Oscillations

(a) It is a non-periodic motion. The motion of the swimmer is to and fro, but its nature is such that it cannot have a definite period. The swimmer does not repeat the trip after a fixed interval of time.

(b) It is a periodic motion, and is also simple harmonic. The suspended bar-magnet oscillates to and fro about its N-S direction under a restoring torque, repeating its motion after a fixed interval of time.

(c) It is a periodic motion. The hydrogen molecule returns to the same state after every complete rotation about its centre of mass, so the motion repeats itself at regular intervals of time.

(d) It is a non-periodic motion, as the arrow never returns to its initial position.

(a) It is a periodic, but not simple harmonic, motion. The earth repeats its rotation after every 24 hours, but it does not have a to and fro motion about a fixed point, which is the first condition for linear S.H.M.

(b) It is (nearly) simple harmonic motion. When the mercury column is depressed in one limb and released, the weight of the unbalanced column provides a restoring force which is directly proportional to the displacement and is directed towards the mean position.

(c) It is (nearly) simple harmonic motion. For a small displacement from the lowermost position, the ball-bearing behaves like the bob of a simple pendulum whose length equals the radius of curvature of the bowl, and the restoring force is proportional to the displacement.

(d) It is a periodic, but not simple harmonic, motion. A polyatomic molecule has a number of natural frequencies, so its general vibration is a superposition of S.H.M.'s of a number of different frequencies. Such a superposition is periodic, but not simple harmonic.

Fig. (a) does not represent periodic motion, because the particle does not repeat the path of its motion. The displacement goes on increasing continuously with time.

Fig. (b) represents periodic motion. The whole pattern of the motion repeats itself after every 2 s, so the period of motion is 2 s.

Fig. (c) does not represent periodic motion. Apparently it looks to be a periodic motion of period 3 s, but for a motion to be periodic the repetition of merely one position is not enough; instead, the entire motion should be repeated successively.

Fig. (d) represents periodic motion. The pattern repeats itself after every 2 s, so the period of motion is 2 s.

(a) Taking $\sqrt2$ common,

$$\begin{aligned}\sin ω\text t - \cos ω\text t = \sqrt2\left[\frac{1}{\sqrt2}\sin ω\text t - \frac{1}{\sqrt2}\cos ω\text t\right] \\[1em] = \sqrt2\left[\sin ω\text t \cos \frac{π}{4} - \cos ω\text t \sin \frac{π}{4}\right] \\[1em] = \sqrt2 \sin\left(ω\text t - \frac{π}{4}\right)\end{aligned}$$

This is of the form a sin (ωt + φ₀). Hence it represents simple harmonic motion, of time period $\text T = \frac{\text{2π}}{ω}$.

(b) Using the trigonometric identity,

$\sin^3 ω\text t = \frac{1}{4}\left[3\sin ω\text t - \sin 3ω\text t\right]$

Here each term, sin ωt and sin 3ωt, individually represents S.H.M. However, the resultant outcome of the superposition of two simple harmonic motions of different frequencies is periodic but not simple harmonic. Its periodic-time is $\text T = \frac{\text{2π}}{ω}$.

(c) Since cos (− θ) = cos θ,

$3\cos\left(\frac{π}{4} - 2ω\text t\right) = 3\cos\left(2ω\text t - \frac{π}{4}\right)$

This is of the form a cos (ωt + φ₀) with angular frequency 2ω. Hence it represents simple harmonic motion, whose time period is

$\text T = \frac{\text{2π}}{2ω} = \frac{π}{ω}$

(d) The three terms cos ωt, cos 3ωt and cos 5ωt have angular frequencies ω, 3ω and 5ω. Each term separately is simple harmonic, but their sum is not. Since the frequencies are integral multiples of ω, the function repeats itself after $\frac{\text{2π}}{ω}$. Hence it represents periodic but not simple harmonic motion, of time period $\text T = \frac{\text{2π}}{ω}$.

(e) $\text e^{-ω^2 \text t^2}$ is an exponential function which decreases continuously with time and never repeats itself. Hence it represents non-periodic motion.

(f) The function 1 + ωt + ω²t² also represents non-periodic motion, because it tends to ∞ as t → ∞ and hence never repeats its values.

The points A and B, 10 cm apart, are the extreme positions of the particle in S.H.M., and the point O is the mean position. The direction from A to B is taken as positive. In S.H.M. the acceleration α and the force F are always directed towards the mean position O.

(a) At the end A : This is an extreme position, so the velocity v is zero. The acceleration α and the force F are directed from A towards O, that is, along the positive direction. Hence the signs are (0, +, +).

(b) At the end B : This is also an extreme position, so v is zero. The acceleration α and the force F are directed from B towards O, that is, along the negative direction. Hence the signs are (0, −, −).

(c) At the mid-point of AB going towards A : Here the particle is at the mean position O, so α and F are zero. The particle is moving towards A, that is, in the negative direction, so v is negative. Hence the signs are (−, 0, 0).

(d) At 2 cm away from B going towards A : The particle lies between O and B and moves towards A, so v is negative. Being on the B-side of O, α and F are directed towards O, that is, negative. Hence the signs are (−, −, −).

(e) At 3 cm away from A going towards B : The particle lies between A and O and moves towards B, so v is positive. Being on the A-side of O, α and F are directed towards O, that is, positive. Hence the signs are (+, +, +).

(f) At 4 cm away from B going towards A, that is, at E : Since AB = 10 cm, this point is 6 cm from A, that is, 1 cm on the B-side of the mean position O. The particle moves towards A, so v is negative. Being on the B-side of O, α and F are directed towards O, that is, along the negative direction. Hence the signs are (−, −, −).

Note: The textbook prints the answer to this part as (−, +, +), its figure marking the point E at 4 cm from A instead of at 4 cm from B.

Relation (c), α = − 10 x, represents simple harmonic motion.

In S.H.M. the acceleration must be directly proportional to the displacement and directed always towards the mean position, that is,

$α = -ω^2 \text x$

This condition requires the acceleration to be proportional to the first power of x and to carry a negative sign.

• In (a) the sign is positive, so the acceleration is directed away from the mean position.
• In (b) the acceleration is proportional to x², and not to x.
• In (d) the acceleration is proportional to x³ and the sign is also positive.

Only relation (c) satisfies α = − ω²x, with ω² = 10 s^-2.

Given,

• Initial displacement, x₀ = 1 cm
• Initial velocity, v₀ = π cm s^-1
• Angular frequency, ω = π rad s^-1

(a) Using the cosine function : The displacement at any instant t is

$\text x = \text A\cos(ω\text t + φ) \qquad \dots(\text i)$

Differentiating with respect to t, the velocity of the particle is

$\text v = \frac{\text{dx}}{\text{dt}} = -\text Aω\sin(ω\text t + φ) \qquad \dots(\text{ii})$

Putting t = 0 in equations (i) and (ii),

$\text x_0 = \text A\cos φ \qquad \dots(\text{iii})$

$\frac{\text v_0}{ω} = -\text A\sin φ \qquad \dots(\text{iv})$

Squaring and adding equations (iii) and (iv),

$\text A^2 = \text x_0^2 + \frac{\text v_0^2}{ω^2}$

Substituting the given values,

$\text A^2 = (1\ \text{cm})^2 + \left(\frac{π\ \text{cm s}^{-1}}{π\ \text{rad s}^{-1}}\right)^2 = 1 + 1 = 2\ \text{cm}^2$

$\text A = \sqrt2\ \text{cm}$

Dividing equation (iv) by equation (iii),

$\tan φ = -\frac{\text v_0}{ω\text x_0} = -\frac{π\ \text{cm s}^{-1}}{π\ \text{rad s}^{-1} \times 1\ \text{cm}} = -1$

$φ = \tan^{-1}(-1) = \text{2π} - \frac{π}{4} = \frac{\text{7π}}{4}$

Hence, with the cosine function the amplitude is $\sqrt2$ cm and the initial phase angle is $\frac{\text{7π}}{4}$.

(b) Using the sine function : The displacement is

$\text x = \text B\sin(ω\text t + α)$

and the velocity is

$\text v = \frac{\text{dx}}{\text{dt}} = \text Bω\cos(ω\text t + α)$

At t = 0,

$\text x_0 = \text B\sin α \qquad \text{and} \qquad \text v_0 = \text Bω\cos α$

Squaring and adding,

$\text B^2 = \text x_0^2 + \frac{\text v_0^2}{ω^2} = (1)^2 + \frac{π^2}{π^2} = 1 + 1 = 2$

$\text B = \sqrt2\ \text{cm}$

Dividing the two equations,

$\tan α = \frac{\text x_0 ω}{\text v_0} = \frac{1 \times π}{π} = 1$

$α = \frac{π}{4}$

Hence, with the sine function the amplitude is again $\sqrt2$ cm and the initial phase is $\frac{π}{4}$. The amplitude is the same in both cases, since it is a physical property of the motion and does not depend on the function chosen to describe it.

Given,

• Maximum load on the scale = 50 kg
• Length of the scale, y = 20 cm = 0.2 m
• Period of oscillation, T = 0.60 s
• g = 9.8 m s^-2

When a mass of 50 kg hangs from the spring balance, the spring extends through the full length of the scale, that is, by 0.2 m. If k is the force-constant of the spring, then by Hooke's law,

$\text{ky} = \text{mg}$

$\text k = \frac{\text{mg}}{\text y} = \frac{50\ \text{kg} \times 9.8\ \text{N kg}^{-1}}{0.2\ \text m} = 2450\ \text{N m}^{-1}$

The periodic-time of a body of mass M suspended by a spring of force-constant k is

$\text T = \text{2π}\sqrt{\frac{\text M}{\text k}}$

Squaring and solving for M,

$\text M = \frac{\text k \text T^2}{4π^2}$

Substituting the values,

$\text M = \frac{2450\ \text{N m}^{-1} \times (0.60\ \text s)^2}{4 \times (3.14)^2} = \frac{2450 \times 0.36}{39.44} = 22.36\ \text{kg}$

The weight of the body is

$\text W = \text{Mg} = 22.36\ \text{kg} \times 9.8\ \text{m s}^{-2} = 219.13\ \text N$

Hence, the weight of the body is 219.13 N.

Given,

• Force-constant of the spring, k = 1200 N m^-1
• Mass attached, m = 3.0 kg
• Amplitude of oscillation, a = 2.0 cm = 0.02 m

(i) Frequency of oscillations : The mass performs S.H.M. whose angular frequency is

$ω = \sqrt{\frac{\text k}{\text m}} = \sqrt{\frac{1200\ \text{N m}^{-1}}{3.0\ \text{kg}}} = \sqrt{400} = 20\ \text{rad s}^{-1}$

The frequency is

$\text n = \frac{ω}{\text{2π}} = \frac{20\ \text{rad s}^{-1}}{2 \times 3.14} = 3.18\ \text s^{-1} ≈ 3.2\ \text s^{-1}$

(ii) Maximum acceleration : In S.H.M. the acceleration is maximum at the extreme position, where y = a,

$|α_{max}| = ω^2 \text a = (20\ \text s^{-1})^2 \times 0.02\ \text m = 400 \times 0.02 = 8.0\ \text{m s}^{-2}$

(iii) Maximum speed : The speed is maximum at the mean position,

$\text u_{max} = ω\text a = (20\ \text s^{-1}) \times 0.02\ \text m = 0.40\ \text{m s}^{-1}$

Hence, the frequency of oscillation is 3.2 s^-1, the maximum acceleration is 8.0 m s^-2 and the maximum speed is 0.40 m s^-1.

Given,

• Amplitude of oscillation, a = 2.0 cm
• Angular frequency, ω = 20 rad s^-1 (from Exercise 9)

The displacement of the mass along the X-axis at any instant t is

$\text x = \text a\sin(ω\text t + φ) \qquad \dots(\text i)$

where φ is the initial phase.

(a) At the mean position at t = 0 : Here x = 0 at t = 0, so sin φ = 0, giving φ = 0, the mass being taken to move along the positive direction of the X-axis at that instant. From equation (i),

$\text x = 2.0\sin(20\text t)\ \text{cm}$

(b) At the maximum stretched position at t = 0 : Here x = + a at t = 0, so sin φ = 1, giving $φ = \frac{π}{2}$. From equation (i),

$\text x = 2.0\sin\left(20\text t + \frac{π}{2}\right)\ \text{cm} \quad \text{or} \quad \text x = 2.0\cos(20\text t)\ \text{cm}$

(c) At the maximum compressed position at t = 0 : Here x = − a at t = 0, so sin φ = − 1, giving $φ = \frac{\text{3π}}{2}$. From equation (i),

$\text x = 2.0\sin\left(20\text t + \frac{\text{3π}}{2}\right)\ \text{cm} \quad \text{or} \quad \text x = -2.0\cos(20\text t)\ \text{cm}$

Hence, the three functions differ neither in amplitude nor in frequency, but in initial phase only.

Fig. (a) : Here the radius of the reference circle is 3 cm, the period is T = 2 s, and the particle P starts from the lowermost point of the circle and revolves clockwise.

Suppose the particle arrives at a position Q in time t. The radius vector sweeps an angle

$θ = ω\text t = \frac{\text{2π}\text t}{\text T} = \frac{\text{2π}\text t}{2} = π\text t$

Let N be the foot of the perpendicular from Q on the X-axis. As P revolves, the projection N executes S.H.M. along the X-axis. The displacement x of N at time t is

$\text{ON} = \text{OQ}\cos \text{NOQ} = \text{OQ}\cos\left(\frac{π}{2} - θ\right) = \text{OQ}\sin θ$

Here ON = − x cm, OQ = 3 cm and θ = πt. Therefore,

$-\text x = 3\sin π\text t$

$\text x = -3\sin π\text t$

Hence, for Fig. (a) the required equation of S.H.M. is x = − 3 sin πt, where x is in cm.

Fig. (b) : Here the radius of the reference circle is 2 m, the period is T = 4 s, and the particle P starts from the extreme left of the circle and revolves anticlockwise.

The radius vector sweeps an angle

$θ = ω\text t = \frac{\text{2π}\text t}{\text T} = \frac{\text{2π}\text t}{4} = \frac{π\text t}{2}$

As above, the displacement x of the X-projection of Q is

$\text{ON} = \text{OQ}\cos θ$

Here ON = − x m, OQ = 2 m and $θ = \frac{π\text t}{2}$. Therefore,

$-\text x = 2\cos\left(\frac{π\text t}{2}\right)$

$\text x = -2\cos\left(\frac{π\text t}{2}\right)$

Hence, for Fig. (b) the required equation of S.H.M. is $\text x = -2\cos\left(\frac{π\text t}{2}\right)$, where x is in metre.

Each equation is first brought to the standard form x = a cos (ωt + φ₀), from which the radius a of the reference circle, the angular speed ω of the revolving particle and the initial phase φ₀ can be read directly.

(a) Using $-\sin θ = \cos\left(θ + \frac{π}{2}\right)$,

$\text x = -2\sin\left(3\text t + \frac{π}{3}\right) = 2\cos\left(3\text t + \frac{π}{3} + \frac{π}{2}\right) = 2\cos\left(3\text t + \frac{\text{5π}}{6}\right)$

Comparing with x = a cos (ωt + φ₀),

$\text a = 2\ \text{cm}, \qquad ω = 3\ \text{rad s}^{-1}, \qquad φ_0 = \frac{\text{5π}}{6}$

The reference circle is of radius 2 cm, and the initial (t = 0) position P of the particle makes an angle of $\frac{\text{5π}}{6}$ with the positive X-axis.

(b) Using cos (− θ) = cos θ,

$\text x = \cos\left(\frac{π}{6} - \text t\right) = \cos\left(\text t - \frac{π}{6}\right)$

Comparing with x = a cos (ωt + φ₀),

$\text a = 1\ \text{cm}, \qquad ω = 1\ \text{rad s}^{-1}, \qquad φ_0 = -\frac{π}{6}$

The reference circle is of radius 1 cm, with the initial position P at an angle of $\frac{π}{6}$ below the positive X-axis.

(c) Using $\sin θ = \cos\left(θ - \frac{π}{2}\right)$,

$\text x = 3\sin\left(\text{2π}\text t + \frac{π}{4}\right) = 3\cos\left(\text{2π}\text t + \frac{π}{4} - \frac{π}{2}\right) = 3\cos\left(\text{2π}\text t - \frac{π}{4}\right)$

Comparing with x = a cos (ωt + φ₀),

$\text a = 3\ \text{cm}, \qquad ω = \text{2π}\ \text{rad s}^{-1}, \qquad φ_0 = -\frac{π}{4}$

The reference circle is of radius 3 cm, with the initial position P at an angle of $\frac{π}{4}$ below the positive X-axis.

(d) The equation x = 2 cos πt is already in the standard form. Comparing with x = a cos (ωt + φ₀),

$\text a = 2\ \text{cm}, \qquad ω = π\ \text{rad s}^{-1}, \qquad φ_0 = 0$

The reference circle is of radius 2 cm, with the initial position P on the positive X-axis itself.

(i) Maximum extension of the spring

In Fig. (a) : Let x be the maximum extension produced in the spring. By Hooke's law,

$\text F = \text{kx} \quad \Rightarrow \quad \text x = \frac{\text F}{\text k}$

In Fig. (b) : Here the forces F, F are applied on the two masses m, m in opposite directions. Hence the middle point O of the spring remains stationary at its position.

Therefore the spring of length l may be considered to be made up of two parts, each of length $\frac{\text l}{2}$, and hence of force-constant 2k, joined to each other at the point O. Since extension is proportional to the length of the spring, each part undergoes an extension given by

$\frac{\text x}{2} = \frac{\text F}{2\text k}$

The total extension in the spring is

$\text x = \frac{\text x}{2} + \frac{\text x}{2} = \frac{\text F}{2\text k} + \frac{\text F}{2\text k} = \frac{\text F}{\text k}$

Hence the maximum extension of the spring is $\frac{\text F}{\text k}$ in both the cases.

(ii) Period of oscillation

In Fig. (a) : The mass m oscillates under a spring of force-constant k, so

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

In Fig. (b) : The spring is equivalent to two springs, each of length $\frac{\text l}{2}$ and force-constant 2k, fixed at the point O. Hence the period of oscillation of each mass m is

$\text T = \text{2π}\sqrt{\frac{\text m}{2\text k}}$

Given,

• Stroke of the piston = 2a = 1.0 m
• Angular frequency, ω = 200 rad min^-1

The amplitude of the motion is

$\text a = \frac{1.0\ \text m}{2} = 0.5\ \text m$

Converting the angular frequency into SI units,

$ω = 200\ \text{rad min}^{-1} = \frac{200}{60}\ \text{rad s}^{-1}$

The maximum speed of a particle in S.H.M. occurs at the mean position and is given by

$\text u_{max} = ω\text a$

Substituting the values,

$\text u_{max} = \left(\frac{200}{60}\ \text s^{-1}\right) \times 0.5\ \text m = \frac{100}{60} = 1.67\ \text{m s}^{-1}$

Hence, the maximum speed of the piston is 1.67 m s^-1.

Given,

• Acceleration due to gravity on the moon, g_m = 1.7 m s^-2
• Acceleration due to gravity on the earth, g_e = 9.8 m s^-2
• Time period on the earth, T_e = 3.5 s

Let l be the length of the simple pendulum. Its time periods on the earth and on the moon are

$\text T_e = \text{2π}\sqrt{\frac{\text l}{\text g_e} } \qquad \text{and} \qquad \text T_m = \text{2π}\sqrt{\frac{\text l}{\text g_m}}$

Dividing the second by the first, the length l cancels out,

$\frac{\text T_m}{\text T_e} = \sqrt{\frac{\text g_e}{\text g_m}}$

$\text T_m = \text T_e \sqrt{\frac{\text g_e}{\text g_m}}$

Substituting the values,

$\text T_m = (3.5\ \text s)\sqrt{\frac{9.8\ \text{m s}^{-2}}{1.7\ \text{m s}^{-2}}} = (3.5\ \text s) \times 2.4 = 8.4\ \text s$

Hence, the time period of the simple pendulum on the moon is 8.4 s. Since g is smaller on the moon, the pendulum oscillates more slowly there.

Given,

• Length of the simple pendulum = l
• Mass of the bob = M
• Radius of the circular track = R
• Uniform speed of the car = v

The bob is suspended in a car moving on a circular track. Hence, in addition to its weight Mg acting vertically downwards, the bob is acted upon by a centripetal force $\frac{\text M\text v^2}{\text R}$ directed horizontally towards the centre of the circle.

These two forces are mutually perpendicular, so the effective force on the bob is their resultant,

$\text{Mg}' = \sqrt{(\text{Mg})^2 + \left(\frac{\text M\text v^2}{\text R}\right)^2}$

Therefore the effective value of the acceleration due to gravity is

$\text g' = \sqrt{\text g^2 + \left(\frac{\text v^2}{\text R}\right)^2} = \sqrt{\text g^2 + \frac{\text v^4}{\text R^2}}$

The time period of the pendulum is obtained by replacing g by g' in the formula for a simple pendulum,

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g'}} = \text{2π}\sqrt{\frac{\text l}{\left(\text g^2 + \frac{\text v^4}{\text R^2}\right)^{\frac{1}{2}}}}$

Hence, the time period of the pendulum is $\text T = \text{2π}\sqrt{\frac{\text l}{\left(\text g^2 + \frac{\text v^4}{\text R^2}\right)^{\frac{1}{2}}}}$. Since g' > g, the time period is reduced on account of the radial acceleration $\frac{\text v^2}{\text R}$.

Given,

• Base area of the cork = A
• Height of the cork = h
• Density of the cork = ρ
• Density of the liquid = ρ₁

Let h₁ be the length of the cork submerged in the liquid at equilibrium. By the law of floatation, the upthrust on the cork equals its weight,

$ρ_1 \text A \text h_1 \text g = ρ \text A \text h \text g \quad \Rightarrow \quad \text h_1 = \frac{ρ}{ρ_1}\text h \qquad \dots(\text i)$

Now let the cork be pushed down further through a small distance x and released. An additional volume Ax of the liquid is displaced, so the increase in upthrust is ρ₁gAx. This acts as the net upward force, that is, the restoring force,

$\text F = -ρ_1 \text g \text A \text x$

The negative sign is taken because the restoring force is directed opposite to the displacement x. By Newton's second law of motion, if α is the instantaneous acceleration of the cork,

$\text m α = -ρ_1 \text g \text A \text x \quad \Rightarrow \quad α = -\left(\frac{ρ_1 \text g \text A}{\text m}\right)\text x \qquad \dots(\text{ii})$

But the mass of the cylindrical piece of cork is

$\text m = \text{volume} \times \text{density} = ρ \text A \text h$

Substituting this in equation (ii),

$α = -\left(\frac{ρ_1 \text g \text A}{ρ \text A \text h}\right)\text x = -\left(\frac{ρ_1 \text g}{ρ \text h}\right)\text x \qquad \dots(\text{iii})$

Here $\frac{ρ_1 \text g}{ρ \text h}$ is a constant. Hence α ∝ − x, that is, the acceleration of the cork is directly proportional to its displacement and is directed towards the mean position. Therefore the motion of the cork is simple harmonic.

