NCERT Solutions for Class 11 Physics Chapter 14 - Oscillations

14.1. Which of the following examples represent periodic motion?

(a) A swimmer completing one (return) trip from one bank of a river to the other and back.

(b) A freely suspended bar magnet displaced from its N-S direction and released.

(c) A hydrogen molecule rotating about its centre of mass.

(d) An arrow released from a bow.

Ans. (a) It isn’t a periodic motion. A swimmer’s motion is to and fro, although it does not have a defined duration.

(b) It is a periodic motion because a freely suspended magnet, when shifted from the N-S direction and released, oscillates around this position.

(c) The rotation of a hydrogen molecule around its mass centre is periodic. This is due to the fact that when a hydrogen molecule spins about its centre of mass, it returns to the same location after an equal time.

(d) When an arrow is released from a bow, it only moves forward. It does not come back or go in the opposite direction. As a result, this motion is not regular i.e. periodic.

14.2. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) the rotation of earth about its axis.

(b) motion of an oscillating mercury column in a U-tube.

(c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point.

(d) general vibrations of a polyatomic molecule about its equilibrium position.

Ans. (a) The earth’s rotation is not a to-and-fro motion around a fixed point. As a result, it is regular but not Simple harmonic motion.

(b) Simple harmonic motion

(d) A polyatomic molecule’s general vibration around its equilibrium position are periodic, but Simple harmonic motion is not. The inherent frequencies of a polyatomic molecule are numerous. As a result, its vibration is a combination of simple harmonic motions at various frequencies.

14.3. Fig. 14.23 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion)?

Ans. (a) It isn’t a periodic motion because it doesn’t repeat itself after a particular amount of time.

(b) The given graph depicts a periodic motion that repeats itself after every two seconds.

(c) Because the motion is repeated in only one position, the depicted graph does not illustrate periodic motion. The full motion during one period must be repeated successively for a periodic motion.

(d) The given graph depicts a periodic motion that repeats itself after every 2 seconds.

14.4. Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (w is any positive constant) :

(a) sin wt – cos wt

(b) sin³ wt

(c) 3 cos (p/4 – 2wt)

(d) cos wt + cos 3wt + cos 5wt

(e) exp (– w²t²)

(f) 1 + wt + w²t²

Ans. The function will represent a periodic motion, if it is identically repeated after a fixed interval of time and will represent S.H.M if it can be written uniquely in the form of

$$a\space\text{cos}\bigg(\frac{2\pi t}{\text{T}}+\phi\bigg)\text{or a sin}\bigg(\frac{2\pi t}{\text{T}}+\phi\bigg),\text{where T is the time period.}\\\text{(a) sin ωt – cos ωt}\\=\sqrt{2}\bigg[\frac{1}{\sqrt{2}}\text{sin}\omega t-\frac{1}{\sqrt{2}}\text{cos}\omega t\bigg]\\=\sqrt{2}\bigg[\text{sin}\omega t\text{cos}\frac{\pi}{4}-\text{cos}\omega t\text{sin}\frac{\pi}{4}\bigg]\\=\sqrt{2}\text{sin}\bigg(\omega t-\frac{\pi}{4}\bigg)$$

It is a S.H.M. and its period is 2π/ω

$$\text{(b)\space sin}^{3}\omega t=\frac{1}{3}[3\space\text{sin}\omega t-\text{sin}\space3\omega t]$$

Here each term sin ωt and sin 3ωt individually represents S.H.M. But (ii) which is the outcome of the superposition of two S.H.M. will only be periodic but not SHMs. Its time period is 2π/ω.

$$\text{(c)}\space 3\space\text{cos}\bigg(\frac{\pi}{4}-2\omega t\bigg)=3\space\text{cos}\bigg(2\omega t-\frac{\pi}{4}\bigg)\\\lbrack\because(-\theta)=\text{cos}\space\theta\rbrack$$

Clearly, it represents SHM and its time period is 2π/2ω.

(d) cos ωt + cos 3 ωt + cos 5ωt.

It represents the periodic but not S.H.M. Its time period is 2π/ω.

