Question

NCERT Solutions Class 11 Physics Chapter 14 Oscillations is based on the latest syllabus proposed by the CBSE. These solutions are made by experts on the subject.

NCERT Solutions Class 10 Physics students get to learn about all kinds of different topics in-depth. Our NCERT solutions also provide in-text question solutions, practical-based, and back-of-chapter questions. All the important topics mentioned in the chapter are: 

• Oscillations – Oscillation means the back and forth movement of particles from the equilibrium position.
• Periodic motion – such motion which is repeated in equal intervals of time is called periodic motion. Example: bouncing ball, rocking chair, the pendulum of the watch, planet’s orbit around the sun.
• Oscillatory motion – oscillatory motion is a type of periodic motion. Oscillating motion is a mechanical system is swinging from one side to another side.
• Simple harmonic motion – simple harmonic motion is such motion that is defined as restoring force that is directly proportional to the displacement of the body from its mean position.
• Uniform circular motion – such motion where the particle moves along the circle with constant velocity.
• Velocity and acceleration in simple harmonic motion – a motion is said to be accelerated when its velocity keeps changing.
• Force law for simple harmonic motion - The force which causes the motion is always directed toward the center of motion and is directly proportional to the distance between them.
• Energy in simple harmonic motion – the energy of simple harmonic motion varies from zero to maximum values. The velocity of the particle is zero at the extreme position and maximum at the mean position. It is mathematically defined as KE = 1/2mv². Where ‘m’ is the mass of the object, ‘v’ is the velocity of velocity and KE stands for Kinetic Energy.
• Some systems execute Simple Harmonic Motion.
• Damped simple harmonic motion – an oscillator and its motion are said to be damped when the motion of the oscillator is reduced to an external force.
• Forced oscillations and resonance – such oscillation is done by the periodic force and is outside the oscillating system.

Our NCERT Solutions Class 11 Physics provides the holistic learning of the chapter. These solutions can be used for learning the basics of physics, preparing for board exams, and also for cracking exams like JEE Mains, JEE Advance, NTSE, and many more.

Answer 1

NCERT Solutions Class 11 Physics Chapter 14 Oscillations

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 center of mass. 

d) An arrow released from a bow. 

Solution:

Correct answer is (b) and (c) 

The swimmer’s motion is not periodic. The motion of the swimmer between the banks of a river is back and forth. However, it does not have a definite period. This is because the time taken by the swimmer during his back and forth journey may not be the same. 

The motion of a freely-suspended magnet, if displaced from its N-S direction and released, is periodic. This is because the magnet oscillates about its position with a definite period of time. 

When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such motion is periodic. 

An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic. 

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. 

Solution:

(b) and (c) are SHMs and (d) are periodic, but not SHMs 

During its rotation about its axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and from motion about its axis. 

An oscillating mercury column in a U-tube is simple harmonic. This is because the mercury moves to and from on the same path, about the fixed position, with a certain period of time. 

The ball moves to and from about the lowermost point of the bowl when released. Also, the ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic. 

A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic. 

3. Figure 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)? 

(a)

Solution: 

(b) and (d) are periodic 

It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case. 

In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s. 

It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time. 

In this case, the motion of the particle repeats itself after 2 s. Hence, it is a periodic motion, having a period of 2 s. 

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 (ω is any positive constant): 

a) sin ωt – cos ωt 

b) sin3 ωt 

c) 3 cos (π/4 – 2ωt) 

d) cos ωt + cos 3ωt + cos 5ωt 

e) exp (–ω² t² ) 

f) 1 + ωt + ω² t² 

Solution:

SHM 

The given function is:

This function represents SHM as it can be written in the form: a sing (ωt + ϕ)

Its period is: \(\frac {2 \pi}{ω}\)

Periodic, but not SHM 

The given function is:

sin³ ωt 

= \(\frac 12[3 sin\, ωt - sin\, 3ωt]\) 

The terms sin ωt and sin ωt individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.

