NCERT Solutions for Class 11 Physics Chapter 14 Oscillations

 

NCERT Solutions for Class 11 Physics Chapter 14 Oscillations: Problems on motion along a straight line, motion in a plane, projectile motion, etc were discussed in the previous chapters. Solutions of NCERT class 11 physics chapter 14 oscillations explains problems on periodic and oscillatory motions. The motion which repeats after a certain interval of time is called periodic motion. For example, the motion of a planet around the sun, the motion of pendulum of a wall clock etc are periodic. To and fro periodic motion about a mean position is known as oscillatory motion. CBSE NCERT solutions for class 11 physics chapter 14 oscillations have questions on simple harmonic motion (SHM). SHM is the simplest form of oscillatory motion. In SHM the force on the oscillating body is directly proportional to the displacement about the mean position and is directed towards the mean position. NCERT solutions are an important tool to score well in the exams and also they are useful if you want to study other subjects of other classes as well. Concept studied in the chapter becomes easy to understand with the help of NCERT solutions for class 11 physics chapter 14 oscillations.

The main topics discussed in NCERT class 11 physics chapter 14 oscillations:

14.1 Introduction

14.2 Periodic and oscillatory motions

14.3 Simple harmonic motion

14.4 Simple harmonic motion and uniform circular motion

14.5 Velocity and acceleration in simple harmonic motion

14.6 Force law for simple harmonic motion

14.7 Energy in simple harmonic motion

14.8 Some systems executing simple harmonic motion

14.9 Damped simple harmonic motion

14.10 Forced oscillations and resonance

Other types of oscillatory motions that are discussed in the NCERT class 11 Physics Chapter 14 are damped oscillations and forced oscillations. In damped oscillation, as the name indicates the oscillation gets damped after an interval of time. And if the damping is small the motion will be approximately periodic, the position-time graph of such damping oscillation is shown below.

We have seen the oscillation of a pendulum, this oscillation will be damped unless an external force is applied to maintain the oscillation. Such maintained oscillation due to an external agency is called forced or driven oscillations.

NCERT solutions for class 11 physics chapter 14 oscillations exercise:

Q .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)?

               

 

Answer:

The x-t plots for linear motion of a particle in Fig. 14.23 (b) and (d) represent periodic motion with both having a period of motion of two seconds.

Q. 14.4 (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 (\omega is any positive constant):

(b) sin^{3}\omega t

Answer:

\\sin3\omega t =3sin\omega t -4sin^{3}\omega t \\sin^{3}\omega t=\frac{1}{4}\left ( 3sin\omega t - sin3\omega t \right )\\

The two functions individually represent SHM but their superposition does not give rise to SHM but the motion will definitely be periodic with a period of \frac{2\pi }{\omega }

Q. 14.6 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=-200x^2

(c) a=-10x

(d) a=100x^3        

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. 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.

Answer:

(a) Let the required function be x(t)=asin(\pm \omega t+\phi ) 

Amplitude = 3 cm = 0.03 m

T = 2 s

\\\omega =\frac{2\pi }{T}\\ \omega =\pi rad\ s

Since initial position x(t) = 0, \phi =0

As the sense of revolution is clock wise

\\x(t)=0.03sin(-\omega t)\\ x(t)=-0.03sin(\pi t)

Here x is in metres and t is in seconds.

(b)Let the required function be x(t)=asin(\pm \omega t+\phi ) 

Amplitude = 2 m

T = 4 s

\\\omega =\frac{2\pi }{T}\\ \omega =\frac{\pi }{2} rad\ s

Since initial position x(t) = -A, \phi =\frac{3\pi }{2}

As the sense of revolution is anti-clock wise

\\x(t)=2sin(\omega t+\frac{3\pi }{2})\\ x(t)=-2cos(\frac{\pi }{2} t)

Here x is in metres and t is in seconds.

Q. 14.12 (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=-2\; sin(3t+\pi /3)

Answer:

\\x=-2\sin(3t+\pi /3)\\ x=2cos(3t+\frac{\pi }{3}+\frac{\pi }{2})\\ x=2cos(3t+\frac{5\pi }{6})

The initial position of the particle is x(0)

\\x(0)=2cos(0+\frac{5\pi }{6})\\ x(0)=2cos(\frac{5\pi }{6})\\ x(0)=-\sqrt{3}cm

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

The angular speed of the rotating particle is \omega =3rad\ s^{-1}

Initial phase is 

\\\phi =\frac{5\pi }{6}\\ \phi =150^{o}

The reference circle for the given simple Harmonic motion is

Reference plot for SHM

Q. 14.12 (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(\pi /6-t)

Answer:

\\x(t)=cos(\frac{\pi }{6}-t)\\ x(t)=cos(t-\frac{\pi }{6})