Comparing equation (iii) with α = − ω²x,

$ω^2 = \frac{ρ_1 \text g}{ρ \text h}$

The time period is

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{ρ \text h}{ρ_1 \text g}}$

Hence, the cork oscillates up and down simple harmonically with a period $\text T = \text{2π}\sqrt{\frac{\text h ρ}{ρ_1 \text g}}$.

Initially the mercury stands at the same height in both the limbs of the U-tube, say at B and C. On working the suction pump, the level C is pressed down through a distance y up to D, and consequently the level B rises up by the same distance y to E. Thus, in the left limb of the U-tube there is now an additional mercury column D′E of length 2y.

When the pressed level of the mercury is left free, the whole mercury column oscillates up and down due to the restoring force provided by the weight of this unbalanced mercury-column D′E.

Let m be the mass of mercury per unit length of the tube. Then the mass of the column D′E is (m × 2y), and the restoring force acting on the mercury is

$\text F = -(\text m \times 2\text y)\text g$

The negative sign is taken because this force acts in the direction opposite to the displacement of the liquid.

If 2h is the length of the whole liquid column, the mass of the whole liquid filled in the tube is (m × 2h). Hence the acceleration of the liquid is

$α = \frac{\text{force}}{\text{mass}} = -\frac{(\text m \times 2\text y)\text g}{\text m \times 2\text h}$

$α = -\frac{\text g}{\text h}\text y = -ω^2 \text y, \qquad \text{where } ω^2 = \frac{\text g}{\text h}$

Here $\frac{\text g}{\text h}$ is a constant for a given liquid column. Hence the acceleration α is directly proportional to the displacement y and is directed opposite to it, towards the mean position.

Hence, the column of mercury in the U-tube executes simple harmonic motion, whose time period is

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{\text h}{\text g}}$

Given,

• Mass suspended, m = 0.1 kg
• Length of the wire, L = 1 m
• Area of cross-section, A = 4.9 × 10^-7 m²
• Angular frequency, ω = 140 rad s^-1
• Young's modulus, Y = n × 10⁹ N m^-2

The Young's modulus of the material of the wire is

$\text Y = \frac{\text F/\text A}{\text l/\text L} = \frac{\text{FL}}{\text{Al}} \quad \Rightarrow \quad \text F = \left(\frac{\text{YA}}{\text L}\right)\text l$

If the elongation of the wire on suspending the mass m is l, then the restoring force acting on the mass due to elasticity is

$\text F = \text k \text l$

Comparing the two equations, the force-constant of the wire is

$\text k = \frac{\text{YA}}{\text L}$

The angular frequency of the resulting S.H.M. is

$ω = \sqrt{\frac{\text k}{\text m}} = \sqrt{\frac{\text{YA}}{\text{mL}}}$

Substituting the values,

$140 = \left[\frac{\text n \times 10^9 \times 4.9 \times 10^{-7}}{0.1 \times 1}\right]^{\frac{1}{2}}$

$140 = \left[\frac{4.9 \times 10^2\text n}{0.1}\right]^{\frac{1}{2}} = \left[4900\text n\right]^{\frac{1}{2}} = 70\sqrt{\text n}$

$\sqrt{\text n} = \frac{140}{70} = 2 \quad \Rightarrow \quad \text n = 4$

Hence, the value of n is 4.

(i) U₀ > E > 0

The particle will perform periodic motion (and not escape to infinity) only if a restoring force continues to act upon it. This is possible only when the particle remains confined within the potential well, that is, when its total energy is less than the constant value U₀ attained by the potential energy outside the well,

$\text E \lt \text U_0$

Since the potential energy is positive everywhere, E must also be positive. Hence the condition is U₀ > E > 0.

(ii) $\frac{1}{\text A}\sqrt{\frac{\text m}{α}}$

The dimensions of α follow from U(x) = αx⁴,

$[α] = \frac{\text U(\text x)}{\text x^4} = \frac{\text{energy}}{(\text{length})^4} = \frac{[\text{ML}^2\text T^{-2}]}{[\text L^4]} = [\text{ML}^{-2}\text T^{-2}]$

Now checking the dimensions of the option $\frac{1}{\text A}\sqrt{\frac{\text m}{α}}$,

$\frac{1}{[\text L]}\sqrt{\frac{[\text M]}{[\text{ML}^{-2}\text T^{-2}]}} = \frac{1}{[\text L]}\sqrt{[\text L^2 \text T^2]} = \frac{[\text{LT}]}{[\text L]} = [\text T]$

This has the dimensions of time. Hence the time period is proportional to $\frac{1}{\text A}\sqrt{\frac{\text m}{α}}$.

(iii) Zero

For | x | > X₀, the potential energy is a constant, U = U₀. Therefore the force acting on the particle is

$\text F = -\frac{\text{dU}}{\text{dx}} = 0 \qquad [\because \text U = \text{constant}]$

Since no force acts on the particle in this region, its acceleration is zero.

$\frac{\text{7A}}{3}$

Given,

• Original amplitude of the motion = A
• Displacement at which the speed is trebled, $\text x = \frac{2\text A}{3}$

The velocity of a particle in S.H.M. at displacement x is

$\text u = ω\sqrt{\text A^2 - \text x^2}$

At $\text x = \frac{2\text A}{3}$,

$\text u = ω\sqrt{\text A^2 - \left(\frac{2\text A}{3}\right)^2} = ω\sqrt{\text A^2 - \frac{4\text A^2}{9}} = ω\sqrt{\frac{5\text A^2}{9}} = \frac{\sqrt5ω\text A}{3}$

The speed at this position is now trebled, so the new speed is

$\text u_{new} = 3\text u = \sqrt5ω\text A$

The displacement is unchanged at this instant, so if A_new is the new amplitude,

$\text u_{new} = ω\sqrt{\text A_{new}^2 - \text x^2} = \sqrt5ω\text A$

Squaring both sides and cancelling ω²,

$\text A_{new}^2 - \text x^2 = 5\text A^2$

$\text A_{new}^2 = 5\text A^2 + \left(\frac{2\text A}{3}\right)^2 = 5\text A^2 + \frac{4\text A^2}{9} = \frac{45\text A^2 + 4\text A^2}{9} = \frac{49}{9}\text A^2$

$\text A_{new} = \frac{7}{3}\text A$

Hence, the new amplitude of the motion is $\frac{\text{7A}}{3}$.

$\text{2π}\sqrt{\frac{\text x_2^2 - \text x_1^2}{\text v_1^2 - \text v_2^2}}$

The velocity of a particle in S.H.M. at a displacement x from the mean position is

$\text v = ω\sqrt{\text a^2 - \text x^2}$

where a is the amplitude. Squaring, at the two given displacements,

$\text v_1^2 = ω^2(\text a^2 - \text x_1^2) \qquad \dots(\text i)$

$\text v_2^2 = ω^2(\text a^2 - \text x_2^2) \qquad \dots(\text{ii})$

Subtracting equation (ii) from equation (i), the amplitude a is eliminated,

$\text v_1^2 - \text v_2^2 = ω^2(\text x_2^2 - \text x_1^2)$

$ω = \sqrt{\frac{\text v_1^2 - \text v_2^2}{\text x_2^2 - \text x_1^2}}$

But $ω = \frac{\text{2π}}{\text T}$, therefore

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{\text x_2^2 - \text x_1^2}{\text v_1^2 - \text v_2^2}}$

time period of oscillations is 6τ

Since the particle starts from rest, it starts from the extreme position. Hence its displacement is measured from the extreme position as

$\text x = \text A\cos ω\text t, \qquad \text{so that at } \text t = 0,\ \text x = \text A$

At t = τ : The particle has travelled a distance a, so x = A − a,

$\text A - \text a = \text A\cos ω\text{τ} \quad \Rightarrow \quad \text a = \text A(1 - \cos ω\text{τ}) \qquad \dots(\text i)$

At t = 2τ : The particle has travelled a further 2a, so the total distance travelled is 3a and x = A − 3a,

$\text A - 3\text a = \text A\cos 2ω\text{τ} \quad \Rightarrow \quad 3\text a = \text A(1 - \cos 2ω\text{τ}) \qquad \dots(\text{ii})$

Dividing equation (i) by equation (ii),

$\frac{1}{3} = \frac{1 - \cos ω\text{τ}}{1 - \cos 2ω\text{τ}} = \frac{1 - \cos ω\text{τ}}{2\sin^2 ω\text{τ}}$

Let x = cos ωτ. Then sin²ωτ = 1 − x², so

$\frac{1}{3} = \frac{1 - \text x}{2(1 - \text x^2)} = \frac{1 - \text x}{2(1 - \text x)(1 + \text x)} = \frac{1}{2(1 + \text x)}$

$2(1 + \text x) = 3 \quad \Rightarrow \quad \text x = \frac{1}{2} = \cos ω\text{τ}$

Therefore,

$ω\text{τ} = \frac{π}{3} \quad \Rightarrow \quad \frac{\text{2π}}{\text T}\text{τ} = \frac{π}{3}$

$\text T = 6\text{τ}$

Hence, the time period of oscillations is 6τ.

$\sqrt{50}$ m/s

Given,

• Unstretched length of the spring = 4.9 m
• Amplitude of oscillation, a = 0.2 m
• Angular frequency, $ω = \frac{π}{3}$ rad s^-1
• Horizontal distance OP = 10 m
• Angle of projection = 45°, time of flight = 1 s, g = 10 m s^-2

Position of the block at t = 1 s : The time period of the block is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π} \times 3}{π} = 6\ \text s$

The block is released from the stretched (extreme) position, so its displacement from the mean position is

$\text x = \text a\cos ω\text t = 0.2\cos\left(\frac{\text{2π}}{\text T}\text t\right)$

At t = 1 s,

$\text x = 0.2\cos\left(\frac{\text{2π}}{6} \times 1\right) = 0.2\cos\frac{π}{3} = \frac{0.2}{2} = 0.1\ \text m$

Hence the position of the block measured from the wall at t = 1 s is

$4.9\ \text m + 0.1\ \text m = 5\ \text m$

Range of the pebble : Since P is at 10 m from O, the horizontal distance covered by the pebble is

$\text R = 10 - 5 = 5\ \text m$

The horizontal component of the velocity of the pebble remains constant, so

$\text v\cos 45° \times \text t = 5\ \text m$

$\text v \times \frac{1}{\sqrt2} \times 1 = 5 \quad \Rightarrow \quad \text v = 5\sqrt2 = \sqrt{50}\ \text{m s}^{-1}$

Hence, the value of v is $\sqrt{50}$ m/s.

$\text A, \frac{\text{4π}}{3}$

Given,

• x₁(t) = A sin ωt
• $\text x_2(\text t) = \text A\sin\left(ω\text t + \frac{\text{2π}}{3}\right)$

The two displacements x₁ and x₂ have equal amplitudes A and a phase difference of $\frac{\text{2π}}{3}$. Their resultant amplitude is

$\text R = \sqrt{\text A^2 + \text A^2 + 2\text A \cdot \text A\cos\frac{\text{2π}}{3}}$

Since $\cos\frac{\text{2π}}{3} = -\frac{1}{2}$,

$\text R = \sqrt{2\text A^2 + 2\text A^2\left(-\frac{1}{2}\right)} = \sqrt{2\text A^2 - \text A^2} = \text A$

Since the two amplitudes are equal, this resultant bisects the angle between them, so its phase is $\frac{π}{3}$.

The third displacement x₃(t) = B sin (ωt + φ) will bring the mass to complete rest only when it is equal and opposite to this resultant. Hence,

$\text B = \text A \qquad \text{and} \qquad φ = π + \frac{π}{3} = \frac{\text{4π}}{3}$

Hence, B = A and $φ = \frac{\text{4π}}{3}$.

For a particle executing S.H.M. with amplitude a, the kinetic energy and the potential energy at a displacement d are

$\text K = \frac{1}{2}\text m ω^2(\text a^2 - \text d^2) \qquad \text{and} \qquad \text U = \frac{1}{2}\text m ω^2 \text d^2$

From these expressions,

• The kinetic energy is maximum at the mean position (d = 0) and becomes minimum (zero) at the extreme positions (d = ± a).
• The potential energy is minimum (zero) at the mean position and becomes maximum at the extreme positions.

Both graphs are parabolas, the KE curve opening downwards and the PE curve opening upwards, and their sum is the constant total energy. Only graph (b) shows the KE curve with its maximum at d = 0 and the PE curve with its minimum at d = 0.

$\frac{6}{5}$

Given,

• Vertical distance covered by the point of suspension, y = kt²
• k = 1 m s^-2
• g = 10 m s^-2

The upward acceleration of the point of suspension is obtained by differentiating y twice with respect to time,

$\frac{\text{dy}}{\text{dt}} = \frac{\text d}{\text{dt}}(\text k \text t^2) = 2\text k \text t$

$\text a_y = \frac{\text d^2\text y}{\text{dt}^2} = \frac{\text d}{\text{dt}}(2\text k \text t) = 2\text k = 2 \times 1 = 2\ \text{m s}^{-2}$

When the support moves upward with acceleration a_y, the effective value of the acceleration due to gravity becomes (g + a_y). Hence the two time periods are

$\text T_1 = \text{2π}\sqrt{\frac{\text l}{\text g}} \qquad \text{and} \qquad \text T_2 = \text{2π}\sqrt{\frac{\text l}{\text g + \text a_y}}$

Squaring and dividing,

$\frac{\text T_1^2}{\text T_2^2} = \frac{\text g + \text a_y}{\text g} = \frac{(10 + 2)\ \text{m s}^{-2}}{10\ \text{m s}^{-2}} = \frac{12}{10} = \frac{6}{5}$

Hence, $\frac{\text T_1^2}{\text T_2^2} = \frac{6}{5}$.

$\frac{π}{3}$

Given,

• Amplitude of each particle = A
• Angular frequency of each particle = ω
• Separation between the mean positions = X₀
• Maximum separation = X₀ + A

Let the displacements of the two particles be

$\text x_1 = \text A\sin ω\text t \qquad \text{and} \qquad \text x_2 = \text X_0 + \text A\sin(ω\text t + φ)$

The separation between them at any instant is

$\text x_2 - \text x_1 = \text X_0 + \text A\sin(ω\text t + φ) - \text A\sin ω\text t$

Using the identity $\sin \text C - \sin \text D = 2\cos\left(\frac{\text C + \text D}{2}\right)\sin\left(\frac{\text C - \text D}{2}\right)$,

$\text x_2 - \text x_1 = \text X_0 + 2\text A\sin\left(\frac{φ}{2}\right)\cos\left(ω\text t + \frac{φ}{2}\right)$

The maximum value of the cosine term is 1, so the maximum separation is

$(\text x_2 - \text x_1)_{max} = \text X_0 + 2\text A\sin\frac{φ}{2}$

But it is given that the maximum separation is X₀ + A. Therefore,

$\text X_0 + 2\text A\sin\frac{φ}{2} = \text X_0 + \text A$

$\sin\frac{φ}{2} = \frac{1}{2} \quad \Rightarrow \quad \frac{φ}{2} = \frac{π}{6} \quad \Rightarrow \quad φ = \frac{π}{3}$

Hence, the phase difference between their motions is $\frac{π}{3}$.

$\frac{\text a}{2\text g \text k}$

Given,

• Equation of the parabolic wire, y = kx²
• Horizontal acceleration of the wire = a

Let the bead settle at a point where the tangent to the parabola makes an angle θ with the horizontal. In the frame of the wire, the bead is in equilibrium under its weight mg, the normal reaction of the smooth wire, and the pseudo force ma acting horizontally opposite to the acceleration.

Resolving along and perpendicular to the wire, the condition of equilibrium gives

$\text{ma} = \text{mg}\tan θ \quad \Rightarrow \quad \text a = \text g\tan θ$

The slope of the parabola at the point (x, y) is

$\tan θ = \frac{\text{dy}}{\text{dx}} = \frac{\text d}{\text{dx}}(\text k \text x^2) = 2\text k \text x$

Substituting this value,

$\text a = \text g \times 2\text k \text x$

$\text x = \frac{\text a}{2\text g \text k}$

Hence, the new equilibrium position of the bead is at a distance $\frac{\text a}{2\text g \text k}$ from the Y-axis.

$\frac{1}{\text{2π}}\sqrt{\frac{\text A^2 γ \text P_0}{\text M \text V_0}}$

Given,

• Mass of the piston = M
• Area of cross-section = A
• Equilibrium volume of the gas = V₀
• Equilibrium pressure of the gas = P₀

Since the system is completely isolated from its surroundings, the process is adiabatic. Hence, by Poisson's law,

$\text{PV}^γ = \text{a constant}$

Taking logarithms and differentiating,

$\frac{\text{dP}}{\text P} + γ\frac{\text{dV}}{\text V} = 0 \quad \Rightarrow \quad \text{dP} = -\frac{γ\text P\text{dV}}{\text V}$

If the piston is displaced through a small distance x, the change in volume is dV = Ax. The net restoring force on the piston is

$\text F_{net} = \text{dP} \times \text A = -\frac{γ\text P_0 (\text{Ax})}{\text V_0} \times \text A = -\left(\frac{γ\text P_0 \text A^2}{\text V_0}\right)\text x$

By Newton's second law of motion, F_net = Mα, so the acceleration of the piston is

$α = -\left(\frac{γ\text P_0 \text A^2}{\text M \text V_0}\right)\text x \qquad \dots(\text i)$

Since α ∝ − x, the motion of the piston is simple harmonic. Comparing equation (i) with α = − ω²x,

$ω^2 = \frac{γ\text P_0 \text A^2}{\text M \text V_0}$

The frequency of oscillation is

$\text n = \frac{ω}{\text{2π}} = \frac{1}{\text{2π}}\sqrt{\frac{\text A^2 γ \text P_0}{\text M \text V_0}}$

a²T² + 4π²v² and $\frac{\text a \text T}{\text x}$

For a particle executing S.H.M. with amplitude A,

$\text x = \text A\sin ω\text t, \qquad \text v = \text Aω\cos ω\text t, \qquad \text a = -\text Aω^2\sin ω\text t$

Checking a²T² + 4π²v² : Substituting the values and using $\text T = \frac{\text{2π}}{ω}$,

$\text a^2\text T^2 + 4π^2\text v^2 = \text A^2ω^4\sin^2 ω\text t \times \frac{4π^2}{ω^2} + 4π^2 \text A^2ω^2\cos^2 ω\text t$

$= 4π^2\text A^2ω^2\sin^2 ω\text t + 4π^2\text A^2ω^2\cos^2 ω\text t$

$= 4π^2\text A^2ω^2(\sin^2 ω\text t + \cos^2 ω\text t) = 4π^2\text A^2ω^2 = \text{a constant}$

Checking $\frac{\text a \text T}{\text x}$ :

$\frac{\text a\text T}{\text x} = \frac{-\text Aω^2\sin ω\text t \times \frac{\text{2π}}{ω}}{\text A\sin ω\text t} = -\text{2π}ω = \text{a constant}$

The remaining two expressions contain sin ωt and cos ωt separately and therefore change with time.

Hence, the quantities a²T² + 4π²v² and $\frac{\text a \text T}{\text x}$ do not change with time.

Note: The question is worded as though only one option is correct, but two of the given expressions, options 2 and 3, are both independent of time. The textbook answer key also marks both (b) and (c) as correct.

Options 1 and 4

Since the particle starts from the equilibrium position, its displacement is

$\text x = \text a\sin ω\text t \quad \Rightarrow \quad \text v = \text aω\cos ω\text t = \text v_0\cos ω\text t$

Checking option 1 : The collision with the rigid wall is elastic, so no mechanical energy is lost. Hence, when the particle returns to its equilibrium position it again has the full speed v₀. Option 1 is correct.

Time of collision : At the instant of collision the speed is 0.5 v₀,

$\frac{\text v_0}{2} = \text v_0\cos ω\text t_1 \quad \Rightarrow \quad \cos ω\text t_1 = \frac{1}{2} \quad \Rightarrow \quad ω\text t_1 = \frac{π}{3} \quad \Rightarrow \quad \text t_1 = \frac{π}{3ω}$

Checking option 2 : After the elastic collision the particle retraces its path, so it returns to the equilibrium position after a further time t₁. Hence

$\text t_2 = 2\text t_1 = \frac{\text{2π}}{3ω} = \frac{\text{2π}}{3} \times \frac{1}{\text{2π}}\text T = \frac{\text{2π}}{3}\sqrt{\frac{\text m}{\text k}}$

using $ω = \sqrt{\frac{\text k}{\text m}}$. This is not $π\sqrt{\frac{\text m}{\text k}}$, so option 2 is incorrect.

Checking option 3 : After crossing the equilibrium position, the spring is compressed and the compression is maximum a quarter of a period later,

$\text t_3 = \text t_2 + \frac{\text T}{4} = \frac{\text{2π}}{3}\sqrt{\frac{\text m}{\text k}} + \frac{\text{2π}}{4}\sqrt{\frac{\text m}{\text k}} = \frac{\text{7π}}{6}\sqrt{\frac{\text m}{\text k}}$

This is not $\frac{\text{4π}}{3}\sqrt{\frac{\text m}{\text k}}$, so option 3 is incorrect.

Checking option 4 : The particle passes the equilibrium position for the second time a further quarter period later,

$\text t_4 = \text t_3 + \frac{\text T}{4} = \frac{\text{7π}}{6}\sqrt{\frac{\text m}{\text k}} + \frac{\text{2π}}{4}\sqrt{\frac{\text m}{\text k}} = \frac{\text{5π}}{3}\sqrt{\frac{\text m}{\text k}}$

Option 4 is correct.

Hence, options 1 and 4 are correct.

Options 2 and 4

Given,

• $\frac{\text a}{\text b} = \text n^2$ and $\frac{\text a}{\text R} = \text n$
• Both oscillators have the same mass m

From the figures, the first oscillator gives an ellipse with semi-axes a (along x) and b (along p), while the second gives a circle of radius R.

First oscillator : The amplitude is a and the maximum momentum is b, so

$\text p_1 = \text m\text a ω_1 = \text b \qquad \dots(\text i)$

Second oscillator : The amplitude is R and the maximum momentum is also R, so

$\text p_2 = \text m \text R ω_2 = \text R \quad \Rightarrow \quad \text m ω_2 = 1 \qquad \dots(\text{ii})$

From equation (i), $\frac{\text b}{\text a} = \text m ω_1$. Dividing this by equation (ii),

$\frac{ω_1}{ω_2} = \frac{\text b}{\text a} = \frac{1}{\text n^2} \quad \Rightarrow \quad \frac{ω_2}{ω_1} = \text n^2$

Hence option 2 is correct.

Total energies : For a harmonic oscillator of mass m, amplitude A and angular frequency ω, the total energy is $\text E = \frac{1}{2}\text m ω^2 \text A^2$. Therefore

$\text E_1 = \frac{1}{2}\text m ω_1^2 \text a^2 \qquad \text{and} \qquad \text E_2 = \frac{1}{2}\text m ω_2^2 \text R^2$

Dividing,

$\frac{\text E_1}{\text E_2} = \frac{ω_1^2}{ω_2^2} \times \left(\frac{\text a}{\text R}\right)^2 = \frac{ω_1^2}{ω_2^2} \times \text n^2$

Substituting $\text n^2 = \frac{ω_2}{ω_1}$,

$\frac{\text E_1}{\text E_2} = \frac{ω_1^2}{ω_2^2} \times \frac{ω_2}{ω_1} = \frac{ω_1}{ω_2}$

$\frac{\text E_1}{ω_1} = \frac{\text E_2}{ω_2}$

Hence options 2 and 4 are correct.