SHM

The given function is:

This function represents simple harmonic motion because it can be written in the form:

a cos (ωt + ϕ)

Its period is:

\(\frac {2\pi}{2ω} = \frac {\pi}{ω}\)

Periodic, but not SHM 

The given function is cos ωt + cos 3ωt + cos 5ωt. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.

Non-periodic motion

The given function exp (-w²t²) is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion. 

The given function 1 + ωt + ω² t² is non-periodic.

Answer 2

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, 

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

Solution:

a) Zero, Positive, Positive 

b) Zero, Negative, Negative 

c) Negative, Zero, Zero 

d) Negative, Negative, Negative 

e) Zero, Positive, Positive 

f) Negative, Negative, Negative

The given situation is shown in the following figure. Points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path.

A particle is in linear simple harmonic motion between the end points 

At the extreme point A, the particle is at rest momentarily. Hence, its velocity is zero at this point. 

Its acceleration is positive as it is directed along AO. 

Force is also positive in this case as the particle is directed rightward. 

At the extreme point B, the particle is at rest momentarily. Hence, its velocity is zero at this point.

Its acceleration is negative as it is directed along B. 

Force is also negative in this case as the particle is directed leftward.

(c)

The particle is executing a simple harmonic motion. O is the mean position of the particle. Its velocity at the mean position O is the maximum. The value for velocity is negative as the particle is directed leftward. The acceleration and force of a particle executing SHM is zero at the mean position.

(d)

The particle is moving toward point O from the end B. This direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, the particle’s velocity and acceleration, and the force on it are all negative.

(e)

The particle is moving toward point O from the end A. This direction of motion is from A to B, which is the conventional positive direction. Hence, the values for velocity, acceleration, and force are all positive.

(f)

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

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

a) a = 0.7x 

b) a = –200x² 

c) a = –10x 

d) a = 100x³

Solution:

A motion represents simple harmonic motion if it is governed by the force law:

F = –kx 

ma = –k

∴ a = - \(\frac kmx\)

Where, F is the force m is the mass (a constant for a body) x is the displacement a is the acceleration k is a constant.

Among the given equations, only equation a = –10 x is written in the above form with \(\frac km=10\). Hence, this relation represents SHM.

7. The motion of a particle executing simple harmonic motion is described by the displacement 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.

Solution:

Initially, at t = 0: 

Displacement, x = 1 cm 

Initial velocity, v = ω cm/sec. 

Angular frequency, ω = π rad/s –1 

It is given that:

Squaring and adding equations (i) and (ii), we get:

Dividing equation (ii) by equation (i), we get:

tan ϕ = -1

∴ ϕ  = \(\frac {3 \pi}{4}, \frac {7 \pi}{4}....\)

SHM is given as:

x = B sin (ωt + a)

Putting the given values in this equation, we get:

1 = B sin [ω x 0 x a]

B sin a = 1 ...(iii)

Velocity, 

v = ωB cos (ωt + a)

Substituting the given values, we get:

π = πB sin a

B sin a = 1 ....(iv)

Squaring and adding equations (iii) and (iv), we get:

B² [sin² a + cos² a] = 1+1

B² = 2

∴ B = √2 cm

Dividing equation (iii) by equation (iv), we get:

\(\frac {B\, sin\, a}{B \, cos\, a} = \frac 11\)

tan a = 1 = tan \(\frac \pi 4\)

∴ a = \(\frac \pi 4\), \(\frac \pi 5\),.....

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? 

Solution:

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

∴ Spring constant, k = \(\frac Fl = \frac {490}{0.2}\) = 2450 Nm^-1

Mass m, is suspended from the balance. 

Time period, T = \(2\pi \sqrt{\frac mk}\)

∴ m = \((\frac {T}{2\pi})^2 \times k = (\frac {0.6}{2\times 3.14})\) x 2450 = 22.36 kg

∴ Weight of the body = mg = 22.36 × 9.8 = 219.167 N

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

Answer 3

 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.

