22 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. Theambulance passes, and then moves away from the observer with no change in velocity.initial positionof ambulancefinal positionof ambulanceobserverWhich overall change in observed frequency takes place between the times at which theambulance 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

This question is about the Doppler effect, which is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.

First, we need to calculate the frequency observed as the ambulance approaches. We can use the formula for the Doppler effect:

f' = f * (v + vo) / v

where: f' is the observed frequency, f is the source frequency, v is the speed of sound, and vo is the speed of the observer relative to the source.

In this case, the ambulance is moving towards the observer, so we consider the speed of the ambulance as negative. Therefore, vo = -25 m/s. Substituting the given values:

f' = 600 Hz * (330 m/s + (-25 m/s)) / 330 m/s = 550 Hz

Next, we calculate the frequency observed as the ambulance moves away. Now, the ambulance is moving away from the observer, so vo = 25 m/s. Substituting the given values:

f'' = 600 Hz * (330 m/s - 25 m/s) / 330 m/s = 650 Hz

The overall change in observed frequency is the difference between the frequency observed when the ambulance is moving away and when it is approaching:

Δf = f'' - f' = 650 Hz - 550 Hz = 100 Hz

Therefore, none of the given options (A 49 Hz, B 84 Hz, C 91 Hz, D 98 Hz) is correct. The overall change in observed frequency is 100 Hz.