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NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

Edited By Vishal kumar | Updated on Apr 03, 2025 04:33 PM IST

Have you seen a swing going back and forth or a ball bouncing? These are oscillations. In this chapter, you will study how objects move in repeating patterns, such as a pendulum or the vibration of a guitar string. Oscillations are everywhere, from the ticking to sound waves' vibrations.

Oscillations is a crucial topic in the Class 11 NCERT syllabus, covering fundamental concepts. The Oscillations chapter in Class 11 Physics is crucial as it forms the foundation for wave motion, SHM (Simple Harmonic Motion), and resonance. It is widely applied in pendulums, springs, sound waves, AC circuits, and quantum mechanics. If you struggle to grasp the concept of Oscillations or solve NCERT textbook questions, Careers360 provides detailed NCERT solutions to assist you in your studies.

This Story also Contains
  1. NCERT Solutions for Class 11 Physics Chapter 13: Oscillations
  2. Additional Questions
  3. Oscillation Class 11 Physics
  4. NCERT Solutions for Class 11 Physics Chapter-Wise
  5. NCERT Solutions for Class 11 Physics Chapter 12: Download PDF
  6. Additional Questions
  7. Kinetic Theory Chapter 12 Physics NCERT Solutions: Concepts and Important Formulas
  8. Kinetic Theory Class 11 Physics NCERT Topics
  9. NCERT Solutions for Class 11 Physics Chapter Wise
  10. NCERT Solutions for Class 11 Subject Wise
  11. Subject wise NCERT Exemplar solutions
  12. Importance of NCERT Solutions for Class 11 Physics Chapter 13 Oscillations:
  13. Smart Tips to learn Class 11 Oscillations:
  14. Subject wise NCERT Exemplar solutions

Through these problems, the students gain a robust concept of wave motion, mechanical vibrations, and oscillatory systems. All of these topics are crucial for CBSE exams, JEE Main, and NEET since oscillation is an important chapter.

NCERT Solutions for Class 11 Physics Chapter 13: Oscillations

Searching for a simple and free solution to crack your Class 11 Physics exams? You can download the Chapter 13 NCERT Solutions PDF. It's full of step-by-step solutions to help you master waves and vibrations and make complex material easy and simple to understand. Ideal for last-minute revisions and to shine in your exam.

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NCERT Solutions For Class 11 Physics Chapter 13 Oscillations: Exercise Solution

Q. 13.1 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

Answer:

(a) The motion is not periodic, though it is to and fro.

(b) The motion is periodic.

(c) The motion is periodic.

(d) The motion is not periodic.

Q. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?

(a) the rotation of the 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 lowermost point.

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

Answer:

(a) Periodic but not S.H.M.

(b) S.H.M.

(c) S.H.M.

(d) Periodic but not S.H.M.M [A polyatomic molecule has a number of natural frequencies, so its vibration is a superposition of SHM’s of a number of different frequencies. This is periodic but not SHM]

Q. 13.3 Fig. 13.18 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)?

1650449930218

Fig 13.18

Answer:

The x-t plots for linear motion of a particle in Fig. 13.18 (b) and (d) represent periodic motion with both having a period of motion of two seconds.

Q. 13.4 (a) Which of the following functions of time represent

(a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion ( ω is any positive constant):

(a) sinωtcosωt

Answer:

sinωtcosωt=2(12sinωt12cosωt)=2(cosπ4sinωtsinπ4cosωt)=2sin(ωtπ4)

Since the above function is of form Asin(ωt+ϕ) it represents SHM with a time period of 2πω

Q. 13.4 (b) 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):

(b) sin3ωt

Answer:

sin3ωt=3sinωt4sin3ωtsin3ωt=14(3sinωtsin3ωt)

The two functions individually represent SHM but their superposition does not give rise to SHM the motion will definitely be periodic with a period of 2πω

Q. 13.4 (c) Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give a period for each case of periodic motion ( ω is any positive constant):

(c) 3cos(π/42ωt)

Answer:

The function represents SHM with a period of πω

Q. 14.4 (c)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 ( is any positive constant): (c)

Q.13.4 (d) 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):

(d) cosωt+cos3ωt+cos5ωt

Answer:

Here, each individual function is SHM. But superposition is not SHM. The function represents periodic motion but not SHM.

period=LCM(2πω,2π3ω,2π5ω)=2πω

Q. 13.5 (c) A particle is in linear simple harmonic motion between two points, A and B, 10cm 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

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

Answer:

Velocity is negative, that is, towards A, and its magnitude is maximum. Acceleration and force are zero.