Solution:

Spring constant, k = 1200 N m^–1 

Mass, m = 3 kg 

Displacement, A = 2.0 cm = 0.02 cm 

Frequency of oscillation v, is given by the relation:

v= \(\frac 1T = \frac 1{2\pi}\sqrt {\frac km}\) 

Where, T is the time period

∴ v = \(\frac {1}{2\times 3.14} \sqrt{\frac {1200}{3}}\) = 3.18 m/s

Hence, the frequency of oscillations is 3.18 m/s. 

Maximum acceleration (a) is given by the relation: 

a = ω² A 

Where,

ω = Angular frequency = \(\sqrt{\frac km}\)

A = Maximum displacement

∴ a = \(\frac km\) = 0.02 x \(\sqrt{\frac {1200}{3}}\) = 0.4 m/s

Hence, the maximum velocity of the mass is 0.4 m/s.

10. In question number 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 at the mean position, at the maximum stretched position, and 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?

Solution:

x = 2sin 20t 

x = 2cos 20t

x = –2cos 20t 

The functions have the same frequency and amplitude, but different initial phases. 

Distance travelled by the mass sideways, A = 2.0 cm 

Force constant of the spring, k = 1200 N m^–1 

Mass, m = 3 kg 

Angular frequency of oscillation:

 ω = \(\sqrt{\frac km}\) 

=  \(= \sqrt {\frac {1200}{3}} = \sqrt{400}\) = 20 rad s^–1

When the mass is at the mean position, initial phase is 0. 

Displacement, x = A sin ωt 

= 2sin 20t 

At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is \(\frac \pi2\)

Displacement, x = A sin (ωt + \(\frac \pi2\))

= 2 sin (20t + \(\frac \pi2\))

= 2cos 20t

At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is \(\frac {3\pi}2\) 

Displacement, x = A sin (ωt  + \(\frac {3\pi}2\))

= 2 sin (20t + \(\frac {3\pi}2\))

= –2cos 20t

The functions have the same frequency (\(\frac {20}{2\pi} Hz\)) and amplitude (2 cm), but different initial phases \((0, \frac \pi2, \frac {3\pi}{2})\)

11. Figures 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 anticlockwise) 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.

Solution:

Time period, T = 2 s 

Amplitude, A = 3 cm 

At time, t = 0, the radius vector OP makes an angle \(\frac \pi2\) with the positive x-axis, i.e., phase angle ϕ = + \(\frac \pi2\)

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given by the displacement equation:

∴ x = -3 sin π t cm

Time period, T = 4 s 

Amplitude, a = 2 m 

At time t = 0, OP makes an angle π with the x-axis, in the anticlockwise direction. Hence, phase angle, Φ = + π

Therefore, the equation of simple harmonic motion for the x-projection of OP, at time t, is given as:

x = a cos \((\frac {2\pi t}{T} + ϕ) = 2 cos (\frac {2\pi t}{4} + \pi)\)

∴ x = -2 cos (\(\frac \pi2\)t)m

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). 

x = –2 sin (3t + π/3) 

x = cos (π/6 – t)

x = 3 sin (2πt + π/4) 

x = 2 cos πt 

Solution:

If this equation is compared with the standard SHM equation x = A cos (\(\frac {2\pi}T t+ ϕ)\) , then we get:

Amplitude, A = 2 cm

Phase angle, ϕ = \(\frac {5\pi}6 \) = 150°

Angular velocity, ω = \(\frac {2\pi}T\) = 3 rad/sec.

The motion of the particle can be plotted as shown in the following figure.

x = cos (\(\frac \pi6\) - t) = cos (t - \(\frac \pi6\))

If this equation is compared with the standard SHM equation  x = A cos (\(\frac {2\pi}T t+ ϕ)\) , then we get:

Amplitude, A = 1

Phase angle, ϕ - \(\frac \pi6\) = 300°

Angular velocity, ω = \(\frac {2\pi}T\) = 1 rad/sec.

