Class 11th

Volume of the air chamber = V

Area of cross-section of the neck = a

Mass of the ball = m

The pressure inside the chamber is equal to atmospheric pressure.

Let the ball be depressed by x units. As a result of depression, there would be decrease in volume and an increase of pressure inside the cylinder.

Decrease in the volume,  $?V$ = ax

Volumetric strain = $\frac{?V}{V}$ = $\frac{ax}{V}$

Bulk modulus of air, B = $\frac{Stress}{Strain}$ = $\frac{-p}{\mathrm{}\frac{ax}{V}}$ : here stress is the increase in pressure. The negative sign indicates that pressure increases with a decrease in volume.

So p = $-\frac{Bax}{V}$

The restoring force acting on the ball, F = p $*a$ = $-\frac{B{a}^{2}x}{V}$ …. (i)

In SHM, the equation of restoring force, F = -kx …. (ii)

Combining equation (i) and (ii), we get k = $\frac{B{a}^2}{V}$

Time period, T = 2 $\pi \sqrt{\frac{m}{k}}$ = 2 $\pi \sqrt{\frac{Vm}{B{a}^2}}$

Base area of the cork = A

Height of the cork = h

Density of the liquid = ${\rho }_l$

Density of the cork = $\rho$

In equilibrium, Weight of the cork = Weight of the liquid displaced by the floating cork

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

Up-thrust (Restoring force) = weight of the extra water displaced

F = mg = ${\rho }_{l}Vg$

Volume = Area $*$ distance through which the cork is depressed

V = Ax

F = A ${\rho }_{l}gx$ ….(i)

According to force law, F= kx, where k is constant

k = $\frac{F}{x}$ = A ${\rho }_{l}g$ …(ii)

Time period of the oscillation of the cork, T = 2 $\pi \sqrt{\frac{m}{k}}$ ….(iii)

Where m = mass of the cork = Base area of the cork $*\mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{k}\mathrm{}*\mathrm{d}\mathrm{e}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y}$

m = Ah $\rho$

So T = 2 $\pi \sqrt{\frac{\mathrm{A}\mathrm{h}\rho }{\mathrm{A}{\rho }_{l}g\mathrm{}}}$ = 2 $\pi \sqrt{\frac{\mathrm{h}\rho }{{\rho }_{l}g}}$

(a) The time period of a simple pendulum, T = 2 $\pi \sqrt{\frac{m}{k}}$

For a simple pendulum, k is expressed in terms of mass, m as : k $\propto m$ or $\frac{m}{k}$ = constant

Hence, the time period of a simple pendulum is independent of the mass of the bob. In the case of a simple pendulum, the restoring force acting on bob is given as F = -mg $\sin ?\theta$ , where

F = restoring force

m = mass of the bob

g = acceleration due to gravity

$\theta =\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}$

(b) For small $\theta ,$ sin $\theta$ $~\theta$ . For larger $\theta ,$ sin $\theta$ is greater than $\theta$ . This decreases the effective value of g.

Hence the time period increase as : T = 2 $\pi \sqrt{\frac{l}{g}}$ , where l is the length of the simple pendulum.

(c) The time shown by the wristwatch of a man falling from the top of the tower is not affected by the fall. Wrist watch works on spring action, independent of acceleration due to gravity.

(d) During free fall under gravity, the acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

(a) For figure (a) : When a force F is applied to the free end of the spring, an extension l is produced. For the maximum extension, it can be written as:

F – kl, where k is the spring constant.

For maximum =extension of the spring, l = $\frac{F}{k}$

For figure (b): The displacement (x) produced in this case is x = $\frac{1}{2}$

Net force F = +2kx = 2k $\frac{1}{2}$ . So l = $\frac{F}{k}$

(b) For figure (a) : For mass (m) of the block, force is written as : F = ma = m $\frac{d^{2}x}{d{t}^2}$ ,

where x is the displacement of the block in time t, then

m $\frac{d^{2}x}{d{t}^2}=-kx$ , it is negative because the direction of the elastic force is opposite to the direction of displacement.