Q. 13.6 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

Answer:

Only the relation given in (c) represents simple harmonic motion as the acceleration is proportional in magnitude to the displacement from the midpoint, and its direction is opposite to that of the displacement from the mean position.

Q. 13.7 The motion of a particle executing a simple harmonic motion is described by the displacement function, x(t)=Acos(ωt+ϕ).

If the initial (t=0) position of the particle is 1cm and its initial velocity is ωcm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is πs1. If instead of the cosine function, we choose the sine function to describe the SHM: x=Bsin(ωt+α), what are the amplitude and initial phase of the particle with the above initial conditions?

Answer:

ω=π rad s1

x(t)=Acos(πt+ϕ)

at t = 0

x(0)=Acos(π×0+ϕ)1=Acosϕ               (i)

v=dx(t)dtv(t)=Aπsin(πt+ϕ)

at t = 0

v(0)=Aπsin(π×0+ϕ)ω=Aπsinϕ 1=Asinϕ                      (ii)

Squaring and adding equation (i) and (ii), we get

12+12=(Acosϕ)2+(Asinϕ)22=A2cos2ϕ+A2sin2ϕ2=A2A=2

Dividing equation (ii) by (i) we get

tanϕ=1ϕ=3π4,7π4,11π4......

x(t)=Bsin(πt+α)

at t = 0

x(0)=Bsin(π×0+α)1=Bsinα               (iii)

v=dx(t)dtv(t)=Bπcos(πt+α)

at t = 0

v(0)=Bπcos(π×0+α)ω=Bπcosα 1=Bcosα                      (iv)

Squaring and adding equation (iii) and (iv), we get

12+12=(Bsinα)2+(Bcosα)22=B2sin2α+B2cos2α2=B2B=2

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

tanα=1α=π4,5π4,9π4......

Q. 13.8 A spring balance has a scale that reads from 0 to 50kg. The length of the scale is 20cm, A body suspended from this balance, when displaced and released, oscillates with a period of 0.6s. What is the weight of the body?

Answer:

Spring constant of the spring is given by

k=Weight of Maximum mass the scale can readMaximum displacement of the scalek=50×9.820×102k=2450 Nm1

The time period of a spring attached to a body of mass m is given by

T=2πmkm=T2k4π2m=(0.6)2×24504π2m=22.34 kgw=mgw=22.34×9.8w=218.95 N

Q. 13.9 (i) A spring having with a spring constant 1200Nm1 is mounted on a horizontal table as shown in Figure 13.19. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

1424

Fig 13.19

Determine

(i) the frequency of oscillations,

Answer:

The frequency of oscillation of an object of mass m attached to a spring of spring constant k is given by

ν=12πkmν=12π×12003ν=3.183 Hz

Q. 13.9 (ii) A spring having with a spring constant 1200Nm1 is mounted on a horizontal table as shown in Figure 13.19. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

1424

Fig 13.19

Determine

(ii) maximum acceleration of the mass, and

Answer:

A body executing S.H.M experiences maximum acceleration at the extreme points

amax=FAmamax=kAmamax=1200×0.23amax=8ms2 (F A = Force experienced by body at displacement A from mean position)

Q. 13.9 (iii) A spring having with a spring constant 1200Nm1 is mounted on a horizontal table as shown in Figure 13.9. A mass of 3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0cm and released.

1424

Fig 13.19

Determine

(iii) the maximum speed of the mass.

Answer:

Maximum speed occurs at the mean position and is given by

vmax=Aωvmax=0.02×2π×3.18vmax=0.4ms1

Q. 13.10 (a) In Exercise 13.9, let us take the position of mass when the spring is unstretched as x=0, and the direction from left to right as the positive direction of the 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,

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

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20 rad/s

(a) At t = 0 the mass is at mean position i.e. at t = 0, x = 0

x(t)=0.02sin(20t)

Here x is in metres and t is in seconds.

Q. 13.10 (b) In Exercise 13.9, let us take the position of mass when the spring is unstretched 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

(b) at the maximum stretched position,

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

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20 rad/s

(b) At t = 0 the mass is at the maximum stretched position.

x(0) = A

ϕ=π2

x(t)=0.02sin(20t+π2)x(t)=0.02cos(20t)

Here x is in metres and t is in seconds.

