Section 24.5 The Doppler Effect andElectromagnetic Waves A speeder is pulling di

Question

Section 24.5 The Doppler Effect andElectromagnetic Waves
A speeder is pulling directly away and increasing his distance froma police car that is moving at 25 m/s withrespect to the ground. The radar gun in the police car emits anelectromagnetic wave with a frequency of 6.00 x 10^0 Hz. The wave reflects from thespeeder's car and returns to the police car, where its frequency ismeasured to be 322 Hz less than theemitted frequency. Find the speeder's speed with respect to theground. 1 m/s Section 24.5 The Doppler Effect andElectromagnetic Waves

Explanation / Answer

Given: -- fo = Frequency observed = (9 x 10^9) - 322 = 8999999678
fs = Frequency of source
V(rel) = relative velocity
C = speed of light (3 x 10^8)
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The frequency drops once as seen from the point of viewof the speeding car: --
f(car) = fs(1 - V(rel)/C)
--
And then, the radar reflects off the back of the car atfrequency f1. Essentially, the speeding car becomes another frequency source emitting a lower frequency. Because of this, asthe radar reflects back to the police car, the doppler effect isapplied a second time, and the frequency observed is in thepolice car is:

fo = f(car)(1 - V(rel)/C)
fo = (fs(1-V(rel)/C))(1-Vrel/C) ----> Plugging in f(car)above
fo = fs(1-V(rel)/C)² -- Thus, 8999999678 = (9 x 10^9) (1-V(rel)/(3 x 10^8))^2 -- Sovle for V(rel): V(rel) = 5.37 m/s -- Add to moving police car velocity: 25 + 5.37 = 30.4 m/s -- Hope this helps.

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