NCERT Solutions For Class 11 Physics Chapter 14 : Oscillations

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations is provided here which is one of the most common topics in Class 11 Physics examination. The Solutions of Chapter 14 Oscillations  for Class 11 Physics are created by subject experts in accordance with the CBSE syllabus.

The NCERT Solutions for Class 11 Physics Chapter 14 is prepared in such a way that students can easily understand the concepts of Oscillation. Get the NCERT Solutions for Class 11 Physics Chapter 14 PDF here.

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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 Fig. (b) is stretched by the same force F.


image name
Volume of the air chamber = V

Cross-sectional area of the neck = A

Mass of the ball = m

The ball is fitted in the neck at position C

The pressure of the air below the ball in the chamber is equal to the atmospheric pressure.

The ball is pressed down a little by increasing the pressure by a small amount p, so the ball moves down to position D.

Increase the pressure on the ball by a little amount p, so that the ball is depressed to position D

The distance C=y

The volume of the air chamber decreases and the pressure increases.

There will be a 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
Volumetric strain = change in volume/ original volume

ΔV/V = Ay/V

Bulk Modulus of elasticity, B = Stress/ volumetric strain

= -p/(Ay/V)

= -pV/Ay

p = – BAy/V

The restoring force on the ball due to the excess pressure

F = p x A = (- BAy/V) x A = – (BA2/V).y ——(1)

F ∝ y and the negative sign indicates that the force is directed towards the equilibrium position.

If the increased pressure is removed the ball will execute simple harmonic motion in the neck of the chamber with C as the mean position.
In S.H.M., the restoring force, F=ky ———(2)
Comparing (1) and (2),

-ky = – (BA2/V).y

k = (BA2/V)

Inertia factor = mass of ball =m

Time period, T = 2π√inertia factor/√spring factor

T = 2π √m/√k

T=2πmEA2V=2πAmVET = 2pi sqrt{frac{m}{frac{EA^{2}}{V}}}=frac{2pi }{A}sqrt{frac{mV}{E}}

Frequency, ν=1T=A2πEmV u =frac{1}{T}= frac{A}{2pi }sqrt{frac{E}{mV}}

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.


(a) Mass of the automobile = 3000 kg

The suspension sags by 15 cm

Decrease in amplitude =50% during one complete oscillation

Let k be the spring constant of each spring, then the spring constant of the four springs in parallel is

K= 4k

Since F = 4kx

Mg = 4kx

⇒ k = Mg/4x = (3000 x 10)/(4 x 0.15) = 5 x 104 N

(b) Each wheel supports 750 kg weight

t = 2π√m/√k = 2 x 3.14 x (√750/√5 x 104) = 0.77 sec

Using, x=x0ebt2mx = x_{0}e^{-frac{bt}{2m}},

we get


50100x0=x0eb×0.772×750frac{50}{100}x_{0} = x_{0}e^{-frac{b imes 0.77}{2 imes 750}}


loge2=(b×0.77)/(1500)logeelog_{e}2 = (b imes 0.77)/(1500)log_{e}e


b=(1500)loge20.77b= frac{(1500)log_{e}2 }{0.77}

b = (1500 x 0.6931)/0.77 = 1350.2 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.


Let m be the mass of the particle executing simple harmonic motion. The displacement of the particle at an instant t is given by

x = A sin ωt

Velocity of the particle, v= dx/dt = Aωcos ωt

Instantaneous Kinetic Energy, K = (1/2) mv2

= (1/2) m (Aωcos ωt)2

= (1/2) m (A2ω2cos2ωt)