Q. 13.10 (c) In Exercise 13.9, let us take the position of mass when the spring is unstretched 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

(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?

Answer:

Amplitude is A = 0.02 m

Time period is ω

ω=kmω=12003ω=20 rad/s

(c) At t = 0 the mass is at the maximum compressed position.

x(0) = -A

ϕ=3π2

x(t)=0.02sin(20t+3π2)x(t)=0.02cos(20t)

Here x is in metres and t is in seconds.

The above functions differ only in the initial phase and not in amplitude or frequency.

Q. 13.11 Figures 13.20 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.

1650450107919

Fig 13.20

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

Answer:

(a) Let the required function be x(t)=asin(±ωt+ϕ)

Amplitude = 3 cm = 0.03 m

T = 2 s

ω=2πTω=πrad s

Since initial position x(t) = 0, ϕ=0

As the sense of revolution is clockwise

x(t)=0.03sin(ωt)x(t)=0.03sin(πt)

Here x is in metres and t is in seconds.

(b)Let the required function be x(t)=asin(±ωt+ϕ)

Amplitude = 2 m

T = 4 s

ω=2πTω=π2rad s

Since initial position x(t) = -A, ϕ=3π2

As the sense of revolution is anti-clockwise

x(t)=2sin(ωt+3π2)x(t)=2cos(π2t)

Here x is in metres and t is in seconds.

Q. 13.12 (a) 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=2sin(3t+π/3)

Answer:

x=2sin(3t+π/3)x=2cos(3t+π3+π2)x=2cos(3t+5π6)

The initial position of the particle is x(0)

x(0)=2cos(0+5π6)x(0)=2cos(5π6)x(0)=3cm

The radius of the circle i.e. the amplitude, is 2 cm

The angular speed of the rotating particle is ω=3rad s1

The initial phase is

ϕ=5π6ϕ=150o

The reference circle for the given simple Harmonic motion is

1650450164265

Q. 13.12 (b) 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).

(b) x=cos(π/6t)

Answer:

x(t)=cos(π6t)x(t)=cos(tπ6)

The initial position of the particle is x(0)

x(0)=cos(0π6)x(0)=cos(π6)x(0)=32cm

The radius of the circle i.e. the amplitude is 1 cm

The angular speed of the rotating particle is ω=1rad s1

Initial phase is

ϕ=π6ϕ=30o

The reference circle for the given simple Harmonic motion is

1650450237789

Q. 13.12 (d) 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).

(d) x=2cosπt

Answer:

x(t)=2cos(πt)

The initial position of the particle is x(0)

x(0)=2cos(0)x(0)=2cm

The radius of the circle i.e. the amplitude is 2 cm

The angular speed of the rotating particle is ω=πrad s1

Initial phase is

ϕ=0o

The reference circle for the given simple Harmonic motion is

1650450280990

Q. 13.13 (a) Figure 13.21 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 13.21 (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. 13.21

(b) is stretched by the same force F.

1650450344807

Fig 13.21

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

Answer:

(a) Let us assume the maximum extension produced in the spring is x.

At maximum extension

F=Kxx=Fk

(b) Let us assume the maximum extension produced in the spring is x. That is x/2 due to force towards left and x/2 due to force towards right

F=kx2+kx2x=Fk

Answer:

Amplitude of SHM = 0.5 m

angular frequency is

ω=200 rad/minω=3.33 rad/s

If the equation of SHM is given by

x(t)=Asin(ωt+ϕ)

The velocity would be given by

v(t)=dx(t)dtv(t)=d(Asin(ωt+ϕ))dtv(t)=Aωcos(ωt+ϕ)

The maximum speed is therefore

vmax=Aωvmax=0.5×3.33vmax=1.67ms1

Q. 13.15 The acceleration due to gravity on the surface of moon is 1.7ms2 What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5s ? (g on the surface of earth is 9.8 m s–2)

Answer:

The time period of a simple pendulum of length l executing S.H.M is given by

T=2πlg

g e = 9.8 m s -2

g m = 1.7 m s -2

The time period of the pendulum on the surface of Earth is T e = 3.5 s

The time period of the pendulum on the surface of the moon is T m

TmTe=gegmTm=Te×gegmTm=3.5×9.81.7Tm=8.4s

Q .13.16 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?