(i) In the case of a simple pendulum, the restoring force acting on the bob is

$\text F = -\left(\frac{\text{mg}}{\text l}\right)\text x = -\text k \text x$

so the force-constant is $\text k = \frac{\text{mg}}{\text l}$. Substituting this value in the relation $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$,

$\text T = \text{2π}\sqrt{\frac{\text m}{\text{mg}/\text l}} = \text{2π}\sqrt{\frac{\text l}{\text g}}$

The mass m cancels out, because for a pendulum the force-constant itself is proportional to the mass. Hence the time period of a pendulum does not depend upon the mass of its bob.

(ii) For larger angles θ of oscillation, the restoring force on the bob of a simple pendulum is

$\text F = -\text{mg}\sin θ = -\text m \text g_e'$

where g_e′ (= g sin θ) is the effective value of g. For a small angle we take sin θ ≈ θ, so the effective value is g_e = gθ.

But sin θ < θ for every value of θ. Therefore

$\text g_e' \lt \text g_e$

that is, the effective value of g decreases for a larger value of θ. Since $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$, a reduction in g implies an increase in T. Hence for larger angles the time period is greater than $\text{2π}\sqrt{\frac{\text l}{\text g}}$.

(iii) Yes, the wrist watch gives correct time during the free fall. The reason is that a wrist watch works on spring action; it has nothing to do with the acceleration due to gravity.

(iv) In free fall the effective value of the acceleration due to gravity becomes zero, that is, g = 0. Hence the frequency of oscillation of the simple pendulum,

$\text n = \frac{1}{\text{2π}}\sqrt{\frac{\text g}{\text l}}$

becomes zero. The pendulum does not oscillate at all.

Let a particle of mass m perform S.H.M. with time period T and angular frequency $ω\left(= \frac{\text{2π}}{\text T}\right)$. Its instantaneous displacement from the mean position is

$\text y = \text a\sin ω\text t$

The instantaneous kinetic energy and potential energy are

$\text K = \frac{1}{2}\text m ω^2(\text a^2 - \text y^2) = \frac{1}{2}\text m ω^2(\text a^2 - \text a^2\sin^2 ω\text t) = \frac{1}{2}\text m ω^2 \text a^2\cos^2 ω\text t$

$\text U = \frac{1}{2}\text m ω^2 \text y^2 = \frac{1}{2}\text m ω^2 \text a^2\sin^2 ω\text t$

Average kinetic energy over one oscillation :

$\text K_{av} = \frac{1}{\text T}\int_0^{\text T} \frac{1}{2}\text m ω^2 \text a^2\cos^2 ω\text t\ \text{dt}$

Using $\cos^2 ω\text t = \frac{1 + \cos 2ω\text t}{2}$,

$\text K_{av} = \frac{\text m ω^2 \text a^2}{2\text T}\int_0^{\text T}\frac{1 + \cos 2ω\text t}{2}\ \text{dt} = \frac{\text m ω^2 \text a^2}{4\text T}\left[\text t + \frac{\sin 2ω\text t}{2ω}\right]_0^{\text T}$

Since $\sin 2ω\text T = \sin \text{4π} = 0$, the second term vanishes at both the limits,

$\text K_{av} = \frac{\text m ω^2 \text a^2}{4\text T}[\text T] = \frac{1}{4}\text m ω^2 \text a^2 \qquad \dots(\text i)$

Average potential energy over one oscillation :

$\text U_{av} = \frac{1}{\text T}\int_0^{\text T}\frac{1}{2}\text m ω^2 \text a^2\sin^2 ω\text t\ \text{dt}$

Using $\sin^2 ω\text t = \frac{1 - \cos 2ω\text t}{2}$,

$\text U_{av} = \frac{\text m ω^2 \text a^2}{2\text T}\int_0^{\text T}\frac{1 - \cos 2ω\text t}{2}\ \text{dt} = \frac{\text m ω^2 \text a^2}{4\text T}\left[\text t - \frac{\sin 2ω\text t}{2ω}\right]_0^{\text T}$

$\text U_{av} = \frac{\text m ω^2 \text a^2}{4\text T}(\text T) = \frac{1}{4}\text m ω^2 \text a^2 \qquad \dots(\text{ii})$

From equations (i) and (ii),

$\text K_{av} = \text U_{av}$

Hence, for a particle in linear S.H.M. the average kinetic energy over a period of oscillation is equal to the average potential energy over the same period, each being one-half of the total energy.

Given,

• Mass of the disc, M = 10 kg
• Radius of the disc, R = 15 cm = 0.15 m
• Period of torsional oscillation, T = 1.5 s

The torsional oscillations of the disc are angular simple harmonic oscillations, whose time period is given by

$\text T = \text{2π}\sqrt{\frac{\text I}{\text c}}$

where I is the moment of inertia of the disc about the wire as axis and c is the torsional rigidity of the wire.

The moment of inertia of a circular disc about an axis passing through its centre and perpendicular to its plane is

$$\begin{aligned}\text I = \frac{1}{2}\text{MR}^2 \\[1em] = \frac{1}{2} \times 10\ \text{kg} \times (0.15\ \text m)^2 \\[1em] = \frac{1}{2} \times 10 \times 0.0225 \\[1em] = 0.1125\ \text{kg m}^2 \\[1em]\end{aligned}$$

Squaring the expression for T and solving for c,

$\text c = \frac{4π^2 \text I}{\text T^2}$

Substituting the values,

$$\begin{aligned}\text c = \frac{4 \times (3.14)^2 \times 0.1125\ \text{kg m}^2}{(1.5\ \text s)^2} \\[1em] = \frac{4 \times 9.8596 \times 0.1125}{2.25} \\[1em] = \frac{4.4368}{2.25}\end{aligned}$$

$\text c = 1.97\ \text{N m}$

Hence, the torsional spring constant of the wire is 1.97 N m.

Given,

• Amplitude, a = 5 cm = 0.05 m
• Time period, T = 0.2 s

The angular frequency of the motion is

$ω = \frac{\text{2π}}{\text T} = \frac{\text{2π}}{0.2\ \text s} = 10π\ \text s^{-1}$

For a body in S.H.M. at displacement y, the acceleration and the velocity are

$α = -ω^2 \text y \qquad \text{and} \qquad \text u = ω\sqrt{\text a^2 - \text y^2}$

(i) When y = 5 cm = 0.05 m :

$α = -(10π\ \text s^{-1})^2(0.05\ \text m) = -5π^2\ \text{m s}^{-2}$

$\text u = (10π\ \text s^{-1})\sqrt{(0.05\ \text m)^2 - (0.05\ \text m)^2} = 0$

(ii) When y = 3 cm = 0.03 m :

$α = -(10π\ \text s^{-1})^2(0.03\ \text m) = -3π^2\ \text{m s}^{-2}$

$\text u = (10π\ \text s^{-1})\sqrt{(0.05\ \text m)^2 - (0.03\ \text m)^2} = (10π\ \text s^{-1})\sqrt{0.0025 - 0.0009}$

$\text u = (10π\ \text s^{-1})(0.04\ \text m) = 0.4π\ \text{m s}^{-1}$

(iii) When y = 0 :

$α = -ω^2(0) = 0$

$\text u = ω\sqrt{\text a^2 - 0} = ω\text a = (10π\ \text s^{-1})(0.05\ \text m) = 0.5π\ \text{m s}^{-1}$

Hence, at y = 5 cm the acceleration is − 5π² m s^-2 and the velocity is zero; at y = 3 cm the acceleration is − 3π² m s^-2 and the velocity is 0.4π m s^-1; and at y = 0 the acceleration is zero and the velocity is 0.5π m s^-1.

Note: The velocity at the mean position is printed in the textbook as 0.05π m s^-1. Substituting ω = 10π s^-1 and a = 0.05 m in u = ωa gives 0.5π m s^-1, so the printed value carries a decimal-place slip.

Given,

• Initial displacement at t = 0 = x₀
• Initial velocity at t = 0 = v₀
• Angular velocity of the oscillation = ω

Let the general equation for the displacement of the mass be

$\text x = \text a\cos(ω\text t + θ)$

where a is the amplitude and θ the initial phase. Differentiating with respect to time, the velocity of the particle is

$\text v = \frac{\text{dx}}{\text{dt}} = -\text aω\sin(ω\text t + θ)$

At t = 0, x = x₀ and v = v₀. Therefore,

$\text x_0 = \text a\cos θ \quad \Rightarrow \quad \frac{\text x_0}{\text a} = \cos θ \qquad \dots(\text i)$

$\text v_0 = -\text aω\sin θ \quad \Rightarrow \quad \frac{\text v_0}{\text aω} = -\sin θ \qquad \dots(\text{ii})$

Squaring and adding equations (i) and (ii), and using sin²θ + cos²θ = 1,

$\left(\frac{\text x_0}{\text a}\right)^2 + \left(\frac{\text v_0}{\text aω}\right)^2 = 1$

$\text a^2 = \text x_0^2 + \left(\frac{\text v_0}{ω}\right)^2$

$\text a = \sqrt{\text x_0^2 + \left(\frac{\text v_0}{ω}\right)^2}$

Hence, the amplitude of the resulting oscillation is $\text a = \sqrt{\text x_0^2 + \left(\frac{\text v_0}{ω}\right)^2}$.

Given,

• Mass of the particle = m
• Force-constant of each of the three springs = k
• The springs B and C are inclined at 45° on either side of the line of the spring A

Let the particle at O be pushed through a small distance y in the direction of the spring A. Then the spring A is compressed by y, while the springs B and C are each stretched by

$\text y' = \text y\cos 45°$

The restoring forces due to B and C act along their own lengths; their components along the line OA are F_B cos 45° and F_C cos 45°. Hence the total restoring force acting on m along AO is

$\text F = -(\text F_A + \text F_B\cos 45° + \text F_C\cos 45°)$

$= -(\text{ky} + \text{ky}'\cos 45° + \text{ky}'\cos 45°) = -(\text{ky} + 2\text{ky}'\cos 45°)$

Substituting y′ = y cos 45°,

$\text F = -\bigg\lbrace\text{ky} + 2\text k(\text y\cos 45°)\cos 45°\bigg\rbrace = -\bigg\lbrace\text{ky} + 2\text{ky}\left(\frac{1}{\sqrt2}\right)\left(\frac{1}{\sqrt2}\right)\bigg\rbrace$

$\text F = -(\text{ky} + \text{ky}) = -2\text{ky}$

By Newton's second law of motion, F = mα, so

$\text m α = -2\text{ky} \quad \Rightarrow \quad α = -\left(\frac{2\text k}{\text m}\right)\text y$

Since α ∝ − y, the motion of the particle is simple harmonic. Comparing with α = − ω²y,

$ω^2 = \frac{2\text k}{\text m}$

The time period of oscillation is

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{\text m}{2\text k}}$

Hence, the time period of oscillation of the particle is $\text T = \text{2π}\sqrt{\frac{\text m}{2\text k}}$.

opposite to the direction of velocity

Reason — In S.H.M. the acceleration is given by α = − ω²y, so it is always directed towards the mean position and opposite to the displacement. Its magnitude changes with displacement, so it is not constant, and it is zero only at the mean position. Among the given options, only the third one describes a directional property of the acceleration in S.H.M.

Note: Strictly, the acceleration in S.H.M. is always directed towards the mean position, that is, opposite to the displacement. It is opposite to the velocity only while the particle moves from the mean position towards an extreme position; while returning from the extreme position to the mean position the two are in the same direction. Option 3 is the intended answer as the other three options are definitely incorrect.

repeats itself at regular intervals of time

Reason — A function f(t) is said to be periodic if it repeats its values at regular intervals of the independent variable, that is,

$\text f(\text t + \text T) = \text f(\text t) \quad \text{for all values of t}$

where T is called the period of the function. Sine and cosine functions of fixed time period are the simplest examples of periodic functions.

mass of the bob

Reason — The periodic-time of a simple pendulum is

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$

This formula does not contain m. Hence the periodic-time of the pendulum does not depend upon the mass of the bob. It does depend on the effective length l and on the acceleration due to gravity g.

$ω = \frac{\text{2π}}{\text T}$

Reason — The angular frequency of an oscillation represents how rapidly the phase of oscillation changes with time. In one complete oscillation the phase changes by 2π radian in a time T, so

$ω = \text{2π}\text n = \frac{\text{2π}}{\text T}$

Its SI unit is radian per second (rad s^-1).

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

Reason — For a body of mass m attached to a spring of force-constant k, the restoring force is F = − kx, so the acceleration is $α = -\left(\frac{\text k}{\text m}\right)\text x$. Comparing with α = − ω²x gives $ω = \sqrt{\frac{\text k}{\text m}}$, and therefore

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{\text m}{\text k}} = \text{2π}\sqrt{\frac{\text{Inertia Factor}}{\text{Spring Factor}}}$

displacement

Reason — A particle is said to execute simple harmonic motion if it moves to and fro along the same straight line about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is directed always towards the mean position,

$\text F \propto -\text y \quad \text{or} \quad \text F = -\text{ky}$

Here k is the force-constant, defined as restoring force per unit displacement.

sinusoidal

Reason — Sine and cosine functions satisfy sin (θ + 2π) = sin θ and cos (θ + 2π) = cos θ, so they repeat themselves after every 2π radian. Exponential, linear and quadratic functions do not repeat their values at regular intervals and are therefore non-periodic.

the length of the string

Reason — The frequency of a simple pendulum is the reciprocal of its periodic-time,

$\text n = \frac{1}{\text T} = \frac{1}{\text{2π}}\sqrt{\frac{\text g}{\text l}}$

It depends upon the effective length l of the pendulum and on the acceleration due to gravity g, but not upon the mass of the bob.

The initial phase of the particle

Reason — In the equation y(t) = A sin (ωt + φ), the quantity φ is the value of the phase of the oscillation at time t = 0. It is therefore called the initial phase or epoch. It tells us the initial state of motion, that is, the position of the oscillating particle and its direction of motion when observation begins. Here A is the amplitude and ω the angular frequency.

zero

Reason — The velocity of a particle in S.H.M. at displacement y is

$\text u = ω\sqrt{\text a^2 - \text y^2}$

This is maximum when y = 0, that is, when the particle passes through its equilibrium position, where u_max = aω. At the maximum displacement (y = a) the velocity is zero.

displacement

Reason — When the body attached to a light horizontal spring is displaced through a small distance x, the spring exerts a restoring force on it given by Hooke's law,

$\text F = -\text{kx}$

where k is the force-constant of the spring. The negative sign indicates that the force F acts opposite to the displacement.

π radian

Reason — If the displacement is y = a sin ωt, then the acceleration is

$α = -\text aω^2\sin ω\text t = \text aω^2\sin(ω\text t + π)$

Thus the acceleration is always directed opposite to the displacement, so the two are 180° (π radian) out of phase.

remains unchanged

Reason — The time period of S.H.M. is determined only by the inertia factor and the spring factor of the system, as in $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$. The amplitude does not appear in this expression, so the time period of S.H.M. is independent of the amplitude of oscillation.

the mean position

Reason — In S.H.M. the speed is $\text u = ω\sqrt{\text a^2 - \text y^2}$, which is greatest where y = 0. For a pendulum this mean position is the lowest point of the swing, at which the whole of the energy is in the form of kinetic energy and the potential energy is zero.

Note: For a simple pendulum the mean position is the lowest point of the swing, so options 2 and 4 name the same position. Option 2 is the answer because 'mean position' is the general term used for S.H.M., the lowest point being its particular form for a pendulum.

constant

Reason — The total energy of a body in S.H.M. is

$\text E = \text U + \text K = \frac{1}{2}\text m ω^2 \text y^2 + \frac{1}{2}\text m ω^2(\text a^2 - \text y^2) = \frac{1}{2}\text m ω^2 \text a^2$

The displacement y has cancelled out, so the total energy remains the same during the motion of the body. The potential energy and kinetic energy change continuously, but their sum is constant.

the mean position

Reason — The acceleration in S.H.M. is α = − ω²y. It is therefore zero when y = 0, that is, at the mean position, where the velocity is maximum. The acceleration is maximum in the position of maximum displacement.

both 1 and 2

Reason — In the equilibrium position of a body of mass m hanging from a spring of force-constant k, the weight of the body is balanced by the restoring force of the stretched spring,

$\text{mg} = \text k \text l \quad \Rightarrow \quad \text l = \frac{\text{mg}}{\text k}$

Hence the extension l which fixes the equilibrium position depends upon both the mass of the object and the spring constant.

mean position

Reason — The kinetic energy of a body in S.H.M. at displacement y is

$\text K = \frac{1}{2}\text m ω^2(\text a^2 - \text y^2)$

Clearly K decreases with increase of displacement y, and is maximum in the equilibrium position, where y = 0, its value being $\frac{1}{2}\text m ω^2 \text a^2$.

f

Reason — The maximum velocity of a particle in S.H.M. occurs at the mean position and is given by

$\text u_{max} = \text a ω = \text a \times \text{2π}\text f$

For a given amplitude a, the quantity 2πa is a constant. Hence u_max ∝ f.

Circular motion of a car on a track

Reason — For linear S.H.M. the motion of the particle must be in a straight line, to and fro about a fixed point, under a restoring force proportional to the displacement. The circular motion of a car on a track is not a to and fro motion along a straight line, so it is not simple harmonic, although it is periodic. The other three are standard examples of S.H.M.

it never repeats itself

Reason — Non-periodic functions are functions that do not repeat their values at regular intervals, that is, they do not satisfy the condition F(t) = F(t + T) for all t. A classic example is the exponential function F(t) = e^t, which never repeats itself.

increase

Reason — The periodic-time of a simple pendulum is inversely proportional to the square-root of the acceleration due to gravity,

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$

On the moon's surface the value of g is $\frac{1}{6}$ th of that on the earth's surface. A smaller g therefore gives a larger T, so the pendulum oscillates more slowly on the moon.

A ball dropped from a height

Reason — A motion is said to be periodic if it repeats itself at regular intervals of time. The motion of a ball dropped from a height is non-repetitive, since the ball does not return to its starting point after a fixed interval. The remaining three are oscillatory, and hence periodic, motions.

$\sqrt2$ times the original time period

Reason — The time period of a spring-mass system is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} \quad \Rightarrow \quad \text T \propto \sqrt{\text m}$

If the mass is doubled while the same spring is used,

$\frac{\text T'}{\text T} = \sqrt{\frac{2\text m}{\text m}} = \sqrt2 \quad \Rightarrow \quad \text T' = \sqrt2\text T$

− Aω sin ωt

Reason — Velocity is the derivative of displacement with respect to time. Differentiating y = A cos ωt,

$\text u = \frac{\text{dy}}{\text{dt}} = \frac{\text d}{\text{dt}}(\text A\cos ω\text t) = -\text Aω\sin ω\text t$

0.5 Hz

Reason — Frequency is the number of complete oscillations completed per unit time,

$\text n = \frac{\text{number of oscillations}}{\text{time}} = \frac{30}{60\ \text s} = 0.5\ \text{Hz}$

$\frac{π}{2}$

Reason — If the displacement is y = a sin ωt, then the velocity is

$\text u = \frac{\text{dy}}{\text{dt}} = \text a ω\cos ω\text t = \text a ω\sin\left(ω\text t + \frac{π}{2}\right)$

Hence the displacement and the velocity are 90°, that is, $\frac{π}{2}$ radian, out of phase.

− ky

Reason — The restoring force is obtained from the potential energy as the negative of its gradient,

$\text F = -\frac{\text{dU}}{\text{dy}} = -\frac{\text d}{\text{dy}}\left(\frac{1}{2}\text k \text y^2\right) = -\frac{1}{2} \times 2\text{ky} = -\text{ky}$

The negative sign shows that the force is directed opposite to the displacement, towards the mean position.

3 : 1

Reason — At a displacement y, the kinetic energy and the potential energy are

$\text K = \frac{1}{2}\text k(\text A^2 - \text y^2) \qquad \text{and} \qquad \text U = \frac{1}{2}\text k \text y^2$

Substituting $\text y = \frac{\text A}{2}$,

$\frac{\text K}{\text U} = \frac{\frac{1}{2}\text k\left(\text A^2 - \frac{\text A^2}{4}\right)}{\frac{1}{2}\text k \times \frac{\text A^2}{4}} = \frac{\frac{3\text A^2}{4}}{\frac{\text A^2}{4}} = 3$

Hence at $\text y = \frac{\text A}{2}$ the kinetic energy is 3 times the potential energy.

1.5 Hz

Reason — In one complete oscillation a particle crosses the mean position twice, once while moving in one direction and again while moving back. Hence the number of oscillations completed in 4 seconds is

$\text{Number of oscillations} = \frac{12}{2} = 6$

Therefore the frequency of oscillation is

$\text n = \frac{\text{number of oscillations}}{\text{time}} = \frac{6}{4\ \text s} = 1.5\ \text{Hz}$

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In S.H.M. the acceleration is α = − ω²y. At the extreme position the displacement has its maximum value y = a, so the acceleration attains its maximum magnitude α_max = ω²a.

Reason (R) is also correct: The acceleration in S.H.M. is directly proportional to the displacement from the mean position and is directed opposite to it, towards the mean position.

Since the acceleration is proportional to the displacement, it must be greatest where the displacement is greatest, that is, at the extreme position. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Adding the potential energy and the kinetic energy of a body in S.H.M.,

$\text E = \frac{1}{2}\text m ω^2 \text y^2 + \frac{1}{2}\text m ω^2(\text a^2 - \text y^2) = \frac{1}{2}\text m ω^2 \text a^2$

The displacement y cancels out, so the total energy is free from the displacement and remains the same during the motion.

Reason (R) is also correct: During oscillation the two energies convert into each other continuously — the potential energy is maximum at the extreme position and the kinetic energy is maximum at the mean position — but their sum stays the same.

The continuous interchange of the two forms of energy, with no loss to friction, is precisely why the total remains constant. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: From $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$, at a given place g is constant, so the periodic-time of a pendulum is directly proportional to the square-root of its effective length l.

Reason (R) is incorrect: The formula for the periodic-time does not contain the mass of the bob, so T is independent of the mass. It is also independent of the amplitude, provided the amplitude is small.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The velocity of a particle in S.H.M. is $\text u = ω\sqrt{\text a^2 - \text y^2}$. At the extreme position y = a, so u = 0.

Reason (R) is also correct: The kinetic energy at displacement y is $\text K = \frac{1}{2}\text m ω^2(\text a^2 - \text y^2)$, which becomes zero at y = a. At the extreme position the whole of the energy is in the form of potential energy.

Since $\text K = \frac{1}{2}\text m \text u^2$, the kinetic energy being zero requires the velocity also to be zero. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: From $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$, doubling the mass with the same spring gives $\text T' = \sqrt2\text T$, so the time-period increases.

