Question

Two identical sound sources are placed at A and B -one in air and one submerged in water Fig. Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times than that of B, what is the ratio between the speeds of sound in air and water?

Answer 1

Given:

Distance traveled by the sound from both sources is identical (let the round-trip distance be $2d$).

Time taken for sound to return to source A in air ($t_A$) is 4.5 times the time taken for sound to return to source B in water ($t_B$).

$$t_A = 4.5 t_B$$

Formulas:

Speed ($v$) = Distance/Time

Speed of sound in air ($v_A$) = $\frac{2d}{t_A}$

Speed of sound in water ($v_B$) = $\frac{2d}{t_B}$

To find the ratio between the speeds of sound in air and water:

$$\frac{v_A}{v_B} = \frac{\frac{2d}{t_A}}{\frac{2d}{t_B}}$$

$$\frac{v_A}{v_B} = \frac{t_B}{t_A}$$

Substituting the given time relationship $t_A = 4.5 t_B$:

$$\frac{v_A}{v_B} = \frac{t_B}{4.5 t_B}$$

$$\frac{v_A}{v_B} = \frac{1}{4.5}$$

$$\frac{v_A}{v_B} = \frac{2}{9}$$