(e) exp (– ω2t2). It is an exponential function which never repeats itself. Therefore, it represents non-periodic motion.

(f) 1 + wt + w²t² also represents a non periodic motion.

14.5. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is

(a) at the end A,

(b) at the end B,

(c) at the mid-point of AB going towards A,

(d) at 2 cm away from B going towards A,

(e) at 3 cm away from A going towards B, and

(f) at 4 cm away from B going towards A.

Ans. (a) Zero, Positive, Positive

(b) Zero, Negative, Negative

(d) Negative, Negative, Negative

(e) Positive, Positive, Positive

(f) Negative, Negative, Negative

Explanation :

(a) and (b) The following diagram depicts the current scenario. The two end points are A and B, with AB = 10 cm and ‘O’ being the path’s halfway.

A particle moves in a linear simple harmonic motion between the two end points. At the extreme point A, the particle is temporarily at rest. As a result, its velocity is zero at this place. Its acceleration is positive because it is travelling along AO. Force is positive in this circumstance since the particle is pointed rightward.

The particle is temporarily at rest at the extreme point B. As a result, at this point, its velocity is zero.

(c) A basic harmonic motion is being performed by the particle. The particle’s mean location is denoted by the letter ‘O.’ Its highest velocity is at the mean position O. Because the particle is moving leftward, the velocity value is negative. At the mean position, the acceleration and force of a particle performing SHM are both zero.

(d) The particle is travelling from end B to point O. This motion is in the opposite direction of the traditional positive direction, which is from A to B. The particle’s velocity, acceleration, and force are all negative as a result.

(e) The particle is travelling from end A to point O. This motion is going from point A to point B, which is the traditional positive direction. As a result, the velocity, acceleration, and force measurements are all positive.

(f) This case is similar to one given in (d).

14.6. Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

(a) α = 0.7x

(b) α = – 200 x²

(c) α = – 10x

(d) α = 100 x³

Ans. In the condition of simple Harmonic Motion, acceleration is directly proportional to negative of displacement of particle. If ‘a’ is acceleration, and x is displacement.

Then, for Simple Harmonic Motion, = – kx where k is constant

(a) a = 0.7x

This is not in the form of a = – kx

Hence, this is not SHM

(b) a = – 200x2
Clearly, it is not SHM

This is in the form of a = – kx

Hence, this is SHM.

(d) a = 100x³

It’s clear it is not SHM.

14.7. The motion of a particle executing simple harmonic motion is described by the displace-ment function,

x(t) = A cos (ωt + Φ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s^–1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Ans. The given displacement function is :

x(t) = A cos (wt + f) ...(i)

t = 0, x(0) = 1 cm.

Also,

ω = π s^–1

∴ 1 = A cos (π × 0 + Φ)

⇒ A cos Φ = 1 ...(ii)

Also, differentiating eqn. (i) w.r.t. ’t’.

$$v=\frac{\text{d}}{\text{dt}}x(t)$$

= – A ω sin (ωt + Φ) ...(iii)

Now at t = 0, v = ω

∴ from eqn. (iii),

ω = – A ω sin (π × 0 + Φ)

or A sin Φ = – 1 ...(iv)

Squaring and adding eqns. (ii) and (iv),

A² cos² Φ + A² sin² Φ

= 1² + 1² or A = 2 cm

Dividing eqns. (ii) and (iv),

$$\frac{\text{A sin}\space\phi}{\text{A cos}\space\phi}=\frac{\normalsize-1}{1}\\\therefore\space\text{tan}\phi=\normalsize-1\\\Rarr\space\phi=\frac{3\pi}{4}\\\text{It instead we use the sine function, i.e.,}\\\text{x = B sin} (\omega t + \alpha),\\\text{Then}\space \text{v}=\frac{d}{dt}\text{B}\omega\text{cos}(\omega t+\alpha)$$

∴ At t = 0, using x = 1

and v = ω,

We get 1 = B sin (ω × 0 + α)

B sin α = 1 ...(v)

and

ω = Bω cos (w × 0 + α)

B cos α = 1 ...(vi)

Dividing (v) by (vi),

$$\text{tan}\space\alpha=1\space\text{or}\space\alpha=\frac{\pi}{4}\text{or}\frac{5\pi}{4}$$

Squaring (v) and (vi), we get

B² sin² α + β² cos²α = 1² + 1²

$$\Rarr\space\text{B}=\sqrt{2}\space\text{cm}.$$

14.8. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ?

