Chapter 13: Oscillations

Ans:

(a) In the figure (a) it is given that:

Time period is \[t\text{ }=\text{ }2\text{ }s\].

Amplitude is \[A\text{ }=\text{ }3\text{ }cm\].

When \[t\text{ }=\text{ }0\] , the radius vector OP makes an angle $\pi /2$ with the positive x-axis.

\[Phase\text{ }angle=\pi /2\].

Hence, the equation of simple harmonic motion for the x-projection of OP, at the time t, and is given by the displacement equation:

$x=A\cos \left[ \frac{2\pi t}{T}+\phi  \right]=3\cos \left[ \frac{2\pi t}{2}+\frac{\pi }{2} \right]=-3\sin \left( \frac{2\pi t}{2} \right)cm$

$\Rightarrow x=-3\sin \pi tcm$, which represents the SHM equation here.

(b) In the figure (b) it is given that:

Time Period, \[t\text{ }=\text{ }4\text{ }s\].

Amplitude, \[a\text{ }=\text{ }2\text{ }m\].

At time \[t\text{ }=\text{ }0\], OP makes an angle $\pi $ with the x-axis, in the anticlockwise direction, 

\[Phase\text{ }angle=\pi \]

Thus, the equation of simple harmonic motion for the x-projection of OP, at the time t, is

$x=A\cos \left[ \frac{2\pi t}{T}+\phi  \right]=2\cos \left[ \frac{2\pi t}{4}+\pi  \right]$

$\Rightarrow x=-2\cos \left( \frac{\pi }{2}t \right)m$, which represents the SHM equation here.

Ans: Given equation is 

$x=-2\sin \left( 3t+\pi /3 \right)=-2co\operatorname{s}\left( 3t+\pi /3+\pi /2 \right)=-2co\operatorname{s}\left( 3t+5\pi /6 \right)$

If this equation is compared with the standard SHM equation 

$x=A\cos \left[ \frac{2\pi t}{T}+\phi  \right]$

Therefore, $A=2cm$,

$\phi =5\pi /6$,

$\omega =2\pi /t=3rad/s$

The motion of the particle is shown in fig (a).

1. $x=\cos \left( \pi /6-t \right)$

Ans: Given equation is 

$x=\cos \left( \pi /6-t \right)=\cos \left( t-\pi /6 \right)$

If this equation is compared with the standard SHM equation 

$x=A\cos \left[ \frac{2\pi t}{T}+\phi  \right]$

Therefore, $A=1cm$,

$\phi =-\pi /6$,

$\omega =2\pi /t=1rad/s$

The motion of the particle is shown in fig (b).

1. $x=3\sin \left( 2\pi t+\pi /4 \right)$

Ans: Given equation is 

$x=3\sin \left( 2\pi t+\pi /4 \right)=-3\cos \left( 2\pi t+\pi /4+\pi /2 \right)=-3\cos \left( 2\pi t+3\pi /4 \right)$

If this equation is compared with the standard SHM equation 

$x=A\cos \left[ \frac{2\pi t}{T}+\phi  \right]$

Therefore, $A=3cm$,

$\phi =3\pi /4$,

$\omega =2\pi /t=2rad/s$

The motion of the particle is shown in fig (c).

1. $x=2\cos \pi t$

Ans: Given equation is 

$x=2\cos \pi t$

If this equation is compared with the standard SHM equation 

$x=A\cos \left[ \frac{2\pi t}{T}+\phi  \right]$

Therefore, $A=2cm$,

$\phi =0$,

$\omega =\pi rad/s$

The motion of the particle is shown in fig (d).

Ans: Consider one block system.

If a force F, is applied to the free end of the spring, then extension l, is produced. For the maximum extension,

i.e., \[F=kl\]

Where, k is the spring constant.

Thus, the maximum extension produced in the spring is $l=\frac{F}{k}$.

Now consider the two-block system:

In this case the displacement (x) produced is $x=\frac{l}{2}$.

Therefore, net force$=2kx=2k\left( \frac{l}{2} \right)$

$\Rightarrow l=\frac{F}{k}$

Thus, the maximum extension of the spring in both cases will be $\frac{F}{k}$.

1. 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: In figure (a) if x is the extension in the spring, when mass m is returning to its mean position after being released free, then restoring force on the mass is $F=-kx$, i.e., $F\propto x$. As, this F is directed towards the mean position of the mass, hence the mass attached to the spring will execute SHM.

