Ever wondered why a pendulum swings or how sound travels in waves? The NCERT Solutions for Class 11 Physics Chapter 13 Oscillations help you to understand all this in a very simple and clear way. This chapter is very important because oscillations are found everywhere from clocks and swings to sound waves and even in electricity. These NCERT solution for class 11 Physics are created by expert faculty as per the latest NCERT syllabus for 2025-26. You can read the answers online or download the PDF for free using the given below link.
This NCERT solution also includes extra questions from old NCERT textbooks and Higher Order Thinking Skills (HOTS) questions to boost your understanding and reasoning skills. It covers all important concepts like wave motion, vibrations and different types of oscillations that are useful for school exams and competitive tests like JEE and NEET. These NCERT solutions make learning easier and help you feel more confident during your exams.
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Searching for a simple and free solution to crack your Class 11 Physics exams? You can download the Chapter 13 NCERT Solutions as a PDF. It is full of step-by-step answers to help you master Oscillations and makes complex material easy and simple to understand. Ideal for last-minute revisions and to shine in your exam.
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) The 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 [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)?

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 ( $\omega$ is any positive constant):
(a) $sin\; \omega t-cos\; \omega t$
Answer:
$
\begin{aligned}
& \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}\left(\cos \frac{\pi}{4} \sin \omega t-\sin \frac{\pi}{4} \cos \omega t\right) \\
& =\sqrt{2} \sin \left(\omega t-\frac{\pi}{4}\right)
\end{aligned}
$
Since the above function is of form $Asin(\omega t+\phi )$ it represents SHM with a time period of $\frac{2\pi }{\omega }$
(b) $sin^{3}\omega t$
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 the motion will definitely be periodic with a period of $\frac{2\pi }{\omega }$
(c) $3\; cos(\pi /4-2\omega t)$
Answer:
The function represents SHM with a period of $\frac{\pi }{\omega }$
(d) $cos\; \omega t+cos\; 3\omega t+cos\; 5\omega t$
Answer:
Here, each individual function is 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) $exp(-\omega ^{2}t^{2})$
Answer:
The given function is exponential and, therefore, does not represent periodic motion.
(f) $1+\omega t+\omega^{2}t^{2}$
Answer:
The given function does not represent periodic motion.
(a) at the end A,
Answer:
Velocity is zero. Force and acceleration are in the positive direction.
(b) at the end $B$,
Answer:
Velocity is zero. Acceleration and force are negative.
(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.
(d) at $2cm$ away from $B$ going towards $A$ ,
Answer:
The velocity is negative. Acceleration and force are also negative.
(e) at $3cm$ away from $A$ going towards $B$ , and
Answer:
Velocity is positive. Acceleration and force are also positive.
(f) at $4cm$ away from $B$ going towards $A.$
Answer:
Velocity, acceleration and force all are negative
(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.
If the initial $(t=0)$ position of the particle is $1\; cm$ and its initial velocity is $\omega \; cm/s,$ what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi s^{-1}.$ If instead of the cosine function, we choose the sine function to describe the SHM: $x=B\; sin(\omega t+\alpha ),$ what are the amplitude and initial phase of the particle with the above initial conditions?