Ans. Maximum mass that the scale can read,

M = 50 kg

Maximum displacement of the spring

= Length of the scale,

l = 20 cm

= 0.2 m

Time period,

T = 0.6 s

Maximum force exerted on the spring, F = mg Where,

g = acceleration due to gravity

= 9.8 m/s²

F = 50 × 9.8 = 490 N

Hence, spring constant,

$$k=\text{F/I}\\=\frac{490}{0.2}$$

= 2450 N m^–1

Mass m is suspended from the balance.

$$\text{Time period},\space\text{T}=2\pi\frac{\sqrt{m}}{K}\\\text{Therfore}, m=\bigg(\frac{\text{T}}{2\pi}\bigg)^{2}×k\\=\bigg(\frac{0.6}{2×3.14}\bigg)^{2}×2450$$

= 22.36 kg

Hence, weight of the body

= mg = 22.36 × 9.8 N

On calculation, we get,

= 218.54 N

Therefore, the weight of the body is about 219N.

14.9. A spring having with a spring constant 1200 N m^–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Ans. Given,

Spring constant,

k = 1200 Nm^–1

Mass, m = 3 kg

Displacement, A = 2.0 cm

= 0.02 m

(i) Frequency of oscillation ‘n’ is given by the relation :

$$v=\frac{1}{\text{T}}\\=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\text{Where, T is the time period of oscillation}\\\text{So},\space v=\frac{1}{2×3.14}\sqrt{\frac{1200}{3}}$$

= 3.18 m/s^{– 1}

Therefore, the frequency of oscillations is 3.18 cycles per second.

(ii) Maximum acceleration (a) is given by the relation:

a=ω²A

Where,

$$ω = \text{Angular frequency} =\sqrt{\frac{k}{m}}\\\text{and, A = Maximum displacement}\\\text{Hence},\\a=\bigg(\frac{k}{m}\bigg)\text{A}\\a=\frac{1200×0.02}{3}$$

= 8 ms^–2

Therefore, the maximum acceleration of the mass is 8.0 m/s².

(iii) Maximum velocity,

v_max = Aω

On substituting, we get,

$$v_{\text{max}}=\text{A}\sqrt{\frac{k}{m}}\space\space\space\bigg[\because\omega=\sqrt{\frac{k}{m}}\bigg]\\=0.02×\sqrt{\frac{1200}{3}}$$

= 0.02 × 20

= 0.4 m/s

Therefore, the maximum velocity of the mass is 0.4 m/s

14.10. In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is :

(a) at the mean position,

(b) at the maximum stretched position, and

(c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase ?

Ans. a = 2 cm,

$$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}}\space\text{s}^{\normalsize-1}$$

= 20 s^–1

(a) Since time is measured from mean position, x = a sin ωt = 2 sin 20t

(b) At the maximum stretched position, the body is at the extreme right position. The initial phase is

$$\frac{\pi}{2}$$.

$$\therefore\space x=a\space\text{sin}\bigg(\omega t+\frac{\pi}{2}\bigg)$$

= a cos ωt

= 2 cos 20t

$$\frac{3\pi}{2}$$

$$\therefore\space x=a\space\text{sin}\bigg(\omega t+\frac{3\pi}{2}\bigg)\\=-a\text{cos}\space\omega t\\=-2\text{cos}\space20t$$

14.11. Figures 14.25 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure.

Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case.

Ans. (a) Let A be any point on the circle of reference of the fig. (a) From A, draw AM per- pendicular on x-axis.