Reason (R) is also correct: For a given spring, k is constant, so $\text T \propto \sqrt{\text m}$, that is, the time period is directly proportional to the square-root of the mass.

This proportionality is exactly what makes the time-period increase when the mass is doubled. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: For small angular displacement θ of the bob, sin θ ≈ θ = $\frac{\text x}{\text l}$, so the restoring force is

$\text F = -\text{mg}\sin θ = -\left(\frac{\text{mg}}{\text l}\right)\text x$

which is directly proportional to the displacement x.

Reason (R) is also correct: The above expression also contains g, so the restoring force is proportional to the acceleration due to gravity.

However, the proportionality of F to displacement follows from the small-angle approximation sin θ ≈ θ, not from the presence of g. The Reason states a separate fact and does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: From $\text u = ω\sqrt{\text a^2 - \text y^2}$, the speed is greatest at y = 0, that is, at the equilibrium position, where u_max = aω.

Reason (R) is incorrect: The potential energy $\text U = \frac{1}{2}\text m ω^2 \text y^2$ is zero at the equilibrium position. It is the kinetic energy, and not the potential energy, that is maximum there.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a body of mass m hanging from a spring, the weight mg is balanced by the restoring force kl at the equilibrium position, so the net force on the body when displaced through y is F″ = − ky. Gravity therefore drops out of the equation of motion, and $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$ contains no g.

Reason (R) is also correct: The only effect of the weight is to stretch the spring by $\text l = \frac{\text{mg}}{\text k}$, which shifts the equilibrium position downward. It does not enter the expression for the time period.

Since gravity only relocates the mean position and cancels out of the restoring force, the oscillation itself is unaffected by it. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For y = a sin ωt, the acceleration is α = − aω² sin ωt = aω² sin (ωt + π), so the displacement and the acceleration differ in phase by π radian.

Reason (R) is also correct: In S.H.M. α = − ω²y, so the acceleration always points opposite to the displacement, towards the mean position.

Two quantities that are always opposite in direction and vary with the same frequency are exactly π radian out of phase. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The periodic-time of a simple pendulum is inversely proportional to the square-root of the acceleration due to gravity, so an increase in g decreases T.

Reason (R) is also correct: The time period of a simple pendulum is indeed given by $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$.

Since g appears in the denominator under the square-root, the formula itself displays the inverse relation stated in the Assertion. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The maximum velocity in S.H.M. is u_max = aω. For a given system ω is fixed, so doubling the amplitude a doubles u_max.

Reason (R) is incorrect: The maximum velocity is directly proportional to the first power of the amplitude, and not to its square.

Therefore, assertion is true but reason is false.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The potential energy $\text U = \frac{1}{2}\text m ω^2 \text y^2$ is zero at the equilibrium position (y = 0), not maximum. It is maximum at the extreme positions.

Reason (R) is correct: The potential energy increases with increase of displacement y and attains its maximum value $\frac{1}{2}\text m ω^2 \text a^2$ in the extreme position of oscillation, where y = a.

Therefore, assertion is false but reason is true.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The frequency of a mass-spring system is $\text n = \frac{1}{\text{2π}}\sqrt{\frac{\text k}{\text m}}$, which contains no g. Hence the frequency is the same on the Moon as on the Earth.

Reason (R) is correct: The acceleration due to gravity on the Moon's surface is $\frac{1}{6}$ th of that on the Earth's surface, so it is indeed less.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: On plotting t along the X-axis and the corresponding values of y along the Y-axis, the displacement curve obtained for S.H.M. is the 'sine curve'.

Reason (R) is also correct: The displacement of a particle in S.H.M. is $\text y = \text a\sin\left(\frac{\text{2π}\text t}{\text T}\right)$, which is a sine function of time; if the motion is timed from an extreme position it becomes a cosine function.

Since the displacement is itself a sine or cosine function of time, its graph must have the sinusoidal shape. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The periodic-time of a pendulum does not depend upon the mass of the bob. If a girl is swinging on a swing and another girl is seated with her on the swing, then the periodic-time of the swing would not change.

Reason (R) is correct: The formula $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$ is the standard expression for the periodic-time of a simple pendulum, and it clearly does not contain m.

Therefore, assertion is false but reason is true.

If assertion and reason both are false.

Explanation

Assertion (A) is incorrect: The frequency of a spring-mass system is

$\text n = \frac{1}{\text{2π}}\sqrt{\frac{\text k}{\text m}}$

which contains k. The frequency therefore depends on the spring constant, increasing as the spring is made stiffer.

Reason (R) is treated as incorrect in the textbook, on the ground that the frequency depends on both the spring constant and the mass, and cannot be described by the mass alone.

Therefore, assertion and reason both are false.

Note: Read strictly, Reason (R) is a correct statement, since for a given spring k is a constant and $\text n \propto \frac{1}{\sqrt{\text m}}$. On that reading the answer would be option 4. The textbook answer key marks option 5, treating (R) as an incomplete and hence unacceptable statement of the dependence of frequency.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a pendulum-clock is taken up to a hill, its periodic-time increases, that is, the clock is slowed down.

Reason (R) is also correct: The value of the acceleration due to gravity g decreases as we go to higher altitudes.

Since $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$, the periodic-time is inversely proportional to the square-root of g, so a smaller g at higher altitudes gives a larger T. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The total energy of a body in S.H.M. is $\text E = \frac{1}{2}\text m ω^2 \text a^2 = 2π^2\text m \text n^2 \text a^2$, so E ∝ a².

Reason (R) is also correct: Adding the two forms of energy,

$\text E = \text K + \text U = \frac{1}{2}\text m ω^2(\text a^2 - \text y^2) + \frac{1}{2}\text m ω^2 \text y^2 = \frac{1}{2}\text m ω^2 \text a^2$

Both K and U involve the square of the displacement, and on adding them the y² terms cancel, leaving an expression containing a² only. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion and reason both are false.

Explanation

Assertion (A) is incorrect: From $\text u = ω\sqrt{\text a^2 - \text y^2}$, the velocity is maximum, and not zero, at the mean position (y = 0). It is zero at the extreme positions.

Reason (R) is also incorrect: The mean position is the position of zero displacement. The displacement is maximum at the extreme positions, where it equals the amplitude.

Therefore, assertion and reason both are false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a simple pendulum executing small oscillations, the restoring force is $\text F = -\left(\frac{\text{mg}}{\text l}\right)\text x$, which is directly proportional to the displacement x and directed towards the mean position.

Reason (R) is also correct: The restoring force is in general F = − mg sin θ. If the angular displacement θ is small and measured in radian, then sin θ ≈ θ = $\frac{\text x}{\text l}$.

It is precisely this small-angle approximation that converts the sine dependence into a linear dependence on x. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: In a damped oscillation, frictional or resistive forces continuously withdraw energy from the system, so the amplitude falls off exponentially with time.

Reason (R) is incorrect: For the usual case of damped harmonic motion, the damping force is directly proportional to the velocity of the particle, and not to the square of the velocity.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: When an external periodic force is applied to a system and the frequency of the external force matches the natural frequency of the system, resonance occurs, resulting in a large increase in amplitude.

Reason (R) is incorrect: In practical systems damping is always present, and it limits the amplitude at resonance. The amplitude becomes large but never infinitely large.

Therefore, assertion is true but reason is false.

If assertion is false but reason is true.

Explanation

Assertion (A) is incorrect: The two energies are equal only at the particular displacement $\text y = \frac{\text a}{\sqrt2}$. At the mean position the energy is wholly kinetic and at the extreme positions it is wholly potential, so the division between them changes continuously as the particle oscillates.

Reason (R) is correct: The total energy $\text E = \frac{1}{2}\text m ω^2 \text a^2$ is free from the displacement and remains constant throughout the motion.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The motion of a simple pendulum is only approximately simple harmonic, and this holds true only for small amplitudes. For large amplitudes the periodic-time itself begins to increase, showing that the motion is no longer simple harmonic.

Reason (R) is also correct: For large angular displacements the approximation sin θ ≈ θ fails, so the restoring force F = − mg sin θ is no longer proportional to the displacement.

Since proportionality of the restoring force to the displacement is the defining condition of S.H.M., its failure at large angles is exactly why the motion ceases to be simple harmonic. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The time period of a spring-mass system is independent of the amplitude of oscillation.

Reason (R) is also correct: The time period is $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$, which contains the mass of the block attached and the force-constant of the spring, and nothing else.

Since the amplitude does not appear anywhere in this expression, the time period cannot depend on it. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Given,

• Equation of S.H.M., y = 0.5 sin 100πt (y in metre, t in second)

The general equation of S.H.M. is

$\text y = \text a\sin ω\text t$

Comparing the given equation with the general equation,

$\text a = 0.5\ \text m \qquad \text{and} \qquad ω = 100π\ \text{rad s}^{-1}$

The frequency is

$\text n = \frac{ω}{\text{2π}} = \frac{100π}{\text{2π}} = 50\ \text{Hz}$

Hence, the amplitude is 0.5 m and the frequency is 50 Hz.

Given,

• Periodic-time of a second's pendulum, T = 2 s
• g = 9.8 m s^-2

The periodic-time of a simple pendulum is

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$

Squaring both sides and solving for l,

$\text l = \frac{\text g \text T^2}{4π^2} = \frac{\text g}{π^2} \qquad [\because \text T = 2\ \text s]$

Substituting the values,

$\text l = \frac{9.8\ \text{m s}^{-2}}{(3.14)^2} = \frac{9.8}{9.8596} = 0.994\ \text m$

Hence, the length of a second's pendulum on the earth is 0.994 m, that is, 99.4 cm (nearly 1 metre).

Given,

• Length of the simple pendulum, $\text l = \frac{\text g}{π^2}$ m

The periodic-time of a simple pendulum is

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$

Substituting the given length,

$\text T = \text{2π}\sqrt{\frac{\text g/π^2}{\text g}} = \text{2π}\sqrt{\frac{1}{π^2}} = \frac{\text{2π}}{π} = 2\ \text s$

Hence, the time-period of the pendulum is 2 s. Such a pendulum is called a second's pendulum.

For a particle executing S.H.M., the acceleration is related to the displacement by

$α = -ω^2 \text y$

Since the angular frequency is related to the frequency n by ω = 2πn, substituting this value gives

$α = -4π^2 \text n^2 \text y$

Hence, the required relation is α = − 4π²n²y, where α is the acceleration, y the displacement from the mean position and n the frequency. The negative sign shows that the acceleration is always directed towards the mean position.

For a particle executing simple harmonic motion of amplitude a and angular frequency ω, at a displacement y from the mean position,

Velocity :

$\text u = ω\sqrt{\text a^2 - \text y^2}$

Acceleration :

$α = -ω^2 \text y$

The velocity is maximum (u_max = aω) at the mean position and zero at the extreme positions, while the acceleration is zero at the mean position and maximum (α_max = ω²a) at the extreme positions.

For a particle executing S.H.M., $ω^2 = \frac{α}{\text y} = \frac{\text{acceleration}}{\text{displacement}}$. Hence the time-period is

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{\text{displacement}}{\text{acceleration}}} = \text{2π}\sqrt{\frac{\text y}{α}}$

where y is the displacement of the particle from the mean position and α is its instantaneous acceleration. Both are taken in magnitude here, since α = − ω²y shows that the acceleration is always directed opposite to the displacement.

The given equation of motion is

$\frac{\text d^2\text x}{\text{dt}^2} = -\text b \text x$

Here $\frac{\text d^2\text x}{\text{dt}^2}$ is the acceleration of the particle. Since b is a constant, the acceleration is directly proportional to the displacement x and is directed opposite to it, towards the mean position. Hence the particle will execute simple harmonic motion.

Comparing the given equation with the differential equation of S.H.M.,

$\frac{\text d^2\text x}{\text{dt}^2} = -ω^2 \text x$

we get

$ω^2 = \text b \quad \Rightarrow \quad ω = \sqrt{\text b}$

The time period of oscillation is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{\sqrt{\text b}}$

Squaring the given relation,

$\text T^2 = \frac{4π^2}{\text g}\text l$

Graph between l and T : From $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$, we have $\text T \propto \sqrt{\text l}$, or l ∝ T². Since l varies as the square of T, the graph between l and T will be a parabola.

Graph between l and T² : Here T² ∝ l with the constant of proportionality $\frac{4π^2}{\text g}$. Since the relation is linear and passes through the origin, the graph between l and T² will be a straight line.

Given,

• Acceleration, $\vec{\text a} = -4π^2 \vec{\text x}$m s^-2

Comparing the given relation with the standard relation for S.H.M.,

$\vec{\text a} = -ω^2 \vec{\text x}$

we get

$ω^2 = 4π^2 \quad \Rightarrow \quad ω = \text{2π}\ \text{rad s}^{-1}$

The time period is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{\text{2π}} = 1\ \text s$

Hence, the time period of the particle is 1 s.

(i) Velocity : The velocity of the particle at displacement y is $\text u = ω\sqrt{\text a^2 - \text y^2}$.

• The velocity is zero at y = ± a, that is, in the position of maximum displacement.
• The velocity is maximum at y = 0, that is, in the equilibrium position, where u_max = aω.

(ii) Acceleration : The acceleration of the particle at displacement y is α = − ω²y.

• The acceleration is zero at y = 0, that is, in the equilibrium position.
• The acceleration is maximum at y = ± a, that is, in the position of maximum displacement, where α_max = ω²a.

A particle is said to execute simple harmonic motion if it moves to and fro along the same straight line about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is directed always towards the mean position.

If y is the displacement of a particle of mass m at any instant, the restoring force acting on it is

$\text F \propto -\text y \quad \text{or} \quad \text F = -\text{ky}$

where k is the proportionality constant called the force constant, defined as restoring force per unit displacement. Its SI unit is N m^-1. The negative sign indicates that the restoring force F is opposite to the displacement y.

If a particle is moving with uniform speed along the circumference of a circle, then the straight-line motion of the foot of the perpendicular drawn from the particle on the diameter of the circle is called 'simple harmonic motion'.

The circle is called the 'reference circle' of the simple harmonic motion. This is the kinematical definition of simple harmonic motion.

The following three conditions should be fulfilled for the linear S.H.M. of a particle :

(i) The motion of the particle should be in a straight line to and fro about a fixed point.

(ii) The restoring force (or acceleration) acting on the particle should always be proportional to the displacement of the particle from that point.

(iii) The force (or acceleration) should always be directed towards that point.

The force which tends to bring a vibrating body from its displaced position to the equilibrium position is called restoring force.

It is always directed towards the mean position and, for simple harmonic motion, is directly proportional to the displacement of the body from the mean position.

Example : In the motion of a simple pendulum, when the bob is displaced through an angle θ, then the component of the force due to gravity, mg sin θ, acts along the arc towards the mean position. This is the restoring force.

Given,

• First mass, m₁ = 1 kg, with frequency n₁ = 4 s^-1
• Second mass, m₂ = 4 kg

The frequency of oscillation of a body of mass m suspended by a spring of force-constant k is

$\text n = \frac{1}{\text{2π}}\sqrt{\frac{\text k}{\text m}}$

The same spring is used in both cases, so k is the same. Hence

$\frac{\text n_1}{\text n_2} = \sqrt{\frac{\text m_2}{\text m_1}}$

Substituting the values,

$\frac{4\ \text s^{-1}}{\text n_2} = \sqrt{\frac{4\ \text{kg}}{1\ \text{kg}}} = 2$

$\text n_2 = \frac{4}{2} = 2\ \text s^{-1}$

Hence, the 4 kg weight will perform 2 oscillations per second.

Given,

• Time-period with mass M, T = 1 s
• Additional mass suspended = 3 kg
• New time-period, T′ = 2 × 1 = 2 s

The time-period of a body of mass m suspended by a spring of force-constant k is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

For the two cases,

$\text T = \text{2π}\sqrt{\frac{\text M}{\text k}} \qquad \text{and} \qquad \text T' = \text{2π}\sqrt{\frac{\text M + 3}{\text k}}$

Dividing the second by the first,

$\frac{\text T'}{\text T} = \sqrt{\frac{\text M + 3}{\text M}}$

Substituting the values,

$\frac{2\ \text s}{1\ \text s} = \sqrt{\frac{\text M + 3}{\text M}} \quad \Rightarrow \quad 4 = \frac{\text M + 3}{\text M}$

$4\text M = \text M + 3 \quad \Rightarrow \quad 3\text M = 3 \quad \Rightarrow \quad \text M = 1\ \text{kg}$

Hence, the value of M is 1 kg.

Second's pendulum : If the periodic-time of a pendulum is 2 seconds, then it is called a 'second pendulum'.

Given,

• Periodic-time, T = 2 s
• g on the earth, g_e = 10 m s^-2
• g on the planet, $\text g_p = \frac{1}{5} \times 10 = 2$ m s^-2

Putting T = 2 s in the formula for the time period of a simple pendulum,

$2 = \text{2π}\sqrt{\frac{\text l}{\text g}} \quad \Rightarrow \quad \text l = \frac{\text g}{π^2}$

Length on the earth :

$\text l_e = \frac{10\ \text{m s}^{-2}}{π^2} = \frac{10}{π^2}\ \text m$

Length on the planet :

$\text l_p = \frac{2\ \text{m s}^{-2}}{π^2} = \frac{10}{5π^2}\ \text m$

Hence, the length of a second's pendulum is $\frac{10}{π^2}$ m on the earth and $\frac{10}{5π^2}$ m on the other planet. Since g is smaller on the planet, the second's pendulum there is shorter.

Given,

• Force-constant of each spring, k = 147 N m^-1
• Mass suspended, m = 5 kg
• g = 9.8 m s^-2

The weight of the suspended body is

$\text{mg} = 5\ \text{kg} \times 9.8\ \text{m s}^{-2} = 49\ \text N$

By Hooke's law, the elongation produced in a spring by a force F is $\text y = \frac{\text F}{\text k}$.

(a) The two springs are in parallel : Here the load is shared equally between the two springs, so the force on each spring is

$\text F = \frac{49\ \text N}{2} = 24.5\ \text N$

$\text y = \frac{24.5\ \text N}{147\ \text{N m}^{-1}} = \frac{1}{6}\ \text m$

(b) The two springs are in series : In a series combination the same force is transmitted to all the springs connected in series. Hence each spring carries the full weight of 49 N,

$\text y = \frac{49\ \text N}{147\ \text{N m}^{-1}} = \frac{1}{3}\ \text m$

so each of the two springs elongates by $\frac{1}{3}$ m.

(c) The body hangs from a single spring : The spring carries the whole weight,

$\text y = \frac{49\ \text N}{147\ \text{N m}^{-1}} = \frac{1}{3}\ \text m$

Hence, the elongations are $\frac{1}{6}$ m in case (a), $\frac{1}{3}$ m in each spring in case (b), and $\frac{1}{3}$ m in case (c).

Given,

• Force-constant of the original spring = k
• Periodic-time of the mass with the original spring = T

Force-constant of each part : The force-constant of a spring is inversely proportional to its length,

$\text k \propto \frac{1}{\text{length of spring}}$

If the spring of length l is cut into three equal pieces, the length of each piece becomes $\frac{\text l}{3}$. Hence, if k′ is the force-constant of each piece,

$\text{k}\text l = \text k'\left(\frac{\text l}{3}\right) \quad \Rightarrow \quad \text k' = 3\text k$

Periodic-time with one piece : The periodic-time of a mass m suspended by a spring of force-constant k is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

With the piece of force-constant 3k,

$\text T' = \text{2π}\sqrt{\frac{\text m}{3\text k}} = \frac{1}{\sqrt3}\left(\text{2π}\sqrt{\frac{\text m}{\text k}}\right) = \frac{\text T}{\sqrt3}$

Hence, the force-constant of each part is 3k and the new periodic-time is $\frac{\text T}{\sqrt3}$.

Given,

• Length of the simple pendulum = l
• Horizontal acceleration of the train = a

When the frame of reference is accelerating horizontally, the bob experiences the vertical gravitational acceleration g downward together with a horizontal acceleration a. These two are mutually perpendicular, so the effective acceleration due to gravity is their resultant,

$\text g_{eff} = \sqrt{\text a^2 + \text g^2} = (\text a^2 + \text g^2)^{\frac{1}{2}}$

Replacing g by g_eff in the formula for the time period of a simple pendulum,

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g_{eff}}} = \text{2π}\sqrt{\frac{\text l}{(\text a^2 + \text g^2)^{\frac{1}{2}}}}$

Hence, the time period of the pendulum is $\text T = \text{2π}\sqrt{\frac{\text l}{(\text a^2 + \text g^2)^{\frac{1}{2}}}}$. Since g_eff > g, the time period is smaller than that in a stationary train.

For a particle in S.H.M. of amplitude a, at displacement y,

$\text K = \frac{1}{2}\text m ω^2(\text a^2 - \text y^2), \qquad \text U = \frac{1}{2}\text m ω^2 \text y^2, \qquad \text E = \frac{1}{2}\text m ω^2 \text a^2$

(i) Kinetic energy is zero : Putting K = 0,

$\text a^2 - \text y^2 = 0 \quad \Rightarrow \quad \text y = \pm\text a$

that is, at the extreme positions.

(ii) Potential energy is zero : Putting U = 0,

$\text y^2 = 0 \quad \Rightarrow \quad \text y = 0$

that is, at the mean position.

(iii) Potential energy is one-fourth of the total energy :

$\frac{1}{2}\text m ω^2 \text y^2 = \frac{1}{4}\left(\frac{1}{2}\text m ω^2 \text a^2\right)$

$\text y^2 = \frac{\text a^2}{4} \quad \Rightarrow \quad \text y = \pm\frac{\text a}{2}$

(iv) Potential energy and kinetic energy are equal :

$\frac{1}{2}\text m ω^2 \text y^2 = \frac{1}{2}\text m ω^2(\text a^2 - \text y^2)$

$\text y^2 = \text a^2 - \text y^2 \quad \Rightarrow \quad 2\text y^2 = \text a^2 \quad \Rightarrow \quad \text y = \pm\frac{\text a}{\sqrt2}$

When the pendulum is at rest in the stationary lift, the net force on it is zero. Hence the tension in its string is

$\text T_{tension} = \text{mg}$

and its period of oscillation is $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$.

(i) When the lift is ascending with acceleration a : The net upward force on the bob is

$\text T_{tension} - \text{mg} = \text{ma} \quad \Rightarrow \quad \text T_{tension} = \text m(\text g + \text a)$

Thus the effective value of g is (g + a), and the time period becomes

$\text T' = \text{2π}\sqrt{\frac{\text l}{\text g + \text a}}$

This is $\sqrt{\frac{\text g}{\text g + \text a}}$ times the period in the stationary lift. That is, the time period decreases.

(ii) When the lift is descending with acceleration a : Proceeding similarly, the effective value of g becomes (g − a), and the time period is

$\text T'' = \text{2π}\sqrt{\frac{\text l}{\text g - \text a}}$

This is $\sqrt{\frac{\text g}{\text g - \text a}}$ times the period in the stationary lift. That is, the time period increases.