Answer:

Acceleration due to gravity = g (in downward direction)

Centripetal acceleration due to the circular movement of the car = a c

ac=v2R (in the horizontal direction)

Effective acceleration is

g=g2+ac2g=g2+v4R2

The time period is T'

T=2πlgT=2πlg2+v4R2

Q. 13.17 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρı . The cork is depressed slightly and then released

Show that the cork oscillates up and down simply harmonically with a period T=2πhρρ1g where ρ is the density of the cork. (Ignore damping due to the viscosity of the liquid).

Answer:

Let the cork be displaced by a small distance x in downwards direction from its equilibrium position where it is floating.

The extra volume of fluid displaced by the cork is Ax

Taking the downwards direction as positive we have

ma=ρ1gAxρAha=ρ1gAxd2xdt2=ρ1gρhx

Comparing with a=-kx we have

k=ρ1gρhT=2πkT=2πρhρ1g

Q. 13.18 One end of a U-tube containing mercury is connected to a suction pump and the other end to the 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 a simple harmonic motion.

Answer:

Let the height of each mercury column be h.

The total length of mercury in both columns = 2h.

Let the cross-sectional area of the mercury column be A.

Let the density of mercury be ρ

When either of the mercury columns dips by a distance x, the total difference between the two columns becomes 2x.

The weight of this difference is 2Axρg

This weight drives the rest of the entire column to the original mean position.

Let the acceleration of the column be a. Since the force is restoring

2hAρ(a)=2xAρga=ghx

d2xdt2=ghx which is the equation of a body executing S.H.M

The time period of the oscillation would be

T=2πhg

Additional Questions

Q. 1 An air chamber of volume V has a neck area of cross section into which a ball of mass m just fits and can move up and down without any friction (Fig.1). 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.1]

1650450479295

Fig 1

Answer:

Let the initial volume and pressure of the chamber be V and P.

Let the ball be pressed by a distance x.

This will change the volume by an amount ax.

Let the change in pressure be ΔP

Let the Bulk's modulus of air be K.

K=ΔPΔV/VΔP=KaxV

This pressure variation would try to restore the position of the ball.

Since force is restoring in nature, displacement and acceleration due to the force would be in different directions.

F=aΔPmd2xdt2=aΔpd2xdt2=ka2mVx

The above is the equation of a body executing S.H.M.

The time period of the oscillation would be

T=2πamVk

Q. 2 (a) You are riding in an automobile of mass 3000kg. . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50o/o during one complete oscillation. Estimate the values of

(a) the spring constant K

Answer:

Mass of automobile (m) = 3000 kg

There are a total of four springs.

Compression in each spring, x = 15 cm = 0.15 m

Let the spring constant of each spring be k

4kx=mgk=3000×9.84×0.15k=4.9×104 N

Q. 2 (b) You are riding in an automobile of mass 3000kg . Assuming that you are examining the oscillation characteristics of its suspension system. The suspension sags 15cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50o/o during one complete oscillation. Estimate the values of

(b) the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports 750kg. .

Answer:

The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.

T=2πmkT=2π×30004×4.9×104T=0.77 s

For damping factor b we have

x=x0e(bt2m)

x=x 0 /2

t=0.77s

m=750 kg

e0.77b2×750=0.5ln(e0.77b2×750)=ln0.50.77b1500=ln2b=0.693×15000.77b=1350.2287 kg s1

Q. 3 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.

Answer:

Let the equation of oscillation be given by x=Asin(ωt)

Velocity would be given as

v=dxdtv=Aωcost(ωt)

Kinetic energy at an instant is given by

K(t)=12m(v(t))2K(t)=12m(Aωcos(ωt))2K(t)=12mA2ω2cos2ωt

Time Period is given by

T=2πω

The Average Kinetic Energy would be given as follows

Kav=0TK(t)dt0TdtKav=1T0TK(t)dtKav=1T0T12mA2ω2cos2ωt dtKav=mA2ω22T0Tcos2ωt dtKav=mA2ω22T0T(1+cos2ωt2)dt

Kav=mA2ω22T[t2+sin2ωt4ω]0TKav=mA2ω22T[(T2+sin2ωT4ω)(0+sin(0))]Kav=mA2ω22T×T2Kav=mA2ω24

The potential energy at an instant T is given by

U(t)=12kx2U(t)=12mω2(Asin(ωt))2U(t)=12mω2A2sin2ωt

The Average Potential Energy would be given by

Uav=0TU(t)dt0TdtUav=1T0T12mω2A2sin2ωt dtUav=mω2A22T0Tsin2ωt dtUav=mω2A22T0T(1cos2ωt)2dt

Uav=mω2A22T[t2sin2ωt4ω]0TUav=mω2A22T[(T2sin2ωT4ω)(0sin0)]Uav=mω2A22T×T2Uav=mω2A24

We can see K av = U av

Q .4 A circular disc of mass 10kg 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 1.5s. The radius of the disc is 15cm . Determine the torsional spring constant of the wire. (Torsional spring constant a is defined by the relation J=aθ , where J is the restoring couple and θ the angle of twist).