Answer:
$\omega =\pi\ rad\ s^{-1}$
$x(t)=Acos(\pi t+\phi )$
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
$\tan\alpha =1$
$\alpha =\frac{\pi }{4},\frac{5\pi }{4},\frac{9\pi }{4}......$
Answer:
Spring constant of the spring is given by
$\begin{aligned}
k & =\frac{\text { Weight of Maximum mass the scale can read }}{\text { Maximum displacement of the scale }} \\
k & =\frac{50 \times 9.8}{20 \times 10^{-2}} \\
k & =2450 \mathrm{Nm}^{-1}
\end{aligned}$
The time period of a spring attached to a body of mass $m$ is given by
$\begin{aligned}
T & =2 \pi \sqrt{\frac{m}{k}} \\
m & =\frac{T^2 k}{4 \pi^2} \\
m & =\frac{(0.6)^2 \times 2450}{4 \pi^2} \\
m & =22.34 \mathrm{~kg} \\
w & =m g \\
w & =22.34 \times 9.8 \\
w & =218.95 \mathrm{~N}
\end{aligned}$
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
$\begin{aligned}
\nu & =\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \\
\nu & =\frac{1}{2 \pi} \times \sqrt{\frac{1200}{3}} \\
\nu & =3.183 \mathrm{~Hz}
\end{aligned}$
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
$\begin{aligned}
& a_{\max }=\frac{F_A}{m} \\
& a_{\max }=\frac{k A}{m} \\
& a_{\max }=\frac{1200 \times 0.2}{3} \\
& a_{\max }=8 \mathrm{~ms}^{-2}
\end{aligned}$
Fig 13.19
Determine
(iii) the maximum speed of the mass.
Answer:
Maximum speed occurs at the mean position and is given by
$\begin{aligned}
& v_{\max }=A \omega \\
& v_{\max }=0.02 \times 2 \pi \times 3.18 \\
& v_{\max }=0.4 \mathrm{~ms}^{-1}
\end{aligned}$
(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 $\\\omega$
$\begin{aligned}
\omega & =\sqrt{\frac{k}{m}} \\
\omega & =\sqrt{\frac{1200}{3}} \\
\omega & =20 \mathrm{rad} / \mathrm{s}
\end{aligned}$
(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?
Answer:
Amplitude is A = 0.02 m
Time period is $\\\omega$
$\begin{aligned}
&\begin{aligned}
\omega & =\sqrt{\frac{k}{m}} \\
\omega & =\sqrt{\frac{1200}{3}} \\
\omega & =20 \mathrm{rad} / \mathrm{s}
\end{aligned}\\
&\text { (b) At } \mathrm{t}=0 \text { the mass is at the maximum stretched position. }\\
&\begin{aligned}
& x(0)=\mathrm{A} \\
& \phi=\frac{\pi}{2} \\
& x(t)=0.02 \sin \left(20 t+\frac{\pi}{2}\right) \\
& x(t)=0.02 \cos (20 t)
\end{aligned}
\end{aligned}$
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?
Answer:
Amplitude is A = 0.02 m
Time period is $\\\omega$
$\begin{aligned}
&\begin{aligned}
\omega & =\sqrt{\frac{k}{m}} \\
\omega & =\sqrt{\frac{1200}{3}} \\
\omega & =20 \mathrm{rad} / \mathrm{s}
\end{aligned}\\
&\text { (c) At } \mathrm{t}=0 \text { the mass is at the maximum compressed position. }\\
&\begin{aligned}
& \mathrm{x}(0)=-\mathrm{A} \\
& \phi=\frac{3 \pi}{2} \\
& x(t)=0.02 \sin \left(20 t+\frac{3 \pi}{2}\right) \\
& x(t)=-0.02 \cos (20 t)
\end{aligned}
\end{aligned}$
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.

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 $x(t)=a \sin ( \pm \omega t+\phi)$
Amplitude $=3 \mathrm{~cm}=0.03 \mathrm{~m}$
$\mathrm{T}=2 \mathrm{~s}$
$\begin{aligned}
\omega & =\frac{2 \pi}{T} \\
\omega & =\pi \mathrm{rad} \mathrm{~s}
\end{aligned}$
Since initial position $\mathrm{x}(\mathrm{t})=0, \phi=0$
As the sense of revolution is clockwise
$\begin{aligned}
& x(t)=0.03 \sin (-\omega t) \\
& x(t)=-0.03 \sin (\pi t)
\end{aligned}$
Here x is in metres and t is in seconds.