Given,

• Radius of the earth, R_e = 6.4 × 10⁶ m
• g = 9.8 m s^-2

For a simple pendulum of infinite length, the restoring force on the bob is directed towards the mean position and its acceleration works out to be $α = -\left(\frac{\text g}{\text R_e}\right)\text x$. Hence the motion is simple harmonic and its periodic-time is

$\text T = \text{2π}\sqrt{\frac{\text R_e}{\text g}}$

Substituting the values,

$\text T = 2 \times 3.14 \times \sqrt{\frac{6.4 \times 10^6\ \text m}{9.8\ \text{m s}^{-2}}} = 6.28 \times \sqrt{6.531 \times 10^5}$

$\text T = 6.28 \times 808.1 = 5076\ \text s$

Converting into minutes,

$\text T = \frac{5076\ \text s}{60} = 84.6\ \text{minutes}$

Hence, the periodic-time of a simple pendulum of infinite length is 84.6 minutes. This is the maximum limit of the time period of a simple pendulum.

(i) Proportional to displacement and acts in the opposite direction

The restoring force in S.H.M. is directly proportional to the displacement from the mean position and acts in the opposite direction,

$\text F = -\text{ky}$

The negative sign shows that the force is always directed towards the mean position.

(ii) It is maximum

The velocity of a particle in S.H.M. is $\text u = ω\sqrt{\text a^2 - \text y^2}$. At the mean position y = 0, so the velocity attains its maximum value u_max = aω. At this instant the whole of the energy of the particle is kinetic.

(iii) Length of the pendulum and gravitational acceleration

The time period of a simple pendulum is $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$. This formula contains only l and g. It does not depend upon the mass of the bob, nor upon the amplitude, provided the amplitude is small.

(iv) At the extreme positions

The potential energy of a body in S.H.M. is $\text U = \frac{1}{2}\text m ω^2 \text y^2$. It increases with increase of displacement y, and hence is maximum in the extreme position of oscillation, where y = a and the displacement is greatest.

(i) Inversely proportional to k

From $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$ for a given mass the time period is inversely proportional to the square-root of the spring constant. A hard spring (large k) therefore gives a smaller periodic-time than a soft spring.

Note: Strictly, $\text T \propto \frac{1}{\sqrt{\text k}}$ that is, the time period is inversely proportional to the square-root of k and not to k itself, so none of the four options states the relation exactly. Option 2 is the intended answer because it alone carries the inverse dependence.

(ii) The time period increases

Since $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$, the time period is directly proportional to the square-root of the mass. Increasing the mass therefore increases the time period, so a heavier body suspended from a spring oscillates more slowly than a lighter body.

(iii) Due to the weight of the mass

When a body of mass m is suspended from the spring, the spring extends until the restoring force balances the weight of the body,

$\text k \text l = \text{mg} \quad \Rightarrow \quad \text l = \frac{\text{mg}}{\text k}$

It is this weight that causes the shift of the equilibrium position in the vertical mass-spring system.

(iv) Hooke's law

The restoring force exerted by the spring is given by Hooke's law, F = − kx, that is, the force is proportional to the displacement and directed opposite to it.

(i) $\sqrt{\text l}$

From $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$, at a given place g is a constant, so the periodic-time of a pendulum is directly proportional to the square-root of its effective length l.

(ii) The time period increases

Since $\text T \propto \sqrt{\text l}$, an increase in the effective length increases the time period. If the length of the pendulum is increased to four times, the periodic-time will become twice.

(iii) When the angular displacement is large

For small angular displacements sin θ ≈ θ, which makes the restoring force proportional to the displacement. For large angular displacements this approximation fails, so the restoring force is no longer proportional to displacement and the motion deviates from simple harmonic motion.

(iv) The acceleration due to gravity

In the formula for the time period of a simple pendulum, g stands for the acceleration due to gravity at the place where the pendulum is oscillating.

(v) The time period will increase

The periodic-time of a simple pendulum is inversely proportional to the square-root of g. On the Moon g is $\frac{1}{6}$ th of its value on the Earth, so the pendulum oscillates more slowly there, leading to a longer time period.

A. (i) Underdamped

In underdamped systems the oscillations occur with gradually decreasing amplitude, since the damping force removes energy slowly enough for the system to continue oscillating.

A. (ii) Overdamped

In overdamped systems the damping force is so large that the system returns to equilibrium without oscillating at all.

A. (iii) Resonance

When the frequency of the external periodic force matches the natural frequency of the system, resonance occurs, causing a large increase in amplitude.

A. (iv) Critically damped

A critically damped system returns to equilibrium in the shortest possible time without oscillating.

B. (i)

Given,

• Time period, T = 2 s
• Displacement required, $\text y = \frac{\text a}{2}$

The displacement equation of S.H.M. is

$\text y = \text a\sin\frac{\text{2π}}{\text T}\text t$

Let t′ be the interval after t = 0 at which $\text y = \frac{\text a}{2}$. Then,

$\frac{\text a}{2} = \text a\sin\left(\frac{\text{2π}}{2} \times \text t'\right)$

$\frac{1}{2} = \sin(π\text t')$

Taking π = 180°,

$30° = 180° \times \text t' \quad \Rightarrow \quad \text t' = \frac{30°}{180°} = \frac{1}{6}\ \text s$

Hence, the displacement becomes half of the amplitude after an interval of $\frac{1}{6}$ s.

B. (ii)

Given,

• Amplitude, a = 10 cm
• Time period, T = 1 s

The maximum value of acceleration in S.H.M. occurs at the extreme position and is

$α_{max} = \text a ω^2 = \text a\left(\frac{\text{2π}}{\text T}\right)^2$

Substituting the values,

$α_{max} = 10\ \text{cm} \times \left(\frac{2 \times 3.14}{1\ \text s}\right)^2 = 10 \times 39.44 = 394.4\ \text{cm s}^{-2}$

Hence, the maximum value of the acceleration of the particle is 394.4 cm s^-2.

B. (iii)

The displacement equation of S.H.M. is

$\text y = \text a\sin ω\text t$

The velocity is the time-derivative of the displacement. Differentiating with respect to t,

$\text v = \frac{\text{dy}}{\text{dt}} = \frac{\text d}{\text{dt}}(\text a\sin ω\text t) = \text a ω\cos ω\text t$

Hence, the equation for the velocity of a particle executing S.H.M. is v = aω cos ωt.

The same result may also be written in terms of the displacement. Using cos ωt = $\sqrt{1 - \sin^2 ω\text t}$,

$\text v = \text a ω\sqrt{1 - \sin^2 ω\text t} = \text a ω\sqrt{1 - \frac{\text y^2}{\text a^2}} = ω\sqrt{\text a^2 - \text y^2}$

This second form gives the magnitude of the velocity at a displacement y, the direction being settled by the sign of cos ωt.

Conditions for simple harmonic motion : The following three conditions should be fulfilled for the linear S.H.M. of a particle :

(i) The motion of the particle should be in a straight line to and fro about a fixed point.

(ii) The restoring force (or acceleration) acting on the particle should always be proportional to the displacement of the particle from that point.

(iii) The force (or acceleration) should always be directed towards that point.

Displacement equation of S.H.M. :

Suppose a particle P is moving with uniform speed along the circumference of a circle of radius a and centre O. This circle is called the reference circle. Let the particle start from the point B and rotate through an angle θ radian in t second.

If the angular velocity of the particle is ω, then

$ω = \frac{θ}{\text t} \quad \Rightarrow \quad θ = ω\text t$

Let N be the foot of the perpendicular drawn from P on the diameter AA′. As P moves along the circumference, N moves in a straight line to and fro about the point O. This straight-line motion of N is simple harmonic motion.

In t second the displacement of N from O is ON. Writing this displacement as y,

$\text y = \text{ON} = \text{OP}\sin \text{NPO}$

But OP = a and ∠NPO = ∠POB = θ = ωt. Therefore,

$\text y = \text a\sin ω\text t \qquad \dots(\text i)$

This is the displacement equation of simple harmonic motion.

If the time is measured from an instant when the particle on the reference circle was at a point P₀ different from B, where ∠P₀OB = φ, then the displacement of N after time t is

$\text y = \text a\sin(ω\text t + φ)$

Here φ is called the initial phase or epoch, and (ωt + φ) is the phase of the motion. This is the general equation of simple harmonic motion.

Since $ω = \frac{\text{2π}}{\text T}$, equation (i) may also be written as

$\text y = \text a\sin\left(\text{2π}\frac{\text t}{\text T}\right)$

Let a particle P move with uniform speed v along the circumference of a circle of radius a, and let N be the foot of the perpendicular drawn from P on the diameter. The straight-line motion of N is simple harmonic motion.

Since P moves on the circumference of a circle, its motion has a centripetal acceleration $\frac{\text v^2}{\text a}$ directed towards the centre O of the circle. This acceleration can be resolved into two components :

• a component $\left(\frac{\text v^2}{\text a}\right)\cos θ$ perpendicular to the direction of motion of N, and
• a component $\left(\frac{\text v^2}{\text a}\right)\sin θ$ parallel to the direction of motion of N.

The component $\left(\frac{\text v^2}{\text a}\right)\sin θ$ is parallel to the line of motion of N, but its direction is opposite to the motion of N, that is, it is directed towards the equilibrium position O. Therefore the acceleration of N is

$α = -\left(\frac{\text v^2}{\text a}\right)\sin θ$

The negative sign indicates that the acceleration of N is directed opposite to its displacement, that is, always towards the equilibrium position O. From the figure,

$\sin θ = \frac{\text{ON}}{\text{OP}} = \frac{\text y}{\text a}$

where y (= ON) is the displacement of N at this instant. Therefore,

$α = -\left(\frac{\text v^2}{\text a^2}\right)\text y$

But v = aω, so $\frac{\text v^2}{\text a^2} = ω^2$, giving

$α = -ω^2 \text y \qquad \dots(\text i)$

This is the relation between the acceleration and the displacement. Since ω² is a constant,

$α \propto -\text y$

Thus, the acceleration of N is proportional to the displacement y and its direction is opposite to the direction of displacement.

Formula for the time period : From equation (i),

$ω^2 = \frac{α}{\text y} = \frac{\text{acceleration}}{\text{displacement}}$

$ω = \sqrt{\frac{\text{acceleration}}{\text{displacement}}}$

The periodic-time of N is $\text T = \frac{\text{2π}}{ω}$. Substituting the value of ω,

$\text T = \text{2π}\sqrt{\frac{\text{displacement}}{\text{acceleration}}}$

This is the formula for the periodic-time of S.H.M.

Second pendulum : If the periodic-time of a pendulum is 2 seconds, then it is called a 'second pendulum'.

Putting T = 2 second in the formula for the time period of a simple pendulum,

$2 = \text{2π}\sqrt{\frac{\text l}{\text g}} \quad \Rightarrow \quad \text l = \frac{\text g}{π^2}$

Taking the value of g as 9.8 metre/second² on the earth's surface, the length of a second pendulum is

$\text l = \frac{9.8}{(3.14)^2} = 0.99\ \text{metre} ≈ 1\ \text{metre}$

Hence the length of a second pendulum on the earth's surface is nearly 1 metre. On the moon's surface the value of g is $\frac{\text g}{6}$, so there the length of a second pendulum will be nearly $\frac{1}{6}$ metre, or 16.6 cm.

Measurement of acceleration due to gravity : A small metallic solid sphere is suspended by a cotton thread from a rigid support, and the distance from the point of suspension to the centre of gravity of the bob is taken as the effective length l of the pendulum.

The periodic-time of a simple pendulum is

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$

Squaring both sides and rearranging,

$\text T^2 = \frac{4π^2 \text l}{\text g} \quad \Rightarrow \quad \text g = \frac{4π^2 \text l}{\text T^2}$

The bob is displaced slightly to one side from its mean position and released, so that it oscillates through a small amplitude. The time for a large number of oscillations, say 20, is measured with a stop watch, and the periodic-time T is obtained by dividing this time by the number of oscillations. The experiment is repeated for several different lengths l.

Substituting the measured values of l and T in the above expression, the value of g at that place is calculated.

Alternatively, a graph is plotted between l and T². Since $\text T^2 = \frac{4π^2}{\text g}\text l$, this graph is a straight line passing through the origin. From its slope $\frac{\text T^2}{\text l}$,

$\text g = \frac{4π^2}{\text{slope}}$

which gives a more accurate value of the acceleration due to gravity.

When a body oscillates in simple harmonic motion, it is acted upon by a restoring force which tends to bring it in the equilibrium position. Due to this force there is potential energy in the body. Also, as the body is in motion it has kinetic energy. During oscillation of the body the two energies convert into each other, but their sum remains constant, taking friction negligible.

Potential energy : Let m be the mass of the oscillating body and y its displacement from the equilibrium position at any instant t. Then the displacement equation of S.H.M. is

$\text y = \text a\sin ω\text t$

where a is the amplitude of oscillation and ω the angular velocity of the body. The linear acceleration of the body is

$α = \frac{\text{du}}{\text{dt}} = -\text a ω^2\sin ω\text t = -ω^2 \text y$

The restoring force acting on the body at this instant is

$\text F = \text{mass} \times \text{acceleration} = \text m α = -\text m ω^2 \text y$

If the body undergoes a further infinitesimally small displacement dy, the work done against the restoring force is

$\text{dW} = (-\text F)\text{dy} = \text m ω^2 \text y\ \text{dy}$

The total work done for displacing the body through y is obtained by integrating the right hand side between the limits y = 0 to y = y,

$\text W = \int_0^{\text y} \text m ω^2 \text y\ \text{dy} = \text m ω^2\left[\frac{\text y^2}{2}\right]_0^{\text y} = \frac{1}{2}\text m ω^2 \text y^2$

This work done on the body appears as the potential energy U of the body at that instant. Thus,

$\text U = \frac{1}{2}\text m ω^2 \text y^2 \qquad \dots(\text i)$

Clearly, the potential energy of the body increases with increase of displacement y, and in the extreme position of oscillation (where y = a) it is maximum.

Kinetic energy : When the displacement of the body in S.H.M. is y, its velocity is

$\text u = ω\sqrt{\text a^2 - \text y^2}$

Hence the instantaneous kinetic energy of the body is

$\text K = \frac{1}{2}\text m \text u^2 = \frac{1}{2}\text m ω^2(\text a^2 - \text y^2) \qquad \dots(\text{ii})$

Clearly, the kinetic energy of the body decreases with increase of displacement y, and in the equilibrium position (where y = 0) it is maximum.

Total energy : On adding equations (i) and (ii), the total energy of the body is

$\text E = \text U + \text K = \frac{1}{2}\text m ω^2 \text y^2 + \frac{1}{2}\text m ω^2(\text a^2 - \text y^2)$

$\text E = \frac{1}{2}\text m ω^2 \text a^2 \qquad \dots(\text{iii})$

It is evident from this that the total energy of the body is free from the displacement y, that is, it remains the same during the motion of the body.

We know that ω = 2πn, where n is the frequency. Substituting this in equation (iii),

$\text E = \frac{1}{2}\text m (2π\text n)^2 \text a^2 = 2π^2 \text m \text n^2 \text a^2$

Hence, in S.H.M. the total energy of a particle is directly proportional to the square of the amplitude (a²) and to the square of the frequency (n²).

Suppose a light spring, whose normal length is L, is hanging from a rigid support A. When a body of mass m is suspended from its lower end, then due to the weight of the body the length of the spring is extended, say, by l. The spring, due to its elasticity, exerts a restoring force F on the body. By Hooke's law,

$\text F = -\text k \text l$

where k is the force-constant of the spring. The negative sign indicates that the force F is directed opposite (upwards) to the weight of the body.

The other force acting on the body is its weight mg. Since at this instant the body has no acceleration, the resultant force acting on it must be zero. That is,

$\text F + \text{mg} = 0$

$-\text k \text l + \text{mg} = 0 \quad \Rightarrow \quad \text{mg} = \text k \text l \qquad \dots(\text i)$

If we now pull the body slightly downwards and release it, then it oscillates up and down. Suppose at any instant during oscillation the body is at a distance y below the equilibrium position. At this instant the total extension in the length of the spring is (l + y). Hence the force exerted on the body by the spring is

$\text F' = -\text k(\text l + \text y) = -\text k \text l - \text k \text y$

The weight of the body is still mg. Hence the resultant force acting on the body is

$\text F'' = \text F' + \text{mg} = (-\text k \text l - \text k \text y) + \text{mg}$

But from equation (i), kl = mg. Therefore,

$\text F'' = -\text k \text y$

According to Newton's law of motion, the resultant force acting on a body is equal to the product of the mass and the acceleration of the body. Hence, if the acceleration of the body is α, then

$\text F'' = \text m α = -\text k \text y$

$α = -\left(\frac{\text k}{\text m}\right)\text y \qquad \dots(\text{ii})$

In this equation $\frac{\text k}{\text m}$ is a constant, as the mass m of the body is constant and k is also constant for the spring. Hence,

$α \propto -\text y$

Thus the acceleration α is directly proportional to the displacement y of the body and its direction is opposite to the displacement. Hence the motion of the body is simple harmonic. Its periodic-time is

$\text T = \text{2π}\sqrt{\frac{\text{displacement}}{\text{acceleration}}}$

From equation (ii),

$\frac{\text{displacement}(\text y)}{\text{acceleration}(α)} = \frac{\text m}{\text k}$

Therefore,

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

Hence proved.

Let the normal length of the spiral spring hanging from a rigid support be L. When the mass m is suspended from its lower end, the spring is extended by l. By Hooke's law the restoring force is F = − kl, where k is the force-constant of the spring. Since the body has no acceleration in this position,

$\text{mg} = \text k \text l \qquad \dots(\text i)$

Now the mass is pulled slightly downwards and released. Suppose at any instant during oscillation the mass is at a distance y below the equilibrium position. The total extension of the spring is then (l + y), and the force exerted by the spring on the mass is

$\text F' = -\text k(\text l + \text y) = -\text k \text l - \text k \text y$

The weight of the body is still mg, so the resultant force on it is

$\text F'' = \text F' + \text{mg} = -\text k \text l - \text k \text y + \text{mg}$

Using equation (i), kl = mg, so

$\text F'' = -\text k \text y$

By Newton's law of motion, F″ = mα, where α is the acceleration of the body. Therefore,

$α = -\left(\frac{\text k}{\text m}\right)\text y$

Here $\frac{\text k}{\text m}$ is a constant, so α ∝ − y, that is, the acceleration is directly proportional to the displacement and is directed opposite to it. Hence the up and down motion of the mass is simple harmonic, and its periodic-time is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

Numerical part

Given,

• Mass, m = 500 g = 0.5 kg
• Force-constant of the spring, k = 50 N m^-1

Substituting the values,

$\text T = 2 \times 3.14 \times \sqrt{\frac{0.5\ \text{kg}}{50\ \text{N m}^{-1}}} = 6.28 \times \sqrt{0.01} = 6.28 \times 0.1$

$\text T = 0.628\ \text s$

Hence, the time period of the spring's vibrations is 0.628 s.

Let O be the centre of the earth and AB the tunnel across it. Suppose at any instant a body D of mass m dropped into the tunnel is at a distance y from the centre C of the tunnel, and at a distance r from the centre O of the earth.

At this instant a gravity-force mg′ directed towards the centre O of the earth will act on the body D, where g′ is the acceleration due to gravity inside the earth at the position of the body. The component of this force directed towards the centre C of the tunnel is mg′ cos θ, and this is the restoring force,

$\text F = -\text m \text g'\cos θ \qquad \dots(\text i)$

The negative sign shows that the force F is directed opposite to the displacement, that is, towards the centre C.

The acceleration due to gravity at the location of D is

$\text g' = \frac{\text G \text M'}{\text r^2}$

where M′ is the mass of the sphere of radius r. If ρ is the density of the earth, then

$\text g' = \frac{\text G\left(\frac{4}{3}π\text r^3 ρ\right)}{\text r^2} = \frac{4}{3}\text G π \text r ρ$

Substituting this value in equation (i),

$\text F = -\text m\left(\frac{4}{3}\text G π \text r ρ\right)\cos θ$

Therefore the acceleration of the body is

$\frac{\text F}{\text m} = -\frac{4}{3}\text G π \text r ρ\cos θ$

From the figure, $\cos θ = \frac{\text y}{\text r}$, so

$\frac{\text F}{\text m} = -\frac{4}{3}\text G π ρ \text y = -ω^2 \text y, \qquad \text{where } ω^2 = \frac{4}{3}\text G π ρ$

Since the acceleration is directly proportional to the displacement y and is directed opposite to it, the motion of the body is simple harmonic.

Time period : The period of the body is

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{1}{\frac{4}{3}\text G π ρ}} \qquad \dots(\text{ii})$

At the surface of the earth,

$\text g = \frac{\text G \text M_e}{\text R_e^2} = \frac{\text G\left(\frac{4}{3}π \text R_e^3 ρ\right)}{\text R_e^2} = \frac{4}{3}\text G π \text R_e ρ$

$\frac{4}{3}\text G π ρ = \frac{\text g}{\text R_e}$

Substituting this value in equation (ii),

$\text T = \text{2π}\sqrt{\frac{\text R_e}{\text g}}$

Putting R_e = 6.4 × 10⁶ m and g = 9.8 m s^-2,

$\text T = 2 \times 3.14 \times \sqrt{\frac{6.4 \times 10^6\ \text m}{9.8\ \text{m s}^{-2}}} = 6.28 \times 808.1 = 5075\ \text s$

$\text T = \frac{5075}{60} = 84.6\ \text{minutes}$

Hence, the ball executes simple harmonic motion of time-period 84.6 minutes. The result is independent of the position of the tunnel, so it is the same whether or not the tunnel passes through the centre of the earth.

The two springs are joined end to end, so they form a series combination. Let the displacement of the body of mass m from the equilibrium position be y. As the force-constants of both the springs are different, the extensions in their lengths will also be different. Let the extension in the length of the spring k₁ be y₁ and that of k₂ be y₂. Then,

$\text y = \text y_1 + \text y_2$

The force exerted by the spring k₁ is

$\text F_1 = -\text k_1 \text y_1 \qquad \dots(\text i)$

and that exerted by the spring k₂ is

$\text F_2 = -\text k_2 \text y_2 \qquad \dots(\text{ii})$

Multiplying equation (i) by k₂ and equation (ii) by k₁, and adding,

$\text k_2 \text F_1 + \text k_1 \text F_2 = -\text k_1 \text k_2(\text y_1 + \text y_2) = -\text k_1 \text k_2 \text y$

As both the springs are in series, they apply the same force on the body, that is, F₁ = F₂ = F (say). Then,

$(\text k_1 + \text k_2)\text F = -\text k_1 \text k_2 \text y$

$\text F = -\frac{\text k_1 \text k_2}{\text k_1 + \text k_2}\text y$

If the effective force-constant of the combination is k, then

$\text k = \frac{\text k_1 \text k_2}{\text k_1 + \text k_2} \qquad \text{or} \qquad \frac{1}{\text k} = \frac{1}{\text k_1} + \frac{1}{\text k_2}$

Therefore the periodic-time of the body is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = \text{2π}\sqrt{\frac{\text m(\text k_1 + \text k_2)}{\text k_1 \text k_2}}$

Hence, the period of oscillation of the body is $\text T = \text{2π}\sqrt{\frac{\text m(\text k_1 + \text k_2)}{\text k_1 \text k_2}}$.

