Question

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
(i) before, and

(ii) after the engine passes the observer.
The speed of sound in air is 340 m/s.

Answer 1

The frequency of the note changes due to the Doppler effect according to the formula (stationary receiver):
f = (V_sound / (V_sound +V_source)) f₀

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