# 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.

<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):

72 km/h = 20 m/s

(i) before:

f = (340/(340 - 20))*500 = 531.25 Hz

(ii) after:

f = (340/(340+20))*500 = 472.22 Hz

*f = (V*_{sound}/ (V_{sound}+V_{source})) f_{0}72 km/h = 20 m/s

(i) before:

*V*is positive_{source }f = (340/(340 - 20))*500 = 531.25 Hz

(ii) after:

*V*is negative_{source }f = (340/(340+20))*500 = 472.22 Hz

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