The motion of the particle can be plotted as shown in the following figure.

If this equation is compared with the standard SHM equation  x = A cos (\(\frac {2\pi}T t+ ϕ)\), then we get:

Amplitude, A = 3 cm

Phase angle, ϕ - \(\frac {3\pi}4\) = 135°

Angular velocity, ω = \(\frac {2\pi}T\) = 2π rads.

The motion of the particle can be plotted as shown in the following figure.

x = 2 cos πt

If this equation is compared with the standard SHM equation  x = A cos (\(\frac {2\pi}T t+ ϕ)\), then we get: 

Amplitude, A = 2 cm 

Phase angle, Φ = 0 

Angular velocity, ω = π rad/s 

The motion of the particle can be plotted as shown in the following figure.

Answer 4

13. Figure (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 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Figure(b) is stretched by the same force F.

What is the maximum extension of the spring in the two cases? If the mass in Figure. (a) and the two masses in Figure. (b) are released, what is the period of oscillation in each case?

Solution:

For the one block system: 

When a force F, is applied to the free end of the spring, an extension l, is produced. For the maximum extension, it can be written as: 

F = kl 

Where, k is the spring constant 

Hence, the maximum extension produced in the spring,  l = \(\frac Fk\) 

For the two block system: 

The displacement (x) produced in this case is:

x = \(\frac l2\) 

Net, F = +2 kx = 2k \(\frac l2\) 

∴ l = \(\frac Fk\) 

For the one block system: 

For mass (m) of the block, force is written as:

F = ma = m \(\frac {d^2x}{dt^2}\) 

Where, x is the displacement of the block in time t

∴ m\(\frac {d^2x}{dt^2}\) = -kx

It is negative because the direction of elastic force is opposite to the direction of displacement.

\(\frac {d^2x}{dt^2}\) = - (\(\frac km\)) x = -ω²x

Where, ω² = \(\frac km\) 

ω = \(\sqrt{\frac km}\)

Where, ω is angular frequency of the oscillation

∴ Time period of the oscillation, T = \(\frac {2\pi}ω\) 

For the two block system:

F =  m\(\frac {d^2x}{dt^2}\) 

 m\(\frac {d^2x}{dt^2}\) = -2 kx

It is negative because the direction of elastic force is opposite to the direction of displacement.

\(\frac {d^2x}{dt^2}\) = - [\(\frac {2k}m\)]x = -ω²x

Where,

Angular frequency, ω = \(\sqrt{\frac {2k}m}\)

∴ Time period, T = \(\frac {2k}ω = 2\pi \sqrt{\frac m{2k}}\)

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?

Solution:

Angular frequency of the piston, ω = 200 rad/ min. 

Stroke = 1.0 m 

Amplitude, A = \(\frac {1.0}{2}= 0.5m\)

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

v_max= Aω 

= 200 x 0.5 = 100 m/min

15. The acceleration due to gravity on the surface of moon is 1.7 ms^–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 ms^–2) 

Solution:

Acceleration due to gravity on the surface of moon, g'= 1.7 m s^–2 

Acceleration due to gravity on the surface of earth, g = 9.8 m s^–2 

Time period of a simple pendulum on earth, T = 3.5 s

T = \(2\pi \sqrt {\frac lg}\)

Where, l is the length of the pendulum

The length of the pendulum remains constant.

On moon’s surface, time period, T' = \(2\pi \sqrt {\frac l{g'}}\)

= \(2\pi \sqrt{\frac {\frac {(3.5)^2}{4\times(3.14)^2} \times 9.8}{1.7}} = 8.4 s\) 

Hence, the time period of the simple pendulum on the surface of moon is 8.4 s.

16. Answer the following questions: 

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

 T = \(2\pi \sqrt {\frac mk}\)

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

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  \(2\pi \sqrt {\frac lg}\). . Think of a qualitative argument to appreciate this result.