If ∠POA = θ, then

∠OAM = θ = ωt

∴ In triangle OAM,

$$\frac{\text{OM}}{\text{OA}}=\text{sin}\space\theta\\\therefore\space\frac{-x}{3}=\text{sin}\space\omega t=\text{sin}\frac{2\pi}{\text{T}}t\\\lbrack x\space \text{is –ve in fourth quadrant}\rbrack\\\therefore\space x=-3\text{sin}\frac{2\pi}{2}t$$

or x = – 3 sin πt

which is the equation of SHM.

(b) Let B be any point on the circle of reference of fig. (b). From B, draw BN perpendicular on x-axis.

Then ∠BON = θ = ωt

∴ In ΔONB,

$$\text{cos}\space\theta=\frac{\text{ON}}{\text{OB}}\\\text{or}\space\text{ON=OB}\text{cos}\theta\\\therefore\space -x=2\text{cos}\omega t\\\Rarr\space x=-2\text{cos}\frac{2\pi}{\text{T}}t\\=-2\space\text{cos}\frac{2\pi}{4}t\\\therefore x=-2\text{cos}\frac{\pi}{2}t,\text{which is equation of SHM.}$$

14.12. Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t = 0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case : (x is in cm and t is in s).

(a) x = – 2 sin (3t + π/3)

(b) x = cos (π/6 – t)

(c) x = 3 sin (2πt + π/4)

(d) x = 2 cos πt

Ans.

$$\text{(a)}\space\text{x = 2 cos}\bigg(3t+\frac{\pi}{3}+\frac{\pi}{2}\bigg)\\\text{Radius of the reference circle, r = amplitude of SHM = 2 cm,}\\\text{At}\space t=0, x=-2\text{sin}\frac{\pi}{3}\\=\frac{-2\sqrt{3}}{2}=-\sqrt{3}\space\text{cm}\\\text{Also},\omega t=3t\\\space\therefore\omega=3\text{rad/s}\\\text{cos}\phi_0=-\frac{\sqrt{3}}{2},\phi_0=150\degree
$$

The reference circle is, thus, as plotted below.

$$\text{(b)}\space x=\text{cos}\bigg(t-\frac{\pi}{6}\bigg)\\\text{Radius of circle,}\\\text{r = amplitude of SHM = 1 cm}\\\text{At}\space t=0,x=\text{cos}\frac{\pi}{6}=\frac{\sqrt{3}}{2}\text{cm}\\\text{Also,}\spaceωt=1t\Rarr\omega=1\text{rad/s}\\\text{cos}\phi_0=\frac{\sqrt{3}}{2},\phi_0=-\frac{\pi}{6}$$

The reference circle is, this as plotted below

$$\text{(c)}\space\text{x = 3 cos}\bigg(2\pi t+\frac{\pi}{4}+\frac{\pi}{2}\bigg)\\\text{Here, radius of reference circle,}\\\text{r = 3 cm and at t = 0, x = 3 sin}\frac{\pi}{4}\\=\frac{3}{\sqrt{2}}\text{cm}\\\omega t=2\pi t\Rarr\omega=2\pi\text{rad/s}\\\text{cos}\phi_{0}=\frac{\frac{3}{\sqrt{2}}}{3}=\frac{1}{\sqrt{2}}$$

Therefore, the reference circle is being shown below.

(d) x = 2 cos πt

Radius of reference circle,

r = 2 cm and at t = 0, x = 2 cm

∴ ωt = πt, or ω = π rad/s

cos Φ₀ = 1, Φ₀ = 0

The reference circle is plotted below.

14.13. Figure 14.26 (a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 14.26 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 14.26(b) is stretched by the same force F.

(a) What is the maximum extension of the spring in the two cases ?

(b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ?

Ans. (a) Let y be the maximum extension produced in the spring in Fig. (a)

Then F = ky (in magnitude)

$$\therefore\space y=\frac{\text{F}}{\text{k}}$$

If fig. (b), the force on one mass acts as the force of reaction due to the force on the other mass. Therefore, each mass behaves
as if it is fixed with respect to other.

