NCERT Solutions for Class 11 Physics Chapter 14

Q.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 center of mass.
(d) An arrow released from a bow.

Ans.

(a) The motion of the swimmer is not periodic as the swimmer’s motion is only back and forth. Its time period is also not definite as the time taken by the swimmer during his back and forth journey may not be the same.

(b) The motion of a freely-suspended magnet, if displaced from its N-S direction and released represents periodic motion as a freely suspended magnet oscillates about its position with a specific time period.

(c) When the hydrogen molecule rotating about its centre of mass represents a periodic motion as it reaches the same position after a fixed time again and again.

(d) An arrow released from a bow travels only in the forward direction so it does not represent a periodic motion.

Q.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion?
(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.

Ans.

(a) During the rotation of earth about its axis, the earth reaches the same position after fixed duration of time. Therefore, it represents a periodic motion. But, it is not a simple harmonic motion, because earth does not have a to and fro motion about its axis.

(b) The motion of an oscillating mercury column in a U-tube represents a simple harmonic motion. This is due to the fact that the mercury moves to and fro on the same path, about the fixed position, with a definite time period.

(c) The ball comes back to its initial position in the same period of time, again and again moving in a to and fro motion. Therefore, its motion is a periodic and simple harmonic

(d) The polyatomic molecule has many natural frequencies of vibrations. Its general motion is the superposition of individual simple harmonic motions of different molecules. Thus, it represents the periodic motion but not the simple harmonic motion.

Q.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)?

Ans.

(a) It does not represent a periodic motion, because there is no repetition of motion in this case.

(b) In the given X-t plot, the motion of the particle repeats itself after every 2 s. Therefore, it is surely a periodic motion, whose period is 2 s.

(c) The given X-t plot does not represent a periodic motion. This is due to the fact that the particle repeats the motion in one position only, whereas, in case of a periodic motion, the entire motion of the particle must be repeated periodically.

(d) In the given X-t plot, the motion of the particle repeats itself after 2 s. Therefore, it represents a periodic motion having a time period of 2 s.

Q.4

\begin{array}{l}\text{Which of the following functions of time}\text{represent}\left(\text{a}\right)\text{simple harmonic,}\left(\text{b}\right)\text{periodic}\text{but not simple harmonic, and}\\ \left(\text{c}\right)\text{non-periodic}\text{motion? Give period}\text{for each case of periodic}\text{motion (}\omega \text{is any positive constant):}\\ \left(\text{a}\right)\text{sin}\omega \text{t}–\text{cos}\omega \text{t}\\ \left(\text{b}\right){\text{sin}}^3\omega \text{t}\\ \left(\text{c}\right)\text{3 cos}\left(\frac{\pi }{\text{4}}–\text{2}\omega \text{t}\right)\end{array} \begin{array}{l}\left(\text{d}\right)\text{cos}\omega \text{t}+\text{cos 3}\omega \text{t}+\text{cos 5}\omega \text{t}\\ \left(\text{e}\right)\text{exp}\left(–{\omega }^{\text{2}}{\text{t}}^{\text{2}}\right)\\ \left(\text{f}\right)\text{1}+\omega \text{t}+{\omega }^{\text{2}}{\text{t}}^{\text{2}}\end{array}

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{the}\text{given}\text{function is:}\\ \text{sin}\omega t-\text{cos}\omega \text{t}=\sqrt{2}\left[\frac{1}{\sqrt{2}}\text{sin}\omega \text{t}-\frac{1}{\sqrt{2}}\text{cos}\omega \text{t}\right]\\ =\sqrt{2}\text{sin}\left(\omega \text{t}-\frac{\pi }{\text{4}}\right)\\ \text{As the}\text{given function}\text{can be written in the form:}\\ \text{a}\text{sin}\left(\omega t+\varphi \right)\text{,}\text{therefore,}\text{it represents SHM}\\ \text{Period of}\text{the}\text{given function}\text{=}\frac{2\pi }{\omega }\\ \left(\text{b}\right)\text{Here,}\text{the given function is:}\\ {\sin }^3\omega t=\frac{1}{4}\left[3\text{sin}\omega t-\text{sin}3\omega t\right]\\ \text{Here,}\text{the terms sinωt}\text{and sin}\text{3}\omega \text{t}\text{represent simple}\text{harmonic motion}\left(\text{SHM}\right)\text{individually}.\text{But},\text{the}\\ \text{superposition of two SHM is periodic and not simple}\text{harmonic}\text{.}\\ \left(\text{c}\right)\text{Here,}\text{the given function is:}\\ 3\text{cos}\left[\frac{\pi }{4}-2\omega t\right]=3\text{cos}\left[2\omega t-\frac{\pi }{4}\right]\\ \text{The}\text{given function represents SHM}\text{.}\\ \text{Period of}\text{the}\text{given function}\text{=}\frac{2\pi }{2\omega }=\frac{\pi }{\omega }\\ \left(\text{d}\right)\text{The given function}\text{=}\text{cos}\omega t+\cos 3\omega t+\cos 5\omega t\\ \text{In}\text{this}\text{case,}\text{each individual cosine function}\text{represents}\text{SHM}\text{.}\\ \text{Here, the superposition of three SHM is}\text{periodic,}\text{but not simple harmonic}\text{.}\\ \left(\text{e}\right)\text{The given function}\text{=}\text{exp}\left(-{\omega }^{\text{2}}{\text{t}}^2\right)\\ \text{It}\text{is an exponential function}.\text{As}\text{the}e\text{xponential}\text{functions}\text{do}\text{not repeat themselves,}\text{therefore},\text{it does}\\ \text{not}\text{represent}\text{a}\text{periodic}\text{motion}.\\ \left(\text{f}\right)\text{The given function}\text{=}\text{1}\text{+}\omega t+{\omega }^2{t}^2\\ \text{It}\text{represents}\text{the}\text{non}-\text{periodic}\text{motion}\text{.}\end{array}

Q.5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm 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
(a) at the end A,
(b) at the end B,
(c) at the mid-point of AB going towards A,
(d) at 2 cm away from B going towards A,
(e) at 3 cm away from A going towards B, and
(f) at 4 cm away from B going towards A

Ans.