(b) Let the required function be $x(t)=a \sin ( \pm \omega t+\phi)$
Amplitude $=2 \mathrm{~m}$
$\mathrm{T}=4 \mathrm{~s}$
$\begin{aligned}
& \omega=\frac{2 \pi}{T} \\
& \omega=\frac{\pi}{2} \mathrm{rad} \mathrm{~s}
\end{aligned}$
Since initial position $x(t)=-A, \phi=\frac{3 \pi}{2}$
As the sense of revolution is anti-clockwise
$\begin{aligned}
& x(t)=2 \sin \left(\omega t+\frac{3 \pi}{2}\right) \\
& x(t)=-2 \cos \left(\frac{\pi}{2} t\right)
\end{aligned}$
Here x is in metres and t is in seconds.
(a) $x=-2\; sin(3t+\pi /3)$
Answer:
$\begin{aligned}
& x=-2 \sin (3 t+\pi / 3) \\
& x=2 \cos \left(3 t+\frac{\pi}{3}+\frac{\pi}{2}\right) \\
& x=2 \cos \left(3 t+\frac{5 \pi}{6}\right)
\end{aligned}$
The initial position of the particle is $x(0)$
$\begin{aligned}
& x(0)=2 \cos \left(0+\frac{5 \pi}{6}\right) \\
& x(0)=2 \cos \left(\frac{5 \pi}{6}\right) \\
& x(0)=-\sqrt{3} \mathrm{~cm}
\end{aligned}$
The radius of the circle i.e. the amplitude is 2 cm
The angular speed of the rotating particle is $\omega=3 \mathrm{rad} \mathrm{s}{ }^{-1}$
Initial phase is
$\begin{aligned}
\phi & =\frac{5 \pi}{6} \\
\phi & =150^{\circ}
\end{aligned}$
The reference circle for the given simple Harmonic motion is

(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

(c) $x=3\; sin(2\pi t+\pi /4)$
Answer:
$\begin{aligned}
x&=3 \sin (2 \pi t+\pi / 4) \\
x&=-3 \cos \left(2 \pi t+\frac{\pi}{4}+\frac{\pi}{2}\right) \\
& =3 \cos \left(2 \pi t+\pi+\frac{\pi}{4}+\frac{\pi}{2}\right) \\
& =3 \cos \left(2 \pi t+\frac{3 \pi}{2}+\frac{\pi}{4}\right) \\
& =3 \cos \left(2 \pi t+\frac{7 \pi}{4}\right)
\end{aligned}$
At t= 0
$phase=\frac{7\pi}{4}$
The reference circle is as follows

(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

(b) is stretched by the same force F.

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
$\\F=Kx$
$x=\frac{F}{k}$
(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
$\\F=k\frac{x}{2}+k\frac{x}{2}$
$\Rightarrow x=\frac{F}{k}$
Answer:
Amplitude of SHM = 0.5 m
angular frequency is
$\begin{aligned}
& \omega=200 \mathrm{rad} / \mathrm{min} \\
& \omega=3.33 \mathrm{rad} / \mathrm{s}
\end{aligned}$
If the equation of SHM is given by
$x(t)=A \sin (\omega t+\phi)$
The velocity would be given by
$\begin{aligned}
& v(t)=\frac{\mathrm{d} x(t)}{\mathrm{d} t} \\
& v(t)=\frac{\mathrm{d}(A \sin (\omega t+\phi))}{\mathrm{d} t} \\
& v(t)=A \omega \cos (\omega t+\phi)
\end{aligned}$
The maximum speed is therefore
$\begin{aligned}
& v_{\max }=A \omega \\
& v_{\max }=0.5 \times 3.33 \\
& v_{\max }=1.67 \mathrm{~ms}^{-1}
\end{aligned}$
Answer:
The time period of a simple pendulum of length l executing S.H.M is given by
$\begin{aligned}
& T=2 \pi \sqrt{\frac{l}{g}} \\
& \mathrm{~g}_{\mathrm{e}}=9.8 \mathrm{~m} \mathrm{~s}^{-2} \\
& \mathrm{~g}_{\mathrm{m}}=1.7 \mathrm{~m} \mathrm{~s}^{-2}
