Chapter 13: Oscillations

Ans: As motion of the swimmer between the banks of the river is to and fro, it does not have a definite period. The time taken by the swimmer during his back-and-forth journey may not be the same. Hence, the swimmer's motion is not periodic.

1. A freely suspended bar magnet displaced from its N-S direction and released.

Ans: If a magnet is displaced from its N-S direction and released, then the motion of the freely-suspended magnet is periodic. This is because the magnet oscillates about its position with a definite period of time.

1. A hydrogen molecule rotating about its centre of mass.

Ans: If we consider a hydrogen molecule rotating about its centre of mass, it is observed that it comes to the same position after an equal interval of time. This type of motion is periodic motion.

1.  An arrow released from a bow.

Ans: When an arrow is released from a bow, it moves only in the forward direction. There is no motion repeated in equal intervals of time. Therefore, this motion is not periodic.

Ans: When the earth rotates about its axis, it comes to the same position in fixed intervals of time. Hence, it is a periodic motion. However, earth does not have a to and fro motion about its axis. Hence, it is not a simple harmonic.

1. motion of an oscillating mercury column in a U-tube.

Ans: In an oscillating mercury column in a U-tube, mercury moves to and from on the same path, about the fixed position, with a certain period of time. Hence, it is a simple harmonic motion.

1. motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lowermost point.

Ans: When a ball is released from a point slightly above the lowermost point, it moves to and from about the lowermost point of the bowl. Also, the ball comes back to its initial position in the fixed interval of time, again and again. Thus, this motion is periodic as well as simple harmonic.

1. general vibrations of a polyatomic molecule about its equilibrium position.

Ans: A polyatomic molecule possesses many natural frequencies of oscillation. Its vibration is the superposition of individual simple harmonic motions of a number of different molecules. Thus, it is not simple harmonic, but periodic.

Ans: Graph in figure (a) is not a periodic motion. It is a unidirectional, linear uniform motion. Also, there is no repetition of motion in this case.

Ans: Graph in figure (b) is a periodic motion. The motion of the particle repeats itself after 2s. Thus, it is a periodic motion with a period of 2s.

Ans: Graph in figure (c) is not a periodic motion as 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.

Ans: Graph in figure (d), the motion of the particle repeats itself after 2s. Thus, it is a periodic motion with a period of 2s.

Ans: The given function is: 

$\sin \omega t-\cos \omega t$

$=\sqrt{2}\left[ \frac{1}{\sqrt{2}}\sin \omega t-\frac{1}{\sqrt{2}}\cos \omega t \right]$

$=\sqrt{2}\left[ \sin \omega t\times \cos \frac{\pi }{4}-\cos \omega t\sin \frac{\pi }{4} \right]$

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

This function is SHM as it can be written in the form of $a\sin \left( \omega t+\phi  \right)$. Its period is $2\pi /\omega $.

1.  ${{\sin }^{3}}\omega t$

Ans: The given function is: 

${{\sin }^{3}}\omega t=\frac{1}{4}\left[ 3\sin \omega t-\sin 3\omega t \right]$

From the above equation, it is clear that $\sin \omega t$and $\sin 3\omega t$represent simple harmonic motion individually and hence does their combination. Hence, the given function is periodic but not simple harmonic.

1. $3\cos \left( \pi /4-2\omega t \right)$

Ans: The given function is: 

$3\cos \left( \pi /4-2\omega t \right)=3\cos \left( 2\omega t-\pi /4 \right)$

This function represents simple harmonic motion because it can be written in the form $aco\operatorname{s}\left( \omega t+\phi  \right)$

Period $=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$

1.  $\cos \omega t+\cos 3\omega t+\cos 5\omega t$

Ans: The given function is: 

$\cos \omega t+\cos 3\omega t+\cos 5\omega t$

From the above equation it is clear that $co\operatorname{s}\omega t$, $co\operatorname{s}2\omega t$and $co\operatorname{s}3\omega t$represent simple harmonic motion individually and hence does their combination. Hence, given function is periodic but not simple harmonic

1.  $\exp \left( -{{\omega }^{2}}{{t}^{2}} \right)$

Ans: The given function is: 

$\exp \left( -{{\omega }^{2}}{{t}^{2}} \right)$ which is an exponential function. As exponential functions do not repeat themselves, hence it is a non-periodic motion.

1. $1+\omega t+{{\omega }^{2}}{{t}^{2}}$

Ans: The given function is:

$1+\omega t+{{\omega }^{2}}{{t}^{2}}$ 

It is non-periodic.

Ans: Consider the figure given in the question. The two extreme positions, A and B are of a SHM. The positive direction of velocity is considered from A to B. The acceleration and the force along AP are positive and BP are negative.

1. at the end A,

$\Rightarrow $The particle is executing SHM is momentarily at rest being its extreme position of motion at the end A. Hence, its velocity is zero. Acceleration is positive as it is directed along AP, force is also positive as the force is directed along AP.

1. at the end B,

$\Rightarrow $Velocity is zero at the end B. As acceleration and force are directed along BP, hence they are negative.

1. at the mid-point of AB going towards A,

$\Rightarrow $Along the direction towards A, at the midpoint of AB, the particle is at its mean position P and has a tendency to move along PA. Thus, velocity is positive. Both acceleration and force are zero.

1. at 2 cm away from B going towards A,

$\Rightarrow $The position of particle at 2 cm away from B going towards A is at Q. At this position it has the tendency to move along QP, which is negative direction. Therefore, velocity, acceleration and force all are positive.

1. at 3 cm away from A going towards B, and

$\Rightarrow $The position of the particle at 3 cm away from A going towards B is at R. It has a tendency to move along RP, which is the positive direction. Here, velocity, acceleration all are positive.

