A source of sound A emitting waves of frequency 1800 Hz is falling towards ground with a terminal speed v. The observer B on the ground directly beneath the source receives waves of frequency 2150 Hz. The source A receives waves, reflected from ground of frequency nearly: (speed of sound = 343 m/s) (in Hz)

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answer is 2500.

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Detailed Solution

Given    $f_{A} = 1800 H z$
$v_{t} = v$
$f_{B} = 2150 H z$
Reflected wave frequency received by A,    ${f_{A}}^{'} = ?$
Applying doppler’s effect of sound,
$f^{'} = \frac{v_{s} f}{v_{s} - v_{t}}$
here,    $v_{t} = v_{s} (1 - \frac{f_{A}}{f_{B}})$
$= 343 (1 - \frac{1800}{2150})$
$v_{t} = 55.8372 m / s$
Now, for the reflected wave,
$\begin{array}{l} ∴ {f_{A}}^{'} = (\frac{v_{s} + v_{t}}{v_{s} - v_{t}}) f_{A} \\ = (\frac{343 + 55.83}{343 - 55.83}) \times 1800 \\ = 2499.44 \approx 2500 H z \end{array}$

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