\end{aligned}$
The time period of the pendulum on the surface of Earth is $\mathrm{T}_{\mathrm{e}}=3.5 \mathrm{~s}$
The time period of the pendulum on the surface of the moon is $T_m$
$\begin{aligned}
& \frac{T_m}{T_e}=\sqrt{\frac{g_e}{g_m}} \\
& T_m=T_e \times \sqrt{\frac{g_e}{g_m}} \\
& T_m=3.5 \times \sqrt{\frac{9.8}{1.7}} \\
& T_m=8.4 \mathrm{~s}
\end{aligned}$
Answer:
Acceleration due to gravity = g (in downward direction)
Centripetal acceleration due to the circular movement of the car = a c
$a_{c}=\frac{v^{2}}{R}$ (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 simply harmonically with a period $T=2\pi \sqrt{\frac{h\rho }{\rho _{ 1}g}}$ where $\rho$ 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 downward 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
$\begin{aligned}
&\begin{aligned}
& m a=-\rho_1 g A x \\
& \rho A h a=-\rho_1 g A x \\
& \frac{\mathrm{~d}^2 x}{\mathrm{~d} t^2}=-\frac{\rho_1 g}{\rho h} x
\end{aligned}\\
&\text { Comparing with } \mathrm{a}=-\mathrm{kx} \text { we have }\\
&\begin{aligned}
k & =\frac{\rho_1 g}{\rho h} \\
T & =\frac{2 \pi}{\sqrt{k}} \\
T & =2 \pi \sqrt{\frac{\rho h}{\rho_1 g}}
\end{aligned}
\end{aligned}$
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 $\rho$
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 $2Ax\rho g$
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
$T=2\pi \sqrt{\frac{h}{g}}$

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 $\Delta P$
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.
$\begin{aligned}
& F=a \Delta P \\
& -m \frac{\mathrm{~d}^2 x}{\mathrm{~d} t^2}=a \Delta p \\
& \frac{\mathrm{~d}^2 x}{\mathrm{~d} t^2}=-\frac{k a^2}{m V} x
\end{aligned}$
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 $K$
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
$\\4kx=mg$
$ 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 $750 \; kg.$ .
Answer:
The amplitude of oscillation decreases by 50 % in one oscillation i.e. in one time period.
$\begin{aligned}
T & =2 \pi \sqrt{\frac{m}{k}} \\
T & =2 \pi \times \sqrt{\frac{3000}{4 \times 4.9 \times 10^4}} \\
T & =0.77 \mathrm{~s}
\end{aligned}$
For damping factor b we have
$\begin{aligned}
& x=x_0 e^{\left(-\frac{b t}{2 m}\right)} \\
& \mathrm{x}=\mathrm{x}_0 / 2 \\
& \mathrm{t}=0.77 \mathrm{~s} \\
& \mathrm{~m}=750 \mathrm{~kg} \\
& e^{-\frac{0.75 \%}{2 \times 750}}=0.5 \\
& \ln \left(e^{-\frac{0.775}{2 \times 780}}\right)=\ln 0.5 \\
& \frac{0.77 b}{1500}=\ln 2 \\
& b=\frac{0.693 \times 1500}{0.77} \\
& b=1350.2287 \mathrm{~kg} \mathrm{~s}^{-1}
\end{aligned}$
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
$
\begin{aligned}
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} m A^2 \omega^2 \cos ^2 \omega t
\end{aligned}
$
Time Period is given by
$T=\frac{2\pi }{\omega }$