Given,

• Displacement equation, y = 0.50 sin (500 t + 0.5), with y in cm and t in second

The general equation of S.H.M. is

$\text y = \text a\sin(ω\text t + φ_0)$

Comparing the given equation with the general equation,

Amplitude :

$\text a = 0.50\ \text{cm}$

Angular frequency :

$ω = 500\ \text{rad s}^{-1}$

Frequency :

$\text n = \frac{ω}{\text{2π}} = \frac{500}{\text{2π}}\ \text s^{-1}$

Initial phase :

$φ_0 = 0.5\ \text{rad}$

Hence, the amplitude is 0.50 cm, the angular frequency is 500 rad s^-1, the frequency is $\frac{500}{\text{2π}}$ s^-1 and the initial phase is 0.5 rad.

Given,

• $\text x = 0.06\cos\left(\frac{π}{2}\text t + \frac{π}{4}\right)$ metre

Comparing with the general equation x = a cos (ωt + φ₀),

$\text a = 0.06\ \text m, \qquad ω = \frac{π}{2}\ \text{rad s}^{-1}, \qquad φ_0 = \frac{π}{4}$

(i) Amplitude :

$\text a = 0.06\ \text m$

(ii) Maximum velocity : In S.H.M. the velocity is maximum at the mean position,

$\text u_{max} = \text a ω = 0.06\ \text m \times \frac{π}{2}\ \text s^{-1} = 0.03π\ \text{m s}^{-1}$

(iii) Maximum acceleration : In S.H.M. the acceleration is maximum at the extreme position,

$α_{max} = \text a ω^2 = 0.06\ \text m \times \left(\frac{π}{2}\ \text s^{-1}\right)^2 = 0.06 \times \frac{π^2}{4} = 0.015π^2\ \text{m s}^{-2}$

(iv) Initial displacement : Putting t = 0 in the given equation,

$\text x = 0.06\cos\frac{π}{4} = 0.06 \times \frac{1}{\sqrt2} = \frac{0.06}{\sqrt2}\ \text m$

Hence, the amplitude is 0.06 m, the maximum velocity is 0.03π m s^-1, the maximum acceleration is 0.015π² m s^-2 and the initial displacement is $\frac{0.06}{\sqrt2}$ m.

Given,

• Mass of the body, m = 0.1 kg
• $\text x = 0.5\cos\left(100\text t + \frac{3π}{4}\right)$ metre

Comparing with x = a cos (ωt + φ₀),

$\text a = 0.5\ \text m, \qquad ω = 100\ \text{rad s}^{-1}, \qquad φ_0 = \frac{3π}{4}$

(i) Frequency of oscillation :

$\text n = \frac{ω}{\text{2π}} = \frac{100}{\text{2π}} = \frac{50}{π}\ \text s^{-1}$

(ii) Initial phase :

$φ_0 = \frac{3π}{4}$

(iii) Maximum velocity :

$\text u_{max} = \text a ω = 0.5\ \text m \times 100\ \text s^{-1} = 50\ \text{m s}^{-1}$

(iv) Maximum acceleration :

$α_{max} = \text a ω^2 = 0.5\ \text m \times (100\ \text s^{-1})^2 = 0.5 \times 10^4 = 5000\ \text{m s}^{-2}$

(v) Total energy :

$\text E = \frac{1}{2}\text m ω^2 \text a^2 = \frac{1}{2} \times 0.1\ \text{kg} \times (100\ \text s^{-1})^2 \times (0.5\ \text m)^2$

$\text E = \frac{1}{2} \times 0.1 \times 10^4 \times 0.25 = 125\ \text J$

Hence, the frequency is $\frac{50}{π}$ s^-1, the initial phase is $\frac{3π}{4}$, the maximum velocity is 50 m s^-1, the maximum acceleration is 5000 m s^-2 and the total energy is 125 J.

Given,

• Amplitude, a = 5 cm = 5 × 10^-2 m
• Time-period, T = 2 s

The angular frequency of the motion is

$ω = \frac{\text{2π}}{\text T} = \frac{\text{2π}}{2\ \text s} = π\ \text{rad s}^{-1}$

The maximum acceleration of a particle in S.H.M. occurs at the extreme position and is

$α_{max} = \text a ω^2$

Substituting the values,

$α_{max} = (5 \times 10^{-2}\ \text m) \times (π\ \text s^{-1})^2 = 5 \times 10^{-2}π^2\ \text{m s}^{-2}$

Hence, the maximum value of the acceleration of the particle is 5 × 10^-2π² m s^-2.

Given,

• Force-constant of the spring, k = 50 N m^-1
• Mass of the body, m = 500 g = 0.5 kg

The period of oscillation of a body of mass m suspended by a spring of force-constant k is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

Substituting the values,

$\text T = 2 \times 3.14 \times \sqrt{\frac{0.5\ \text{kg}}{50\ \text{N m}^{-1}}} = 6.28 \times \sqrt{0.01} = 6.28 \times 0.1$

$\text T = 0.628\ \text s$

Hence, the period of oscillation is 0.628 s.

Given,

• Mass suspended, m = 1.2 kg
• Extension of the spring, y = 2 cm = 0.02 m
• g = 10 m s^-2

(i) Force-constant of the spring : If a force F produces an increase of y in the length of a spring, then

$\text k = \frac{\text F}{\text y} = \frac{\text{mg}}{\text y}$

Substituting the values,

$\text k = \frac{1.2\ \text{kg} \times 10\ \text{m s}^{-2}}{0.02\ \text m} = \frac{12\ \text N}{0.02\ \text m} = 600\ \text{N m}^{-1}$

(ii) Time-period of oscillation :

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = 2 \times 3.14 \times \sqrt{\frac{1.2\ \text{kg}}{600\ \text{N m}^{-1}}}$

$\text T = 6.28 \times \sqrt{0.002} = 6.28 \times 0.0447 = 0.28\ \text s$

Hence, the force-constant of the spring is 600 N m^-1 and the time-period of oscillation is 0.28 s.

Given,

• Force-constant of the spring, k = 16 N m^-1
• Mass of the body, m = 1.0 kg
• Amplitude of oscillation, a = 5 cm = 0.05 m

(i) Period of oscillation :

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = 2 \times 3.14 \times \sqrt{\frac{1.0\ \text{kg}}{16\ \text{N m}^{-1}}}$

$\text T = 6.28 \times 0.25 = 1.57\ \text s$

(ii) Maximum potential energy of the spring : The potential energy of the spring is maximum at the extreme position, where the extension equals the amplitude,

$\text U_{max} = \frac{1}{2}\text k \text a^2 = \frac{1}{2} \times 16\ \text{N m}^{-1} \times (0.05\ \text m)^2$

$\text U_{max} = 8 \times 0.0025 = 0.02\ \text J = 2 \times 10^{-2}\ \text J$

Hence, the period of oscillation is 1.57 s and the maximum potential energy of the spring is 2 × 10^-2 J.

Given,

• Mass of the first body, m = 2.0 kg
• Additional mass suspended = 300 g = 0.3 kg
• Further extension produced, y = 2.0 cm = 0.02 m

Force-constant of the spring : The additional weight of 0.3 kg produces a further extension of 0.02 m, so

$\text k = \frac{\text F}{\text y} = \frac{0.3\ \text{kg} \times 9.8\ \text{m s}^{-2}}{0.02\ \text m} = \frac{2.94\ \text N}{0.02\ \text m} = 147\ \text{N m}^{-1}$

Time period of the first body : When the second body is removed, only the 2.0 kg body oscillates,

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = 2 \times 3.14 \times \sqrt{\frac{2.0\ \text{kg}}{147\ \text{N m}^{-1}}}$

$\text T = 6.28 \times \sqrt{0.01361} = 6.28 \times 0.1166 = 0.73\ \text s$

Hence, the time period of the 2.0 kg body is 0.73 s.

Given,

• Extension of the spring, l = 20 cm = 0.20 m
• g = 9.8 m s^-2

In the equilibrium position, the weight of the body is balanced by the restoring force of the spring,

$\text{mg} = \text k \text l \quad \Rightarrow \quad \frac{\text m}{\text k} = \frac{\text l}{\text g}$

Substituting this in the expression for the period of oscillation,

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = \text{2π}\sqrt{\frac{\text l}{\text g}}$

Substituting the values,

$$\begin{aligned}\text T = 2 \times 3.14 \times \sqrt{\frac{0.20\ \text m}{9.8\ \text{m s}^{-2}}} \\[1em] = 6.28 \times \sqrt{0.02041} \\[1em] = 6.28 \times 0.1429 \\[1em]\end{aligned}$$

$\text T = 0.898\ \text s$

Hence, the period of oscillation of the body is 0.898 s.

Given,

• Distance fallen by the freely-falling body, h = 125 m
• Time of fall, t = 5 s
• Length of the pendulum, l = 2.5 m

Value of g at that place : The body falls freely from rest, so using $\text h = \frac{1}{2}\text g \text t^2$,

$125\ \text m = \frac{1}{2} \times \text g \times (5\ \text s)^2 = \frac{25\text g}{2}$

$\text g = \frac{2 \times 125}{25} = 10\ \text{m s}^{-2}$

Period of the pendulum :

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g}} = 2 \times 3.14 \times \sqrt{\frac{2.5\ \text m}{10\ \text{m s}^{-2}}}$

$\text T = 6.28 \times \sqrt{0.25} = 6.28 \times 0.5 = 3.14\ \text s$

Hence, the period of the pendulum at that place is 3.14 s.

Given,

• Mass of the mercury poured = 10 g
• Mass of the test tube = 8 g
• External diameter of the test tube = 2 cm, so radius r = 1 cm = 0.01 m
• Density of water, ρ_l = 1000 kg m^-3

The total mass of the floating system is

$\text m = 10\ \text g + 8\ \text g = 18\ \text g = 0.018\ \text{kg}$

The area of cross-section of the test tube is

$\text A = π\text r^2 = 3.14 \times (0.01\ \text m)^2 = 3.14 \times 10^{-4}\ \text m^2$

Proof that the motion is simple harmonic : By the law of floatation, in the equilibrium position the weight of the water displaced by the immersed part of the tube equals the total weight of the tube.

Suppose the tube is pressed down through a small distance y and left. Then an additional volume Ay of water is displaced, and the extra upthrust provides the restoring force,

$\text F = -(\text A \text y)ρ_l \text g$

The negative sign is taken because F acts in the direction opposite to the displacement of the tube. By Newton's law of motion, if α is the instantaneous acceleration of the tube,

$α = \frac{\text F}{\text m} = -\left(\frac{\text A ρ_l \text g}{\text m}\right)\text y = -ω^2 \text y, \qquad \text{where } ω = \sqrt{\frac{\text A ρ_l \text g}{\text m}}$

Here $\frac{\text A ρ_l \text g}{\text m}$ is a constant, so the acceleration α is proportional to the displacement y and is directed opposite to it. Hence the motion of the test tube is simple harmonic.

Time period :

$\text T = \frac{\text{2π}}{ω} = \text{2π}\sqrt{\frac{\text m}{\text A ρ_l \text g}}$

Substituting the values,

$\text T = 2 \times 3.14 \times \sqrt{\frac{0.018\ \text{kg}}{(3.14 \times 10^{-4}\ \text m^2)(1000\ \text{kg m}^{-3})(9.8\ \text{m s}^{-2})}}$

$$\begin{aligned}\text T = 6.28 \times \sqrt{\frac{0.018}{3.077}} \\[1em] = 6.28 \times \sqrt{5.85 \times 10^{-3}} \\[1em] = 6.28 \times 0.0765 \\[1em]\end{aligned}$$

$\text T = 0.48\ \text s\ \text{(approx.)}$

Hence, the motion of the test tube is simple harmonic with a time period of about 0.48 s.

Given,

• Time period, T = 2 s
• Amplitude, a = 5 cm
• The particle starts from the mean position

Equation of motion : Since the motion starts from the mean position, the initial phase is zero, so

$\text y = \text a\sin ω\text t = \text a\sin\frac{\text{2π}}{\text T}\text t$

Substituting a = 5 cm and T = 2 s,

$\text y = 5\sin\left(\frac{\text{2π}}{2}\text t\right) = 5\sin π\text t\ \text{cm}$

Displacement after 1.5 s :

$$\begin{aligned}\text y = 5\sin(π \times 1.5) \\[1em] = 5\sin\left(\frac{\text{3π}}{2}\right) \\[1em] = 5 \times (-1) \\[1em] = -5\ \text{cm} \\[1em]\end{aligned}$$

Velocity after 1.5 s : Differentiating the displacement equation with respect to time,

$\text u = \frac{\text{dy}}{\text{dt}} = \frac{\text d}{\text{dt}}(5\sin π\text t) = 5π\cos π\text t$

At t = 1.5 s,

$\text u = 5π\cos\left(\frac{\text{3π}}{2}\right) = 5π \times 0 = 0$

Hence, the equation of motion is y = 5 sin πt cm, the displacement after 1.5 s is − 5 cm and the velocity is zero. The particle is at an extreme position at that instant, which is why its velocity vanishes.

Given,

• Mass of the body, m = 0.10 kg
• Amplitude, a = 1.0 m
• Time period, T = 0.20 s

The angular frequency of the motion is

$ω = \frac{\text{2π}}{\text T} = \frac{2 \times 3.14}{0.20\ \text s} = 31.4\ \text{rad s}^{-1}$

Maximum force : The force is maximum at the extreme position, where the acceleration is maximum,

$\text F_{max} = \text m α_{max} = \text m ω^2 \text a$

Substituting the values,

$\text F_{max} = 0.10\ \text{kg} \times (31.4\ \text s^{-1})^2 \times 1.0\ \text m = 0.10 \times 986.0 = 98.6\ \text N$

Force-constant of the spring : For a spring-mass system,

$ω = \sqrt{\frac{\text k}{\text m}} \quad \Rightarrow \quad \text k = \text m ω^2$

$\text k = 0.10\ \text{kg} \times 986.0\ \text s^{-2} = 98.6\ \text{N m}^{-1}$

Hence, the maximum value of the force acting on the body is 98.6 N and the force-constant of the spring is 98.6 N m^-1.

Given,

• Mass of the body, m = 8 kg
• Amplitude, a = 30 cm = 0.30 m
• Force at maximum displacement, F = 60 N
• Displacement at which the quantities are required, y = 12 cm = 0.12 m

Force-constant : At the maximum displacement the restoring force is F = ka, so

$\text k = \frac{\text F}{\text a} = \frac{60\ \text N}{0.30\ \text m} = 200\ \text{N m}^{-1}$

(i) Time period : The angular frequency is

$ω = \sqrt{\frac{\text k}{\text m}} = \sqrt{\frac{200\ \text{N m}^{-1}}{8\ \text{kg}}} = \sqrt{25} = 5\ \text{rad s}^{-1}$

$\text T = \frac{\text{2π}}{ω} = \frac{2 \times 3.14}{5\ \text s^{-1}} = 1.256\ \text s$

(ii) At y = 0.12 m :

Acceleration :

$|α| = ω^2 \text y = (5\ \text s^{-1})^2 \times 0.12\ \text m = 25 \times 0.12 = 3.0\ \text{m s}^{-2}$

Potential energy :

$\text U = \frac{1}{2}\text k \text y^2 = \frac{1}{2} \times 200\ \text{N m}^{-1} \times (0.12\ \text m)^2 = 100 \times 0.0144 = 1.44\ \text J$

Kinetic energy : The total energy of the body is

$\text E = \frac{1}{2}\text k \text a^2 = \frac{1}{2} \times 200 \times (0.30)^2 = 100 \times 0.09 = 9.0\ \text J$

Therefore,

$\text K = \text E - \text U = 9.0\ \text J - 1.44\ \text J = 7.56\ \text J$

Hence, the time period is 1.256 s, and at 12 cm displacement the acceleration is 3.0 m s^-2, the potential energy is 1.44 J and the kinetic energy is 7.56 J.

Given,

• Change in the length of the spring, x = 0.1 m
• Change in the potential energy, U = 0.5 J

The potential energy stored in a spring stretched or compressed through x is

$\text U = \frac{1}{2}\text k \text x^2$

Solving for the force-constant,

$\text k = \frac{2\text U}{\text x^2}$

Substituting the values,

$\text k = \frac{2 \times 0.5\ \text J}{(0.1\ \text m)^2} = \frac{1.0}{0.01} = 100\ \text{N m}^{-1}$

Hence, the force-constant of the spring is 100 N m^-1.

Given,

• Time period with mass m, T₁ = 2 s
• Time period with mass (m + 2), T₂ = 3 s

The time period of a body of mass m suspended by a spring of force-constant k is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$

For the two cases,

$\text T_1 = \text{2π}\sqrt{\frac{\text m}{\text k}} \qquad \text{and} \qquad \text T_2 = \text{2π}\sqrt{\frac{\text m + 2}{\text k}}$

Dividing the second by the first and squaring,

$\frac{\text T_2^2}{\text T_1^2} = \frac{\text m + 2}{\text m}$

Substituting the values,

$\frac{(3\ \text s)^2}{(2\ \text s)^2} = \frac{\text m + 2}{\text m} \quad \Rightarrow \quad \frac{9}{4} = \frac{\text m + 2}{\text m}$

$9\text m = 4\text m + 8 \quad \Rightarrow \quad 5\text m = 8 \quad \Rightarrow \quad \text m = 1.6\ \text{kg}$

Hence, the value of the mass m is 1.6 kg.

Given,

• Mass of the body, m = 1 kg
• Increase in the length of the spring, l = 20 cm = 0.20 m
• g = 9.8 m s^-2

Force-constant of the spring :

$\text k = \frac{\text{mg}}{\text l} = \frac{1\ \text{kg} \times 9.8\ \text{m s}^{-2}}{0.20\ \text m} = 49\ \text{N m}^{-1}$

Period of vertical oscillations :

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = 2 \times 3.14 \times \sqrt{\frac{1\ \text{kg}}{49\ \text{N m}^{-1}}} = 6.28 \times \frac{1}{7} = 0.9\ \text s$

Period on a frictionless horizontal surface : The time period of a spring-mass system depends only on the mass of the block attached and the force-constant of the spring, and not on the acceleration due to gravity. Hence the time period remains the same,

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = 0.9\ \text s$

Hence, the force-constant of the spring is 49 N m^-1, and the time period is 0.9 s both for the vertical oscillations and for the oscillations on the frictionless horizontal surface.

Given,

• Mass suspended, m = 0.2 kg
• Increase in the length of the spring, y = 4.9 cm = 0.049 m
• g = 9.8 m s^-2

Force-constant of the spring :

$\text k = \frac{\text{mg}}{\text y} = \frac{0.2\ \text{kg} \times 9.8\ \text{m s}^{-2}}{0.049\ \text m} = \frac{1.96\ \text N}{0.049\ \text m} = 40\ \text{N m}^{-1}$

(i) Period of oscillation with a 0.4 kg body :

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = 2 \times 3.14 \times \sqrt{\frac{0.4\ \text{kg}}{40\ \text{N m}^{-1}}}$

$\text T = 6.28 \times \sqrt{0.01} = 6.28 \times 0.1 = 0.628\ \text s$

(ii) Change in the accumulated energy : The energy stored in a spring extended through x is $\text U = \frac{1}{2}\text k \text x^2$. Here the extension changes from x₁ = 4.9 cm = 0.049 m to x₂ = 9.8 cm = 0.098 m,

$Δ\text U = \frac{1}{2}\text k(\text x_2^2 - \text x_1^2)$

$Δ\text U = \frac{1}{2} \times 40\ \text{N m}^{-1} \times \left[(0.098\ \text m)^2 - (0.049\ \text m)^2\right]$

$Δ\text U = 20 \times (0.009604 - 0.002401) = 20 \times 0.007203 = 0.144\ \text J$

Hence, the period of oscillation with the 0.4 kg body is 0.628 s and the change in the accumulated energy of the spring is 0.144 J.

Given,

• m₁ = 300 g = 0.3 kg and m₂ = 200 g = 0.2 kg
• Increase in the length of the spring, y = 5 cm = 0.05 m
• g = 10 m s^-2

Force-constant of the spring : With both the masses suspended,

$(\text m_1 + \text m_2)\text g = \text k \text y$

$\text k = \frac{(0.3 + 0.2)\ \text{kg} \times 10\ \text{m s}^{-2}}{0.05\ \text m} = \frac{5\ \text N}{0.05\ \text m} = 100\ \text{N m}^{-1}$

Angular frequency of the 200 g body :

$ω = \sqrt{\frac{\text k}{\text m_2}} = \sqrt{\frac{100\ \text{N m}^{-1}}{0.2\ \text{kg}}} = \sqrt{500} = 10\sqrt5\ \text{rad s}^{-1}$

Amplitude of oscillation : When the 300 g body is removed, the equilibrium position of the spring rises. The shift in the equilibrium position is exactly the extension that the removed weight had produced, and this becomes the amplitude of the resulting oscillation,

$\text a = \frac{\text m_1 \text g}{\text k} = \frac{0.3\ \text{kg} \times 10\ \text{m s}^{-2}}{100\ \text{N m}^{-1}} = \frac{3\ \text N}{100\ \text{N m}^{-1}} = 0.03\ \text m = 3\ \text{cm}$

Hence, the angular frequency is $10\sqrt5$ rad s^-1 and the amplitude of oscillation is 3 cm.

Given,

• Period of oscillation with the original spring, T = 1.5 s

The force-constant of a spring is inversely proportional to its length. Hence, if the spring of force-constant k is cut into three equal parts, the force-constant of each part becomes

$\text k' = 3\text k$

(i) Body suspended by any one part :

$\text T_1 = \text{2π}\sqrt{\frac{\text m}{3\text k}} = \frac{1}{\sqrt3}\left(\text{2π}\sqrt{\frac{\text m}{\text k}}\right) = \frac{\text T}{\sqrt3}$

$\text T_1 = \frac{1.5\ \text s}{1.732} = 0.866\ \text s$

(ii) Body suspended by all the three parts in parallel : For springs in parallel the effective force-constant is the sum of the individual force-constants,

$\text k_p = 3\text k' = 3 \times 3\text k = 9\text k$

$\text T_2 = \text{2π}\sqrt{\frac{\text m}{9\text k}} = \frac{1}{3}\left(\text{2π}\sqrt{\frac{\text m}{\text k}}\right) = \frac{\text T}{3}$

$\text T_2 = \frac{1.5\ \text s}{3} = 0.5\ \text s$

Hence, the new periods are 0.866 s in the first case and 0.5 s in the second case.

Given,

• Length of the pendulum, l = 20 cm = 0.20 m
• Upward acceleration of the lift, a = 3.0 m s^-2
• g = 9.8 m s^-2

When the lift is rising up with an accelerated motion, the effective value of the acceleration due to gravity is increased to

$\text g_{eff} = \text g + \text a = 9.8 + 3.0 = 12.8\ \text{m s}^{-2}$

Hence the time period of the pendulum is

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g + \text a}}$

Substituting the values,

$\text T = 2 \times 3.14 \times \sqrt{\frac{0.20\ \text m}{12.8\ \text{m s}^{-2}}} = 6.28 \times \sqrt{0.015625} = 6.28 \times 0.125$

$\text T = 0.79\ \text s$

Hence, the time period of the pendulum is 0.79 s. It is smaller than the value in a stationary lift, because the effective value of g is increased.

