You have attempted this problem 6 times Police use radar guns and the Doppler ef

Question

You have attempted this problem 6 times 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 x 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

f = actual frequency emitted = 8 x 109 Hz

f' = frequency observed by the speeder

f'' = frequency emitted by the speeder

V = speed of speeder = 25 m/s

Vs = speed of sound = 343 m/s

f' = (V + Vs) f / Vs = (25 + 343) (8 x 109 ) /343 = 8.6 x 109 Hz

f'' = Vsf' / (Vs - V) = 343 (8.6 x 109 ) / (343 - 25) = 9.3 x 109 Hz

difference = f'' - f = 9.3 x 109 - 8 x 109 = 1.3 x 109

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