The situation of the problem is shown in the given figure. The particle is in linear SHM. Here, points A and B are the two end points, with AB = 10 cm. O is the midpoint of the path AB.

(a) In this case, the particle is at the extreme point A, at rest momentarily. Therefore, its velocity is zero at this point.

The acceleration of the particle is positive as its direction is along AO.

As the particle is directed rightward, therefore, force is also positive in this case.

(b) In this case, the particle is at the extreme point B, at rest momentarily. Therefore, the velocity of the particle is equal to zero at this point.

As it is directed along B, therefore, its acceleration is negative.

As the particle is directed leftward, therefore, force is also negative in this case.

(c)

In this case, the particle is in simple harmonic motion. Here, O represents the mean position of the particle. At the mean position O, the velocity of the particle is the maximum. As the particle is directed leftward, therefore, the velocity is negative. At the mean position O, the acceleration and force of a particle executing SHM is zero.

(d)

The particle is traveling toward point O from the end B. In this case, the direction of motion is opposite to the conventional positive direction, which is from A to B. Hence, in this case, the velocity and acceleration of the particle are negative. The force on it is also negative.

(e)

In this case, the particle is traveling towards point O from the end A. Here, the direction of motion is from A to B, which is the positive direction conventionally. Therefore, the velocity, acceleration, and force are all positive in this case.

(f)

This case is similar to the case given in (d). Therefore, velocity, acceleration and force are all negative in this case.

Q.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

Ans.

In the SHM, acceleration is related to displacement as:

a = –kx

Out of the given relations, only relation (c) is written in the above form. Therefore, relation (c) represents the SHM.

Q.7 The motion of a particle executing simple harmonic motion is described by the displacement function,
x (t) = A cos (ωt + Φ).
If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cms^-1, what are its amplitude and initial phase angle? The angular frequency of the particle is πs^–1. If instead of the cosine function, we choose the sine function to describe the SHM: x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

Ans.

\begin{array}{l}\text{Here,}\text{initially, at}\text{t}\text{=}\text{0:}\\ \text{Displacement,}\text{x}\text{=}\text{1 cm}\\ \text{Initial velocity,}\text{v}\text{=}\omega \text{cm}{\text{sec}}^{\text{-1}}\\ \text{Angular frequency,}\omega =\pi {\text{rads}}^{–\text{1}}\\ \text{Given that:}\\ x\left(t\right)=\text{A}\text{cos}\left(\omega t+\varphi \right)\\ 1=\text{A}\text{cos}\left(\omega \times 0+\varphi \right)=A\text{cos}\varphi \\ \text{A}\text{cos}\varphi =1\to \text{(i)}\\ \text{As}\text{velocity},v=\frac{\text{dx}}{\text{dt}}\\ \therefore \omega =-\text{A}\omega \text{sin}\left(\omega t+\varphi \right)\\ 1=-\text{A}\text{sin}\left(\omega \times 0+\varphi \right)=-A\text{sin}\varphi \\ \text{A}\text{sin}\varphi =-1\to \text{(ii)}\\ \text{Squaring and adding equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we have:}\\ \text{tan}\varphi =-1\\ \therefore \varphi =\frac{3\pi }{4},\frac{7\pi }{4},\mathrm{\dots ..}\\ \text{SHM}\text{}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ x=B\text{sin}\left(\omega t+\alpha \right)\\ \text{Putting}\text{the}\text{given}\text{values}\text{in}\text{the}\text{above}\text{relation,}\text{we}\text{obtain:}\\ 1=\text{Bsin}\left[\omega \times 0+\alpha \right]\\ B\sin \alpha =1\to (iv)\\ \text{Squaring}\text{and}\text{adding}\text{equation}\text{(iii)}\text{and}\text{(iv),}\text{we}\text{have:}\\ B^2\left[{\sin }^2\alpha +{\cos }^2\alpha \right]=1+1\\ B^2=2\\ \therefore B=\sqrt{2}\text{cm}\\ \text{Dividing equation}\left(\text{iii}\right)\text{by equation}\left(\text{iv}\right)\text{, we obtain:}\\ \frac{\text{B}\text{sin}\alpha }{\text{B}\text{cos}\alpha }=\frac{1}{1}\\ \therefore \text{tan}\alpha =1=\text{tan}\frac{\pi }{4}\\ \therefore \alpha =\frac{\pi }{4},\frac{5\pi }{4},\mathrm{\dots \dots ..}\end{array}

Q.8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Ans.

\begin{array}{l}\text{Here,}\text{maximum mass that the scale can read,}\text{M = 50 kg}\\ \text{Maximum extension of spring = Length of scale, l = 20 cm = 0}\text{.2 m}\\ \text{Time period}\text{of}\text{oscillation,}\text{T = 0}\text{.6 s}\\ \text{Maximum force applied on the spring, F = Mg}\\ \text{Here,}\text{g = acceleration due to gravity = 9}{\text{.8 ms}}^{\text{-2}}\\ \therefore \text{F = 50 kg× 9}{\text{.8 ms}}^{\text{-2}}\text{= 490 N}\\ \therefore \text{Spring constant,}\text{k =}\frac{\text{F}}{\text{l}}\text{=}\frac{\text{490 N}}{\text{0}\text{.2 m}}\text{= 2450}{\text{Nm}}^{\text{-1}}\\ \text{Let}\text{mass suspended from the balance be m}\text{.}\\ \text{Time period,T = 2π}\sqrt{\frac{\text{m}}{\text{k}}}\\ \therefore \text{m =}{\left(\frac{\text{T}}{\text{2π}}\right)}^{\text{2}}\text{×k =}{\left(\frac{\text{0}\text{.6 s}}{\text{2×3}\text{.14}}\right)}^{\text{2}}{\text{×2450 Nm}}^{\text{-1}}\text{= 22}\text{.36}\text{kg}\\ \therefore \text{Weight of body,}\text{W = mg = 22}\text{.36×9}\text{.8 = 219}\text{.167 N}\\ \therefore \text{The weight of the body is approximately 219 N}\text{.}\end{array}

Q.9 A spring having with a spring constant 1200 Nm^–1 is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.