The Average Kinetic Energy would be given as follows
$
\begin{aligned}
& K_{a v}=\frac{\int_0^T K(t) d t}{\int_0^T d t} \\
& K_{a v}=\frac{1}{T} \int_0^T K(t) d t \\
& K_{a v}=\frac{1}{T} \int_0^T \frac{1}{2} m A^2 \omega^2 \cos ^2 \omega t d t \\
& K_{a v}=\frac{m A^2 \omega^2}{2 T} \int_0^T \cos ^2 \omega t d t \\
& K_{a v}=\frac{m A^2 \omega^2}{2 T} \int_0^T\left(\frac{1+\cos 2 \omega t}{2}\right) d t \\
& K_{a v}=\frac{m A^2 \omega^2}{2 T}\left[\frac{t}{2}+\frac{\sin 2 \omega t}{4 \omega}\right]_0^T \\
& K_{a v}=\frac{m A^2 \omega^2}{2 T}\left[\left(\frac{T}{2}+\frac{\sin 2 \omega T}{4 \omega}\right)-(0+\sin (0))\right] \\
& K_{a v}=\frac{m A^2 \omega^2}{2 T} \times \frac{T}{2} \\
& K_{a v}=\frac{m A^2 \omega^2}{4}
\end{aligned}
$
The potential energy at an instant T is given by
$
\begin{aligned}
& U(t)=\frac{1}{2} k x^2 \\
& U(t)=\frac{1}{2} m \omega^2(A \sin (\omega t))^2 \\
& U(t)=\frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t
\end{aligned}
$
The Average Potential Energy would be given by
$
\begin{aligned}
& U_{a v}=\frac{\int_0^T U(t) d t}{\int_0^T d t} \\
& U_{a v}=\frac{1}{T} \int_0^T \frac{1}{2} m \omega^2 A^2 \sin ^2 \omega t d t \\
& U_{a v}=\frac{m \omega^2 A^2}{2 T} \int_0^T \sin ^2 \omega t d t \\
& U_{a v}=\frac{m \omega^2 A^2}{2 T} \int_0^T \frac{(1-\cos 2 \omega t)}{2} d t \\
& U_{a v}=\frac{m \omega^2 A^2}{2 T}\left[\frac{t}{2}-\frac{\sin 2 \omega t}{4 \omega}\right]_0^T \\
& U_{a v}=\frac{m \omega^2 A^2}{2 T}\left[\left(\frac{T}{2}-\frac{\sin 2 \omega T}{4 \omega}\right)-(0-\sin 0)\right] \\
& U_{a v}=\frac{m \omega^2 A^2}{2 T} \times \frac{T}{2} \\
& U_{a v}=\frac{m \omega^2 A^2}{4}
\end{aligned}
$
We can see K av = U av
Answer:
$J=-a\; \theta$
Moment of Inertia of the disc about the axis passing through its centre and perpendicular to it is
$
\begin{aligned}
& I=\frac{M R^2}{2} \\
& J=I \frac{\mathrm{~d}^2 \theta}{\mathrm{~d} t^2} \\
& -a \theta=\frac{M R^2}{2} \frac{\mathrm{~d}^2 \theta}{\mathrm{~d} t^2} \\
& \frac{\mathrm{~d}^2 \theta}{\mathrm{~d} t^2}=-\frac{2 a}{M R^2} \theta
\end{aligned}
$
The period of Torsional oscillations would be
$
\begin{aligned}
& T=2 \pi \sqrt{\frac{M R^2}{2 a}} \\
& a=\frac{2 \pi^2 M R^2}{T^2} \\
& a=\frac{2 \pi^2 \times 10 \times(0.15)^2}{(1.5)^2} \\
& a=1.97 \mathrm{~N} \mathrm{~m} \mathrm{rad}{ }^{-1}
\end{aligned}
$
(a) $5\; cm$
Answer:
$\begin{aligned}
& \mathrm{A}=5 \mathrm{~cm}=0.05 \mathrm{~m} \\
& \mathrm{~T}=0.2 \mathrm{~s} \\
& \omega=\frac{2 \pi}{T} \\
& \omega=\frac{2 \pi}{0.2} \\
& \omega=10 \pi \mathrm{rad} \mathrm{~s}^{-1}
\end{aligned}$
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
$
\begin{aligned}
& v=10 \pi \sqrt{(0.05)^2-(0.05)^2} \\
& v=0 \\ \\
& a=-(10 \pi)^2 \times 0.05 \\
& a=-49.35 m s^{-2}
\end{aligned}
$
(b) $3\; cm$
Answer:
$
\begin{aligned}
