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In the figure shown, an observer $O_{1}$ floats (static) on water surface with ears in air while another observer $O_{2}$ is moving upwards with constant velocity $V_{1} = V / 5$ in water. The source moves down with constant velocity $V_{2} = V / 5$ and emits sound of frequency $f$ . The velocity of sound in air is V and that in water is 4V. For this situation the wavelength of the sound received by $O_{2}$ is $\frac{16 V}{x f}$ . Find x

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answer is 5.

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Detailed Solution

For observer $O_{1}$
$λ_{4} = \frac{V - V_{S}}{f} = \frac{V - V / 5}{f} = \frac{4 V}{5 f}$
For $O_{2}$ , there is change of medium. Hence, at the surface of water, keeping frequency unchanged
$\frac{V}{λ_{a}} = \frac{4 V}{λ_{w}}$

$\Rightarrow λ_{w} = 4 λ_{a} = \frac{16 V}{5 f}$

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