Spring factor = spring constant = k

inertia factor = mass of the given mass = m

As time period,

$T=2\pi \sqrt{\frac{\text{inertia factor}}{\text{spring factor}}}=2\pi \sqrt{\frac{m}{k}}$

If we consider figure (b), we have a two body system of spring constant k and reduced mass,

$\mu =\frac{m\times m}{m+m}=\frac{m}{2}$

$Inertia\text{ }factor\text{ }=\text{ }m/2$

$Spring\text{ }factor\text{ }=\text{ }k$

Hence, time period, $T=2\pi \sqrt{\frac{m/2}{k}=}2\pi \sqrt{\frac{m}{2k}}$.

Ans: In the above question it is given that:

Angular frequency of the piston is \[200\text{ }rad/min\].

\[Stroke\text{ }=\text{ }1.0\text{ }m\]

Amplitude is \[A\text{ }=\text{ }1.0/2\text{ }=\text{ }0.5\text{ }m\].

The maximum speed $\left( {{v}_{\max }} \right)$ of piston is given by:

${{v}_{\max }}=A\omega =200\times 0.5=100m/\min $

Ans: In the above question it is given that:

Acceleration due to gravity on the surface of the moon is $g'=1.7m/{{s}^{2}}$.

Acceleration due to gravity on the surface of earth is $g=9.8m/{{s}^{2}}$.

Time period of a simple pendulum on earth is \[T\text{ }=\text{ }3.5\text{ }s\].

$T=2\pi \sqrt{\frac{l}{g}}$

\[l=\frac{{{T}^{2}}}{4{{\pi }^{2}}}g=\frac{{{\left( 3.5 \right)}^{2}}}{4{{\left( 3.14 \right)}^{2}}}\left( 9.8 \right)\]

Where,

l is the length of the pendulum.

The length of pendulum remains constant

On moon’s surface, time period, $T'=2\pi \sqrt{\frac{l}{g'}}$

$\Rightarrow T'=2\pi \sqrt{\frac{\frac{{{\left( 3.5 \right)}^{2}}}{4{{\left( 3.14 \right)}^{2}}}\left( 9.8 \right)}{1.7}}=8.4s$

Therefore, Hence, the time period of the simple pendulum on the surface of moon is $8.4s$.

Ans: We know that:

The time period of the simple pendulum is $T=2\pi \sqrt{\frac{m}{k}}$

For a simple pendulum, k is expressed in terms of mass m, as: $k\propto m$,

Where, $\frac{m}{k}$ is a constant.

Thus, the time period T, of a simple pendulum is independent of the mass of the bob.

1. 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.

Ans: If we consider a simple pendulum, the restoring force acting on the bob of the pendulum is given by: \[F=-mg\sin \theta \]

Where,

\[\text{F = Restoring force}\]

\[\text{m = Mass of the bob}\]

\[\text{g = Acceleration due to gravity}\]

\[\theta \text{=Angle of displacement}\]

For small $\theta $, $\sin \theta \approx \theta $

For large $\theta $, $\sin \theta >\theta $

This decreases the effective value of g.

Thus, the time period increases as:

$T=2\pi \sqrt{\frac{l}{g}}$ where l is the length of the simple pendulum.

1. 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?

Ans: The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. As a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. The working of wristwatch depends on spring action.

1. What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

Ans: If a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Thus, the frequency of the oscillation of this pendulum is zero.

Ans: From the above question it is clear that the bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car.

\[\text{Acceleration due to gravity = g}\]

\[\text{Centripetal acceleration }{{\text{a}}_{c}}={{v}^{2}}/r\]

where,

v is the uniform speed of the car

R is the radius of the track

Hence, effective acceleration is given as: 

$g'=\sqrt{{{g}^{2}}+{{a}_{c}}^{2}}$

$g'=\sqrt{{{g}^{2}}+{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}}$

Time period, 

$T=2\pi \sqrt{\frac{l}{g'}}\frac{dy}{dx}$

$\Rightarrow T=2\pi \sqrt{\frac{l}{\sqrt{{{g}^{2}}+\frac{{{v}^{4}}}{{{r}^{2}}}}}}$

$\Rightarrow T=2\pi \sqrt{\frac{l}{\sqrt{{{g}^{2}}+{{\left( \frac{{{v}^{2}}}{r} \right)}^{2}}}}}$, which represents the time period here.