Ans.

\begin{array}{l}\text{Here,}\text{spring constant,}{\text{k = 1200 Nm}}^{\text{–1}}\\ \text{Mass}\text{attached,}\text{m = 3 kg}\\ \text{Displacement}\text{of}\text{the}\text{mass,}\text{A = 2}\text{.0 cm = 0}\text{.02 m}\\ \left(\text{i}\right)\text{Frequency of oscillation, ν =}\frac{\text{1}}{\text{T}}\text{=}\frac{\text{1}}{\text{2π}}\sqrt{\frac{\text{k}}{\text{m}}}\\ \text{Here,T}\text{= Time period}\\ \therefore \text{ν =}\frac{\text{1}}{\text{2×3}\text{.14}}\sqrt{\frac{{\text{1200 Nm}}^{\text{-1}}}{\text{3 kg}}}\text{= 3}\text{.18}{\text{ms}}^{\text{-1}}\\ \therefore \text{The frequency of oscillations is 3}{\text{.18 ms}}^{\text{-1}}\text{.}\\ \left(\text{ii}\right){\text{Maximum acceleration, a = ω}}^{\text{2}}\text{A}\\ \text{Here, ω = Angular frequency =}\sqrt{\frac{\text{k}}{\text{m}}}\\ \text{A = Maximum displacement}\\ \therefore \text{a =}\frac{\text{k}}{\text{m}}\text{A =}\frac{{\text{1200 Nm}}^{\text{-1}}\text{×0}\text{.02 m}}{\text{3 kg}}\text{= 8}{\text{ms}}^{\text{-2}}\\ \therefore \text{The maximum acceleration of the mass is 8}{\text{.0 ms}}^{\text{-2}}\text{.}\\ \left(\text{iii}\right){\text{Maximum velocity, v}}_{\text{max}}\text{= Aω = A}\sqrt{\frac{\text{k}}{\text{m}}}\\ {\text{v}}_{\text{max}}\text{= 0}\text{.02 m×}\sqrt{\frac{{\text{1200 Nm}}^{\text{-1}}}{\text{3 kg}}}\text{= 0}\text{.4}{\text{ms}}^{\text{-1}}\\ \therefore \text{The maximum velocity of the mass is 0}{\text{.4 ms}}^{\text{-1}}.\end{array}

Q.10 In Exercise 14.9, 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
(a) at the mean position,
(b) at the maximum stretched position, and
(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?

Ans.

\begin{array}{l}\text{Here,}\text{distance travelled by the mass,}\text{a}\text{=}\text{2}\text{.0 cm}\\ \text{Spring}\text{constant,}\text{k}\text{=}{\text{1200 Nm}}^{\text{–1}}\\ \text{Mass}\text{attached,}\text{m}\text{=}\text{3 kg}\\ \text{Angular frequency of oscillation}\text{is}\text{given}\text{as:}\\ \omega =\sqrt{\frac{\text{k}}{\text{m}}}=\sqrt{\frac{1200}{3}}=\sqrt{400}\\ =\text{2}0{\text{rad s}}^{\text{–1}}\\ \left(\text{a}\right)\text{When the mass is at the mean position,}\text{Initial phase}\text{=}0\\ \text{Displacement}\text{is}\text{given}\text{as:}\\ \text{x}=\text{A}\text{sin}\omega \text{t}\\ =\text{2}\text{sin 2}0\text{t cm}\\ \left(\text{b}\right)\text{At the maximum stretched position, the mass is}\text{at the extreme right}\text{.}\\ \therefore \text{Initial phase}\text{=}\frac{\pi }{2}\\ \text{Displacement}\text{is}\text{given}\text{as:}\\ x=\text{A}\text{sin}\left(\omega \text{t}+\frac{\pi }{2}\right)\\ =2\text{sin}\left(\text{20t}+\frac{\pi }{2}\right)\\ =\text{2}\text{cos 2}0\text{t}\\ \left(\text{c}\right)\text{At the maximum compressed position, the mass is at the extreme left}\text{.}\\ \therefore \text{Initial phase}\text{=}\frac{3\pi }{2}\\ \text{Displacement}\text{is}\text{given}\text{as:}\\ \text{x}=\text{A}\text{sin}\left(\omega \text{t}+\frac{3\pi }{2}\right)\\ =2\text{sin}\left(\text{20}\text{t}+\frac{3\pi }{2}\right)\\ \text{=}–\text{2cos 2}0\text{t}\\ \therefore \text{The functions have the same frequency}\left(\frac{20}{2\pi }\right)\text{and}\text{amplitude}\text{(2}\text{cm), but they}\text{differ}\text{in initial phase}\text{.}\end{array}

Q.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.

