A man standing next to a stationary train hears sound of frequency 400 Hz emitted from the train’s horn.

Question 25

A man standing next to a stationary train hears sound of frequency 400 Hz emitted from the train’s horn. The train then moves directly away from the man and sounds its horn when it has a speed of 50 m s^-1. The speed of sound is 340 m s^-1.

What is the difference in frequency of the sound heard by the man on the two occasions?

A 51 Hz                       B 69 Hz                       C 349 Hz                     D 469 Hz

Reference: Past Exam Paper – November 2016 Paper 12 Q25

Solution:

Answer: A.

The formula for the Doppler effect is as follows (given in the list of formula):

f_o = f_s v / (v ± v_s)

where  fo is the observed frequency

            fs is the frequency of the source

            v is the speed of sound

            v_s is the speed of motion of the source (train)

When the source moves towards the observer, the observed wavelength decreases and so, the observed frequency increases. The negative sign is used.

When the source moves away from the observer, the observed wavelength increases and so, observed frequency decreases. The positive sign is used in the formula.

When stationary, the sound from the train has a frequency of 400 Hz. This is the frequency of the source, f_s (= 400 Hz).

Speed of motion of the train, v_s = 50 m s^-1

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

For a receding source (going away),

f_o = f_s v / (v + v_s)

f_o = 400 × 340 / (340 + 50)

f_o = 348.7 = 349 Hz

Difference in frequency = 400 – 349 = 51 Hz