Oscillations

Class 11 NCERT Physics

NCERT

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.

Solution :

(a) The swimmer's motion is not periodic. The motion of the swimmer between the banks of the river is to and fro. However, it does not have a definite period. This is because 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, is periodic because the magnet oscillates about its position with a definite period of time.$\\$ (c) When a hydrogen molecule rotates about its centre of mass, it comes to the same position again and again after an equal interval of time. Such a motion is periodic.$\\$ (d) An arrow released from a bow moves only in the forward direction. It does not come backward. Hence, this motion is not a periodic.

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.

Solution :

ball comes back to its initial position in the same period of time, again and again. Hence, its motion is periodic as well as simple harmonic.$\\$ (d) A polyatomic molecule has many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Hence, it is not simple harmonic, but periodic.

(a)During its rotation about the axis, earth comes to the same position again and again in equal intervals of time. Hence, it is a periodic motion. However, this motion is not simple harmonic. This is because earth does not have a to and fro motion about its axis.$\\$ (b)An oscillating mercury column in a U-tube is a simple harmonic motion because the mercury moves to and from on the same path, about the fixed position, with a certain period of time.$\\$ (c) the ball moves to and from about the lowermost point of the bowl when released. Also, the

3   Figure 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)?

Solution :

(a) It is not a periodic motion. This represents a unidirectional, linear uniform motion. There is no repetition of motion in this case.$\\$ (b) In this case, the motion of the particle repeats itself after $2 s$ . Hence, it is a periodic motion, having a period of $2 s.$$\\$ (c) It is not a periodic motion. This is because the particle repeats the motion in one position only. For a periodic motion, the entire motion of the particle must be repeated in equal intervals of time.$\\$ (d) In this case, the motion of the particle repeats itself after $2 s$. Hence, it is a periodic motion, having a period of $2 s$.

4   Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion ($\omega $ is any positive constant):$\\$ (a)$\sin \omega t-\cos \omega t$$\\$ (b)$\sin^3 \omega t$$\\$ (c)$3 \cos \omega t+\cos 3 \omega t+\cos 5 \omega t$$\\$ (d)$\cos \omega t + \cos 3 \omega t+\cos 5 \omega t $$\\$ (e)exp$(-\omega^2 t^2)$$\\$ (f)$1+\omega t+\omega^2 t^2$

Solution :

(a) SHM$\\$ The given function is:$\\$ $\sin \omega t -\cos omega t\\ =\sqrt{2}[\dfrac{1}{\sqrt{2}}\sin \omega t -\dfrac{1}{\sqrt{2}}\cos \omega t]\\ =\sqrt{2}[\sin \omega t*\cos \dfrac{\pi}{4}\\ -\cos \omega t* \sin \dfrac{\pi}{4}]\\ =\sqrt{2}\sin(\omega t-\dfrac{\pi}{4})$$\\$ This function represents SHM as it can be written in the form:$\\$ $a \sin(\omega t+\Phi)$$\\$ Its period is: $2\pi / \omega$$\\$ (b) Periodic but not SHM$\\$ The given function is:$\\$ $\sin^3 \omega t=1/4[3 \sin \omega t -\sin 3 \omega t]$ The terms $\sin \omega t$ and $\sin \omega t$ individually represent simple harmonic motion (SHM). However, the superposition of two SHM is periodic and not simple harmonic.$\\$ This function represents simple harmonic motion because it can be written in the form:$\\$$ a \cos(\omega t+\Phi)$ Its period is :$ 2\pi / 2 \omega =\pi / \omega$$\\$ (d) Periodic, but not SHM$\\$ The given function is $\cos \omega t + \cos 3 \omega t + \cos 5 \omega t $. Each individual cosine function represents SHM. However, the superposition of three simple harmonic motions is periodic, but not simple harmonic.$\\$ (e) Non-periodic motion$\\$ The given function exp $(-\omega^2 t^2 )$ is an exponential function. Exponential functions do not repeat themselves. Therefore, it is a non-periodic motion.$\\$ (f) The given function $1+ \omega t +\omega^2 t^2$ is non-periodic.

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.

Solution :

From above figure, where A and B represent the two extreme positions of a SHM. For velocity, the direction from A to B is taken to b positive. The acceleration and the force, along AP are taken as positive and also BP are taken as negative.$\\$ (a) At the end A, the particle executing SHM is momentarily at rest being its extreme position of motion. Therefore, its velocity is zero. Acceleration is positive because it is directed along AP, Force is also Positive since the force is directed along AP.$\\$ (b) At the end B, velocity is zero. Here, acceleration and force are negative as they are directed along BP.$\\$ (c) At the mid point of AB going towards A, the particle is at its mean position P, with a tendency to move along PA. Hence, velocity is positive. Both acceleration and force are zero.$\\$ (d) At $2 cm$ away from B going towards A, the particle is at Q, with a tendency to move along QP, which is negative direction. Therefore, velocity, acceleration and force all are positive.$\\$ (e) At $3 cm$ away from A going towards B, the particle is at R, with a tendency to move along RP, which is positive direction. Here, velocity, acceleration all are positive.$\\$ (f) At $4 cm$ away from A going towards A, the particles is at S, with a tendency to move along SA, which is negative direction. Therefore, velocity is negative but acceleration is directed towards mean position, along SP. Hence it is positive and also force is positive similarly.

6   Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion?

Solution :

In SHM, acceleration a is related to displacement by the relation of the form $a=- kx$ , which is for relation (c).

7   The motion of a particle executing simple harmonic motion is described by the displacement function, $x(t)=A\cos(\omega+\varphi)$.$\\$ 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.