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{time period}\text{of}\text{revolution},\text{T}=\text{2}\text{s}\\ \text{Amplitude,}\text{A}\text{=}\text{3 cm}\\ \text{At time,}\text{t}=0,\text{the radius vector OP makes an angle}\frac{\pi }{2}\\ \text{with}\text{the positive}\text{direction}\text{of}\text{x}-\text{axis}\text{.}\\ \therefore \text{Phase angle,}\varphi \text{=}+\frac{\pi }{2}\\ \therefore \text{The equation of SHM for the}\text{x}-\text{projection of OP,}\text{at time}\text{t, is given as:}\\ \text{x}\text{=}\text{A}\text{cos}\left[\frac{2\pi \text{t}}{\text{T}}+\varphi \right]\\ =3\text{cos}\left[\frac{2\pi t}{2}+\frac{\pi }{2}\right]=-3\text{sin}\left(\frac{2\pi \text{t}}{2}\right)\\ \therefore \text{x}=-3\text{sin}\pi \text{t}\text{cm}\\ \left(\text{b}\right)\text{Here,}\text{time period}\text{of}\text{oscillation,}\text{T}=\text{4 s}\\ \text{Amplitude}\text{of}\text{oscillation,a}=\text{2 m}\\ \text{At time}\text{t}=0,\text{the}\text{radius}\text{vector}\text{OP makes an angle}\pi \text{with the}\text{x}-\text{axis, in the anticlockwise direction}\text{.}\\ \therefore \text{Phase angle},\varphi =+\pi \\ \therefore \text{The equation of SHM for the}\text{x}–\text{projection of the}\text{radius}\text{vector}\text{OP, at time}\text{t, is given by}\text{the}\text{relation:}\\ \text{x}=a\text{cos}\left[\frac{2\pi t}{T}+\varphi \right]=2\text{cos}\left[\frac{2\pi \text{t}}{4}+\pi \right]\\ \therefore x=-2\text{cos}\left(\frac{\pi \text{t}}{2}\right)\text{m, is the required equation of SHM}\end{array}

Q.12 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 + p /3)
(b) x = cos (p/6 – t)
(c) x = 3 sin (2pt + p/4)
(d) x = 2 cos pt

Ans.

\begin{array}{l}\text{Standard}\text{SHM equation is}\text{given}\text{as:}\\ x=\text{A}\text{cos}\left(\frac{2\pi }{\text{T}}\text{t}+\varphi \right)\\ \left(\text{a}\right)\text{Here,}\text{x=}-\text{2}\text{sin}\left(\text{3t}+\frac{\pi }{\text{3}}\right)=+\text{2}\text{cos}\left(\text{3t}+\frac{\pi }{\text{3}}+\frac{\pi }{\text{2}}\right)\\ =\text{2}\text{cos}\left(3\text{t}+\frac{5\pi }{6}\right)\\ \text{Comparing}\text{the}\text{above equation with the standard}\text{SHM equation,}\text{we}\text{obtain:}\\ \text{Amplitude,}\text{A}\text{=}\text{2}\text{cm}\\ \text{Phase}\text{angle},\varphi =\frac{5\pi }{6}={\text{150}}^{\text{0}}\\ \text{Angular}\text{velocity},\omega =\frac{2\pi }{\text{T}}=3\text{rad}{\text{sec}}^{-1}\\ \text{The motion of the particle can be represented}\text{as}\text{shown in the given figure}\text{.}\end{array}

\begin{array}{l}\left(\text{b}\right)\text{Here,}\text{x}\text{=}\text{cos}\left(\frac{\pi }{6}-t\right)=\text{cos}\left(\text{t}-\frac{\pi }{6}\right)\\ \text{Comparing the}\text{above equation with the standard SHM}\text{equation,}\text{we obtain:}\\ \text{Amplitude,}\text{A}\text{=1}\text{cm}\\ \text{Phase}\text{angle},\varphi =-\frac{\pi }{6}=-{30}^0\\ \text{Angular}\text{velocity}\text{is}\text{given}\text{as:}\\ \omega =\frac{2\pi }{T}=1\text{rad}{\text{sec}}^{-1}\\ \text{The motion of the particle can}\text{be}\text{represented}\text{as}\text{shown in the}\text{given figure}\text{.}\end{array}

\begin{array}{l}\left(\text{c}\right)\text{Here,}\text{x}=\text{3}\text{sin}\left(\text{2}\pi \text{t}+\frac{\pi }{\text{4}}\right)\\ =-\text{3}\text{cos}\left[\left(2\pi \text{t}+\frac{\pi }{\text{4}}\right)+\frac{\pi }{\text{2}}\right]=\text{3}\text{cos}\left(2\pi \text{t}-\frac{\pi }{\text{4}}\right)\\ \text{Comparing}\text{the}\text{above equation with the standard}\text{SHM}\text{equation ,we obtain:}\\ \text{Amplitude,}\text{A}=\text{3 cm}\\ \text{Phase angle,}\varphi =-\frac{\pi }{4}=-\text{45}°\\ \text{Angular velocity},\omega =\frac{2\pi }{T}=2\pi \text{rad}{\text{s}}^{-1}\\ \text{The motion of the particle can be represented as shown in the given figure}\text{.}\end{array}

\begin{array}{l}\left(\text{d}\right)\text{Here,}\text{x}\text{=}\text{2 cos}\pi \text{t}\\ \text{If we}\text{compare}\text{the}\text{above equation with the standard}\text{SHM equation ,we have:}\\ \text{Amplitude,}\text{A}\text{=}\text{2 cm}\\ \text{Phase angle},\varphi =0\\ \text{Angular velocity},\omega =\pi \text{rad}{\text{s}}^{\text{-1}}\\ \text{Motion of the particle can be represented as shown}\text{in the given figure}\text{.}\end{array}

Q.13 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.

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

Ans.

