An engine moving away from a vertical cliff b | vedclass

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Question

(A) Let $v$ be the speed of sound in air. The speed of the engine is $v_s = 0.005v$.First,the sound travels from the moving engine to the stationary c

Vedclass · Accepted Answer

$0.990$

Vedclass · Answer

(A) Let $v$ be the speed of sound in air. The speed of the engine is $v_s = 0.005v$.First,the sound travels from the moving engine to the stationary cliff. The frequency $f_1$ received by the cliff is given by the Doppler effect formula for a moving source and stationary observer:$f_1 = f \left( \frac{v}{v + v_s} ight) = f \left( \frac{v}{v + 0.005v} ight) = f \left( \frac{1}{1.005} ight)$Next,the cliff acts as a stationary source reflecting the sound of frequency $f_1$ back to the moving engine (observer). The frequency $f_2$ received by the engine is given by the Doppler effect formula for a stationary source and moving observer:$f_2 = f_1 \left( \frac{v - v_o}{v} ight)$Since the engine is moving away from the cliff,the observer speed $v_o = v_s = 0.005v$ is taken as negative relative to the direction of the reflected sound:$f_2 = f_1 \left( \frac{v - 0.005v}{v} ight) = f_1 \left( \frac{0.995v}{v} ight) = 0.995 f_1$Substituting $f_1$ into the equation for $f_2$:$f_2 = 0.995 	imes f \left( \frac{1}{1.005} ight) \approx 0.995 	imes 0.995 f \approx 0.990 f$Thus,the correct option is $A$.

An engine moving away from a vertical cliff blows a horn at a frequency $f$. Its speed is $0.5 \%$ of the speed of sound in air. The frequency of the reflected sound received at the engine is ............ $f$.

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