Answer:

J=aθ

Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is

I=MR22

J=Id2θdt2aθ=MR22d2θdt2d2θdt2=2aMR2θ

The period of Torsional oscillations would be

T=2πMR22aa=2π2MR2T2a=2π2×10×(0.15)2(1.5)2a=1.97 N m rad1

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

(a) 5cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

ω=2πTω=2π0.2ω=10π rad s1

At displacement x acceleration is a=ω2x

At displacement x velocity is v=ωA2x2

(a)At displacement 5 cm

v=10π(0.05)2(0.05)2v=0

a=(10π)2×0.05a=49.35ms2

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

(b) 3cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

ω=2πTω=2π0.2ω=10π rad s1

At displacement x acceleration is a=ω2x

At displacement x velocity is v=ωA2x2

(a)At displacement 3 cm

v=10π(0.05)2(0.03)2v=10π0.0016v=10π×0.04v=1.257ms1

a=(10π)2×0.03a=29.61ms2

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

(c) 0cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

ω=2πTω=2π0.2ω=10π rad s1

At displacement x acceleration is a=ω2x

At displacement x velocity is v=ωA2x2

(a)At displacement 0 cm

v=10π(0.05)2(0)2v=10π×0.05v=1.57ms1

a=(10π)2×0a=0

Q. 6 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 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 . [Hint : Start with the equation x=acos(ωt+θ) and note that the initial velocity is negative.]

Answer:

At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring.

Let the amplitude be A

12kA2=12mv02+12kx02A=x02+mkv02

The angular frequency of a spring-mass system is always equal to km

Therefore

A=x02+v02ω2

Oscillation Class 11 Physics

NCERT answers for oscillations allow students to have a thorough understanding of the various factors that affect an object's motion. Period, frequency, simple harmonic motion, and uniform circular motion are a few of these factors. Chapter 13 of Physics in Class 11 NCERT Solutions offers a thorough explanation of every important concept involved, such as force laws, velocity, forced oscillations, systems that undergo harmonic motion (such as spring- mass systems), etc.

NCERT Solutions for Class 11 Physics Chapter-Wise

Have you ever wondered why gases appear to expand or how they produce pressure? The Kinetic Theory of Gases explains it by imagining that gases consist of small particles moving about rapidly. Scientists such as Boyle, Newton, Maxwell, and Boltzmann contributed significantly to this theory, which explains gas behavior, their response to changes in temperature and pressure, and even how they flow and mix together.

NCERT solutions for important chapter kinetic theory class 11 chapter 12 physics are created by subject matter experts to offer accurate and clear answers to each and every NCERT exercise problem. Students can build a strong concept by practicing these Kinetic Theory Class 11 Physics solutions. Step-by-step solutions make learning convenient and are provided below.

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NCERT Solutions for Class 11 Physics Chapter 12 Kinetic Theory - Exercise Questions

Q12.1 Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V actual is

Vactual=NA43π(d2)3
Vactual=6.023×1023×43π×(3×10102)3
Vactual=8.51×106m3
Vactual=8.51×103litres

The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres

VactualVmolar=8.51×10322.4

VactualVmolar=3.8×104

Q12.2 Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 00C ). Show that it is 22.4Litres .

Answer:

As per the ideal gas equation

PV=nRTV=nRTP

For one mole of a gas at STP we have

V=1×8.314×2731.013×105
V=0.0224m3

V=22.4 litres

Q12.3 Figure 13.8 shows plot of PV/T versus P for 1.00×103 kg of oxygen gas at two different temperatures.

