physics ncert solutions 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 $\pi \sqrt{\frac{\mathrm{h}\rho }{{\rho }_{l}g}}$