A highway patrol is parked at the roadside on the 405 and sends out an audio-sig

Question

A highway patrol is parked at the roadside on the 405 and sends out an audio-signal at 4.5 kHz to measure your speed as you are passing by at a constant velocity. The cop measures you from behind as you move away from the police car. He measures a frequency of 3.5 kHz coming back after the wave is reflected off your car.

a. How fast were you driving?

b. What is the frequency of the signal that you hear in your car?

c. The policeman tickets you for violating the 70 mi/hour speed limit. You object and argue that the measurement was hampered by the cold air temperature near the freezing point and thus the fact the speed of sound was smaller (Vsound=330m/s). Cound this different sound velocity really explain the measured frequency if you were not exceeding the speed limit?

Explanation / Answer

a) when the police car is at rest and you drive away, frequency heard by you is

f car = (v - vcar)fo/ v

when the police receives the signal, your car is still moving away from the stationary police car;

so frequency heard by the police is Less than that sent from your car

f police = (v) fcar /(v+vcar); v= speed of sound = 340m/s

now solve these two equations to get an equation for v car.

v car = v (fo/fpolice -1) /(fo/fpolice+1) = 340 (4.5/3.5 - 1)/ (4.5/3.5 +1) = 42.5 m/s= 95 mi/hr

b.) fcar = (v-vcar)fo/v = (340-42.5)(4.5)/340 = 3.9 kHz

c) If v=330 m/s , the frequency measured by the police from the car would be slightly greater than 3.5kHz, which can justify your argument that you have not exceeded the speed limit.

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