$\frac{d^{2}x}{d{t}^2}=-\left(\frac{k}{m}\right)x$ = ${-\omega }^2$ xwhere, ${\omega }^2$ = $\frac{k}{m}$ , $\omega \mathrm{i}\mathrm{s}\mathrm{}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{o}\mathrm{s}\mathrm{c}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}.$

Time period of the oscillation, T = $\frac{2\pi }{\omega }$ = $\frac{2\pi }{\surd \frac{k}{m}}$ = 2 $\pi \sqrt{\frac{m}{k}}$

For figure (b) :

F = m $\frac{d^{2}x}{d{t}^2}$ = -2kx. It is negative because the direction of elastic force is opposite to the direction of displacement.

$\frac{d^{2}x}{d{t}^2}=-\left[\frac{2kx}{m}\right]x$ = - ${\omega }^2$ x, where angular frequency $\omega$ = $\sqrt{\frac{2k}{m}}$

Time period, T = $\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{2k}}$

(a) X = -2 sin( 3t + $\frac{\pi }{3}$ ) = +2cos ( 3t + $\frac{\pi }{3}$ + $\frac{\pi }{2})$ = 2 cos (3t + $\frac{5\pi }{6}$ )

when we compare this equation with standard SHM equation

x = Acos ( $\omega$ t + $\phi$ ), then we get

Amplitude A = 2 cm. Phase angle $\phi =\frac{5\pi }{6}$ = 150 $°$ , angular velocity $\omega$ = 3 rad/s

(b) X= cos ( $\frac{\pi }{6}-$ t) = cos ( $t-\frac{\pi }{6}$ )

when we compare this equation with standard SHM equation

x = Acos( $\omega$ t + $\phi$ ), then we get

Amplitude A = 1 cm. Phase angle $\phi =-\frac{\pi }{6}$ = - 30 $°$ , angular velocity $\omega$ = 1 rad/s

(c) X = 3sin (2 $\pi$ t + $\frac{\pi }{4}$ ) = -3cos $\left[\left(2\pi \mathrm{t}\mathrm{}+\mathrm{}\frac{\pi }{4}\right)+\mathrm{}\frac{\mathrm{\pi }}{2}\right]=-3\mathrm{c}\mathrm{o}\mathrm{s}?(2\pi \mathrm{t}+\mathrm{}\frac{3\pi }{4})$

when we compare this equation with standard SHM equation

x = Acos( $\omega$ t + $\phi$ ), then we get

Amplitude A = 3 cm. Phase angle $\phi =\frac{3\pi }{4}$ = 135 $°$ , angular velocity $\omega$ = $2\pi$ rad/s

(d) X = 2cos $\pi$ t

when we compare this equation with standard SHM equation

x = Acos( $\omega$ t + $\phi$ ), then we get

Amplitude A = 2 cm. Phase angle $\phi =0°$ , angular velocity $\omega$ = $\pi$ rad/s

(a) Time period, T = 2 s, Amplitude A = 3 cm

At time, t = 0, the radius vector makes an angle $\frac{\pi }{2}$ with the positive x-axis, i.e. phase angle $\phi$ = + $\frac{\pi }{2}$

Therefore, the equation of simple harmonic motion for the x-projection of the radius vector, at time t is given by the displacement equation:

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

(b) Time period, T = 4 s, Amplitude A = 2 m

At time, t = 0, the radius vector makes an angle $\pi$ with the positive x-axis, i.e. phase angle $\phi$ = + $\pi$

Therefore, the equation of simple harmonic motion for the x-projection of the radius vector, at time t is given by the displacement equation:

x = Acos $\left. [\frac{2\pi t}{T}+\phi \right]$ = 2cos $\left. [\frac{2\pi t}{4}+\pi \right]$ = -2cos ( $\frac{\pi }{2}t)$ m