In one of the first experiments to demonstrate the Doppler effect, a train was filled with trumpeters all playing a note of frequency 440 Hz.

Question 23

In one of the first experiments to demonstrate the Doppler effect, a train was filled with trumpeters all playing a note of frequency 440 Hz. The difference in observed frequency of the note as the train directly approached a stationary observer was 22 Hz. The speed of sound was 340 m s^-1.

At which speed was the train moving?

A 15.4 m s^-1                     B 16.2 m s^-1                     C 17.0 m s^-1                     D 17.9 m s^-1

Reference: Past Exam Paper – June 2019 Paper 12 Q26

Solution:

Answer: B.

Frequency of sound, fs = 440 Hz

Speed of sound, v = 340 m s^-1

Difference in observed frequency = 22 Hz

Speed of train = v_s

For Doppler effect: f_O = fs v / (v ± v_s)

Since the train is approaching, the wavefronts are compressed, causing the observed wavelength to be smaller. Thus, the observed frequency is greater. In the equation, we use the minus sign.

Observed frequency f_O = 440 + 22 = 462 Hz

 f_O = fs v / (v – v_s)

462 = 440×340 / (340 – v_s)

340 – v_s = 440×340 / 462

340 – v_s = 323.8

Speed v_s = 340 – 323.8 = 16.2 m s^-1