Therefore, F = ky(in magnitude)

$$\Rarr\space y=\frac{\text{F}}{\text{k}}$$

If fig. (b), the force on one mass acts as the force of reaction due to the force on the other mass. Therefore, each mass behaves
as if it is fixed with respect to other.

Therefore, F = ky

$$\Rarr\space y=\frac{\text{F}}{\text{k}}$$

(b) In fig. (a) F = – ky

$$\Rarr\space ma=-ky\Rarr a=-\frac{k}{m}y\\\therefore \omega^{2}=\frac{k}{m}i.e,\omega=\sqrt\frac{k}{m}\\ \text{Therefore, period T=}\frac{2\pi}{\omega}\\\text{T}=\sqrt{\frac{m}{k}}$$

In fig. (b), we may consider the centre of the system is O and there are two springs each of length $$\frac{l}{2}$$ attached to the two masses, each m, so that k′ is the spring factor of each of the springs.

Then, K′ = 2k

$$\therefore\space\text{T}=2\pi\sqrt\frac{m}{k'}\\=2\pi\sqrt{\frac{m}{2k}}$$

14.14. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed ?

Ans. Given,

Angular frequency of the piston,

w = 200 rad/min

Stroke = 1.0 m

$$\text{Amplitude, A =}\frac{1.0}{2}$$

= 0.5 m

The maximum speed (v_max) of the piston is given by the relation :

v_max = Aω

= 200 × 0.5

= 100 m/min

Therefore, its maximum speed is 100 m/min

14.15. The acceleration due to gravity on the surface of moon is 1.7 m s^–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s^–2)

Ans. Here, g_m = 1.7 ms^–2; g_e = 9.8 ms^–2;

T_m = ?; Te = 3.5 s^–1

$$\text{Since},\space\text{T}_{e}=2\pi\sqrt\frac{1}{g_e}\text{and T}_m=2\pi\sqrt\frac{1}{g_m}\\\therefore\space\frac{\text{T}_m}{\text{T}_e}=\sqrt{\frac{g_e}{g_m}}\\\Rarr\space\text{T}_m=\text{T}_e×\sqrt{\frac{g_e}{g_m}}\\\text{T}_m=3.5\sqrt{\frac{9.8}{1.7}}=8.4 s.$$

14.16. Answer the following questions :

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle :

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

A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than $$2\pi\sqrt\frac{l}{g}$$.

Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity ?

Ans. (a) In case of a spring, k does not depend upon m. However, in case of a simple pendulum, k is directly proportional to m and hence the ratio $$\frac{m}{k}$$

is a constant quantity.

(b) The restoring force for the bob of the pendulum is given by,

F = – mg sin θ

$$\text{If}\space\theta\space \text{is small, then sin} \theta = \theta =\frac{y}{l}\\\therefore\space\text{F}=-\frac{\text{mg}}{l}y\\\text{i.e., the motion is simple harmonic and time period is }\text{T}=2\pi\sqrt{\frac{l}{g}}.$$

Clearly, the above formula is obtained only if we apply the approximation sin θ = θ.

For large angles, this approximation is not valid and T is greater than

$$2\pi\sqrt\frac{l}{g}.$$

(c) The wristwatch uses an electronic system or spring system to give the time, which does not change with acceleration due to 0gravity. Therefore, the watch will give the correct time.

(d) During free fall of the cabin, the acceleration due to gravity is zero. Therefore, the frequency of oscillations is also zero i.e., the pendulum will not vibrate at all.

14.17. A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ?

Ans. The necessary acceleration that acts upon the car moving in a circular track are :

(i) Acceleration due to gravity ‘g’ acting vertically downward.

$$\text{(ii)\space Centripetal acceleration,}\space a_c=\frac{v^{2}}{\text{R}}\text{acting along the horizontal direction.}\\\therefore\text{Effective acceleration,}\\\text{g'}=\sqrt{g^{2}+a^{2}_c}\\\text{or}\space \text{g}'=\sqrt{g^{2}+\frac{v^{4}}{\text{R}^{4}}}\\\text{Now time period}\\\text{T'}=2\pi\sqrt{\frac{l}{g}}\\=2\pi\sqrt{\frac{l}{\sqrt {g^{2}+\frac{v^{4}}{R^{2}}}}}$$

14.18. A cylindrical piece of cork of density of base area A and height h floats in a liquid of density rl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

$$\text{T}=2\pi\sqrt{\frac{\text{hp}}{p_1 g} }$$

where r is the density of cork. (Ignore damping due to viscosity of the liquid).

