Police use radar guns and the Doppler effect to catch speeders. The figure below

Question

Police use radar guns and the Doppler effect to catch speeders. The figure below illustrates a moving car approaching a stationary police car. A radar gun emits an electromagnetic wave that reflects from the oncoming car. The reflected wave returns to the police car with a frequency (measured by on-board equipment) that is different from the emitted frequency. One such radar gun emits a wave whose frequency is 8.0 times 10^9 Hz. When the speed of the car is 25.0 m/s and the approach is essentially head-on, what is the difference between the frequency of the wave returning to the police car and that emitted by the radar gun?

Explanation / Answer

Use the general formula of Doppler effect:

Let ve = emitter velocity, vr = reflector velocity, +v = receding
ve = +0 m/s, Since the police car is stationary.

And, vr = - 25 m/s
f = f0*((c-vr)/(c+ve))*((c-ve)/(c+vr)) = (8x10^9)*((3x10^8+25)/(3x10^8+0))*((3x10^8-0)/(3x10^8-25))

= 8000001333 Hz

df = f-f0 = 1333.00 Hz

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