(a) What does the dotted plot signify?
(b) Which is true: T1>T2orT1<T2 ?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If we obtained similar plots for 1.00×103 kg of hydrogen, would we get the same
value of PV/T at the point where the curves meet on the y-axis? If not, what mass
of hydrogen yields the same value of PV/T (for low pressure high temperature
region of the plot) ? (Molecular mass of H2=2.02μ , of O2=32.0μ ,
R=8.31Jmol1K1 .)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T 1 is closer to the horizontal line that the one corresponding to T 2 we conclude T 1 is greater than T 2.

(c) As per the ideal gas equation

PVT=nR

The molar mass of oxygen = 32 g

n=132

R = 8.314

nR=132×8.314nR=0.256JK1

(d) If we obtained similar plots for 1.00×103 kg of hydrogen we would not get the same
value of PV/T at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

m=PVTMR=0.256×28.314=5.48×105Kg

Q12.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 270C . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 170C . Estimate the mass of
oxygen taken out of the cylinder ( R=8.31Jmol1K1 , molecular mass of O2=32μ ).

Answer:

Initial volume, V 1 = Volume of Cylinder = 30 l

Initial Pressure P 1 = 15 atm

Initial Temperature T 1 = 27 o C = 300 K

The initial number of moles n 1 inside the cylinder is

n1=P1V1RT1n1=15×1.013×105×30×1038.314×300n1=18.28

Final volume, V 2 = Volume of Cylinder = 30 l

Final Pressure P 2 = 11 atm

Final Temperature T 2 = 17 o C = 290 K

The final number of moles n 2 inside the cylinder is

n2=P2V2RT2n2=11×1.013×105×30×1038.314×290n2=13.86

Moles of oxygen taken out of the cylinder = n 2 -n 1 = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

m=4.42×32m=141.44g

Q12.5 An air bubble of volume 1.0cm3 rises from the bottom of a lake 40m deep at a temperature of 120C . To what volume does it grow when it reaches the surface, which is at a temperature of 350C ?

Answer:

Initial Volume of the bubble, V 1 = 1.0 cm 3

Initial temperature, T 1 = 12 o C = 273 + 12 = 285 K

The density of water is ρw=103 kg m3

Initial Pressure is P 1

Depth of the bottom of the lake = 40 m

P1=Atmospheric Pressure+Pressure due to waterP1=Patm+ρwghP1=1.013×105+103×9.8×40P1=4.93×105Pa

Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure =1.013×105Pa

Let the final volume be V 2

As the number of moles inside the bubble remains constant, we have

P1V1T1=P2V2T2

V2=P1T2V1P2T1

V2=4.93×105×308×11.013×105×285

V2=5.26 cm3

Q12.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m3 at a temperature of 270C and 1atm pressure.

Answer:

The volume of the room, V = 25.0 m 3

Temperature of the room, T = 27 o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

n=PVRTn=1.013×105×258.314×300n=1015.35

Avogadro's Number, NA=6.022×1023

The number of molecules inside the room is N

N=nNAN=1015.35×6.022×1023N=6.114×1026

Q12.7 Estimate the average thermal energy of a helium atom at (i) room temperature ( 270C ), (ii) the temperature on the surface of the Sun ( 6000K ), (iii) the temperature of 10 million kelvin (the typical core temperature in the case of a star).

Answer:

The average energy of a Helium atom is given as 3kT2 since it is monoatomic

(i)

E=3kT2

E=3×1.38×1023×3002

E=6.21×1021 J

(ii)

E=3kT2

E=3×1.38×1023×60002

E=1.242×1019 J

(iii)

E=3kT2

E=3×1.38×1023×1072

E=2.07×1016 J

Q12.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain an equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v rms the largest?

Answer:

As per Avogadro's Hypothesis, under similar conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. Since the volume of the vessels is the same and all vessels are kept at the same conditions of pressure and temperature, they would contain an equal number of molecules.

Root mean square velocity is given as

vrms=3kTm

As we can see, vrms is inversely proportional to the square root of the molar mass; the root mean square velocity will be maximum in the case of Neon as its molar mass is the least.

Q12.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at 200C ? (atomic mass of Ar=39.9μ , of He=4.0μ ).

Answer:

As we know root mean square velocity is given as vrms=3RTM

Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at 200C

3R×T39.9=3R×2534T=2523.7 K

Q12.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0atm and temperature 170C . Take the radius of a nitrogen molecule to be roughly 1.0A . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of N2=28.0μ ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 o C

The radius of the Nitrogen molecule, r=1 Å.

