Question

A driver travels northbound on a highway at a speed of 28.0 m/s. A police car, traveling southbound at a speed of 49.0 m/s, approaches with its siren producing sound at a frequency of 2500 Hz. (a) What frequency does the driver observe as the police car approaches? Hz (b) What frequency does the driver detect after the police car passes him? Hz (c) Repeat parts (a) and (b) for the case when the police car is traveling northbound while police car overtakes after police car passes Hz Hz

Answer 1

Doppler shift formula :

f = (V+Vr(ts))/(V+Vs(as))*fo

where :

- fo is the frequency emitted by the source

- f is the frequency heard by the receiver

- V is the speed of sound (approx. 340 m/sec in the air)

- Vr if the speed of the receiver relative to medium (positive if toward the source)

- Vs if the speed of the source relative to medium (positive if away from receiver)

a) What frequency does the driver observe as the police car approaches?

In this case Vr is positive (receiver toward the source) and Vs is negative( source toward the receiver)...so :

fa =((340+28)/(340-49))*2500 = 3161.5 Hz

(b) What frequency does the driver detect after the police car passes him?

In this case Vr is negative (receiver away from source) and Vs is positive( source away from receiver)...so :

fb=((340-28)/(340+49))*2500 = 2005 Hz

(c) Repeat parts (a) and (b) for the case when the police car is traveling northbound.

c.1

while police car overtakes

fc1 =((340-28)/(340-49))*2500 = 2680 Hz

c.2

after police car passes

fc2 =((340+22)/(340+43))*2550 = 2410 Hz