& \mathrm{A}=5 \mathrm{~cm}=0.05 \mathrm{~m} \\
& \mathrm{~T}=0.2 \mathrm{~s} \\
& \omega=\frac{2 \pi}{T} \\
& \omega=\frac{2 \pi}{0.2} \\
& \omega=10 \pi \mathrm{rad} \mathrm{~s}^{-1}
\end{aligned}
$
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
$
\begin{aligned}
& v=10 \pi \sqrt{(0.05)^2-(0.03)^2} \\
& v=10 \pi \sqrt{0.0016} \\
& v=10 \pi \times 0.04 \\
& v=1.257 \mathrm{~ms}^{-1} \\ \\
& a=-(10 \pi)^2 \times 0.03 \\
& a=-29.61 \mathrm{~ms}^{-2}
\end{aligned}
$
(c) $0 \; cm$
Answer:
$
\begin{aligned}
& \mathrm{A}=5 \mathrm{~cm}=0.05 \mathrm{~m} \\
& \mathrm{~T}=0.2 \mathrm{~s} \\
& \omega=\frac{2 \pi}{T} \\
& \omega=\frac{2 \pi}{0.2} \\
& \omega=10 \pi \mathrm{rad} \mathrm{~s}^{-1}
\end{aligned}
$
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
$
\begin{aligned}
& v=10 \pi \sqrt{(0.05)^2-(0)^2} \\
& v=10 \pi \times 0.05 \\
& v=1.57 \mathrm{~ms}^{-1} \\ \\
& a=-(10 \pi)^2 \times 0 \\
& a=0
\end{aligned}
$
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
$\begin{aligned}
& \frac{1}{2} k A^2=\frac{1}{2} m v_0^2+\frac{1}{2} k x_0^2 \\
& A=\sqrt{x_0^2+\frac{m}{k} v_0^2}
\end{aligned}$
The angular frequency of a spring-mass system is always equal to $\sqrt{\frac{k}{m}}$
Therefore
$A=\sqrt{x_0^2+\frac{v_0^2}{\omega^2}}$
Q1. A particle executes simple harmonic motion with a time period 2 seconds and amplitude 1 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then $\frac{\mathrm{D}}{\mathrm{d}}$ is:-
Answer:
$\mathrm{A}=1 \mathrm{~cm}$
$\mathrm{T} = 2 \mathrm{~sec}$
Now, 0.5 sec = T/4
$\therefore d = 1 \mathrm{~cm} \sin(\frac{2\pi}{T}\frac{T}{4}) = 1 \mathrm{~cm}$
Number of complete cycles in 12.5 sec = 12.5 / 2 = 6.25 cycles.
Distance covered in one cycle = 4A = 4cm
$\therefore D = 6.25 \times 4 = 25 \mathrm{~cm}$
$\frac{\mathrm{D}}{\mathrm{d}}=25$
Hence, the answer is 25.
Q2. With a period of 5s, a particle moves in simple harmonic motion. The particle takes 1/a time to cover a displacement from the mean location equal to half of its amplitude. To the closest integer, the value of "a" is:
Answer:
Time period, T=5sec
Time taken by the particle, $\mathrm{t}=1 / \mathrm{a}$
$x=A \sin (\omega t+\phi)$ [ Where, phase constant $\phi=0$ at mean position, $\omega$ is angular velocity, A is amplitude.]
$ \frac{A}{2}=A \sin \left(\frac{2 \pi}{T}\right) t $
$ \frac{2 \pi}{T} t=\sin ^{-1}\left(\frac{1}{2}\right) $
$\frac{2 \pi}{T} t=\frac{\pi}{6} $
$ t=\frac{1}{6}$
So, $a=6$
Hence, the answer is 6.
13.1 Introduction
13.2 Periodic And Oscillatory Motions
13.3 Simple harmonic motion
13.4 Simple harmonic motion and uniform circular motion
13.5 Velocity and acceleration in simple harmonic motion
13.6 Force law for simple harmonic motion
13.7 Energy In Simple Harmonic Motion
13.8 The simple pendulum
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