The initial position of the particle is x(0)

\\x(0)=cos(0-\frac{\pi }{6})\\ x(0)=cos(\frac{\pi }{6})\\ x(0)=\frac{\sqrt{3}}{2}cm

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

The angular speed of the rotating particle is \omega =1rad\ s^{-1}

Initial phase is 

\\\phi =-\frac{\pi }{6}\\ \phi =-30^{o}

The reference circle for the given simple Harmonic motion is

Oscillations Excercise:

Question:

 

Q. 14.12 (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=2\; cos\; \pi t

 

Answer:

\\x(t)=2cos(\pi t)\\

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 \omega =\pi rad\ s^{-1}

Initial phase is 

\\\phi =0^{o}

The reference circle for the given simple Harmonic motion is

Oscillations Excercise:

Question:

 

Q. 14.13 (b) 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.

                   

(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?

Answer:

(b).(a) In Fig, (a) we have

F=-kx

ma=-kx

a=-\frac{k}{m}x

\\\omega ^{2}=\frac{k}{m}\\ T=\frac{2\pi }{\omega }\\ T=2\pi \sqrt{\frac{m}{k}}

(b) In fig (b) the two equal masses will be executing SHM about their centre of mass. The time  period of the system would be equal to a single object of same mass m attached to a spring of half the length of the given spring (or undergoing half the extension of the given spring while applied with the same force)

Spring constant of such a spring would be 2k

F=-2kx

ma=-2kx

\\a=-\frac{2k}{m}x\\ \omega ^{2}=\frac{2k}{m}\\ T=\frac{2\pi }{\omega }\\ T=2\pi \sqrt{\frac{m}{2k}}\\ T=\pi \sqrt{\frac{2m}{k}}

Q. 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 ?

Answer:

Amplitude of SHM = 0.5 m

angular frequency is

\\\omega =200\ rad/min \\\omega =3.33\ rad/s

If the equation of SHM is given by

\\x(t)=Asin(\omega t+\phi )\\

The velocity would be given by

\\v(t)=\frac{\mathrm{d} x(t)}{\mathrm{d} t}\\ v(t)=\frac{\mathrm{d} (Asin(\omega t+\phi ))}{\mathrm{d} t}\\ v(t)=A\omega cos(\omega t+\phi )

The maximum speed is therefore

\\v_{max}=A\omega \\ v_{max}=0.5\times 3.33 \\v_{max}=1.67ms^{-1}

Q. 14.15 The acceleration due to gravity on the surface of moon is 1.7m\; 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)

Answer:

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

T=2\pi \sqrt{\frac{l}{g}}

g= 9.8 m s-2

gm = 1.7 m s-2

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

The time period of the pendulum on the surface of the moon is Tm

\\\frac{T_{m}}{T_{e}}=\sqrt{\frac{g_{e}}{g_{m}}}\\ T_{m}=T_{e}\times \sqrt{\frac{g_{e}}{g_{m}}}\\ T_{m}=3.5\times \sqrt{\frac{9.8}{1.7}} \\T_{m}=8.4s

Q. 14.16 (a)  Answer the following questions :

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

                        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?

Answer:

In case of spring, the spring constant is independent of the mass attached whereas in case of a pendulum k is proportional to m making k/m constant and thus the time period comes out to be independent of the mass of the body attached.

Q. 14.16 (b) Answer the following questions :

(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.

Answer:

In reaching the result T=2\pi \sqrt{\frac{l}{g}} we have assumed sin(x/l)=x/l. This assumption is only true for very small values of x (or\ \theta ). Therefore it is obvious that once x takes larger values we will have deviations from the above-mentioned value.

Q. 14.16 (c) Answer the following questions :

(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?

Answer:

The watch must be using an electrical circuit or a spring system to tell the time and therefore free falling would not affect the time his watch predicts.

Q. 14.16 (d) Answer the following questions :

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

Answer:

While free falling the effective value of g inside the cabin will be zero and therefore the frequency of oscillation of a simple pendulum would be zero i.e. it would not vibrate at all because of the absence of a restoring force.

Q .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?

Answer:

Acceleration due to gravity = g (in downwards direction)

Centripetal acceleration due to the circular movement of the car = ac

a_{c}=\frac{v^{2}}{R} (in the horizontal direction)

Effective acceleration is

\\g'=\sqrt{g^{2}+a_{c}^{2}}\\ g'=\sqrt{g^{2}+\frac{v^{4}}{R^{2}}}

The time period is T'

\\T'=2\pi \sqrt{\frac{l}{g'}}\\ T'=2\pi \sqrt{\frac{l}{\sqrt{g^{2}+\frac{v^{4}}{R^{2}}}}}

Q. 14.18 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density \rho _{\imath }. 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{h\rho }{\rho _{ 1}g}} where \rho is the density of cork. (Ignore damping due to 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=-\rho _{1}gAx\\ \rho Aha=-\rho _{1}gAx\\ \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}=-\frac{\rho _{1}g}{\rho h}x

Comparing with a=-kx we have

\\k=\frac{\rho _{1}g}{\rho h}\\ T=\frac{2\pi }{\sqrt{k}}\\ T=2\pi \sqrt{\frac{\rho h}{\rho_{1}g }}

Q. 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.

Answer:

Let the height of each mercury column be h.