\begin{array}{l}\left(\text{a}\right)\text{In}\text{case}\text{of the one block system:}\\ \text{When a force}\left(\text{F}\right)\text{is exerted to the free end of the}\text{spring, an extension}\left(\text{l}\right)\text{is developed}\text{.}\\ \text{For the maximum value}\text{of}\text{extension, we}\text{have:}\\ \text{F}\text{=}\text{kl}\\ \text{Here,}\text{k}\text{=}\text{Spring constant}\\ \therefore \text{The maximum extension in the spring:}\\ \text{l}\text{=}\frac{\text{F}}{\text{k}}\\ \text{In}\text{case}\text{of}\text{two block system:}\\ \text{Here, displacement,}\text{x}\text{=}\frac{\text{l}}{\text{2}}\\ \text{Total force,}\text{F}=+\text{2kx}\text{=}\text{2k}\frac{l}{2}\\ \therefore l=\frac{F}{k}\\ \left(\text{b}\right)\text{In}\text{case}\text{of one block system:}\\ \text{Let mass of the block}\text{=}\text{m}\\ \text{Force}\text{is}\text{give}\text{as:}\text{F}\text{=}\text{ma}=m\frac{d^{2}x}{d{t}^2}\\ \text{Here,}\text{x}\text{=}\text{Displacement of the block in time}\text{t}\\ \therefore m\frac{d^{2}x}{d{t}^2}=-kx\\ \text{Its}\text{vaue is negative}\text{as the direction of elastic force is}\text{opposite to the direction of}\text{displacement}\text{.}\\ \frac{d^{2}x}{d{t}^2}=-\left(\frac{k}{m}\right)x=-{\omega }^{2}x\\ \text{Here,}\text{angular frequency},\omega =\sqrt{\frac{k}{m}}\\ \therefore \text{Time period}\text{is}\text{given}\text{as:}\\ \text{T=}\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}\\ \text{For two block system:}\\ \text{Force},F=m‘\frac{d^{2}x}{d{t}^2}\\ \text{When the two masses are released, effective mass,}\\ \text{m’ =}\frac{m_{1}x{m}_2}{m_1+m_2}=\frac{m}{2}\\ \frac{m}{2}\frac{d^{2}x}{d{t}^2}=-kx\\ \text{Its}\text{value is negative as the direction of elastic force is}\text{opposite to the direction of}\text{displacement}\text{.}\\ \frac{d^{2}x}{d{t}^2}=-\left[\frac{2k}{m}\right]x=-{\omega }^{2}x\\ \text{Here,}\text{angular}\text{frequency,}\omega =\sqrt{\frac{2k}{m}}\\ \therefore \text{Time}\text{period}\text{is}\text{given}\text{as:}T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{2k}}\end{array}

Q.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?

Ans.

\begin{array}{l}\text{Here,}\text{stroke}\text{of}\text{piston}=\text{1}.0\text{m}\\ \text{Angular frequency of piston,}\omega =\text{2}00\text{rad}{\text{min}}^{\text{-1}}\\ \text{Amplitude},A=\frac{\text{1}.0}{2}=0.5\text{m}\\ \text{The maximum speed}\left({\text{v}}_{\text{max}}\right)\text{of the piston is given as:}\\ v_{\text{max}}=A\omega =\text{2}00\times 0.5=100\text{m}{\text{min}}^{-1}\end{array}

Q.15 The acceleration due to gravity on the surface of moon is 1.7 ms^–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 ms^–2)

Ans.

\begin{array}{l}\text{Here,}\text{acceleration due to gravity on the moon’s}\text{surface,}\text{g’ = 1}{\text{.7 ms}}^{\text{–2}}\\ \text{Acceleration due to gravity on the earth’s}\text{surface,}\text{g = 9}{\text{.8 ms}}^{\text{–2}}\\ \text{Time period of a simple pendulum on earth,}{\text{T}}_{\text{e}}\text{= 3}\text{.5 s}\\ \text{As}{\text{T}}_{\text{e}}\text{= 2π}\sqrt{\frac{\text{l}}{\text{g}}}\\ \text{Here,}\text{l}\text{= Length of pendulum}\\ \therefore \text{l =}\frac{{\text{T}}_{\text{e}}{}^{\text{2}}}{{\left(\text{2π}\right)}^{\text{2}}}\text{×g =}\frac{{\left(\text{3}\text{.5 s}\right)}^{\text{2}}}{\text{4×}{\left(\text{3}\text{.14}\right)}^{\text{2}}}\text{×9}\text{.8}{\text{ms}}^{\text{–2}}\\ \text{The value}\text{of}\text{the length of the pendulum}\text{remains constant}\text{.}\\ \text{On the surface}\text{of}\text{moon, time period}\text{is}\text{given}\text{as:}\\ \text{T’ = 2π}\sqrt{\frac{\text{l}}{\text{g’}}}\text{= 2π}\sqrt{\frac{\frac{{\left(\text{3}\text{.5 s}\right)}^{\text{2}}}{\text{4×}{\left(\text{3}\text{.14}\right)}^{\text{2}}}\text{×9}{\text{.8 ms}}^{\text{–2}}}{\text{1}{\text{.7 ms}}^{\text{–2}}}}\text{= 8}\text{.4}\text{s}\\ \therefore \text{Time period of the simple pendulum on the surface of moon = 8}\text{.4 s}\end{array}

Q.16

\begin{array}{l}\text{Answer the following questions:}\\ \left(\text{a}\right)\text{Time period of a particle in SHM depends on}\text{the force constant}\text{k}\text{and mass m of the particle:}\\ \text{T}\text{=}\text{2}\pi \sqrt{\frac{\text{m}}{\text{k}}}.\text{A simple pendulum executes SHM}\text{approximately}\text{.}\text{Why then is the time period of}\\ \text{a pendulum independent of the mass of the}\text{pendulum?}\\ \left(\text{b}\right)\text{The motion of a simple pendulum is}\text{approximately simple harmonic for small}\text{angle}\\ \text{oscillations}\text{. For larger angles of oscillation, a}\text{more involved analysis shows that}\text{T}\text{is greater}\\ \text{than}2\pi \sqrt{\frac{\text{l}}{\text{g}}}.\text{Think of a qualitative argument to}\text{appreciate this result}\text{.}\\ \left(\text{c}\right)\text{A man with a wristwatch on his hand falls}\text{from the top of a tower}\text{. Does the watch give}\\ \text{correct}\text{time during the free fall?}\\ \left(\text{d}\right)\text{What is the frequency of oscillation of a}\text{simple pendulum mounted in a cabin that is}\\ \text{freely falling under gravity?}\end{array}

Ans.

