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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)

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2150 Hz

2400 Hz

1800 Hz

2500 Hz

answer is B.

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Detailed Solution

Given $f_{A} = 1800 Hz$

v_t = v

f_B = 2150 Hz

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,

$∴ f_{A}' = (\frac{v_{s} + v_{t}}{v_{s} - v_{t}}) f_{A} = (\frac{343 + 55.83}{343 - 55.83}) \times 1800 = 2499.44 \approx 2500 Hz$

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