The warning signal on an ambulance has a frequency of 600 Hz. The speed of sound is 330 m s-1. The ambulance is travelling with a constant velocity of 25 m s-1 towards an observer.

Question 4

The warning signal on an ambulance has a frequency of 600 Hz. The speed of sound is 330 m s^-1.

The ambulance is travelling with a constant velocity of 25 m s^-1 towards an observer.

Which overall change in observed frequency takes place between the times at which the

ambulance is a long way behind the observer and when it is a long way in front of the observer?

A 49 Hz                       B 84 Hz                       C 91 Hz                       D 98 Hz

Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q26

Solution:

Answer: C.

The equation 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 (ambulance)

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

When approaching the observer (initial position of ambulance),

Observed frequency f_o1 = f_s v / (v – v_s) = 600 × 330 / (330 – 25) = 649 Hz

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

When receding the observer (final position of ambulance),

Observed frequency f_o2 = f_s v / (v + v_s) = 600 × 330 / (330 + 25) = 558 Hz

Overall change in observed frequency = 649 – 558 = 91 Hz