\begin{array}{l}\left(\text{a}\right)\text{The time period of a simple pendulum}\text{is}\text{given}\text{as:}\\ T=2\pi \sqrt{\frac{m}{k}}\\ \text{In}\text{case}\text{of a simple pendulum,}\\ k\propto m\\ \frac{\text{m}}{\text{k}}\text{=}\text{Constant}\\ \therefore \text{The time period of a simple pendulum does}\text{not}\text{depend}\text{on}\text{the mass of the}\text{bob}\text{.}\\ \left(\text{b}\right)\text{For}\text{a simple pendulum, the restoring force acting}\text{on the bob of the}\text{pendulum is given by}\text{the}\text{relation:}\\ \text{F}=–\text{mgsin}\theta \\ \text{For small}\text{values}\text{of}\theta ,\text{sin}\theta \simeq \theta \\ \text{For large}\text{values}\text{of}\theta ,\text{sin}\theta >\theta \\ \text{It decreases the effective value of}\text{acceleration}\text{due}\text{to}\text{gravity}\text{g}\text{.}\\ \therefore \text{The time period increases as:}\\ T=2\pi \sqrt{\frac{l}{g‘}}\\ \text{Here,}\text{l}\text{= Length of the simple pendulum}\\ \left(\text{c}\right)\text{Yes,}\text{because}\text{the}\text{working}\text{of}\text{a}\text{wrist}\text{watch}\text{is not}\text{based on pendulum movement, rather, it works on}\\ \text{spring action}\text{. Hence, its}\text{working}\text{is}\text{not affected by}\text{the acceleration due to gravity during}\text{free fall}\text{.}\\ \text{(d)}\text{When}\text{a}\text{simple}\text{pendulum}\text{falls}\text{freely}\text{under}\text{gravity,}\text{its}\text{acceleration}\text{is}\text{zero}\text{.}\text{Therefore,}\text{the}\text{frequency}\text{of}\\ \text{vibration}\text{of}\text{this}\text{simple}\text{pendulum}\text{is}\text{zero}\text{.}\end{array}

Q.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?

Ans.

\begin{array}{l}\text{Here,}\text{the bob of the simple pendulum will experience}\text{the acceleration due to gravity and}\text{the circular motion}\\ \text{of the car}\text{will}\text{provide}\text{the}\text{centipetal}\text{acceleration}\text{.}\\ \text{Acceleration due to gravity}\text{=}\text{g}\\ \text{Centripetal acceleration}\text{is}\text{given}\text{as:}\frac{{\text{v}}^{\text{2}}}{\text{R}}\\ \text{Here,}\text{v}\text{=}\text{Uniform speed of the car}\\ \text{R}\text{=}\text{Radius of the circular}\text{track}\\ \text{Effective acceleration of}\text{the}\text{pendulum}\left({\text{a}}_{\text{eff}}\right)\text{is given}\text{by}\text{the}\text{relation:}\\ {\text{a}}_{\text{eff}}=\sqrt{g^2+{\left(\frac{v^2}{R}\right)}^2}\\ \text{Time period}\text{of}\text{the}\text{pendulum}\text{is}\text{given}\text{as:}\\ \text{T}\text{=}\text{2}\pi \sqrt{\frac{l}{{\text{a}}_{\text{eff}}}}\\ \text{Here,}\text{l}\text{=}\text{Length of the pendulum}\\ \therefore \text{Time period}\text{of}\text{the}\text{pendulum,}\text{T}\text{=}\text{2}\pi \sqrt{\frac{l}{g^2+\frac{v^4}{R^2}}}\end{array}

Q.18

\begin{array}{l}\text{A cylindrical piece of cork of density of base area}\text{A}\text{and heighthfloats in a liquid of density}\text{. The cork is}\\ \text{depressed slightly and then released}\text{. Show that}\text{the cork oscillates up and down simple}\\ \text{harmonically with a period}\\ T=2\pi \sqrt{\frac{h\rho }{{\rho }_{l}g}}\\ \text{where}\rho \text{is the density of cork}\text{. (Ignore dampingdue to viscosity of the liquid)}\text{.}\end{array}

Ans.

\begin{array}{l}\text{Here,}\text{base area of cork}=\text{A}\\ \text{Height of cork}=\text{h}\\ \text{Density of liquid}={\rho }_l\\ \text{Density of cork}=\rho \\ \text{At equilibrium:}\\ \text{Weight of cork}\text{=}\text{Weight of liquid displaced by}\text{floating cork}\\ \text{Let the cork is depressed slightly by}\text{x}\text{and}\text{released}\text{.}\\ \text{As}\text{a}\text{result}\text{of}\text{it,}\text{some extra water of a definite volume}\text{is displaced}\text{.}\\ \therefore \text{An extra up}-\text{thrust acts in}\text{the}\text{upward}\text{direction and}\text{provides}\text{the}\text{restoring force to the cork}\text{.}\\ \text{Up}-\text{thrust}=\text{Restoring force}\\ \text{Let}\text{F}=\text{Weight of extra water displaced}\\ \text{F}=–\left(\text{Volume}\times \text{Density}\times \text{g}\right)\\ \text{Volume = Area}\times \text{Distance by which the cork is depressed}\\ \text{Volume}=\text{Ax}\\ \therefore \text{F}=–\text{A x}{\rho }_{l}g\to \left(\text{i}\right)\\ \text{As}\text{per the force law:}\\ \text{F}=-\text{kx}\\ \text{Here,k}\text{is a constant}\\ \text{-k}\text{=}\frac{F}{x}=-\text{A}{\rho }_{l}g\to \text{(ii)}\\ \text{Time period of the oscillations of cork}\text{is}\text{given}\text{as:}\\ T=2\pi \sqrt{\frac{m}{k}}\to \text{(iii)}\\ \text{Here},\text{m}=\text{mass of cork}\\ =\text{volume}\text{of}\text{cork}\times \text{density}\\ =\text{Base area of cork}\times \text{Height of cork}\times \text{Density of cork}=\text{Ah}\rho \\ \therefore \text{Time period of}\text{oscillation}\text{of}\text{cork:}\\ T=2\pi \sqrt{\frac{\text{Ah}\rho }{\text{A}{\rho }_{l}g}}=2\pi \sqrt{\frac{\text{h}\rho }{{\rho }_{l}g}}\end{array}

Q.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.

