this is a straight forward doppler shift
just plug the information into your formulas
A siren on the top of the police car moving with a speed of 40 m/s emits sound at frequency of 820Hz. Ahead of the car, there is a truck moving in the same direction as the car but with a speed of 30 m/s. Assume the speed of sound in air to be 340m/s.
a) What frequency does the driver of the truck hears as the police car approaches it?
b) What frequency does the driver of the truck hears after the police car has passed it and is receding from it?
2 answers
a. Fd = ((Vs+Vd)/(Vs-Vc)) * Fc.
Fd = ((340+30)/(340-40)) * 820 =
b. Fd = ((340-30)/(340+40)) * 820 =
So the frequency heard by the driver will be greater than 820 Hz as the police car
approaches but less than 820 Hz as the car passes the truck.
Fd = ((340+30)/(340-40)) * 820 =
b. Fd = ((340-30)/(340+40)) * 820 =
So the frequency heard by the driver will be greater than 820 Hz as the police car
approaches but less than 820 Hz as the car passes the truck.
1 answer