Ans. Say, initially in equilibrium, y height of cylinder is inside the liquid. Then, Weight of the cylinder = upthrust due to liquid displaced

∴ Ahpg=Ayp₁g

When the cork cylinder is depressed slightly by Δy and released, a restoring force, equal to additional upthrust, acts on it. The restoring force is

F = A(y + Δy)ρ₁g – Ayρ₁g

= Aρ₁gΔy

$$\therefore\space\text{Acceleration, a}=\frac{\text{F}}{\text{m}}\\=\frac{\text{A}ρ_1g\Delta y}{Ahp}=\frac{ρ_1g}{hp}.\Delta y$$

and the acceleration is directed in a direction opposite to Dy. Obviously, as a ∝ – Dy, the motion of cork cylinder is SHM, whose time period is give by

$$\text{T}=2\pi\sqrt{\frac{\text{displacement}}{\text{acceleration}}}\\=2\pi\sqrt{\frac{\Delta y}{a}}\\=2\pi\sqrt{\frac{hp}{ρ_1g}}$$

14.19. One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion.

Ans. Area of cross-section of the U-tube = A

Density of the mercury column = ρ

Acceleration due to gravity = g

Restoring force,

F = Weight of the mercury
column of a certain height

F = – (Volume × Density × g)

F = – (A × 2h × r × g)

= – 2Argh

= – k × Displacement in one of
the arms (h)

Where, 2h is the height of the mercury column in the two arms k is a constant, given by

k = – F / h

Then, k = 2Arg

$$\text{Time period}, \text{T}=2\pi\sqrt{\frac{m}{k}}\\\text{On substituting k value, we get,}\\\text{Time period, T}=2\pi\sqrt{\frac{m}{2A\rho g}}\\\text{Where, m is the mass of the mercury column
Let l be the length of the total mercury in the U-tube}\\=\text{volume of mercury × Density of mercury}\\=Al\rho\\\text{Hence,T} = 2\pi\sqrt{\frac{AlP}{2\text{A}\rho g}}\\\text{T}=2\pi\sqrt{\frac{l}{2g}}.$$

Therefore, the mercury column executes simple harmonic motion with time period

$$2\pi\sqrt{\frac{l}{2g}}.$$

14.20. An air chamber of volume V has a neck area of cross section a into which a ball of mass m just fits and can move up and down without any friction (Fig.14.27). Show that when the ball is pressed down a little and released , it executes SHM. Obtain an expression for the time period of oscillations assuming pressure-volume variations of air to be isothermal [see Fig. 14.27].

Ans. Consider an air chamber of volume V with a long neck of uniform area of cross-section A, and a frictionless ball of mass m fitted smoothly in the neck at position C, Fig. The pressure of air below the ball inside the chamber is equal to the atmospheric pressure.

Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D, where CD = y.

There will be decrease in volume and hence increase in pressure of air inside the chamber. The decrease in volume of the air inside the chamber, ΔV = Ay

$$\text{Volumetric strain}=\frac{\text{change in volume}}{\text{original volume}}\\=\frac{\Delta V}{V}=\frac{Ay}{V}\\\therefore\text{Bulk Modulus of elasticity E, will be E}=\frac{\text{stress (or increase in pressure)}}{\text{volumetric strain}}\\=\frac{-p}{Ay/V}=\frac{-pV}{Ay}$$

Here, negative sign shows that the increase in pressure will decrease the volume of air in the chamber.

$$\text{Now},\space p=\frac{-\text{EAy}}{\text{V}}\\\text{Due to this excess pressure, the restoring force acting on the ball is}\\\text{F}=p×\text{A}\\=\frac{-\text{EAy}}{\text{V}}.\text{A}=\frac{-\text{EA}^{2}}{V}y\space\space\text{...(i)}$$