Solution :

Initially, at $t = 0;$$\\$ Displacement,$ x = 1 cm$$\\$ Initial velocity, $v =\omega \ cm / sec .$$\\$ Angular frequency, $\omega =\pi \ rad / s^{-1}$$\\$ . It is given that,$\\$ $x(t)=A \cos(\omega t+ \Phi)\\ 1=A \cos(\omega * 0+\Phi)=A\cos \Phi\\ A \cos \Phi =1 \quad (i)\\ \text{Velocity }, v= dx/dt\\ \omega =-A \omega \sin(\omega t+\Phi)\\ 1=-A \sin(\omega * 0+\Phi)=-A \sin \Phi\\ A \sin \Phi =-1 \quad ....(ii)$$\\$ Squaring and adding equations (i) and (ii), we get:$\\$ $A^2(\sin^2 \Phi +\cos^2 \Phi)=1+1\\ A^2=2\\ \therefore A=\sqrt{2} \ cm$$\\$ Dividing equation (ii) by equation (i), we get:$\\$ $\tan \Phi=-1\\ \therefore \Phi =3 \pi / 4, 7 \pi / 4,.......$$\\$ SHM is given as:$\\$ $x= B\sin(\omega t+\alpha)$$\\$ Putting the given values in this equation, we get:$\\$ $1= B \sin[\omega * 0+\alpha]=1+1\\ B\sin \alpha=1 \quad ....(iii)\\ \text{Velocity}, v=\omega B \cos (\omega t +\alpha)$$\\$ Substituting the given values, we get:$\\$ $\pi=\pi B \sin \alpha\\ B\sin \alpha=1 \quad .......(iv)$$\\$ Squaring and adding equations (iii) and (iv), we get:$\\$ $B^2[\sin^2 \alpha +\cos^2 \alpha]=1+1\\ B^2=2\\ \therefore B=\sqrt{2 }cm$$\\$ Dividing equation (iii) by equation (iv), we get:$\\$ $B \sin \alpha /B \cos \alpha =1/1\\ \tan \alpha=1=\tan \pi/4\\ \therefore \alpha =\dfrac{\pi}{4},\dfrac{5 \pi}{4},.......$$\\$

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?

Solution :

Maximum mass that the scale can read,$ M = 50 kg$$\\$ Maximum displacement of the spring = Length of the scale, $l = 20 cm = 0.2 m$$\\$ Time period, $T = 0.6 s$$\\$ Maximum force exerted on the spring, $F = Mg$$\\$ where, g = acceleration due to gravity =$ 9.8 m/s^2$$\\$ $F = 50 × 9.8 = 490 N$$\\$ $\therefore $ Spring constant, $k = F / l = 490 / 0.2 =2450 N m^{-1}$ .$\\$ Mass $m$, is suspended from the balance.$\\$ Time period,$ T=t=2\pi \sqrt{\dfrac{m}{k}}$$\\$ $\therefore m=(\dfrac{T}{2\pi})^2*k=(\dfrac{0.6}{2*3.14})^2* 2450=22.36 kg$$\\$ $\therefore $ Weight of the body =mg=$22.36*9.8=219.167N$$\\$ Hence,the weight of the body is about $ 219N$$\\$

9   A spring having with a spring constant $1200 N m^{-1}$ is mounted on a horizontal table as shown in figure. 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.

Solution :

Spring constant, $k = 1200 N m^{-1}$$\\$ Mass, $m = 3 kg$$\\$ Displacement, $A = 2.0 cm = 0.02 cm$$\\$ (i) Frequency of oscillation $v$, is given by the relation:$\\$ $v=\dfrac{1}{T}=\dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}$$\\$ Where, $T$ is time period $\\$ $\therefore v=\dfrac{1}{2*3.14}\sqrt{\dfrac{1200}{3}}=3.18 m/s$$\\$ Hence, the frequency of oscillations is $3.18$ cycles per second.$\\$$\\$ (ii) Maximum acceleration (a) is given by the relation:$\\$ $a =\omega ^2 A$$\\$ where, A= maximum displacement $\\$ $\omega =$ Angular frequency=$\sqrt{\dfrac{k}{m}}$$\\$ A=max imum displacement$\\$ $\therefore \alpha =\dfrac{k}{m}A=\dfrac{1200*0.02}{3}=8 ms^{-2}$$\\$ Hence,the maximum acceleration of the mass is $ 8.0 m/s^2.$$\\$ (iii) Maximum velocity, $v _{max }= A \omega .$$\\$ $=A \sqrt{\dfrac{k}{m}} =0.02*\sqrt{\dfrac{1200}{3}}=0.4 m/s.$$\\$ Hence,the maximum velocity of the mass is $ 0.4 m/s.$

10   In Question 9, let us take the position of mass when the spring is unstarched 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?

Solution :

Distance travelled by the mass sideways, $ a = 2.0 cm $$\\$ Angular frequency of oscillation:$\\$ $ \omega =\sqrt{\dfrac{k}{m}}$$\\$ $=\sqrt{\dfrac{1200}{3}}=\sqrt{400}=20 rad s^{-1}$$\\$ (a) As time is noted from the mean position, initial phase is $0$. hence using Displacement, $x = a \sin \omega t ,$$\\$ $x= 2\sin 20 t$$\\$ (b) At maximum stretched position, the body is at the extreme right position, with an initial phase of $\pi / 2 rad $. Then,$\\$ Displacement, $x=\alpha \sin(\omega t+\dfrac{\pi}{2})=\alpha \cos \omega t=2\cos 20 t$$\\$ (c) At maximum compressed position, the body is at left position, with an initial phase of $3 \pi / 2 rad $ . Then, Displacement , $x=\alpha \sin(\omega t+\dfrac{3 \pi}{2})=-\alpha \cos \omega t=-2 \cos 20 t$$\\$ The functions neither differ in amplitude nor in frequency. They differ in initial phase.