Waves Class 11 Physics Ch 14 NCERT Solutions | Clear Your Concepts

Ans. 15.1: Mass of the string, M = 2.5 kg

Tension in the string, T = 200 N

Length of the string, l = 20 m

Mass per unit length,  $\mu$ =  $\frac{M}{l}$ =  $\frac{2.5}{20}$ = 0.125 kg/m

The velocity (v) of the transverse wave in the string is given by the relation:

$v=\sqrt{\frac{T}{\mu }}$ =  $\sqrt{\frac{200}{0.125}}$ = 40 m/s

Therefore, time taken by the disturbance to reach the other end, t =  $\frac{l}{v}$ =  $\frac{20}{40}$ = 0.5 s

Q.15.2 A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s^–1? (g = 9.8 m s^–2)

Ans.15.2: Height of the tower, h = 300 m

Initial velocity of the stone, u = 0

Acceleration, a = g = 9.8 m/  $s^2$

Speed of sound in air, V = 340 m/s

The time taken by the stone (t), to strike the water can be calculated from the relation

s =us +  $\frac{1}{2}$ a  $t^2$ as

300 = 0 +  $\frac{1}{2}\times 9.8\times t^2$ or t = 7.82 s

Time taken by the sound to reach the top of the tower,  $t_1$ =  $\frac{h}{V}$ =  $\frac{300}{340}$ = 0.88 s

Therefore, the time when the splash can be heard = 7.82 + 0.88 = 8.7 s

Q.15.3 A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s^–1.

Ans. 15.3: Length of the steel wire, l = 12 m

Mass of the steel wire, m = 2.1 kg

Velocity of the transverse wave, v = 343 m/s

Mass per unit length,  $\mu$ =  $\frac{m}{l}$ =  $\frac{2.1}{12}$ = 0.175 kg/m

The velocity (v) of the transverse wave in the string is given by the relation:

$v=\sqrt{\frac{T}{\mu }}$ , where T is the tension

T =  $v^2\times \mu$ =  ${343}^2\times 0.175$ = 20588.575 N = 2.06  $\times {10}^4$ N

Q.15.4 Use the formula  $\mathit{v}=\sqrt{\frac{\mathit{\gamma }\mathit{P}}{\mathit{\rho }}}\mathit{}\mathit{}$ to explain why the speed of sound in air

(a) Is independent of pressure

(b) Increases with temperature

(c) Increases with humidity

Ans.15.4: In the equation  $v=\sqrt{\frac{\gamma P}{\rho }}$ ……(i)

$\rho$ Density =  $\frac{Mass}{Volume}$ =  $\frac{M}{V}$ where M = molecular weight of the gas, V = Volume of the gas, so we can write

$v=\sqrt{\frac{\gamma PV}{M}}$ ……..(ii)

For ideal gas equation, PV = nRT, n = 1 so PV = RT

For constant T, PV = constant

In equation (ii), since PV = constant,  $\gamma$ and M constant, v is also constant. Hence, at a constant temperature, the speed of sound in a gaseous medium is independent of the change in the pressure of the gas.

From equation (i)  $v=\sqrt{\frac{\gamma P}{\rho }}$

For 1 mole of an ideal gas, the gas equation can be written as PV = RT or P =  $\frac{RT}{V}$

Substituting in equation (i), we get  $v=\sqrt{\frac{\gamma RT}{\rho V}}$ =  $\sqrt{\frac{\gamma RT}{M}}$

Since  $\gamma$ , R and M are constant, we get v  $\propto \surd T$

Hence, the speed of sound in a gas is directly proportional to the square root of the temperature of the gaseous medium. i.e. the speed of sound increase with an increase in the temperature of the gaseous medium and vice versa.

Let  $v_m$ and  $v_d$ be the speed of sound in moist air and dry air respectively and  ${\rho }_m$ and  ${\rho }_d$ be the corresponding densities

From equation (i)  $v=\sqrt{\frac{\gamma P}{\rho }}$ , we get  $v_m\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_m}}$ and  $v_d\mathrm{}=\sqrt{\frac{\gamma P}{{\rho }_d}}$

$\frac{v_m}{v_d}$ =  $\sqrt{\frac{{\rho }_d}{{\rho }_m}}$

However, the presence of water vapour reduces the density of air, i.e.  ${\rho }_d<$  ${\rho }_m$ , so  $v_m<$  $v_d$