Ans: In the above question it given that:

\[Base\text{ }area\text{ }of\text{ }the\text{ }cork\text{ }=\text{ }A\]

\[Height\text{ }of\text{ }the\text{ }cork\text{ }=\text{ }h\]

Density of the liquid \[={{\rho }_{1}}\]

Density of the cork\[=\rho \]

In equilibrium:

\[Weight\text{ }of\text{ }the\text{ }cork\text{ }=\text{ }Weight\text{ }of\text{ }the\text{ }liquid\text{ }displaced\text{ }by\text{ }the\text{ }floating\text{ }cork\]

If the cork is depressed slightly by x, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.

\[Upthrust\text{ }=\text{ }Restoring\text{ }force,\text{ }F\text{ }=\text{ }Weight\text{ }of\text{ }the\text{ }extra\text{ }water\text{ }displaced\]

\[F\text{ }=\text{ }\text{ }\left( Volume\text{ }\times \text{ }Density\text{ }\times \text{ }g \right)\]

\[Volume\text{ }=\text{ }Area\text{ }\times \text{ }Distance\text{ }through\text{ }which\text{ }the\text{ }cork\text{ }is\text{ }depressed\]

$\Rightarrow F=-Ax{{\rho }_{1}}g$ …… (i)

According to the force law:

$F=kx\Rightarrow k=F/x$

where, k is constant.

$k=F/x=-A{{\rho }_{1}}g$…… (ii)

The time period of the oscillations of the cork: $T=2\pi \sqrt{\frac{m}{k}}$ ..(iii)

where,

\[m=\text{ }Mass\text{ }of\text{ }the\text{ }cork\]

\[m\text{ }=\text{ }Volume\text{ }of\text{ }the\text{ }cork\text{ }\times \text{ }Density\]

\[m\text{ }=\text{ }Base\text{ }area\text{ }of\text{ }the\text{ }cork\text{ }\times \text{ }Height\text{ }of\text{ }the\text{ }cork\text{ }\times \text{ }Density\text{ }of\text{ }the\text{ }cork\]

\[m=Ah\rho \]

Hence, the expression for the time period becomes:

$T=2\pi \sqrt{\frac{Ah\rho }{A{{\rho }_{1}}g}}=2\pi \sqrt{\frac{h\rho }{{{\rho }_{1}}g}}$

Ans: Consider,

\[Area\text{ }of\text{ }cross-section\text{ }of\text{ }the\text{ }U-tube\text{ }=\text{ }A\]

\[Density\text{ }of\text{ }the\text{ }mercury\text{ }column\text{ }=\rho \]

\[\text{Acceleration due to gravity = g}\]

\[\text{Restoring force, F = Weight of the mercury column of a certain height}\]

\[F\text{ }=\text{ }\left( Volume\text{ }\times \text{ }Density\text{ }\times \text{ }g \right)\]

\[F=-\left( 2A\times h\times \rho \times g \right)=-2A\rho gh\]

\[=k\times \text{Displacement in one of the arms (h)}\]

Where,

$2h$ is the height of the mercury column in the two arms

k is a constant, given by

$k=-F/h=2A\rho g$

Time period, $T=2\pi \sqrt{\frac{m}{A\rho g}}$

where,

m is the mass of the mercury column

Let l be the length of the total mercury in the U-tube

\[Mass\text{ }of\text{ }mercury,\text{ }m\text{ }=\text{ }Volume\text{ }of\text{ }mercury\text{ }\times \text{ }Density\text{ }of\text{ }mercury=Al\rho \]

$T=2\pi \sqrt{\frac{Al\rho }{2A\rho g}}=2\pi \sqrt{\frac{l}{2g}}$

Thus, the mercury column executes simple harmonic motion with time period $2\pi \sqrt{\frac{l}{2g}}$.

Ans: In the above question it given that:

\[Volume\text{ }of\text{ }the\text{ }air\text{ }chamber=V\]

\[Area\text{ }of\text{ }cross-section\text{ }of\text{ }the\text{ }neck\text{ }=\text{ }a\]

\[Mass\text{ }of\text{ }the\text{ }ball\text{ }=\text{ }m\]

\[The\text{ }pressure\text{ }inside\text{ }the\text{ }chamber=atmospheric\text{ }pressure.\]

Let the ball be depressed by x units. As a result of this depression, there would be a decrease in

the volume and an increase in the pressure inside the chamber.

Decrease in the volume of the air chamber, $\Delta V=ax$

$\text{Volumetric Strain = }\frac{\text{change in volume}}{\text{original volume}}=\frac{ax}{V}$

Bulk modulus of air, $B=\frac{Stress}{Strain}=\frac{-p}{\frac{ax}{V}}$

Here, stress is the increase in pressure. The negative sign indicates that pressure increases with decrease in volume.

$p=-Bax/V$

The restoring force acting on the ball,

$F=p\times a$

$=-\frac{B{{a}^{2}}x}{V}$ …… (i)

In simple harmonic motion, the equation for restoring force is:

\[F\text{ }=\text{ }-kx\] ..... (ii)

where, k is the spring constant

Comparing equations (i) and (ii), we get:

$k=-\frac{B{{a}^{2}}}{V}$

Time Period, 

\[T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{Vm}{B{{\alpha }^{2}}}}\]