Ans.

\begin{array}{l}\text{Here, cross}-\text{sectional area of the U}-\text{tube}=\text{A}\\ \text{Density of mercury column}\text{=}\rho \\ \text{Acceleration due to gravity}=\text{g}\\ \text{Let}\text{height of the mercury column in the two arms}\text{=}\text{2h}\\ \text{Restoring force,}\text{F}\text{=}\text{Weight of the mercury column of a definite height}\\ \text{F}=–\left(\text{Volume}\times \text{Density}\times \text{g}\right)\\ \text{F}=–\left(\text{A}\times \text{2h}\times \rho \times \text{g}\right)=–\text{2A}\rho \text{gh}\\ F=–\text{k}\times \text{Displacement in one of the arms}\left(\text{h}\right)\\ \text{Here,}\text{k}\text{=}\text{Constant}\\ \text{k=-}\frac{F}{h}=\frac{–\text{2A}\rho \text{gh}}{h}=\text{2A}\rho \text{g}\\ \text{Time period}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{T}\text{=}\text{2}\pi \sqrt{\frac{m}{k}}=\text{2}\pi \sqrt{\frac{m}{\text{2A}\rho \text{g}}}\\ \text{Here,}\text{m}\text{=}\text{Mass of the mercury column}\\ \text{Let}\text{total length of the}\text{mercury column}=l\\ \text{Mass of mercury,}\text{m}=\text{Volume}\times \text{Density}=\text{Al}\rho \\ \therefore \text{Time}\text{period,}T=\text{2}\pi \sqrt{\frac{\text{Al}\rho }{\text{2A}\rho \text{g}}}=\text{2}\pi \sqrt{\frac{\text{l}}{\text{2g}}}\\ \therefore \text{The mercury column undergoes SHM with time}\text{period}\\ \text{2}\pi \sqrt{\frac{\text{l}}{\text{2g}}}.\end{array}

Q.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.14.27). 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. 14.27]

Ans.

\begin{array}{l}\text{Here,}\text{volume of air chamber}\text{=}\text{V}\\ \text{Cross}-\text{sectional}\text{area of the neck}\text{=}\text{a}\\ \text{Let}\text{mass of ball}\text{=}\text{m}\\ \text{Pressure inside the chamber}\text{=}\text{Atmospheric pressure}\\ \text{Let the ball be pressed}\text{through}\text{a}\text{small}\text{distance}\text{x}\text{vertically}\text{downward}\text{.}\\ \text{As a result, there will be a decrease in the volume and an increase in the pressure inside the chamber}\text{.}\\ \text{Decrease in the volume of}\text{air chamber},\Delta \text{V}=\text{ax}\\ \text{Volumetric strain=}\frac{\text{Change}\text{in}\text{volume}}{\text{Original}\text{volume}}=\frac{\Delta \text{V}}{V}=\frac{ax}{V}\\ \text{Bulk Modulus of air}\text{is}\text{given}\text{as:}\\ \text{B}\text{=}\frac{\text{Stress}}{\text{Strain}}=\frac{-p}{\frac{ax}{V}}\\ \text{Here, stress is the increase in pressure}\text{. The negative}\text{sign shows that the}\text{pressure increases with decrease}\\ \text{in volume}\text{.}\\ p=\frac{-Bax}{V}\\ \text{The restoring force acting on the ball}\text{is}\text{given}\text{as:}\\ \text{F}=\text{p}\times \text{a}\\ \text{=}\frac{-Bax}{V}\times \text{a=}\frac{-B{a}^{2}x}{V}\to \text{(i)}\\ \text{For SHM, the equation for restoring}\text{force is}\text{given}\text{as:}\\ \text{F}=–\text{kx}\to \left(\text{ii}\right)\\ \text{Here,}\text{k}\text{= Spring constant}\\ \text{Comparing equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{, we obtain:}\\ k=\frac{B{a}^2}{V}\to \text{(iii)}\\ \text{Time period}\text{is}\text{given}\text{by}\text{the}\text{relation:}\\ \text{T}\text{=}\text{2}\pi \sqrt{\frac{m}{k}}\to \text{(iv)}\\ \text{Substituting}\text{the}\text{value}\text{of}\text{k}\text{from}\text{equation}\text{(iii)}\text{in}\text{equation}\text{(iv),}\text{we}\text{obtain:}\\ \text{Time}\text{period,}\text{T}=\text{2}\pi \sqrt{\frac{Vm}{B{a}^2}}\end{array}

Q.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

Ans.

\begin{array}{l}\left(\text{a}\right)\text{Here,}\text{mass of automobile,m = 3000 kg}\\ \text{Displacement in the suspension system,x = 15 cm = 0}\text{.15 m}\\ \text{Let}\text{spring}\text{constant}\text{of}\text{each}\text{spring be k}\\ \text{As}\text{there are 4 springs in parallel to support the whole}\text{mass of the automobile,}\\ \text{Spring}\text{constant}\text{of}\text{4 springs,}\text{K = 4k}\\ \text{The equation for the restoring force for the system}\text{is}\text{given}\text{as:}\\ \text{F = –4kx = mg}\\ \text{or}\text{k =}\frac{\text{mg}}{\text{4x}}\text{=}\frac{{\text{3000 kg×10 ms}}^{\text{-2}}}{\text{4×0}\text{.15 m}}{\text{= 5×10}}^{\text{4}}{\text{Nm}}^{\text{-1}}\\ \therefore \text{Spring constant,}{\text{k = 5 × 10}}^{\text{4}}{\text{Nm}}^{\text{-1}}\\ \left(\text{b}\right)\text{Here,}\text{mass}\text{supported}\text{by}\text{each wheel,}\text{M =}\frac{\text{3000 kg}}{\text{4}}\text{= 750 kg}\\ \text{For damping factor}{\text{b, the equation for displacement, x = x}}_{\text{0}}{\text{e}}^{\text{-bt/2M}}\\ \text{Since amplitude of oscillation decreases by 50\%,}\\ \therefore \text{x =}\frac{{\text{x}}_{\text{0}}}{\text{2}}\\ \frac{{\text{x}}_{\text{0}}}{\text{2}}{\text{= x}}_{\text{0}}{\text{e}}^{\text{-bt/2M}}\\ {\text{log}}_{\text{e}}\text{2 =}\frac{\text{bt}}{\text{2M}}\\ \therefore \text{b =}\frac{{\text{2Mlog}}_{\text{e}}\text{2}}{\text{t}}\\ \text{Time}\text{period, t = 2π}\sqrt{\frac{\text{m}}{\text{4k}}}\text{= 2π}\sqrt{\frac{\text{3000}\text{kg}}{{\text{4×5×10}}^{\text{4}}{\text{Nm}}^{\text{-1}}}}\text{= 0}\text{.7691}\text{s}\\ \therefore \text{b =}\frac{\text{2×750 kg×0}\text{.693}}{\text{0}\text{.7691 s}}\text{= 1351}{\text{.58 kgs}}^{\text{-1}}\\ \therefore \text{Damping constant of the spring = 1351}{\text{.58 kgs}}^{\text{-1}}\end{array}

