# 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:

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

(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

(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.

(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.

(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]

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.

(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):

(a)  $\inline sin\; \omega t-cos\; \omega t$

$\\sin\omega t-cos\omega t\\ =\sqrt{2}\left ( \frac{1}{\sqrt{2}}sin\omega t-\frac{1}{\sqrt{2}}cos\omega t \right )\\ =\sqrt{2}(cos\frac{\pi }{4}sin\omega t-sin\frac{\pi }{4}cos\omega t)\\ =\sqrt{2}sin(\omega t-\frac{\pi }{4})$

Since the above function is of form $Asin(\omega t+\phi )$ it represents SHM with a time period of $\frac{2\pi }{\omega }$

(b) $sin^{3}\omega t$

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

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 $\frac{2\pi }{\omega }$

(d)   $cos\; \omega t+cos\; 3\omega t+cos\; 5\omega t$

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

$period=LCM(\frac{2\pi}{\omega},\frac{2\pi}{3\omega},\frac{2\pi}{5\omega})=\frac{2\pi}{\omega}$

(e)  $exp(-\omega ^{2}t^{2})$

The given function is exponential and therefore does not represent periodic motion.

(f)  $1+\omega t+\omega^{2}t^{2}$

The given function does not represent periodic motion.

(a) at the end A,

Velocity is zero. Force and acceleration are in the positive direction.

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

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

(a)  $a=0.7x$

(b) $a=-200x^2$

(c) $a=-10x$

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.

If the initial $\inline (t=0)$  position of the particle is $\inline 1\; cm$ and its initial velocity is $\inline \omega \; cm/s,$  what are its amplitude and initial phase angle ? The angular frequency of the particle is  $\inline \pi s^{-1}.$ If instead of the cosine function, we choose the sine function to describe the SHM : $\inline x=B\; sin(\omega t+\alpha ),$  what are the amplitude and initial phase of the particle with the above initial conditions.

$\omega =\pi\ rad\ s^{-1}$

$x(t)=Acos(\pi t+\phi )$

at t = 0

$\\x(0)=Acos(\pi \times 0+\phi )\\ 1=Acos\phi\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (i)$

$\\v=\frac{\mathrm{d}x(t) }{\mathrm{d} t}\\ v(t)=-A\pi sin(\pi t+ \phi )$

at t = 0

$\\v(0) =-A\pi sin(\pi \times 0+ \phi ) \\ \omega =-A\pi sin\phi\\\ 1 =-A sin\phi \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (ii)$

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

$\\1^{2}+1^{2}=(Acos\phi )^{2}+(-Asin\phi )^{2} \\2=A^{2}cos^{2}\phi +A^{2}sin^{2}\phi \\ 2=A^{2}\\ A=\sqrt{2}$

Dividing equation (ii) by (i) we get

$\\tan\phi =-1\\ \phi =\frac{3\pi }{4},\frac{7\pi }{4},\frac{11\pi }{4}......$

$x(t)=Bsin(\pi t+\alpha )$

at t = 0

$\\x(0)=Bsin(\pi \times 0+\alpha )\\ 1=Bsin\alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iii)$

$\\v=\frac{\mathrm{d}x(t) }{\mathrm{d} t}\\ v(t)=B\pi cos(\pi t+ \alpha )$

at t = 0

$\\v(0) =B\pi cos(\pi \times 0+ \alpha ) \\ \omega =B\pi cos\alpha \\\ 1 =B cos\alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (iv)$

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

$\\1^{2}+1^{2}=(Bsin\alpha )^{2}+(Bcos\alpha )^{2} \\2=B^{2}sin^{2}\alpha +B^{2}cos^{2}\alpha \\ 2=B^{2}\\ B=\sqrt{2}$

Dividing equation (iii) by (iv) we get

$\\tan\alpha =1\\ \alpha =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}......$

Spring constant of the spring is given by

$\\k=\frac{Weight\ of\ Maximum\ mass\ the\ scale\ can\ read}{Maximum\ displacement\ of\ the\ scale}\\ k=\frac{50\times 9.8}{20\times 10^{-2}}\\ k=2450\ Nm^{-1}$

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

$\\T=2\pi \sqrt{\frac{m}{k}}\\ m=\frac{T^{2}k}{4\pi ^{2}}\\ m=\frac{(0.6)^{2}\times 2450}{4\pi ^{2}}\\ m=22.34\ kg\\ w=mg\\ w=22.34\times 9.8\\ w=218.95\ N$

Determine

(i) the frequency of oscillations,

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

$\\\nu =\frac{1}{2\pi }\sqrt{\frac{k}{m}}\\ \nu =\frac{1}{2\pi }\times \sqrt{\frac{1200}{3}}\\ \nu =3.183\ Hz$

Determine

(ii) maximum acceleration of the mass, and

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

$\\a_{max}=\frac{F_{A}}{m}\\ a_{max}=\frac{kA}{m}\\ a_{max}=\frac{1200\times 0.2}{3} \\a_{max}=8ms^{-2}$ (FA = Force experienced by body at displacement A from mean position)

Determine

(iii) the maximum speed of the mass.