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? 

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

Solution:

The time period of a simple pendulum,   T = \(2\pi \sqrt {\frac mk}\)

For a simple pendulum, k is expressed in terms of mass m, as: 

k ∝ m 

\(\frac mk\) = Constant 

Hence, the time period T, of a simple pendulum is independent of the mass of the bob. 

In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as: 

F = –mg sinθ 

Where, 

F = Restoring force 

m = Mass of the bob

g = Acceleration due to gravity 

θ = Angle of displacement 

For small θ, sin θ 

For large θ, sin θ is greater than θ. 

This decreases the effective value of g. 

Hence, the time period increases as:

 T = \(2\pi \sqrt {\frac l{g}}\) 

Where, l is the length of the simple pendulum.

The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. Since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action. 

When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero. 

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? 

Solution:

The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. 

Acceleration due to gravity = g 

Centripetal acceleration = \(\frac {v^2}R\)

Where, v is the uniform speed of the car 

R is the radius of the track 

Effective acceleration (a_eff) is given as:

a_eff= \(\sqrt {g^2 + (\frac {v^2}{R^2})^2}\)

Time period, T = \(2\pi \sqrt {\frac {l}{a_{eff}}}\) 

Where, l is the length of the pendulum

∴ Time period, T= \(2\pi \sqrt {\frac {l}{g^2 + \frac {v^4}{R^2}}}\)

Answer 5

18. Cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ₁. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period.

T = \(2\pi \sqrt{\frac{hp}{ρ_1g}}\) 

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

Solution:

Base area of the cork = A 

Height of the cork = h 

Density of the liquid = ρ₁

Density of the cork = ρ 

In equilibrium: 

Weight of the cork = Weight of the liquid displaced by the floating cork 

Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork. 

Up-thrust = Restoring force, 

F = Weight of the extra water displaced 

F = –(Volume × Density × g) 

Volume = Area × Distance through which the cork is depressed 

Volume = Ax 

F = – A x ρ₁g … (i) 

According to the force law:

F = kx 

k = \(\frac Fx\)

Where, k is a constant 

 k = \(\frac Fx\) ρ₁g ...(ii)

The time period of the oscillations of the cork:

T = \(2\pi \sqrt{\frac{m}{k}}\)  ....(iii)

Where, 

m = Mass of the cork 

= Volume of the cork × Density 

= Base area of the cork × Height of the cork × Density of the cork 

= Ahρ 

Hence, the expression for the time period becomes:

T = \(2\pi \sqrt {\frac {Ahρ}{Aρ_1g}} = 2\pi \) \( \sqrt{\frac{hp}{ρ_1g}}\) 

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.

Solution:

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 × ρ ×g) = –2Aρgh = –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 = - \(\frac Fh = 2 Aρg\)

Time period,

Where, m is the mass of the mercury column 

Let l be the length of the total mercury in the U-tube. 

Mass of mercury, m = Volume of mercury × Density of mercury = Alρ

Hence, the mercury column executes simple harmonic motion with time period \(2\pi \sqrt {\frac l{2g}}\)

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 (Figure). 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

Solution:

Volume of the air chamber = V 

Area of cross-section of the neck = a

 Mass of the ball = m 

The pressure inside the chamber is equal to the atmospheric pressure. 

Let the ball be depressed by x units. As a result of this depression, there would be a decrease in the volume and an increase in the pressure inside the chamber. 