Average value of kinetic energy over one complete cycle

Kav=1T0T12mA2ω2cos2ωtdtK_{av}=frac{1}{T}int_{0}^{T}frac{1}{2}mA^{2}omega ^{2}cos^{2}omega tdt =mA2ω22T0Tcos2ωtdt= frac{mA^{2}omega ^{2}}{2T}int_{0}^{T}cos^{2}omega tdt =mA2ω22T0T(1+cos2ωt)2dt= frac{mA^{2}omega ^{2}}{2T}int_{0}^{T}frac{(1+cos2omega t )}{2}dt =mA2ω24T[t+sin2ωt2ω]0T=mA2ω24T[(T0)+(sin2ωtsin02ω)]=14mA2w2egin{array}{l} =frac{m A^{2} omega^{2}}{4 T}left[t+frac{sin 2 omega t}{2 omega} ight]_{0}^{T} =frac{m A^{2} omega^{2}}{4 T}left[(T-0)+left(frac{sin 2 omega t-sin 0}{2 omega} ight) ight] =frac{1}{4} m A^{2} w^{2} end{array}

Average instantaneous potential energy, U = (1/2)kx2 = (1/2) mω2x2= (1/2)mωA2 sin2 ωt

Average value of potential energy over one complete cycle

Uav=1T0T12mω2A2sin2ωt=mω2A22T0Tsin2ωtdt=mω2A22T0T(1cos2ωt)2dt=mω2A24T[tsin2ωt2ω]0T=mω2A24T[(T0)(sin2ωtsin0)2ω]=14mω2A2egin{aligned} U_{a v} &=frac{1}{T} int_{0}^{T} frac{1}{2} m omega^{2} A^{2} sin ^{2} omega t=frac{m omega^{2} A^{2}}{2 T} int_{0}^{T} sin ^{2} omega t d t &=frac{m omega^{2} A^{2}}{2 T} int_{0}^{T} frac{(1-cos 2 omega t)}{2} d t &=frac{m omega^{2} A^{2}}{4 T}left[t-frac{sin 2 omega t}{2 omega} ight]_{0}^{T} &=frac{m omega^{2} A^{2}}{4 T}left[(T-0)-frac{(sin 2 omega t-sin 0)}{2 omega} ight] &=frac{1}{4} m omega^{2} A^{2} end{aligned}

Kinetic energy = Potential energy

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).


Mass of the circular disc = 10 kg

Period of torsional oscillation = 1.5 s

Radius of the disc = 15 cm = 0.15 m

Restoring couple, J = –α θ

Moment of inertia, I = (1/2) mR2

= (1/2) x 10 x (0.15)2

= 0.1125 kgm2

Time period is given by the relation

T=2πIαT = 2pi sqrt{frac{I}{alpha }}

So, α=4π2IT2alpha =frac{4pi^{2}I }{T^{2}} α=4×(3.14)2×0.1125(1.5)2alpha = frac{4 imes (3.14)^{2} imes 0.1125}{(1.5)^{2}}

= 4.44/2.25

= 1.97 Nm/rad

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.


Amplitude = 5 cm = 0.05 m

Time period = 0.2 s

When the displacement is y, then

acceleration, A = -ω2y

Velocity, v=ωr2y2v = omega sqrt{r^{2}-y^{2}}

ω = 2π/T

= 2π/0.2 = 10π rad/s

(a) When the displacement y = 5 cm = 0.05 m

Acceleration, A = – (10π)2(0.05) = 5π2 m/s2

Velocity, V=10π(0.05)2(0.05)2=0V = 10pi sqrt{(0.05)^{2}-(0.05)^{2}}=0

(b) When the displacement y = 3 cm = 0.03 m

Acceleration, A = – (10π)2(0.03) = 3π2 m/s2

Velocity, V=10π(0.05)2(0.03)2=10π×0.04=0.4πm/sV = 10pi sqrt{(0.05)^{2}-(0.03)^{2}}=10pi imes 0.04 = 0.4pi m/s

(c) When the displacement y = 0

Acceleration, A =  – (10π)2(0) = 0

Velocity, V=10π(0.05)2(0)2=10π×0.05=0.5πm/sV = 10pi sqrt{(0.05)^{2}-(0)^{2}}=10pi imes 0.05 = 0.5pi 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. [Hint: Start with the equation x = a cos (ωt+θ) and note that the initial velocity is negative.]