Given,

• Length of the shorter pendulum, l₁ = 100 cm
• Length of the longer pendulum, l₂ = 110.25 cm

The time periods of the two pendulums are

$\text T_1 = \text{2π}\sqrt{\frac{100}{\text g}} \qquad \text{and} \qquad \text T_2 = \text{2π}\sqrt{\frac{110.25}{\text g}}$

Taking the ratio,

$\frac{\text T_1}{\text T_2} = \sqrt{\frac{100}{110.25}} = \frac{10}{10.5} = \frac{20}{21}$

Since the longer pendulum has the larger time period, it completes fewer oscillations in the same time. For the two to be in phase again, if the larger pendulum makes n oscillations then the smaller one must make (n + 1) oscillations in the same time. Hence,

$\text n \text T_2 = (\text n + 1)\text T_1$

$\frac{\text n + 1}{\text n} = \frac{\text T_2}{\text T_1} = \frac{21}{20}$

$20\text n + 20 = 21\text n \quad \Rightarrow \quad \text n = 20$

Hence, the two pendulums will again be in phase after 20 oscillations of the larger pendulum, that is, after 21 oscillations of the smaller one.

Given,

• Time period, T = 0.05 s
• Amplitude, a = 4 cm = 0.04 m

The angular frequency of the motion is

$ω = \frac{\text{2π}}{\text T} = \frac{\text{2π}}{0.05\ \text s} = 40π\ \text{rad s}^{-1}$

Maximum velocity : The velocity is maximum at the mean position,

$\text u_{max} = \text a ω = 0.04\ \text m \times 40π\ \text s^{-1} = 1.6π\ \text{m s}^{-1}$

Maximum acceleration : The acceleration is maximum at the extreme position,

$α_{max} = \text a ω^2 = 0.04\ \text m \times (40π\ \text s^{-1})^2 = 0.04 \times 1600π^2 = 64π^2\ \text{m s}^{-2}$

Hence, the maximum velocity is 1.6π m s^-1 and the maximum acceleration is 64π² m s^-2.

Given,

• Displacement, y = 4 cm
• Acceleration, $α = \frac{π^2}{4}$ cm s^-2

In S.H.M. the magnitude of the acceleration at displacement y is

$α = ω^2 \text y \quad \Rightarrow \quad ω^2 = \frac{α}{\text y}$

Substituting the values,

$ω^2 = \frac{\frac{π^2}{4}\ \text{cm s}^{-2}}{4\ \text{cm}} = \frac{π^2}{16}\ \text s^{-2}$

$ω = \frac{π}{4}\ \text{rad s}^{-1}$

The time period is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{\frac{π}{4}} = \text{2π} \times \frac{4}{π} = 8\ \text s$

Hence, the time period of the particle is 8 s.

Given,

• Maximum velocity, u_max = 10 m s^-1
• Amplitude, a = 2.5 m

The maximum velocity of a body in S.H.M. occurs at the mean position and is

$\text u_{max} = \text a ω$

Solving for the angular velocity,

$ω = \frac{\text u_{max}}{\text a} = \frac{10\ \text{m s}^{-1}}{2.5\ \text m} = 4\ \text{rad s}^{-1}$

Hence, the angular velocity of the body is 4 radian/second.

Given,

• Maximum velocity, u_max = aω = 1.0 m s^-1
• Maximum acceleration, α_max = aω² = 1.57 m s^-2

Dividing the second relation by the first, the amplitude a cancels out,

$\frac{α_{max}}{\text u_{max}} = \frac{\text a ω^2}{\text a ω} = ω$

$ω = \frac{1.57\ \text{m s}^{-2}}{1.0\ \text{m s}^{-1}} = 1.57\ \text{rad s}^{-1}$

The periodic-time is

$\text T = \frac{\text{2π}}{ω} = \frac{2 \times 3.14}{1.57\ \text s^{-1}} = \frac{6.28}{1.57} = 4\ \text s$

Hence, the periodic-time of the particle is 4 s.

Given,

• Mass of the body, m = 0.50 kg
• Time period, T = 0.1 s
• Amplitude, a = 10 cm = 0.10 m
• Displacement, y = 5 cm = 0.05 m

The angular frequency of the motion is

$ω = \frac{\text{2π}}{\text T} = \frac{2 \times 3.14}{0.1\ \text s} = 62.8\ \text{rad s}^{-1}$

$ω^2 = (62.8)^2 = 3943.8\ \text s^{-2}$

Acceleration :

$|α| = ω^2 \text y = 3943.8\ \text s^{-2} \times 0.05\ \text m = 197.2\ \text{m s}^{-2}$

Force acting on the body :

$\text F = \text m α = 0.50\ \text{kg} \times 197.2\ \text{m s}^{-2} = 98.6\ \text N$

Potential energy :

$\text U = \frac{1}{2}\text m ω^2 \text y^2 = \frac{1}{2} \times 0.50\ \text{kg} \times 3943.8\ \text s^{-2} \times (0.05\ \text m)^2$

$\text U = 0.25 \times 3943.8 \times 0.0025 = 2.465\ \text J$

Hence, the acceleration is 197.2 m s^-2, the force acting on the body is 98.6 N and its potential energy is 2.465 J.

Given,

• Mass of the body, m = 8 kg
• Amplitude, a = 0.3 m
• Time period, T = 0.6 s

The angular frequency of the motion is

$ω = \frac{\text{2π}}{\text T} = \frac{2 \times 3.14}{0.6\ \text s} = 10.47\ \text{rad s}^{-1}$

$ω^2 = (10.47)^2 = 109.6\ \text s^{-2}$

Maximum kinetic energy : The kinetic energy is maximum at the mean position,

$\text K_{max} = \frac{1}{2}\text m ω^2 \text a^2 = \frac{1}{2} \times 8\ \text{kg} \times 109.6\ \text s^{-2} \times (0.3\ \text m)^2$

$\text K_{max} = 4 \times 109.6 \times 0.09 = 39.44\ \text J$

Force constant of the spring :

$\text k = \text m ω^2 = 8\ \text{kg} \times 109.6\ \text s^{-2} = 876.40\ \text{N m}^{-1}$

Hence, the maximum kinetic energy of the body is 39.44 J and the force constant of the spring is 876.40 N m^-1.

Given,

• Increase in the length of the spring, l = 0.1 m
• Mass suspended, m = 2.0 kg
• g = 9.8 m s^-2

In the equilibrium position the weight of the load is balanced by the restoring force of the spring,

$\text{mg} = \text k \text l \quad \Rightarrow \quad \frac{\text m}{\text k} = \frac{\text l}{\text g}$

Hence the time period of the up and down oscillations is

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = \text{2π}\sqrt{\frac{\text l}{\text g}}$

Substituting the values,

$\text T = 2 \times 3.14 \times \sqrt{\frac{0.1\ \text m}{9.8\ \text{m s}^{-2}}} = 6.28 \times \sqrt{0.010204} = 6.28 \times 0.101$

$\text T = 0.634\ \text s$

Hence, the time period of the up and down oscillations is 0.634 s.

Given,

• Mass of the body, m = 10 kg
• Force-constant of the spring, k = 250 N m^-1
• Amplitude, a = 30 cm = 0.30 m

(i) Time period :

$\text T = \text{2π}\sqrt{\frac{\text m}{\text k}} = 2 \times 3.14 \times \sqrt{\frac{10\ \text{kg}}{250\ \text{N m}^{-1}}}$

$\text T = 6.28 \times \sqrt{0.04} = 6.28 \times 0.2 = 1.256\ \text s$

(ii) Speed at the mean position : The angular frequency is

$ω = \sqrt{\frac{\text k}{\text m}} = \sqrt{\frac{250}{10}} = \sqrt{25} = 5\ \text{rad s}^{-1}$

The speed is maximum at the mean position,

$\text u_{max} = \text a ω = 0.30\ \text m \times 5\ \text s^{-1} = 1.50\ \text{m s}^{-1}$

(iii) Total energy :

$\text E = \frac{1}{2}\text k \text a^2 = \frac{1}{2} \times 250\ \text{N m}^{-1} \times (0.30\ \text m)^2 = 125 \times 0.09 = 11.25\ \text J$

(iv) Time from the extreme position to the mean position : The body travels from an extreme position to the mean position in one quarter of a complete oscillation,

$\text t = \frac{\text T}{4} = \frac{1.256\ \text s}{4} = 0.314\ \text s$

Hence, the time period is 1.256 s, the speed at the mean position is 1.50 m s^-1, the total energy is 11.25 J and the time taken to reach the mean position from the position of maximum displacement is 0.314 s.

Given,

• Frequency, $\text n = \frac{1}{π}$ Hz
• Total energy, E = 10 J
• Maximum speed, u_max = 0.4 m s^-1

The angular frequency of the motion is

$ω = \text{2π}\text n = \text{2π} \times \frac{1}{π} = 2\ \text{rad s}^{-1}$

Amplitude : Since u_max = aω,

$\text a = \frac{\text u_{max}}{ω} = \frac{0.4\ \text{m s}^{-1}}{2\ \text s^{-1}} = 0.2\ \text m$

Force-constant of the spring : The total energy is $\text E = \frac{1}{2}\text k \text a^2$, so

$\text k = \frac{2\text E}{\text a^2} = \frac{2 \times 10\ \text J}{(0.2\ \text m)^2} = \frac{20}{0.04} = 500\ \text{N m}^{-1}$

Maximum potential energy : At the extreme position the whole of the energy of the particle is in the form of potential energy. Hence,

$\text U_{max} = \text E = 10\ \text J$

Hence, the force-constant of the spring is 500 N m^-1 and the maximum potential energy of the spring during the motion is 10 J.

Given,

• Mass of the body, m = 0.5 kg
• Extension of the spring, l = 7 cm = 0.07 m
• Velocity given at the mean position, u_max = 40 cm s^-1 = 0.40 m s^-1
• g = 9.8 m s^-2

Force-constant of the spring :

$\text k = \frac{\text{mg}}{\text l} = \frac{0.5\ \text{kg} \times 9.8\ \text{m s}^{-2}}{0.07\ \text m} = \frac{4.9\ \text N}{0.07\ \text m} = 70\ \text{N m}^{-1}$

Angular velocity :

$ω = \sqrt{\frac{\text k}{\text m}} = \sqrt{\frac{70\ \text{N m}^{-1}}{0.5\ \text{kg}}} = \sqrt{140} = 11.8\ \text{rad s}^{-1}$

Amplitude : The body is given the velocity at the mean position, where the speed is maximum. Since u_max = aω,

$\text a = \frac{\text u_{max}}{ω} = \frac{0.40\ \text{m s}^{-1}}{11.8\ \text s^{-1}} = 0.034\ \text m = 3.4\ \text{cm}$

Hence, the angular velocity is 11.8 rad s^-1 and the amplitude of the resulting oscillatory motion is 3.4 cm.

Given,

• Periodic-time of a second's pendulum, T = 2 s
• g on the earth, g_e = 9.8 m s^-2
• g on the moon, $\text g_m = \frac{\text g_e}{6}$

The periodic-time of a simple pendulum is

$\text T = \text{2π}\sqrt{\frac{\text l}{\text g}} \quad \Rightarrow \quad \text l = \frac{\text g \text T^2}{4π^2} = \frac{\text g}{π^2} \qquad [\because \text T = 2\ \text s]$

Length on the earth :

$\text l_e = \frac{9.8\ \text{m s}^{-2}}{(3.14)^2} = \frac{9.8}{9.8596} = 0.9939\ \text m = 99.39\ \text{cm}$

Length on the moon : Since $\text l \propto \text g$ for a fixed time period,

$\text l_m = \frac{\text l_e}{6} = \frac{99.39\ \text{cm}}{6} = 16.56\ \text{cm}$

Hence, the length of a second's pendulum is 99.39 cm on the earth and 16.56 cm on the moon.

Given,

• Time period of a second's pendulum on the earth, T = 2 s
• g on the earth's surface, g = 9.8 m s^-2
• g at the height, g′ = 4.36 m s^-2

The length of the pendulum remains unchanged, so from $\text T = \text{2π}\sqrt{\frac{\text l}{\text g}}$,

$\frac{\text T'}{\text T} = \sqrt{\frac{\text g}{\text g'}}$

$\text T' = \text T\sqrt{\frac{\text g}{\text g'}}$

Substituting the values,

$\text T' = 2\ \text s \times \sqrt{\frac{9.8\ \text{m s}^{-2}}{4.36\ \text{m s}^{-2}}} = 2 \times \sqrt{2.2477} = 2 \times 1.499$

$\text T' = 3\ \text s$

Hence, the new time period of the pendulum is 3 seconds. The periodic-time has increased because the value of g is smaller at that height.

$\frac{\text{4π}}{\sqrt5}$

Reason —

Given,

• Amplitude, a = 3 cm
• Displacement, y = 2 cm
• Magnitude of velocity = magnitude of acceleration

For a particle in S.H.M., the velocity and the magnitude of the linear acceleration at displacement y are

$\text u = ω\sqrt{\text a^2 - \text y^2} \qquad \text{and} \qquad |α| = ω^2 \text y$

Equating the two at y = 2 cm,

$ω\sqrt{\text a^2 - \text y^2} = ω^2 \text y$

$\sqrt{\text a^2 - \text y^2} = ω\text y$

Substituting a = 3 cm and y = 2 cm,

$\sqrt{3^2 - 2^2} = ω \times 2 \quad \Rightarrow \quad \sqrt5 = 2ω$

$ω = \frac{\sqrt5}{2}\ \text{rad s}^{-1}$

The time period is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{\frac{\sqrt5}{2}} = \frac{\text{4π}}{\sqrt5}\ \text s$

π s

Reason —

Given,

• Acceleration, |α| = 20 m s^-2
• Displacement, y = 5 m

In S.H.M. the magnitude of the acceleration at displacement y is

$|α| = ω^2 \text y$

Substituting the values,

$20\ \text{m s}^{-2} = ω^2 \times 5\ \text m$

$ω^2 = 4\ \text s^{-2} \quad \Rightarrow \quad ω = 2\ \text{rad s}^{-1}$

The time period of oscillation is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{2} = π\ \text s$

1 : 11

Reason —

Given,

• Force constant of the original spring = k
• The spring is cut into lengths in the ratio 1 : 2 : 3

The force constant of a spring is inversely proportional to its length,

$\text k \propto \frac{1}{\text l}$

If the original length l is cut in the ratio 1 : 2 : 3, the three parts have lengths $\frac{\text l}{6}$, $\frac{\text l}{3}$ and $\frac{\text l}{2}$. Hence their force constants are

$\text k_1 = 6\text k, \qquad \text k_2 = 3\text k, \qquad \text k_3 = 2\text k$

When connected in series :

$\frac{1}{\text k'} = \frac{1}{\text k_1} + \frac{1}{\text k_2} + \frac{1}{\text k_3} = \frac{1}{6\text k} + \frac{1}{3\text k} + \frac{1}{2\text k}$

$\frac{1}{\text k'} = \frac{1 + 2 + 3}{6\text k} = \frac{6}{6\text k} = \frac{1}{\text k} \quad \Rightarrow \quad \text k' = \text k$

This is expected, since joining the three pieces in series simply restores the original spring.

When connected in parallel :

$\text k'' = \text k_1 + \text k_2 + \text k_3 = 6\text k + 3\text k + 2\text k = 11\text k$

Therefore,

$\text k' : \text k'' = \text k : 11\text k = 1 : 11$

$\frac{\text g}{3}, \text g$

Reason —

Given,

• Mass of block A = 3m and mass of block B = m
• The two blocks are joined by a massless inextensible string and the system hangs from a spring

Before the string is cut : The whole system is in equilibrium, so the spring force balances the total weight,

$\text{kx} = 3\text{mg} + \text{mg} = 4\text{mg}$

and the tension in the string is T = mg.

Immediately after the string is cut : The spring cannot change its extension instantaneously, so the spring force is still 4mg. The tension in the string becomes zero.

For block A : The forces acting on it are the spring force 4mg upward and its weight 3mg downward,

$\text a_\text A = \frac{\text{kx} - 3\text{mg}}{3\text m} = \frac{4\text{mg} - 3\text{mg}}{3\text m} = \frac{\text g}{3}\ (\text{upward})$

For block B : The only force acting on it is its own weight, since the string has been cut,

$\text a_\text B = \frac{\text{mg}}{\text m} = \text g\ (\text{downward})$

Graph (a)

Reason —

Given,

• At t = 0 the particle is at its equilibrium position

The velocity is maximum at the equilibrium position and zero at the extreme position. Hence the motion is described by

$\text x = \text A\sin ω\text t \quad \Rightarrow \quad \text v = \frac{\text{dx}}{\text{dt}} = \text Aω\cos ω\text t$

The kinetic energy at any instant is

$\text K = \frac{1}{2}\text m \text v^2 = \frac{1}{2}\text m \text A^2 ω^2\cos^2 ω\text t$

At t = 0 : cos 0 = 1, so the kinetic energy is maximum.

At $\text t = \frac{\text T}{4}$ : the particle is at the extreme position, and

$\text K = \frac{1}{2}\text m \text A^2 ω^2\cos^2\frac{\text{2π}}{4} = 0$

Since K varies as cos²ωt, the kinetic energy completes two full cycles in one time period T, starting from a maximum at t = 0 and touching zero at $\frac{\text T}{4}$, $\frac{\text{3T}}{4}$ and so on. Only graph (a) shows this behaviour.

7.1 N/m

Reason —

Given,

• Frequency of oscillation, n = 10¹² s^-1
• Molecular weight of silver = 108
• Avogadro number = 6.02 × 10²³ per gram mole

Mass of one silver atom :

$\text m = \frac{108}{6.02 \times 10^{23}}\ \text{gm} = \frac{108}{6.02 \times 10^{23}} \times 10^{-3}\ \text{kg}$

Force constant : The frequency of a body of mass m attached to a spring of force constant k is

$\text n = \frac{1}{\text{2π}}\sqrt{\frac{\text k}{\text m}} \quad \Rightarrow \quad \text k = 4π^2 \text m \text n^2$

Substituting the values,

$\text k = 4π^2 \times \frac{108 \times 10^{-3}}{6.02 \times 10^{23}} \times (10^{12})^2$

$\text k = \frac{4 \times 10 \times 108 \times 10^{-3}}{6.02 \times 10^{23}} \times 10^{24} \qquad [\because π^2 ≈ 10]$

$\text k = \frac{4.32}{6.02} \times 10 = 7.1\ \text{N m}^{-1}$

$\frac{1}{3}$

Reason —

Given,

• $\text x(\text t) = \text A\sin\frac{π\text t}{90}$
• t = 210 s

Potential energy :

$\text U = \frac{1}{2}\text m ω^2 \text x^2 = \frac{1}{2}\text m ω^2 \text A^2\sin^2\left(\frac{π\text t}{90}\right)$

At t = 210 s,

$\frac{π\text t}{90} = \frac{π \times 210}{90} = \frac{\text{7π}}{3} = \text{2π} + \frac{π}{3}$

$\text U = \frac{1}{2}\text m ω^2 \text A^2\sin^2\frac{π}{3} = \frac{1}{2}\text m ω^2 \text A^2 \times \frac{3}{4} = \frac{3}{8}\text m ω^2 \text A^2 \qquad \dots(\text i)$

Kinetic energy :

$\text K = \frac{1}{2}\text m ω^2(\text A^2 - \text x^2) = \frac{1}{2}\text m ω^2 \text A^2\cos^2\left(\frac{π\text t}{90}\right)$

At t = 210 s,

$\text K = \frac{1}{2}\text m ω^2 \text A^2\cos^2\frac{π}{3} = \frac{1}{2}\text m ω^2 \text A^2 \times \frac{1}{4} = \frac{1}{8}\text m ω^2 \text A^2 \qquad \dots(\text{ii})$

Dividing equation (ii) by equation (i),

$\frac{\text K}{\text U} = \frac{\frac{1}{8}\text m ω^2 \text A^2}{\frac{3}{8}\text m ω^2 \text A^2} = \frac{1}{3}$

$\frac{\text A}{\sqrt2}$

Reason —

Given,

• Amplitude of the S.H.M. = A
• Condition : potential energy = kinetic energy

At a displacement x from the mean position,

$\text U = \frac{1}{2}\text m ω^2 \text x^2 \qquad \text{and} \qquad \text K = \frac{1}{2}\text m ω^2(\text A^2 - \text x^2)$

Putting K = U,

$\frac{1}{2}\text m ω^2(\text A^2 - \text x^2) = \frac{1}{2}\text m ω^2 \text x^2$

$\text A^2 - \text x^2 = \text x^2 \quad \Rightarrow \quad \text A^2 = 2\text x^2$

$\frac{\text A^2}{2} = \text x^2 \quad \Rightarrow \quad \text x = \frac{\text A}{\sqrt2}$

10 cm, $\frac{2}{3}$ s

Reason —

Given,

• $\text y = 5\left(\sin 3π\text t + \sqrt3\cos 3π\text t\right)$ cm

Taking 2 common inside the bracket,

$\text y = 5 \times 2\left(\frac{1}{2}\sin 3π\text t + \frac{\sqrt3}{2}\cos 3π\text t\right)$

$\text y = 10\left(\cos\frac{π}{3}\sin 3π\text t + \sin\frac{π}{3}\cos 3π\text t\right)$

Using sin (A + B) = sin A cos B + cos A sin B,

$\text y = 10\sin\left(3π\text t + \frac{π}{3}\right) \qquad \dots(\text i)$

Comparing equation (i) with the general equation of S.H.M., y = A sin (ωt + φ),

$\text A = 10\ \text{cm} \qquad \text{and} \qquad ω = 3π\ \text{rad s}^{-1}$

The time period is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{3π} = \frac{2}{3}\ \text s$

$\frac{\text{8π}}{3}$

Reason —

Given,

• Amplitude, a = 5 cm
• Displacement, y = 4 cm
• Magnitude of velocity = magnitude of acceleration

In S.H.M.,

$\text u = ω\sqrt{\text a^2 - \text y^2} \qquad \text{and} \qquad |α| = ω^2 \text y$

Equating the two at y = 4 cm,

$ω\sqrt{5^2 - 4^2} = ω^2 \times 4$

$\sqrt{25 - 16} = 4ω \quad \Rightarrow \quad 3 = 4ω$

$ω = \frac{3}{4}\ \text{rad s}^{-1}$

The periodic time is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{\frac{3}{4}} = \frac{\text{8π}}{3}\ \text s$

$\frac{1}{\text n}$

Reason —

Given,

• Force constant of the original spring = k
• l₁ = nl₂, where n is an integer

Since the restoring force is F = − kl, the force constant of a spring is inversely proportional to its unstretched length,

$\text k \propto \frac{1}{\text l}$

Hence, for the two pieces,

$\frac{\text k_1}{\text k_2} = \frac{\text l_2}{\text l_1} = \frac{\text l_2}{\text n \text l_2} = \frac{1}{\text n}$

Note: The stem asks for $\frac{\text k_2}{\text k_1}$ but then names the constants in the order "k₁ and k₂", and the printed answer key gives $\frac{1}{\text n}$, which is the value of $\frac{\text k_1}{\text k_2}$. Read exactly as printed, $\frac{\text k_2}{\text k_1} = \text n$, that is, option 1. The ratio intended by the question is $\frac{\text k_1}{\text k_2} = \frac{1}{\text n}$.