Decrease in the volume of the air chamber, ΔV = ax

Volumetric strain = \(\frac {Change \,in\,volume}{Original\,volume}\)

= \(\frac {\triangle V}{V} = \frac {ax}V\) 

Bulk Modulus of air, B = \(\frac {Stress}{Strain} = \frac {-p}{\frac {ax}{V}}\)

In this case, stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.

p = \(\frac {-Bax}{V}\) 

The restoring force acting on the ball,

F = p × a

= \(\frac {-Ba^2x}{V}.a\) 

= \(\frac {-Ba^2x}{V} \) ....(i)

In simple harmonic motion, the equation for restoring force is: 

F = –kx … (ii) 

Where, k is the spring constant 

Comparing equations (i) and (ii), we get:

k = \(\frac {-Ba^2}{V} \) 

Time period,

T = \(2\pi\sqrt{\frac mk}\) 

= \(2\pi\sqrt{\frac {Vm}{Ba^2}}\) 

Answer 6

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.

Solution:

Mass of the automobile, m = 3000 kg 

Displacement in the suspension system, x = 15 cm = 0.15 m 

There are 4 springs in parallel to the support of the mass of the automobile. 

The equation for the restoring force for the system: 

F = –4kx = mg 

Where, k is the spring constant of the suspension system

Time period, T = \(2\pi \sqrt{\frac m{4k}}\) And k = \(\frac {mg}{4x} = \frac {3000\times 10}{4 \times 0.15}\) 

= 5000 = 5 × 10⁴ N/m

Spring constant, k = 5 × 10⁴ N/m

Each wheel supports a mass, M = \(\frac {3000}{4} \) = 750 kg

For damping factor b, the equation for displacement is written as:

x = \(x_0e^{-bt/2M}\)

The amplitude of oscillation decreases by 50%.

∴ x = \(\frac {x_0}{2}\)

\(\frac {x_0}{2}\) = \(x_0e^{-bt/2M}\)

Where,

Time, period, 

Therefore, the damping constant of the spring is 1351.58 kg/s.

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. 

Solution:

The equation of displacement of a particle executing SHM at an instant t is given as: 

x = A sin ωt

Where, 

A = Amplitude of oscillation 

ω = Angular frequency  = \(\sqrt {\frac kM}\)

 The velocity of the particle is: 

v = \(\frac {dx}{dt} = A ω\, cos \,ωt\)

The kinetic energy of the particle is: 

The potential energy of the particle is: 

For time period T, the average kinetic energy over a single cycle is given as:

And, average potential energy over one cycle is given as:

It can be inferred from equations (i) and (ii) that the average kinetic energy for a given time period is equal to the average potential energy for the same time period. 

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 α is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist). 

Solution:

Mass of the circular disc, m = 10 kg

Radius of the disc, r = 15 cm = 0.15 m The torsional oscillations of the disc has a time period, T = 1.5 s 

The moment of inertia of the disc is:

l = \(\frac 12\) mr²

= \(\frac 12\) x (10) x (0.15)²

= 0.1125 kg m²

Time period, T = \(2\pi \sqrt{\frac la}_a\)

is the torsional constant.

= 1.972 Nm/rad 

Hence, the torsional spring constant of the wire is 1.972 Nm rad^–1. 

24. A body describes simple harmonic motion with 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

Solution:

Amplitude, A = 5 cm = 0.05 m 

Time period, T = 0.2 s 

For displacement, x = 5 cm = 0.05 m 

Acceleration is given by:

Velocity is given by:

= 0 

When the displacement of the body is 5 cm, its acceleration is –5π² m/s² and velocity is 0. 

For displacement, x = 3 cm = 0.03 m

Acceleration is given by:

Velocity is given by:

= 0.4 π m/s 

When the displacement of the body is 3 cm, its acceleration is –3π m/s² and velocity is 0.4π m/s.

For displacement, x = 0 

Acceleration is given by:

a = -ωx = 0

Velocity is given by:

When the displacement of the body is 0, its acceleration is 0 and velocity is 0.5 π m/s.

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 ω, x₀ and v₀. 

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

Solution:

The displacement equation for an oscillating mass is given by: 

x = A cos (ωt + θ)

Where,

A is the amplitude x 

is the displacement θ 

is the phase constant

Squaring and adding equations (i) and (ii), we get:

Hence, the amplitude of the resulting oscillation is