Class 11th

The equation of displacement of a particle executing SHM at an instant t is given bas:

x = Asin $\omega t$ , where A = Amplitude of oscillation and $\omega$ = angular frequency = $\sqrt{\frac{k}{M}}$

The velocity of the particle, v = $\frac{dx}{dt}$ = A $\omega \cos ?\omega t$

The kinetic energy of the particle $E_k$ = $\frac{1}{2}$ M $v^2$ = $\frac{1}{2}$ M ${\left(\mathrm{A}\omega \cos ?\omega t\right)}^2$ = $\frac{1}{2}$ M $A^2{\omega }^2{cos}^2\omega t$

The potential energy of the particle $E_p$ = $\frac{1}{2}$ k $x^2$ = $\frac{1}{2}$ k $A^2{sin}^2\omega t$

For time period T, the average kinetic energy over a single cycle is given as :

${E_k}_{avg}=\frac{1}{T}{\int }_0^T{E}_{k}dt$ = $\frac{1}{T}{\int }_0^T\frac{1}{2}\mathrm{M}{A}^2{\omega }^2{cos}^2\omega tdt$ = $\frac{\mathrm{M}{A}^2{\omega }^2}{2T}{\int }_0^T{cos}^2\omega tdt$

= $\frac{\mathrm{M}{A}^2{\omega }^2}{2T}{\int }_0^T\frac{(1-cos2\omega t)}{2}dt$ = $\frac{\mathrm{M}{A}^2{\omega }^2}{2T}({t+\frac{sin2\omega t}{2\omega })}_0^T$ = $\frac{\mathrm{M}{A}^2{\omega }^2}{4T}\left(T\right)$ = $\frac{1}{4}\mathrm{}\mathrm{M}{A}^2{\omega }^2$ …….(i)

Average potential energy over one cycle is given as :

${E_p}_{avg}=\frac{1}{T}{\int }_0^T{E}_{p}dt$ = $\frac{1}{T}{\int }_0^T\frac{1}{2}\mathrm{M}{A}^2{\omega }^2\sin ?\omega tdt$

= $\frac{\mathrm{M}{A}^2{\omega }^2}{2T}{\int }_0^T\frac{(1-cos2\omega t)}{2}dt$ = $\frac{\mathrm{M}{A}^2{\omega }^2}{2T}({t-\frac{sin2\omega t}{2\omega })}_0^T$ = $\frac{\mathrm{M}{A}^2{\omega }^2}{4T}\left(T\right)$ = $\frac{1}{4}\mathrm{}\mathrm{M}{A}^2{\omega }^2$ …….(ii)

From equations (i) and (ii), it is proved that average kinetic energy for a given period of time is equal to the average potential energy for the same period of time.

Mass of the automobile, m = 3000 kg

Displacement of the suspension system, x = 15 cm = 0.15 m

There are 4 springs in parallel to the support of the mass of the automobile.

(a) The equation for restoring force for the system, F = -4kx = mg, where k is the spring constant of the suspension system

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

k = $\frac{mg}{4x}$ = $\frac{3000*9.8}{4*0.15}=4.9*{10}^4$ N/m

(b) Each wheel supports a mass, M = 3000/4 = 750 kg

For damping factor b, the equation for displacement is written as

x = $x_0{e}^{-bt/2M}$

The amplitude of oscillation decrease by 50%. Therefore x = $\frac{x_0}{2}$ = $x_0{e}^{-bT/2M}$

${\log }_e?2$ = $\frac{bT}{2M}$

b = $\frac{2M{\log }_e?2}{T}$

T = 2 $\pi \sqrt{\frac{m}{4k}}$ = 2 $\pi \sqrt{\frac{3000}{4*4.9*{10}^4}}$ = 0.7773 s

b = $\frac{2*750*0.693}{0.7773}$ = 1337.32 kg/s

Therefore the damping constant of the spring is 1337.32 kg/s

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.