Maximum speed occurs at the mean position and is given by

$\\v_{max}=A\omega \\ v_{max}=0.02\times 2\pi \times 3.18\\ v_{max}=0.4ms^{-1}$

(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?

Amplitude is A = 0.02 m

Time period is $\\\omega$

$\\\omega =\sqrt{\frac{k}{m}}\\ \omega =\sqrt{\frac{1200}{3}}\\ \omega =20\ rad/s$

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

$\\x(t)=0.02sin\left ( 20t \right )$

Here x is in metres and t is in seconds.

(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?

Amplitude is A = 0.02 m

Time period is $\\\omega$

$\\\omega =\sqrt{\frac{k}{m}}\\ \omega =\sqrt{\frac{1200}{3}}\\ \omega =20\ rad/s$

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

x(0) = A

$\phi =\frac{\pi }{2}$

$\\x(t)=0.02sin\left ( 20t + \frac{\pi }{2}\right )\\ x(t)=0.02cos(20t)$

Here x is in metres and t is in seconds.

(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?

Amplitude is A = 0.02 m

Time period is $\\\omega$

$\\\omega =\sqrt{\frac{k}{m}}\\ \omega =\sqrt{\frac{1200}{3}}\\ \omega =20\ rad/s$

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

x(0) = -A

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

$\\x(t)=0.02sin\left ( 20t + \frac{3\pi }{2}\right )\\ x(t)=-0.02cos(20t)$

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.

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

(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.

(a)  $x=-2\; sin(3t+\pi /3)$

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

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

(b)  $x=cos(\pi /6-t)$

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

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

(c)  $\inline x=3\; sin(2\pi t+\pi /4)$

$\inline x=3\; sin(2\pi t+\pi /4)$

$\\x=-3cos(2\pi t+\frac{\pi}{4}+\frac{\pi}{2})\\ \ =3cos(2\pi t+\pi+\frac{\pi}{4}+\frac{\pi}{2})\\=3cos(2\pi t+\frac{3\pi}{2}+\frac{\pi}{4})\\=3cos(2\pi t+\frac{7\pi}{4})$

At t= 0

$phase=\frac{7\pi}{4}$

Reference circle is as follows

## Oscillations Excercise:

### Question:

(d)  $x=2\; cos\; \pi t$

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

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

Q. 14.13 (a) Figure 14.30

(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 14.30 (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. 14.30

(b) is stretched by the same force F.

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

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

At maximum extension

$\\F=Kx\\ x=\frac{F}{k}$

(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

$\\F=k\frac{x}{2}+k\frac{x}{2}\\\Rightarrow x=\frac{F}{k}$

## Oscillations Excercise:

### Question:

(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?

(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

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

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

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

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?

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.

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?

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?

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.

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'

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

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).

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

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

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:

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

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

(a) the spring constant $\inline K$

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

$\\4kx=mg\\ k=\frac{3000\times 9.8}{4\times 0.15}\\ k=4.9\times 10^{4}\ N$

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

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

$\\e^{-\frac{0.77b}{2\times 750}}=0.5\\ ln(e^{-\frac{0.77b}{2\times 750}})=ln0.5\\ \frac{0.77b}{1500}=ln2\\ b=\frac{0.693\times 1500}{0.77}\\ b=1350.2287\ kg\ s^{-1}$

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

$\inline J=-a\; \theta$

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

$I=\frac{MR^{2}}{2}$

$\\J=I\frac{\mathrm{d}^{2}\theta }{\mathrm{d} t^{2}}\\ -a\theta =\frac{MR^{2}}{2}\frac{\mathrm{d}^{2}\theta }{\mathrm{d} t^{2}}\\ \frac{\mathrm{d}^{2}\theta }{\mathrm{d} t^{2}}=-\frac{2a}{MR^{2}}\theta$

The period of Torsional oscillations would be

$\\T=2\pi \sqrt{\frac{MR^{2}}{2a}}\\ a=\frac{2\pi ^{2}MR^{2}}{T^{2}}\\ a=\frac{2\pi ^{2}\times 10\times (0.15)^{2}}{(1.5)^{2}}\\ a=1.97\ N\ m\ rad^{-1}$

(a)  $5\; cm$

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

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

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

(b)  $\inline 3\; cm$

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

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

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

(c) $0 \; cm$

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

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

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

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

$\\\frac{1}{2}kA^{2}=\frac{1}{2}mv_{0}^{2}+\frac{1}{2}kx_{0}^{2}\\ A=\sqrt{x_{0}^{2}+\frac{m}{k}v_{0}^{2}}$

The angular frequency of a spring-mass system is always equal to $\sqrt{\frac{k}{m}}$

Therefore

$A=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega ^{2}}}$

## 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.