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

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

Let the density of mercury be \rho

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

Weight of this difference is 2Ax\rho g

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

\\2hA\rho (-a)=2xA\rho g\\ a=-\frac{g}{h}x

\frac{\mathrm{d}^{2}x }{\mathrm{d} t^{2}}=-\frac{g}{h}x which is the equation of a body executing S.H.M

The time period of the oscillation would be

T=2\pi \sqrt{\frac{h}{g}}

NCERT solutions for class 11 physics chapter 14 oscillations additional exercise:

Q. 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.14.33). 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. 14.33].

                                

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 \Delta P

Let the Bulk's modulus of air be K.

\\K=\frac{\Delta P}{\Delta V/V}\\ \Delta P=\frac{Kax}{V}

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=a\Delta P\\ -m\frac{\mathrm{d^{2}}x }{\mathrm{d}t^{2}}=a\Delta p\\ \frac{\mathrm{d^{2}}x }{\mathrm{d}t^{2}}=-\frac{ka^{2}}{mV}x

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

The time period of the oscillation would be

T=\frac{2\pi }{a}\sqrt{\frac{mV}{k}}

Q. 14.21 (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 15\; cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50^{o}/_{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 supports750 \; kg..

Answer:

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

\\T=2\pi \sqrt{\frac{m}{k}}\\ T=2\pi \times \sqrt{\frac{3000}{4\times 4.9\times 10^{4}}}\\ T=0.77\ s

For damping factor b we have

x=x_{0}e^{\left ( -\frac{bt}{2m} \right )}

x=x0/2

t=0.77s

m=750 kg

\\e^{-\frac{0.77b}{2\times 750}}=0.5\\ ln(e^{-\frac{0.77b}{2\times 750}})=ln0.5\\ \frac{0.77b}{1500}=ln2\\ b=\frac{0.693\times 1500}{0.77}\\ b=1350.2287\ kg\ s^{-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.

Answer:

Let the equation of oscillation be given by x=Asin(\omega t)

Velocity would be given as 

\\v=\frac{dx}{dt}\\ v=A\omega cost(\omega t)

Kinetic energy at an instant is given by

\\K(t)=\frac{1}{2}m(v(t))^{2}\\ K(t)=\frac{1}{2}m(A\omega cos(\omega t))^{2}\\ K(t)=\frac{1}{2}mA^{2}\omega ^{2}cos^{2}\omega t

Time Period  is given by

T=\frac{2\pi }{\omega }

The Average Kinetic Energy would be given as follows

\\K_{av}=\frac{\int _{0}^{T}K(t)dt}{\int _{0}^{T}dt}\\ K_{av}=\frac{1}{T}\int _{0}^{T}K(t)dt\\ K_{av}=\frac{1}{T}\int_{0}^{T}\frac{1}{2}mA^{2}\omega ^{2}cos^{2}\omega t\ dt\\K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\int_{0}^{T}cos^{2}\omega t\ dt\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\int_{0}^{T}\left ( \frac{1+cos2\omega t}{2} \right )dt

\\K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\left [ \frac{t}{2} +\frac{sin2\omega t}{4\omega }\right ]_{0}^{T}\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\left [ \left ( \frac{T}{2}+\frac{sin2\omega T}{4\omega } \right )-\left ( 0+sin(0) \right ) \right ]\\ K_{av}=\frac{mA^{2}\omega ^{2}}{2T}\times \frac{T}{2}\\ K_{av}=\frac{mA^{2}\omega ^{2}}{4}

The potential energy at an instant T is given by 

\\U(t)=\frac{1}{2}kx^{2}\\ U(t)=\frac{1}{2}m\omega ^{2}(Asin(\omega t))^{2}\\ U(t)=\frac{1}{2}m\omega ^{2}A^{2}sin^{2}\omega t

The Average Potential Energy would be given by

\\U_{av}=\frac{\int_{0}^{T}U(t)dt}{\int_{0}^{T}dt}\\ \\U_{av}=\frac{1}{T}\int_{0}^{T}\frac{1}{2}m\omega ^{2}A^{2}sin^{2}\omega t\ dt\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\int_{0}^{T}sin^{2}\omega t\ dt\\\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\int_{0}^{T}\frac{(1-cos2\omega t)}{2}dt