Since F ∝ y and negative sign shows that the force is directed towards equilibrium position. If the applied increased pressure is removed from the ball, the ball will start executing linear SHM in the neck of chamber with C as mean position
in S.H.M. the restoring force,

F = – ky ...(ii)

Comparing (i) and (ii), we have

Spring factor,

$$k=\frac{\text{EA}^{2}}{\text{V}}\\\text{Here, Inertia factor}\\= \text{mass of ball = m}.\\\text{Period, T}=2\pi\sqrt{\frac{\text{inertia factor}}{\text{spring factor}}}\\=2\pi\sqrt{\frac{m}{\text{EA}^{2}/\text{V}}}=\frac{2\pi}{\text{A}}.\sqrt{\frac{\text{mV}}{\text{E}}}$$

14.21. You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750 kg.

Ans. (a) Here, mass, M = 3000 kg,

displacement,

x = 15 cm = 0.15 m, g = 10 m/s².

There are four spring systems, If k is the spring constant of each spring, then total
spring constant of all the four springs in parallel is

K_p = 4k

∴ M_g = K_px = 4kx

$$\Rarr\space k=\frac{\text{M}_g}{4x}\\=\frac{3000×10}{4×0.15}=5×10^{4}\text{N}$$

(b) For each spring system supporting 750 kg of weight,

$$t=2\pi\sqrt{\frac{m}{k}}\\=2×3.15×\sqrt{\frac{750}{5×10^{4}}}\\=0.77\space\text{sec.}\\\therefore\space\text{Using}x=x_0 e^{-\frac{bt}{2m},}\text{we get}\\\frac{50}{100}x_0=x_0e^{-\frac{b×0.77}{2×750}}\text{or}\space e^{-\frac{0.77b}{1500}}=2\\\text{Taking logarithm of both sides,}\\\frac{0.77 b}{1500}=ln\space2=2.303\space\text{log}\space2\\\therefore\space b=\frac{1500}{0.77}×2.303×0.3010\\=1350.4\space\text{kg s}^{\space^-1}.$$

14.22. Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Ans. Let the particle executing SHM starts oscillating from its mean position. Then displacement equation is

x = A sin ωt

∴ Particle velocity,

v = Aw cos ωt

∴ Instantaneous K.E.,

$$\text{K}=\frac{1}{2}\text{mv}^{2}\\=\frac{1}{2}\text{mA}^{2}\omega^{2}\text{cos}^{2}\omega t\\\therefore\space\text{Average value of K.E. over one complete cycle}\\\text{K}_{av}=\frac{1}{\text{T}}\int^{T}_{0}\frac{1}{2}\text{mA}^{2}\text{cos}^{2}\omega t\space dt\\=\frac{\text{mA}^{2}\omega^{2}}{2T}\int^{T}_{0}\text{cos}^{2}\omega t\space dt\\=\frac{mA^{2}\omega^{2}}{2\text{T}}\int^{T}_{0}\frac{(1+\text{cos 2}\omega t)}{2}\text{dt}\\=\frac{mA^{2}\omega^{2}}{4T}\bigg[t+\frac{\text{sin} 2\omega t}{2 \omega}\bigg]^{T}_{0}$$

$$=\frac{mA^{2}\omega^{2}}{4T}[T]\\=\frac{1}{4}mA^{2}\omega^{2}\space...\text{(i)}\\\text{Again instantaneous P.E.}\\\text{U}=\frac{1}{2}kx^{2}=\frac{1}{2}m\omega^{2}x^{2}\\=\frac{1}{2}m\omega^{2}A^{2}\text{sin}^{2}\omega t\\\therefore\text{Average value of P.E. over one complete cycle}\\\text{U}_{av}=\frac{1}{\text{T}}\int^{T}_{0}\frac{1}{2}m\omega^{2}A^{2}\text{sin}^{2}\omega t \text{dt} \\=\frac{m\omega^{2}A^{2}}{2T}\int^{T}_{0}\text{sin}^{2}\omega t dt$$

