A siren emitting a sound of frequency $800 ext{ Hz}$ moves away from an observer towards a cliff at a speed o

Question

A siren emitting a sound of frequency $800	ext{ Hz}$ moves away from an observer towards a cliff at a speed of $15	ext{ ms}^{-1}$. The frequency of sound that the observer hears in the echo reflected from the cliff will be: (Take, velocity of sound in air = $330	ext{ ms}^{-1}$)

Accepted Answer

The siren is moving towards the cliff, so the cliff acts as a stationary observer receiving sound from a moving source. The apparent frequency $f'$ reaching the cliff is given by the Doppler effect formula:
$f' = f_0 \left(\frac{v}{v - v_s}ight)$
where:
$f_0 = 800	ext{ Hz}$ (actual frequency of the siren)
$v = 330	ext{ ms}^{-1}$ (speed of sound)
$v_s = 15	ext{ ms}^{-1}$ (speed of the source)

$f' = 800 	imes \left(\frac{330}{330 - 15}ight) = 800 	imes \frac{330}{315} = 800 	imes \frac{22}{21} \approx 838.09	ext{ Hz}$

The cliff then acts as a stationary source of sound, reflecting it back to the stationary observer. The frequency of the echo heard by the observer will be the same as the frequency received by the cliff.
Therefore, the frequency of the echo heard by the observer is approximately $838	ext{ Hz}$.

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