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Answer on Acoustics Question for nirmala

Question #1871
A locomotive engine passes a stationary observer at a speed of 72 km/h emitting a note of frequency 500 Hz. Calculate the frequency of sound heard by the observer <br>(i) before, and
<br>(ii) after the engine passes the observer. <br>The speed of sound in air is 340 m/s.
Expert's answer
The frequency of the note changes due to the Doppler effect according to the formula (stationary receiver):
f = (Vsound / (Vsound +Vsource)) f0

72 km/h = 20 m/s
(i) before: Vsource is positive
f = (340/(340 - 20))*500 = 531.25 Hz
(ii) after: Vsource is negative
f = (340/(340+20))*500 = 472.22 Hz

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