A police car moving at 41 m/s is initially behind a truck moving in the same direction at 17 m/s. The natural frequency of the car's siren is 800 Hz.

The change in frequency observed by the truck driver as the car overtakes him is __(Hz)

Take the speed of sound to be 340m/s.

2 answers

Fd = (Vs+Vd)/(Vs-Vp) * Fp.
Fd = (343+17)/(343-41) * 800. = 953.6 Hz. = Freq. heard by driver of the truck.

Change = 953.6 - 800 = 153.6 Hz.
f' = Frequency Perceived (by Truck Driver)
f0 = Natural Frequency
v = Speed of Sound
vs = Speed of Source (Police Car)
vd = Speed of Detector (Truck Driver)

f' before = f0 * [ (v-vd) / (v-vs) ] = 800Hz [ (340-17 m/s) / (340-41m/s) ] = 864.21 Hz
f' after = f0 * [ (v+vd) / (v+vs) ] = 800Hz [ (340+17m/s) / (340+41m/s) ] = 749.61 Hz
Change = | f' before - f' after | = 114.61 Hz

Hope this helps.