Q.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.

Ans.

\begin{array}{l}\text{Consider}\text{a}\text{particle}\text{of}\text{mass}\text{m}\text{be}\text{executing}\text{SHM}\text{with time}\text{period}\text{T}\text{.}\\ \text{The displacement of the particle executing SHM at an instant}\text{t}\text{is given by}\text{the}\text{relation:}\\ \text{x = Asinωt}\\ \text{Here,}\text{A = Amplitude}\\ \text{ω = Angular frequency}\text{=}\sqrt{\frac{\text{k}}{\text{M}}}\\ \text{Velocity of the particle, v =}\frac{\text{dx}}{\text{dt}}\text{= Aωcosωt}\\ \text{Kinetic energy of the particle,}{\text{E}}_{\text{k}}\text{=}\frac{\text{1}}{\text{2}}{\text{Mv}}^{\text{2}}\text{=}\frac{\text{1}}{\text{2}}{\text{MA}}^{\text{2}}{\text{ω}}^{\text{2}}{\text{cos}}^{\text{2}}\text{ωt}\\ \text{Potential energy of the particle,}{\text{E}}_{\text{p}}\text{=}\frac{\text{1}}{\text{2}}{\text{kx}}^{\text{2}}\text{=}\frac{\text{1}}{\text{2}}{\text{Mω}}^{\text{2}}{\text{A}}^{\text{2}}{\text{sin}}^{\text{2}}\text{ωt}\left({\text{Qk=Mω}}^{\text{2}}\right)\\ \text{Let}\text{time}\text{period = T}\\ \text{Average kinetic energy over one cycle,}{\left({\text{E}}_{\text{k}}\right)}_{\text{av}}\text{=}\frac{\text{1}}{\text{T}}{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}{\text{E}}_{\text{k}}\text{dt =}}\frac{\text{1}}{\text{T}}{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}\frac{\text{1}}{\text{2}}{\text{MA}}^{\text{2}}{\text{ω}}^{\text{2}}{\text{cos}}^{\text{2}}\text{ωtdt}}\\ {\left({\text{E}}_{\text{k}}\right)}_{\text{av}}\text{=}\frac{\text{1}}{\text{2T}}{\text{MA}}^{\text{2}}{\text{ω}}^{\text{2}}{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}\frac{\left(\text{1+cos2ωt}\right)}{\text{2}}\text{dt}}\text{=}\frac{\text{1}}{\text{4T}}{\text{MA}}^{\text{2}}{\text{ω}}^{\text{2}}{\left[\text{t+}\frac{\text{sin2ωt}}{\text{2ω}}\right]}_{\text{0}}^{\text{T}}\\ {\left({\text{E}}_{\text{k}}\right)}_{\text{av}}\text{=}\frac{\text{1}}{\text{4T}}{\text{MA}}^{\text{2}}{\text{ω}}^{\text{2}}\left(\text{T}\right)\text{=}\frac{\text{1}}{\text{4}}{\text{MA}}^{\text{2}}{\text{ω}}^{\text{2}}\mathrm{\dots }\text{(i)}\\ \text{Average potential energy over one cycle,}{\left({\text{E}}_{\text{p}}\right)}_{\text{av}}\text{=}\frac{\text{1}}{\text{T}}{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}{\text{E}}_{\text{p}}\text{dt =}}\frac{\text{1}}{\text{T}}{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}\frac{\text{1}}{\text{2}}{\text{Mω}}^{\text{2}}{\text{A}}^{\text{2}}{\text{sin}}^{\text{2}}\text{ωtdt}}\\ {\left({\text{E}}_{\text{p}}\right)}_{\text{av}}\text{=}\frac{\text{1}}{\text{2T}}{\text{Mω}}^{\text{2}}{\text{A}}^{\text{2}}{\displaystyle \underset{\text{0}}{\overset{\text{T}}{\int }}\frac{\left(\text{1-cos2ωt}\right)}{\text{2}}\text{dt}}\text{=}\frac{\text{1}}{\text{4T}}{\text{Mω}}^{\text{2}}{\text{A}}^{\text{2}}{\left[\text{t-}\frac{\text{sin2ωt}}{\text{2ω}}\right]}_{\text{0}}^{\text{T}}\\ {\left({\text{E}}_{\text{p}}\right)}_{\text{av}}\text{=}\frac{\text{1}}{\text{4T}}{\text{Mω}}^{\text{2}}{\text{A}}^{\text{2}}\left(\text{T}\right)\text{=}\frac{\text{1}}{\text{4}}{\text{Mω}}^{\text{2}}{\text{A}}^{\text{2}}\mathrm{\dots }\text{(ii)}\\ \text{From equations}\left(\text{i}\right)\text{and}\left(\text{ii}\right)\text{it}\text{can}\text{be}\text{concluded}\text{that the}\text{average kinetic energy for a}\text{given time period is}\\ \text{equal to the average potential energy for the same time period}\text{.}\end{array}