The molecular mass of N 2 = 28 u

The molar mass of N 2 = 28 g

From the ideal gas equation

PV=nRT

nV=PRT

The above tells us about the number of moles per unit volume; the number of molecules per unit volume would be given as

n=NAnV=6.022×1023×2×1.013×1058.314×(17+273)

n=5.06×1025

The mean free path λ is given as

λ=12πnd2

λ=12×π×5.06×1025×(2×1×1010)2

λ=1.11×107 m

The root mean square velocity v rms is given as

vrms=3RTM

vrms=3×8.314×29028×103

vrms=508.26 m s1

The time between collisions T is given as

T=1Collision Frequency

T=1ν

T=λvrms

T=1.11×107508.26

T=2.18×1010s

Collision time T' is equal to the average time taken by a molecule to travel a distance equal to its diameter

T=dvrms

T=2×1×1010508.26

T=3.935×1013s

The ratio of the average time between collisions to the collision time is

TT=2.18×10103.935×1013TT=554
Thus, we can see that the time between collisions is much larger than the collision time.

Additional Questions

Q 1) A metre-long narrow bore held horizontally (and closed at one end) contains a 76cm long mercury thread, which traps a 15cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P 1 = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm 2

The initial volume of the air column, V 1 = 15x cm 3

Let's assume that once the tube is held vertical, y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P 2 = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V 2 = (24 + y)x cm 3

Since the temperature of the air column does not change

P1V1=P2V276×15x=y×(24+y)x1140=y2+24yy2+24y1140=0

Solving the above quadratic equation, we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore, y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore, once the tube is held vertically, 23.8 cm of Mercury flows out of it, and the length of the air column becomes 47.8 cm

Q 2) From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7cm3s1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2cm3s1. Identify the gas.

Answer:

As per Graham's Law of diffusion, if two gases of Molar Mass M 1 and M 2 diffuse with rates R 1 and R 2 respectively, their diffusion rates are related by the following equation

R1R2=M2M1

In the given question

R 1 = 28.7 cm 3 s -1

R 2 = 7.2 cm 3 s -1

M 1 = 2 g

M2=M1(R1R2)2M2=2×(28.77.2)2M2=31.78g

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q 3) A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n2=n1exp[mg(h2h1)/kbT]

where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for the sedimentation equilibrium of a suspension in a liquid column:

n2=n1exp[mgNA(ρρ)(h2h1)/(ρRT)]
where ρ is the density of the suspended particle, and ρ , that of the surrounding medium. [ NA s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes' principle to find the apparent weight of the suspended particle.]

Answer:

n2=n1exp[mg(h2h1)/kbT] (i)

Let the suspended particles be spherical and have a radius r

The gravitational force acting on the suspended particles would be

FG=43πr3ρg

The buoyant force acting on them would be

FB=43πr3ρg

The net force acting on the particles become

Fnet=FGFBFnet=43πr3ρg43πr3ρgFnet=43πr3g(ρρ)

Replacing mg in equation (i) with the above equation, we get

n2=n1exp[43πr3g(ρρ)(h2h1)/kbT]n2=n1exp[43πr3g(ρρ)(h2h1)RTNA]n2=n1exp[mgNA(ρρ)(h2h1)RTρ]

The above is the equation to be derived

Q 4) Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance

Atomic Mass (u)

Density (10 3 Kg m 3 )

Carbon (diamond)

12.01

2.22

Gold

197

19.32

Nitrogen (liquid)

14.01

1

Lithium

6.94

0.53

Fluorine

19

1.14

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Answer:

Let one mole of a substance of atomic radius r and density ρ have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is NA=6.022×1023

NA43πr3ρ=Mr=(3M4NAπρ)13

For Carbon

NA43πr3ρ=Mr=(3×12.014×6.022×1023×π×2.22×103)13r=1.29 Å.

For gold

NA43πr3ρ=Mr=(3×197.004×6.022×1023×π×19.32×103)13r=1.59 Å.

For Nitrogen

NA43πr3ρ=Mr=(3×14.014×6.022×1023×π×1.00×103)13r=1.77 Å.

For Lithium

NA43πr3ρ=Mr=(3×6.944×6.022×1023×π×0.53×103)13r=1.73 Å.

For Fluorine

NA43πr3ρ=Mr=(3×19.004×6.022×1023×π×1.14×103)13r=1.88 Å.

Class 11 Physics Chapter 12 NCERT solutions help students understand how tiny particles in matter behave. This chapter explains gases and their thermal properties, linking to thermodynamics. Studying it helps in exams and builds a strong foundation in physics.