1. at 4 cm away from B going towards A.

$\Rightarrow $The position of the particle at 4 cm away from A going towards A, is at S. It has a tendency to move along SA, which is the negative direction. Thus, velocity is negative but acceleration is directed towards the mean position, along with SP. Hence it is positive and also force is positive similarly.

Ans: For SHM, acceleration and displacement are related by the relation (c)  $a=-10x$ (of the form $a=-kx$).

Ans: In the above question it is given that:

Displacement at $t=0$, is $x=1cm$.

Initial velocity, $v=\text{ }\!\!\omega\!\!\text{  cm/sec}$.

We have,

$x\left( t \right)=A\cos \left( \omega t+\varphi  \right)$

$1=A\cos \left( \omega \times 0+\varphi  \right)=A\cos \varphi $

$A\cos \varphi =1$ ….. (i)

Velocity, $v=\frac{dx}{dt}$

$\omega =-A\omega \sin \left( \omega t+\varphi  \right)$

$1=A\sin \left( \omega \times 0+\varphi  \right)=-A\sin \varphi $

$A\sin \varphi =-1$ …… (ii)

Squaring and adding equations (i) and (ii), we get:

${{A}^{2}}\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)=1+1$$\Rightarrow {{A}^{2}}=2$$\Rightarrow A=\sqrt{2}cm$

Dividing equation (ii) by equation (i), we get:

$\tan \varphi =-1$

$\Rightarrow \varphi =3\pi /4,7\pi /4.....$

SHM is given by:

$x\left( t \right)=B\sin \left( \omega t+\alpha  \right)$

Substituting the given values in this equation, we get:

$1=B\sin \left[ \omega \times 0+\alpha  \right]=1+1$

$B\sin \alpha =1$ …… (iv)

Squaring and adding equations (iii) and (iv), we get:

${{B}^{2}}\left( {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha  \right)=1+1\Rightarrow {{B}^{2}}=2\Rightarrow B=\sqrt{2}cm$

Dividing equation (iii) by equation (iv), we get:

$\tan \alpha =1\Rightarrow \alpha =\pi /4,5\pi /4.....$

Ans: In the above question it is given that:

Maximum mass that the scale can read is \[M\text{ }=\text{ }50\text{ }kg\].

Maximum displacement of the spring \[=\text{ }Length\text{ }of\text{ }the\text{ }scale,\text{ }l\text{ }=\text{ }20\text{ }cm\text{ }=\text{ }0.2\text{ }m\]

Time period, \[T\text{ }=\text{ }0.6\text{ }s\]

Maximum force exerted on the spring is \[F\text{ }=\text{ }Mg\].

where,

\[g\text{ }=\text{ }acceleration\text{ }due\text{ }to\text{ }gravity\text{ }=\text{ }9.8\text{ }m/{{s}^{2}}\]

\[F\text{ }=\text{ }50\text{ }\times \text{ }9.8\text{ }=\text{ }490\text{ }N\]

$\therefore$ Spring constant, $k=\frac{F}{l}=\frac{490}{0.2}=2450N{{m}^{-1}}$.

Mass m, is suspended from the balance.

Time period, 

$T=t=2\pi \sqrt{\frac{m}{k}}$

$\Rightarrow m=\left( \frac{T}{2\pi } \right)\times k={{\left( \frac{0.6}{2\times 3.14} \right)}^{2}}\times 2450=22.36kg$

Weight of the body \[=\text{ }mg\text{ }=\text{ }22.36\times 9.8\text{ }=\text{ }219.167\text{ }N\]

Therefore, the weight of the body is \[219\text{ }N\].

Ans: In the above question it is given that:

Spring constant is $k=1200N{{m}^{-1}}$

Mass, \[m\text{ }=\text{ }3\text{ }kg\]

Displacement, \[A\text{ }=\text{ }2.0\text{ }cm\text{ }=\text{ }0.02\text{ }cm\]

1. Frequency of oscillation v, is given by:

$\upsilon =\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Here, T is time period.

$\Rightarrow \upsilon =\frac{1}{2\pi }\sqrt{\frac{1200}{3}=3.18m/s}$

Thus, the frequency of oscillations is \[3.18\] cycles per second.

1. Maximum acceleration (a) is given by:

$a={{\omega }^{2}}A$

Where, 

\[\text{A= maximum displacement}\]

$\alpha =\frac{k}{m}A=\frac{1200\times 0.02}{3}=8m{{s}^{-2}}$

Thus, the maximum acceleration of the mass is $8m{{s}^{-2}}$.

1. Maximum velocity is 

${{v}_{\max }}=A\omega =A\sqrt{\frac{k}{m}=0.02\times }\sqrt{\frac{1200}{3}=0.4m/s}$

Hence, the maximum velocity of the mass is 0.4 m/s.

Ans: In the above question it is given that:

Distance travelled by the mass sideways is \[a\text{ }=\text{ }2.0\text{ }cm\].

Angular frequency of oscillation is given by:

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

1. As time is noted from the mean position, initial phase is $0$. Hence, Displacement, \[x\text{ }=\text{ }2\text{ }sin\text{ }20\text{ }t\].

2. When the body is at maximum stretched position, it is at the extreme right position, with an initial phase of $\pi /2$ rad.

Then, displacement $x=\alpha \sin \left( \omega t+\frac{\pi }{2} \right)=\alpha \cos \omega t=2\cos 20t$.

1. The body is at left position at maximum compressed position, with an initial phase of  $3\pi /2$ rad. 

Then,

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

The functions differ in initial phase. They neither differ in amplitude nor in frequency.