$$=\frac{mA^{2}\omega^{2}}{4T}\int^{T}_{0}\frac{(1-\text{cos}2\omega t)}{2}dt\\=\frac{mA^{2}\omega^{2}}{4T}\bigg[t-\frac{\text{sin}2\omega t}{2\omega}\bigg]^{T}_{0}\\=\frac{mA^{2}\omega^{2}}{4\text{T}}[\text{T}]\\=\frac{1}{4}m\omega^{2}A^{2}\space...\text{(ii)}\\\text{Simple comparison of (i) and (ii), show that}\\\text{K}_{av}=\text{U}_{sv}=\frac{1}{4}m\omega^{2}A^{2}$$

14.23. A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J = – α θ , where J is the restoring couple and q the angle of twist).

Ans. $$\text{T}=2\pi\sqrt{\frac{l}{\alpha}}\space\text{or}\space \text{T}^{2}=\frac{4\pi^{2}I}{\alpha}\\\text{or}\space\alpha=\frac{4\pi^{2}I}{T^{2}}\text{or}\space\alpha=\frac{4\pi^{2}}{T^{2}}\bigg(\frac{1}{2}\text{MR}^{2}\bigg)\\\text{or}\space\alpha=\frac{2\pi^{2}\text{MR}^{2}}{\text{T}^{2}}\\\text{or}\space\alpha=\frac{2(3.14)^{2}×10×(0.15)^{2}}{(1.5)^{2}}\text{Nm\space rad}^{\normalsize-1}\\=1.97\text{Nm\space \text{rad}}^{\normalsize-1}.$$

14.24. A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is (a) 5 cm (b) 3 cm (c) 0 cm

$$\textbf{Ans.}\space\text{Here, r = 5 cm = 0.05 m; T = 0.2s;}\space \omega=\frac{2\pi}{T}=\frac{2\pi}{0.2}=10\pi\space\text{rad/s}\\\text{When displacement is y, then acceleration, A =} – \omega^2y\\\text{velocity}, v=\omega\sqrt{r^{2}-y^{2}}\\\text{Case (a) When}\\\text{y = 5 cm = 0.05 m}\\\text{A} = – (10\pi)^2 × 0.05 = – 5\pi^2 m/s^2\\v=10\pi\sqrt{(0.05)^{2}-(0.05)^{2}}=0.\\\text{Case (b) When}\\\text{y = 3 cm = 0.03 m}\\\text{A}=-(10\pi)^{2}×0.03=-3\pi^{2}\text{m/s}^{2}\\v=10\pi\sqrt{(0.05)^{2}-(0.03)^{2}}\\=10\pi×0.04=0.4\pi\space\text{m/s}\\\text{Case (c) When}$$

$$y = 0, A = – (10\pi)^2 × 0 = 0\\v=10\pi\sqrt{(0.05)^{2}-0^{2}}\\=10\pi×0.05=0.5\pi\space\text{m/s}$$

14.25. A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x₀ and pushed towards the centre with a velocity v₀ at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters w, x₀ and v₀.

[Hint : Start with the equation x = a cos (ωt + θ) and note that the initial velocity is negative.]

Ans. x = a cos (ωt + θ)

$$v=\frac{dx}{dt}=-a\omega\space\text{sin}(\omega t+\theta)\\\text{When}\space t=o,x=x_0\space\text{and}\frac{dx}{dt}=-v_0\\\therefore\space x_0=a\text{cos}\theta\space\text{...(i)}\\\text{and}\space-v_0=-a\omega\text{sin}\theta \text{or}\space a\space\text{sin}\space\theta =\frac{v_0}{\omega}\space...(\text{ii})\\\text{Squaring and adding (i) and (ii), we get}\\a^{2}(\text{cos}^{2}\theta+\text{sin}^{2}\theta)\\=x_{0}^{2}+\frac{v_0^{2}}{\omega ^{2}}\\\text{or}\space a=\sqrt{x^{2}_0+\frac{v_{0}^{2}}{\omega^{2}}}$$