Question

Solve the following examples.

a. The speed of sound in air at 0^oC is 332 m/s. If it increases at the rate of 0.6 m/s per degree, what will be the temperature when the velocity has increased to 344 m/s?

b. Nita heard the sound of lightning after 4 seconds of seeing it. What was the distance of the lightning from her?(The velocity of sound in air is 340 m/s?)

c. Sunil is standing between two walls. The wall closest to him is at a distance of 360 m. If he shouts, he hears the first echo after 4 s and another after another 2 seconds.

1. What is the velocity of sound in air?

2. What is the distance between the two walls?

d. Hydrogen gas is filled in two identical bottles, A and B, at the same temperature. The mass of hydrogen in the two bottles is 12 gm and 48 gm respectively. In which bottle will sound travel faster? How may times as fast as the other?

e. Helium gas is filled in two identical bottles A and B. The mass of the gas in the two bottles is 10 gm and 40 gm respectively. If the speed of sound is the same in both bottles, what conclusions will you draw?

Answer 1

a. Let the temperature at which the velocity of air becomes 344 m/s be T^oC.
The increase in velocity of sound in air from 0^oC to T^oC = 344$-$332 = 12 m/s
Given,
The velocity of sound in air increases by 0.6 m/s for 1^oC rise in temperature.
Therefore, for 1 m/s increase in velocity of sound in air, the temperature should rise by $\frac{1}{0.6}{}^{\mathrm{o}}\mathrm{C}$.
For 12 m/s increase in velocity of sound in air, the temperature should rise by = $\frac{12}{0.6}{}^{\mathrm{o}}\mathrm{C}=20{}^{\mathrm{o}}\mathrm{C}$
Thus, the temperature at which the velocity of air becomes 344 m/s = T^oC = 20^oC$-$ 0^oC = 20^oC

b. Let the distance between Nita and the source of lightning be x.
Speed of sound in air = 340 m/s
Time taken by sound to reach Nita = 4 s
x = 340$\times 4$ = 1360 m

c.

1. Time to hear first echo = 4 s
Distance travelled by sound in these 4 s = 360$\times 2$ m
Speed of sound = $\frac{360\times 2}{4}=180\mathrm{m}/\mathrm{s}$
2. Time taken to hear second echo = 4 s + 2 s = 6 s
Distance travelled in these 6 s = 2x
Speed of sound in air = 180 m/s
Therefore, 2x = 180 $\times 6$
x = 540 m
Distance between the walls = 540 + 360 = 900 m

d. The velocity of sound in gas is related to density of gas as
$v\propto \frac{1}{\sqrt{\rho }}$ .....(i)
and the velocity of sound in gas is related to temperature of gas as
$v\propto \sqrt{T}$ .....(ii)
Combining (i) and (ii), we get
$v\propto \frac{\sqrt{T}}{\sqrt{\rho }}$
Now, the two bottles given are identical i.e. the volumes of gases are same. Let the volume of the bottle be V. Let M₁ and M₂ be the masses of the gases in bottles A and B, respectively and v₁ and v₂ be the velocity of the sound in the two bottles, respectively. Since, the temperature of gas in both the bottles are same, let that common temperature be T. Therefore,
$$\begin{gathered} \frac{v_1}{v_2}=\frac{\sqrt{{\displaystyle \frac{M_2}{V}}}\sqrt{T}}{\sqrt{{\displaystyle \frac{M_1}{V}}}\sqrt{T}}=\frac{{\displaystyle \sqrt{48}}}{{\displaystyle \sqrt{12}}}=\sqrt{2} \\ \Rightarrow v_1=2v_2 \end{gathered}$$
Thus, the sound travels 2 times faster in bottle B as compared to bottle A.

e. Mass of the gas in bottle A = 10 g
Mass of gas in bottle B = 40 g
The velocity of sound in gas is related to density of gas as
$v\propto \frac{1}{\sqrt{\rho }}$ .....(i)
and the velocity of sound in gas is related to temperature of gas as
$v\propto \sqrt{T}$ .....(ii)
Combining (i) and (ii), we get
$v\propto \frac{\sqrt{T}}{\sqrt{\rho }}$
The bottle are identical, this means the volumes of the gases are equal. Let the volume of the bottle be V. Let M₁ and M₂ be the masses of gases in bottles A and B, respectively and v₁ and v₂ be the velocity of the sound in the two bottles, respectively. Also, let T₁ and T₂ be their respective temperatures. Therefore,
$\frac{v_1}{v_2}=\frac{\sqrt{{\displaystyle \frac{M_2}{V}}}\sqrt{T_1}}{\sqrt{{\displaystyle \frac{M_1}{V}}}\sqrt{T_2}}$
Now given that, v₁ = v₂
$$\begin{gathered} \Rightarrow \sqrt{M_1}\sqrt{T_2}=\sqrt{M_2}\sqrt{T_1} \\ \mathrm{or},T_2=\frac{M_2{T}_1}{M_2} \\ \mathrm{Given},M_1=10\mathrm{g},M_2=40\mathrm{g} \\ \Rightarrow T_2=\frac{40T_1}{10}=4T_1 \end{gathered}$$
Thus, it can be concluded that the temperature of bottle B is 4 times the temperature of A.