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 from the moving source is given by the Doppler effect formula: $f_1 = f_0 \left( \frac{v

Vedclass · Accepted Answer

$1032$

Vedclass · Answer

(B) The frequency of the sound heard directly by the observer from the moving source is given by the Doppler effect formula: $f_1 = f_0 \left( \frac{v}{v + v_s} \right)$,where $f_0 = 1000 ~Hz$,$v = 330 ~m/s$,and $f_1 = 970 ~Hz$.Substituting the values: $970 = 1000 \left( \frac{330}{330 + v_s} \right)$.Solving for $v_s$: $330 + v_s = \frac{330000}{970} \approx 340.2 ~m/s$,so $v_s \approx 10.2 ~m/s$.The sound reflected from the hill acts as if it is coming from a stationary source (the hill) towards the observer,but the source (van) is moving towards the hill. The hill receives the sound at a frequency $f' = f_0 \left( \frac{v}{v - v_s} \right)$.Since the hill is stationary,it reflects this frequency back to the observer: $f_2 = f' = 1000 \left( \frac{330}{330 - 10.2} \right)$.$f_2 = \frac{330000}{319.8} \approx 1031.89 ~Hz$.Rounding to the nearest integer,the frequency is approximately $1032 ~Hz$.

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