\\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\left [ \frac{t}{2} -\frac{sin2\omega t}{4\omega }\right ]_{0}^{T}\\ \\U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\left [ \left ( \frac{T}{2}-\frac{sin2\omega T}{4\omega } \right )-\left ( 0-sin0 \right ) \right ]\\ U_{av}=\frac{m\omega ^{2}A^{2}}{2T}\times \frac{T}{2}\\ U_{av}=\frac{m\omega ^{2}A^{2}}{4}

We can see Kav = Uav

Q. 14.24 (a)  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

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

\\\omega =\frac{2\pi }{T}\\ \omega =\frac{2\pi }{0.2}\\ \omega =10\pi\ rad\ s^{-1}

At displacement x acceleration is a=-\omega ^{2}x

At displacement x velocity is v=\omega \sqrt{A^{2}-x^{2}}

(a)At displacement 5 cm

\\v=10\pi \sqrt{(0.05)^{2}-(0.05)^{2}}\\ v=0

\\a=-(10\pi )^{2}\times 0.05\\a=-49.35ms^{-2}

Q. 14.24 (b) 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

(b)  3\; cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

\\\omega =\frac{2\pi }{T}\\ \omega =\frac{2\pi }{0.2}\\ \omega =10\pi\ rad\ s^{-1}

At displacement x acceleration is a=-\omega ^{2}x

At displacement x velocity is v=\omega \sqrt{A^{2}-x^{2}}

(a)At displacement 3 cm

\\v=10\pi \sqrt{(0.05)^{2}-(0.03)^{2}}\\ v=10\pi \sqrt{0.0016}\\v=10\pi \times 0.04\\v=1.257ms^{-1}

\\a=-(10\pi )^{2}\times 0.03\\a=-29.61ms^{-2}

Q. 14.24 (c) 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 

(c) 0 \; cm

Answer:

A = 5 cm = 0.05 m

T = 0.2 s

\\\omega =\frac{2\pi }{T}\\ \omega =\frac{2\pi }{0.2}\\ \omega =10\pi\ rad\ s^{-1}

At displacement x acceleration is a=-\omega ^{2}x

At displacement x velocity is v=\omega \sqrt{A^{2}-x^{2}}

(a)At displacement 0 cm

\\v=10\pi \sqrt{(0.05)^{2}-(0)^{2}}\\ v=10\pi \times 0.05\\v=1.57ms^{-1}

\\a=-(10\pi )^{2}\times0\\a=0

NCERT solutions for class 11 physics chapter wise

Chapter 1

NCERT solutions for class 11 physics chapter 1 Physical world

Chapter 2

Solutions of NCERT for class 11 physics chapter 2 Units and Measurement

Chapter 3

CBSE NCERT solutions for class 11 physics chapter 3 physics Motion in a straight line

Chapter 4

NCERT solutions for class 11 physics chapter 4 Motion in a Plane

Chapter 5

Solutions of NCERT for class 11 physics chapter 5 Laws of Motion

Chapter 6

CBSE NCERT solutions for class 11 physics chapter 6 Work, Energy and Power

Chapter 7

NCERT solutions for class 11 physics chapter 7 System of Particles and Rotational motion

Chapter 8

Solutions of NCERT for class 11 physics chapter 8 Gravitation

Chapter 9

CBSE NCERT solutions for class 11 physics chapter 9 Mechanical Properties of Solids

Chapter 10

NCERT solutions for class 11 physics chapter 10 Mechanical Properties of Fluids

Chapter 11

Solutions of NCERT for class 11 physics chapter 11 Thermal Properties of Matter

Chapter 12

CBSE NCERT solutions for class 11 physics chapter 12 Thermodynamics

Chapter 13

NCERT solutions for class 11 physics chapter 13 Kinetic Theory

Chapter 14

NCERT solutions for class 11 physics chapter 14 Oscillations

Chapter 15

CBSE NCERT solutions for class 11 physics chapter 15 Waves

NCERT Solutions for Class 11 Subject wise

CBSE NCERT solutions for class 11 biology

NCERT solutions for class 11 maths

CBSE NCERT solutions for class 11 chemistry

NCERT solutions for class 11 physics

Importance of NCERT solutions for class 11 physics chapter 14 oscillations:

On an average 6.67 % of questions from oscillation and waves are asked for JEE Mains.  Most of the previous JEE mains questions from oscillation asked are from topics SHM and simple pendulum. For NEET exam 2 questions are expected from oscillation. The CBSE NCERT solutions for class 11 physics chapter 14 oscillations will help to score well in class 11 and competitive exams.

 

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