The angular velocity of the spring = ω

x = a cos (ωt+θ)

At t = 0, x = x0

Substituting these values in the above equation we get

x0 = A cos θ —–(1)

Velocity, v= dx/dt = – Aω sin (ωt+θ)

At t = 0, v = – v0

Substituting these values in the above equation we get

– v= – Aω sin θ

Asin θ =  v0/ω———-(2)

Squaring and adding (1) and (2) we get

A2(cos2θ+sin2θ)=x02+v02ω2A^{2}(cos^{2} heta+sin^{2} heta )=x_{0}^{2}+frac{v_{0}^{2}}{omega ^{2}}


In our daily life, we encounter various kinds of motions. The study of oscillatory motion is basic to Physics. Oscillations and Resonance is one of the most interesting chapters. Students will get to know about periodic motion, period, Simple harmonic motion, and other related concepts from this chapter. Some important concepts and key points of Oscillations and Resonance are given below.

  • The period T is the least time after which motion repeats itself. Thus, motion repeats itself after nT where n is an integer.
  • Every periodic motion is not SHM. Only that periodic motion governed by the force law F = – k x is simple harmonic.
  • The motion of a simple pendulum is simple harmonic for small angular displacement
  • Under forced oscillation, the phase of the harmonic motion of the particle differs from the phase of the driving force.


Class 11 Physics NCERT Solutions for  Chapter 14 Oscillations

These NCERT Solutions provide you with the answers to the question from the textbook, important questions from previous year question papers and sample papers. The solution comprises worksheets, exemplary problems, short and long answer questions, MCQs, tips and tricks to help you understand them thoroughly. NCERT Solutions are one of the best tools to prepare Physics for Class 11. Oscillations  is a very important chapter in CBSE Class 11. Students must prepare this Chapter well to score high marks in their board examination. The NCERT Solutions for Class 11 Physics Oscillations is given below so that students can understand the concepts of this Chapter in-depth.

Subtopics of Class 11 Physics Chapter 14 Oscillations

  1. Introduction
  2. Periodic and oscillatory motions
  3. Simple harmonic motion
  4. Simple harmonic motion and uniform circular motion
  5. Velocity and acceleration in simple harmonic motion
  6. Force law for simple harmonic motion
  7. Energy in simple harmonic motion
  8. Some systems executing SHM
  9. Damped simple harmonic motion
  10. Forced oscillations and resonance

Students are suggested to get well versed with the concepts in this chapter to score well in their board examination. The concepts in this topic will also be useful for students when they will prepare for competitive entrance examinations like JEE and NEET.

Students must follow some strategies while preparing for their physics exam. Some effective strategies to prepare are given below.

  • Students must know the syllabus completely during their preparation. It will help them to understand the pattern of the exam.
  • They must follow a timetable during their preparation.
  • Students must thoroughly follow the NCERT books while preparing.
  • Making notes is one of the best ways to retain the concepts for a longer time
  • Practise lots of question papers and sample papers.

CoolGyan’S bring you the best study materials, notes, sample papers, important questions, MCQs, NCERT Books, tips and tricks that help you to face Class 11 examination confidently and score well. CoolGyan’S videos and animations help you to remember the concepts for a long period.

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The NCERT Solutions for Class 11 Physics Chapter 14 from CoolGyan’S are designed with the main aim of helping students to focus on the important concepts. Each and every minute detail is explained with utmost care to improve the conceptual knowledge among students. The solutions also contain various shortcut techniques which can be used to remember the concepts effectively. Students can refer to the solutions while answering the textbook questions and understand the method of answering them without any difficulty.

What are the topics covered under the Chapter 14 of NCERT Solutions for Class 11 Physics?

The topics covered under the Chapter 14 of NCERT Solutions for Class 11 Physics are –
Periodic and oscillatory motions
Simple harmonic motion
Simple harmonic motion and uniform circular motion
Velocity and acceleration in simple harmonic motion
Force law for simple harmonic motion
Energy in simple harmonic motion
Some systems executing SHM
Damped simple harmonic motion
Forced oscillations and resonance

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