Kinetic Theory Chapter 12 Physics NCERT Solutions: Concepts and Important Formulas

12.1 Introduction

The kinetic theory explains the behaviour of gases by assuming they consist of a large number of tiny particles (atoms or molecules) in constant random motion. This theory helps derive gas laws and explain properties like pressure, temperature, and specific heat.


12.2 Molecular Nature of Matter

  • Matter is made up of tiny molecules that are in continuous motion.
  • The molecular interactions vary in solids, liquids, and gases.
  • Gases have negligible interatomic forces, allowing them to expand and fill any container.

12.3 Behavior of Gases

  • Gases obey different laws like Boyle's Law and Charles's Law at low pressure and high temperature.
  • The Ideal Gas Equation is given by:

PV=nRT

where P= pressure, V= volume, n= number of moles, R= universal gas constant, T= temperature (Kelvin).


12.4 Kinetic Theory of an Ideal Gas

  • Gas molecules move in random directions, colliding elastically with each other and with the walls of the container.
  • Pressure of an Ideal Gas:

P=13ρvrms2

where ρ= density of gas, vamxam= root mean square velocity of gas molecules.

  • Root Mean Square Speed ( vmas ):

vrms=3kTm

where k= Boltzmann constant, T= absolute temperature, m= mass of a gas molecule.

12.5 Law of Equipartition of Energy

  • Energy is equally distributed among all degrees of freedom in thermal equilibrium.
  • Energy per molecule in an ideal gas:

E=f2kT

where f= degrees of freedom.


12.6 Specific Heat Capacity

  • Monoatomic gas: Cv=32R,Cp=52R
  • Diatomic gas: Cv=52R,Cp=72R
  • Relation between Cp and Cv :

CpCv=R

12.7 Mean Free Path

  • The mean free path (λ) is the average distance a molecule travels before colliding with another molecule.

λ=kT2πd2P

where d= diameter of the gas molecule, P= pressure .

Importance of NCERT Solutions for Class 11 Physics Chapter 12 – Kinetic Theory

  • Helps in understanding key concepts and equations in the chapter.
  • Useful for class exams and competitive exams like NEET & JEE Mains (1 question is usually asked).
  • Covers important formulas, making problem-solving easier.
  • This chapter holds 2-3% weightage in competitive exams, so mastering it is beneficial.

Kinetic Theory Class 11 Physics NCERT Topics

SectionTopic Name
12.1Introduction
12.2Molecular nature of matter
12.3Behaviour of gases
12.4Kinetic theory of an ideal gas
12.5Law of equipartition of energy
12.6Specific heat capacity
12.7Mean free path

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NCERT Solutions for Class 11 Subject Wise

Also, check NCERT Books and NCERT Syllabus here

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Importance of NCERT Solutions for Class 11 Physics Chapter 13 Oscillations:

For JEE Mains, oscillation and wave questions represent 6.67 percent of the total questions. The majority of the earlier oscillation-related JEE mains problems were from the subjects of SHM and basic pendulum. There will likely be two oscillation-related questions in the NEET test. To perform well in Class 11 and competitive exams, use the CBSE NCERT answers for Class 11 Physics Chapter 13 Oscillations.

Smart Tips to learn Class 11 Oscillations:

1. Understand Basics First- Understand what oscillations and SHM are before diving into formulas. Focus on SHM Concepts. Learn words such as amplitude, frequency, period, and phase correctly.

2. Use Graphs & Diagrams- Visualize SHM using displacement, velocity, and acceleration vs time graphs.

3. Memorize Formulas with Meaning- Don't just memorize—understand what each formula means and when to apply it.

4. Practice NCERT Numericals- Complete all examples and exercises, even for solid conceptual use.

5. Revise Regularly- Create a formula sheet and regularly revise it.


NCERT solutions for class 11 Subject-wise

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Frequently Asked Questions (FAQs)

1. What is the weightage of Oscillations Class 11 for NEET

From the NCERT chapter oscillations, two questions can be expected for NEET exam. For more questions solve NEET previous year papers.

2. Is the chapter oscillation important for JEE Main

Yes, oscillation is important for JEE Main. One or two question can be expected from oscillations for JEE Main. It is one of the important chapter for scholarship exams like KVPY and NSEP. To solve more problems on Oscillations refer to NCERT book, NCERT exemplar and JEE main previous year papers.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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