An observer is standing 500 ,m away from a ve | vedclass

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(B) The frequency of the sound heard directly by the observer moving away from the source is given by the Doppler effect formula: $f_1 = f_0 \left( \f

Vedclass · Accepted Answer

$1032$

Vedclass · Answer

(B) The frequency of the sound heard directly by the observer moving away from the source is given by the Doppler effect formula: $f_1 = f_0 \left( \frac{v}{v + v_s} ight)$.Given $f_1 = 970 \,Hz$, $f_0 = 1000 \,Hz$, and $v = 330 \,m/s$, we have: $970 = 1000 \left( \frac{330}{330 + v_s} ight)$.Solving for $v_s$: $330 + v_s = \frac{1000 	imes 330}{970} \approx 340.2 \,m/s$, so $v_s \approx 10.2 \,m/s$.The sound reflected from the hill acts as if it is coming from a source moving towards the observer. The frequency of the reflected sound $f_2$ is given by: $f_2 = f_0 \left( \frac{v}{v - v_s} ight)$.Substituting the values: $f_2 = 1000 \left( \frac{330}{330 - 10.2} ight) = 1000 \left( \frac{330}{319.8} ight) \approx 1031.89 \,Hz$.Rounding to the nearest integer, the frequency is approximately $1032 \,Hz$.

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