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 = (V*_{sound} / (V_{sound} +V_{source})) f_{0}

72 km/h = 20 m/s

(i) before:*V*_{source }is positive

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

(ii) after:*V*_{source }is negative

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

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

## Comments

## Leave a comment