10^-5 K

Reason —

Given,

• Force constant of the spring, k = 800 N m^-1
• Mass attached, m₁ = 500 g = 0.5 kg, specific heat c₁ = 400 J kg^-1 K^-1
• Mass of water, m₂ = 1 kg, specific heat c₂ = 4184 J kg^-1 K^-1
• Amplitude of oscillation, x = 2 cm = 0.02 m

When the vibrations of the mass stop completely, the energy stored in the spring is dissipated in the form of heat, due to which the temperature of the water rises. Hence, by the law of conservation of energy,

$\frac{1}{2}\text k \text x^2 = (\text m_1 \text c_1 + \text m_2 \text c_2)Δ\text T$

Solving for the rise in temperature,

$Δ\text T = \frac{\frac{1}{2}\text k \text x^2}{\text m_1 \text c_1 + \text m_2 \text c_2}$

Substituting the values,

$Δ\text T = \frac{\frac{1}{2} \times 800 \times (0.02)^2}{(0.5 \times 400) + (1 \times 4184)} = \frac{0.16}{200 + 4184}$

$Δ\text T = \frac{0.16}{4384} = 3.65 \times 10^{-5}\ \text K$

Hence the order of magnitude of the change in the temperature of water is 10^-5 K.

$\frac{1}{\text{2π}}\sqrt{\frac{6\text k}{\text m}}$

Reason —

Given,

• Spring constant of each spring = k
• Length of the rod = l and its mass = m
• The rod is pivoted at its centre O

When the rod is turned through a small angle θ, each end of the rod is displaced through

$\text x = \frac{\text l}{2}θ$

Each spring then exerts a restoring force kx at a perpendicular distance $\frac{\text l}{2}$ from the pivot. Hence the total restoring torque due to the two springs is

$τ = 2\left(\text{kx} \times \frac{\text l}{2}\right)\cos θ = \text{kxl}\cos θ$

For a small deflection, cos θ ≈ 1 and $\text x = \frac{\text l}{2}θ$, so

$τ = \text k\left(\frac{\text l}{2}θ\right)\text l = \frac{\text k \text l^2}{2}θ$

The moment of inertia of the rod about an axis through its centre and perpendicular to its length is

$\text I = \frac{\text m \text l^2}{12}$

The angular acceleration is therefore

$α = \frac{τ}{\text I} = \frac{\frac{\text k \text l^2}{2}θ}{\frac{\text m \text l^2}{12}} = \frac{6\text k}{\text m}θ$

Since the angular acceleration is proportional to the angular displacement and is directed opposite to it, the motion is angular simple harmonic with

$ω^2 = \frac{6\text k}{\text m}$

Hence the frequency of the resulting oscillation is

$\text n = \frac{ω}{\text{2π}} = \frac{1}{\text{2π}}\sqrt{\frac{6\text k}{\text m}}$

$\frac{\text F}{\sqrt{\text{mk}}}$

Reason —

Given,

• Mass of the block = m
• Spring constant = k
• Constant applied force = F

The speed of the block is maximum at the new equilibrium position, that is, at the point where the restoring force of the spring balances the applied force F. If x is the extension at this position, then

$\text F = \text{kx} \quad \Rightarrow \quad \text x = \frac{\text F}{\text k}$

Since the block starts from rest at the natural length of the spring and executes S.H.M. about this new equilibrium position, the amplitude of the motion is

$\text A = \text x = \frac{\text F}{\text k}$

The maximum speed of a particle in S.H.M. is

$\text v_{max} = \text A ω = \frac{\text F}{\text k}\sqrt{\frac{\text k}{\text m}} \qquad \left[\because ω = \sqrt{\frac{\text k}{\text m}}\right]$

$\text v_{max} = \frac{\text F}{\sqrt{\text{mk}}}$

$4\text T\sqrt{\frac{1}{15}}$

Reason —

Given,

• Period of oscillation in air = T
• Density of the liquid, $ρ_{liquid} = \frac{ρ}{16}$, where ρ is the density of the material of the bob

In air : The effective value of the acceleration due to gravity is g, so

$\text T = \text{2π}\sqrt{\frac{\text L}{\text g}} \qquad \dots(\text i)$

In the liquid : The bob experiences an upthrust equal to the weight of the liquid displaced. If V is the volume of the bob, the net downward force on it is

$\text F_{net} = \text Vρ\text g - \text V\left(\frac{ρ}{16}\right)\text g \qquad \dots(\text{ii})$

Writing this net force in terms of the effective acceleration due to gravity,

$\text F_{net} = \text Vρ\text g_{effective} \qquad \dots(\text{iii})$

From equations (ii) and (iii),

$\text Vρ\text g_{effective} = \text Vρ\text g - \frac{\text Vρ\text g}{16}$

$\text g_{effective} = \text g - \frac{\text g}{16} = \frac{15\text g}{16}$

Hence the new time period is

$\text T' = \text{2π}\sqrt{\frac{\text L}{\frac{15\text g}{16}}} = \sqrt{\frac{16}{15}} \times \text{2π}\sqrt{\frac{\text L}{\text g}}$

$\text T' = 4\text T\sqrt{\frac{1}{15}}$

10^-3 rad/s

Reason —

Given,

• Length of the pendulum, l = 1 m
• Angular frequency of the pendulum, ω = 10 rad s^-1
• Angular frequency of the support, ω₁ = 1 rad s^-1
• Amplitude of the support, A = 10^-2 m

The angular frequency of a simple pendulum is

$ω = \frac{\text{2π}}{\text T} = \sqrt{\frac{\text g}{\text l}} \qquad \dots(\text i)$

Differentiating with respect to g,

$\frac{\text dω}{\text{dg}} = \frac{1}{2\sqrt{\text g}\sqrt{\text l}} \quad \Rightarrow \quad \text dω = \frac{\text{dg}}{2\sqrt{\text{gl}}} \qquad \dots(\text{ii})$

Dividing equation (ii) by equation (i),

$\frac{\text dω}{ω} = \frac{\text{dg}}{2\text g} \quad \Rightarrow \quad \frac{Δω}{ω} = \frac{Δ\text g}{2\text g}$

The change in the effective value of g arises from the oscillation of the support, whose maximum acceleration is $ω_1^2 \text A$. Taking the full variation between the two extremes,

$Δ\text g = 2ω_1^2 \text A$

Substituting this value,

$\frac{Δω}{ω} = \frac{2ω_1^2 \text A}{2\text g} = \frac{ω_1^2 \text A}{\text g} = \frac{(1\ \text{rad s}^{-1})^2 \times 10^{-2}\ \text m}{10\ \text{m s}^{-2}}$

$\frac{Δω}{ω} = 10^{-3}$

$\text K_2 = \frac{\text K_1}{2}$

Reason —

Given,

• The amplitude a is the same in both the cases
• The length of the pendulum in the second case, l₂ = 2l₁

In S.H.M. the maximum speed is u_max = aω, so the maximum kinetic energy is

$\text K_{max} = \frac{1}{2}\text m(\text a ω)^2 = \frac{1}{2}\text m \text a^2 ω^2$

For a simple pendulum the angular frequency is

$ω = \frac{\text{2π}}{\text T} = \frac{\text{2π}}{\text{2π}\sqrt{\frac{\text l}{\text g}}} = \sqrt{\frac{\text g}{\text l}} \quad \Rightarrow \quad ω^2 = \frac{\text g}{\text l}$

Hence, for a fixed amplitude,

$\text K_{max} \propto ω^2 \propto \frac{1}{\text l}$

Therefore,

$\frac{(\text K_{max})_1}{(\text K_{max})_2} = \frac{\text l_2}{\text l_1} = \frac{2\text l_1}{\text l_1} = 2$

$\text K_2 = \frac{\text K_1}{2}$

elliptical

Reason —

Let the displacement of the particle be

$\text x = \text A\sin ω\text t$

Differentiating with respect to time, the velocity is

$\text v = \text Aω\cos ω\text t \quad \Rightarrow \quad \text v = \pmω\sqrt{\text A^2 - \text x^2}$

Squaring and rearranging,

$\text v^2 = ω^2(\text A^2 - \text x^2) \quad \Rightarrow \quad \frac{\text v^2}{ω^2} + \text x^2 = \text A^2$

$\frac{\text v^2}{\text A^2ω^2} + \frac{\text x^2}{\text A^2} = 1$

This is the standard equation of an ellipse with semi-axes A along the displacement axis and Aω along the velocity axis. Hence the velocity-displacement graph of a particle in S.H.M. is an ellipse.

$\frac{\text v}{2}$

Reason —

Given,

• Mass of each block, m = 250 g = 0.25 kg
• Spring constant, k = 2 N m^-1
• Each block is given a velocity v in opposite directions

At the instant of maximum elongation the two blocks are momentarily at rest relative to each other, so the whole of the kinetic energy of the two blocks has been converted into the potential energy stored in the spring. By the law of conservation of energy,

$2 \times \frac{1}{2}\text m \text v^2 = \frac{1}{2}\text k \text x_m^2$

Substituting the values,

$2 \times \frac{1}{2} \times 0.25 \times \text v^2 = \frac{1}{2} \times 2 \times \text x_m^2$

$0.25\text v^2 = \text x_m^2$

$\text x_m = \sqrt{\frac{\text v^2}{4}} = \frac{\text v}{2}$

11

Reason —

Given,

• Length of the longer pendulum, L₁ = 121 cm
• Length of the shorter pendulum, L₂ = 100 cm

The time period of a simple pendulum is $\text T = \text{2π}\sqrt{\frac{\text L}{\text g}}$, so

$\frac{\text T_1}{\text T_2} = \sqrt{\frac{\text L_1}{\text L_2}} = \sqrt{\frac{121}{100}} = \frac{11}{10}$

$10\text T_1 = 11\text T_2$

Thus the time taken by the longer pendulum to complete 10 vibrations is exactly equal to the time taken by the shorter pendulum to complete 11 vibrations. At the end of this interval both pendulums are again at the mean position in the same phase.

Hence, the minimum number of vibrations of the shorter pendulum is 11.

5 m, 2 s

Reason —

Given,

• $\text x = 5\sin\left(π\text t + \frac{π}{3}\right)$ m

Comparing with the general equation of S.H.M., x = A sin (ωt + φ),

$\text A = 5\ \text m \qquad \text{and} \qquad ω = π\ \text{rad s}^{-1}$

The time period of the motion is

$\text T = \frac{\text{2π}}{ω} = \frac{\text{2π}}{π} = 2\ \text s$

Since the displacement is expressed in metre, the amplitude is 5 m and the time period is 2 s.

175

Reason —

Given,

• $\text x_1 = \sqrt7\sin 5\text t$ cm and $\text x_2 = 2\sqrt7\sin\left(5\text t + \frac{π}{3}\right)$ cm
• Phase difference, $θ = \frac{π}{3} = 60°$

Resultant amplitude : For two simple harmonic motions of the same angular frequency, the resultant amplitude is

$\text A = \sqrt{\text A_1^2 + \text A_2^2 + 2\text A_1\text A_2\cos θ}$

Substituting $\text A_1 = \sqrt7$ cm, $\text A_2 = 2\sqrt7$ cm and cos 60° = 0.5,

$\text A = \sqrt{(\sqrt7)^2 + (2\sqrt7)^2 + 2 \times \sqrt7 \times 2\sqrt7 \times 0.5}$

$\text A = \sqrt{7 + 28 + 14} = \sqrt{49} = 7\ \text{cm}$

Maximum acceleration : Since both the motions have ω = 5 rad s^-1, the resultant motion also has ω = 5 rad s^-1. Hence

$α_{max} = \text A ω^2 = 7\ \text{cm} \times (5\ \text s^{-1})^2 = 175\ \text{cm s}^{-2}$

$α_{max} = 175 \times 10^{-2}\ \text{m s}^{-2}$

Comparing with the given form x × 10^-2 m s^-2, the value of x is 175.

Graph (b)

Reason —

The angular frequency of a spring-mass system is

$ω = \sqrt{\frac{\text k}{\text m}}$

where k is a constant and m decreases with time as the sand leaks out.

Effect on ω(t) : Since m appears in the denominator under the square-root, a decrease in m makes ω increase with time. The increase becomes slower as the box empties, so ω(t) rises and then levels off.

Effect on A(t) : The leaking sand carries energy away from the oscillating system, which acts like a dissipative effect. Hence the amplitude A(t) must decrease with time.

Checking the four options against these two requirements :

• (a) ω(t) increases and levels off, but A(t) remains constant. The amplitude cannot stay constant while energy is being lost, so this is wrong.
• (b) ω(t) increases and levels off, and A(t) decreases with time. This is correct.
• (c) ω(t) increases, but A(t) increases with time. The amplitude cannot increase with energy loss, so this is wrong.
• (d) ω(t) decreases with time, which is wrong.

Hence, graph (b) depicts the changes correctly.

$\sqrt{\frac{\text k_1}{\text k_2}}$

Reason —

Given,

• The two point masses are identical, so m_P = m_Q = m
• Their maximum speeds are the same

The maximum speed of a particle in S.H.M. is u_max = Aω. Since the maximum speeds are equal,

$\text A_\text P ω_\text P = \text A_\text Q ω_\text Q$

For a spring-mass system, $ω = \sqrt{\frac{\text k}{\text m}}$. Substituting,

$\text A_\text P\sqrt{\frac{\text k_1}{\text m_\text P}} = \text A_\text Q\sqrt{\frac{\text k_2}{\text m_\text Q}}$

Since the masses are equal, they cancel out,

$\frac{\text A_\text Q}{\text A_\text P} = \sqrt{\frac{\text k_1}{\text k_2}}$

$\sqrt{\frac{2}{3}\left(\frac{\text g}{\text R - \text r} + \frac{\text k}{\text m}\right)}$

Reason —

Given,

• Radius of the disk = r and its mass = m
• Radius of the ring = R, with R > r
• Spring constant = k

Let the centre of the disk be displaced through a small arc x along the periphery, so that the line joining the centres makes a small angle θ with the vertical, where

$θ = \frac{\text x}{\text R - \text r}$

Taking the forces along the direction of motion, with f the force of friction and a the linear acceleration of the centre of mass,

$\text{kx} + \text{mg}\sin θ - \text f = \text{ma}$

For a small displacement, sin θ ≈ θ, so

$\text{kx} + \text{mg}\left(\frac{\text x}{\text R - \text r}\right) - \text f = \text{ma} \qquad \dots(\text i)$

Frictional force : The disk rolls without slipping, so taking torques about its own centre,

$\text f \text r = \text I α = \left(\frac{\text m \text r^2}{2}\right)\left(\frac{\text a}{\text r}\right) \quad \Rightarrow \quad \text f = \frac{\text{ma}}{2}$

Substituting this value in equation (i),

$\text{kx} + \frac{\text{mgx}}{\text R - \text r} - \frac{\text{ma}}{2} = \text{ma}$

$\left(\text k + \frac{\text{mg}}{\text R - \text r}\right)\text x = \frac{3\text{ma}}{2}$

$\text a = \frac{2}{3}\left(\frac{\text k}{\text m} + \frac{\text g}{\text R - \text r}\right)\text x$

Since the acceleration is proportional to the displacement and is directed towards the equilibrium position, the motion is simple harmonic with

$ω = \sqrt{\frac{2}{3}\left(\frac{\text g}{\text R - \text r} + \frac{\text k}{\text m}\right)}$

$\sqrt{\frac{5}{2}}$

Reason —

Given,

• Length of the rod = l, mass = M, hinged at O
• Spring constant of each spring = K

The moment of inertia of a uniform rod about an axis through one end and perpendicular to its length is

$\text I = \frac{\text M \text l^2}{3}$

Fig. 1 : One spring is connected at the mid-point $\left(\frac{\text l}{2}\right)$ and the other at the top end (l). For a small angular displacement θ, the corresponding linear displacements are $\frac{\text l}{2}θ$ and lθ. Hence the equation of angular motion is

$\frac{\text M \text l^2}{3}\left(\frac{\text d^2θ}{\text{dt}^2}\right) + \text K \times \frac{\text l}{2}θ \times \frac{\text l}{2} + \text K \times \text lθ \times \text l = 0$

$\frac{\text d^2θ}{\text{dt}^2} + \frac{15}{4} \cdot \frac{\text K}{\text M}θ = 0$

Comparing with $\frac{\text d^2θ}{\text{dt}^2} = -ω^2θ$,

$ω_1 = \sqrt{\frac{15\text K}{4\text M}}$

Fig. 2 : Both the springs are connected at the mid-point $\left(\frac{\text l}{2}\right)$, so

$\frac{\text M \text l^2}{3}\left(\frac{\text d^2θ}{\text{dt}^2}\right) + 2\text K \times \frac{\text l}{2}θ \times \frac{\text l}{2} = 0$

$\frac{\text d^2θ}{\text{dt}^2} + \frac{3}{2} \cdot \frac{\text K}{\text M}θ = 0$

$ω_2 = \sqrt{\frac{3\text K}{2\text M}}$

Ratio of the frequencies : Since $\text n = \frac{ω}{\text{2π}}$,

$\frac{\text n_1}{\text n_2} = \frac{ω_1}{ω_2} = \sqrt{\frac{15\text K}{4\text M} \times \frac{2\text M}{3\text K}} = \sqrt{\frac{15 \times 2}{4 \times 3}} = \sqrt{\frac{5}{2}}$

Given,

• Mass of the bob, m = 0.1 kg
• Natural length of the spring, l₀ = 0.1 m
• Spring constant when l > l₀, k₁ = 0.009 N m^-1
• Spring constant when l < l₀, k₂ = 0.016 N m^-1

The bob oscillates about the natural length of the spring. During one full oscillation it spends half the period on the stretched side, where the spring constant is k₁, and the other half on the compressed side, where the spring constant is k₂. The angular frequencies for the two halves are

$ω_1 = \sqrt{\frac{\text k_1}{\text m}} \qquad \text{and} \qquad ω_2 = \sqrt{\frac{\text k_2}{\text m}}$

Hence the time period of the full oscillation is the sum of the two half-periods,

$\text T = \frac{1}{2}\left(\text{2π}\sqrt{\frac{\text m}{\text k_1}}\right) + \frac{1}{2}\left(\text{2π}\sqrt{\frac{\text m}{\text k_2}}\right) = π\sqrt{\frac{\text m}{\text k_1}} + π\sqrt{\frac{\text m}{\text k_2}}$

Substituting the values,

$\text T = π\sqrt{\frac{0.1}{0.009}} + π\sqrt{\frac{0.1}{0.016}}$

$\text T = \frac{π}{0.3} + \frac{π}{0.4} = π\left(\frac{1}{0.3} + \frac{1}{0.4}\right)$

$\text T = π\left(\frac{4 + 3}{12}\right) \times 10 = \frac{70π}{12} = 5.83π\ \text s$

Comparing with T = (nπ) s,

$\text n = 5.83$

Hence, the value of n is 5.83.

Note: The question asks for "the integer closest to n". Since n = 5.83, the closest integer is 6. The textbook answer key records the computed value 5.83 rather than the nearest integer.

Given,

• Spring constants of the two springs = k and 2k
• Period of oscillation in figure (a), T_a = 3 s
• Period of oscillation in figure (b), $\text T_b = \sqrt{\text x}$ s

For case (a) : The two springs are joined end to end, so they form a series combination. The effective spring constant is

$\text k_a = \frac{\text k \times 2\text k}{\text k + 2\text k} = \frac{2\text k^2}{3\text k} = \frac{2\text k}{3}$

For case (b) : The two springs support the mass side by side, so they form a parallel combination. The effective spring constant is

$\text k_b = \text k + 2\text k = 3\text k$

Ratio of the time periods : Since $\text T = \text{2π}\sqrt{\frac{\text m}{\text k}}$, for the same mass,

$\frac{\text T_a}{\text T_b} = \sqrt{\frac{\text k_b}{\text k_a}} = \sqrt{\frac{3\text k}{\frac{2\text k}{3}}} = \sqrt{\frac{3\text k \times 3}{2\text k}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt2}$

Substituting T_a = 3 s,

$\frac{3}{\text T_b} = \frac{3}{\sqrt2} \quad \Rightarrow \quad \text T_b = \sqrt2\ \text s$

Comparing with $\text T_b = \sqrt{\text x}$ s,

$\text x = 2$

Hence, the value of x is 2.

Given,

• Amplitude, A = 4 cm
• Velocity at the mean position, u_max = 10 cm s^-1
• Speed at the required position, v = 5 cm s^-1

Angular frequency : The velocity is maximum at the mean position, where u_max = Aω,

$10\ \text{cm s}^{-1} = 4\ \text{cm} \times ω \quad \Rightarrow \quad ω = \frac{10}{4}\ \text{rad s}^{-1}$

Distance from the mean position : At a displacement y the velocity is

$\text v = ω\sqrt{\text A^2 - \text y^2}$

Substituting the values,

$5 = \frac{10}{4}\sqrt{4^2 - \text y^2}$

$\frac{5 \times 4}{10} = \sqrt{16 - \text y^2} \quad \Rightarrow \quad 2 = \sqrt{16 - \text y^2}$

Squaring both sides,

$4 = 16 - \text y^2 \quad \Rightarrow \quad \text y^2 = 12$

$\text y = \sqrt{12}\ \text{cm} = 2\sqrt3\ \text{cm}$

Writing this in the form $\text y = \sqrt{α}$ cm,

$α = 12$

Hence, the value of α is 12.

Note: As printed, the stem reads "is α cm", which would give α = $2\sqrt3$ ≈ 3.46. The printed answer key value of 12 corresponds to the form $\sqrt{α}$ cm, so the stem should read "is $\sqrt{α}$ cm".

Given,

• Mass of the object, m = 50 kg
• $\text y = 8\left[1 + \sin\left(\frac{\text{2π}\text t}{\text T}\right)\right]$ metre
• T = 40π s
• g = 10 m s^-2

Nature of the motion : Expanding the given relation,

$\text y = 8 + 8\sin\left(\frac{\text{2π}\text t}{\text T}\right)$

This is of the form y = y₀ + A sin ωt, so the elevator is performing S.H.M. of amplitude A = 8 m about a mean height of 8 m, with angular frequency

$ω = \frac{\text{2π}}{\text T} = \frac{\text{2π}}{40π} = \frac{1}{20}\ \text{rad s}^{-1}$

Maximum acceleration of the elevator :

$α_{max} = \text A ω^2 = 8\ \text m \times \left(\frac{1}{20}\ \text s^{-1}\right)^2 = \frac{8}{400} = 0.02\ \text{m s}^{-2}$

Maximum variation of weight : The apparent weight of the object varies between m(g − α_max) and m(g + α_max). Hence the maximum variation is

$Δ\text W = \text m(\text g + α_{max}) - \text m(\text g - α_{max}) = 2\text m α_{max}$

Substituting the values,

$Δ\text W = 2 \times 50\ \text{kg} \times 0.02\ \text{m s}^{-2} = 2\ \text N$

Hence, the maximum variation of the object's weight is 2 N.