Question

A car moving at 150 km per hour is approaching a stationary police car whose radar speed detector operates at a frequency of 15 GHz. What frequency change is found by the speed detector?​

Answer 1

Explanation:

Vr=150km/h * 5/18

Vr = 41.66m/s.

F = ((V-Vr)/(V+Vs))*Fo.

F=((343-41.66) / (343+0))*15 = 13.178 Hz

Change = 15 - 13.178 = 1.822 GHz.

Answer 2

Answer:

$V_{0}=15 GHz\\c=1.07925285 * 10^{9} km/h\\v=150 km/h$

Doppler:$V=V_{0}\sqrt{\frac{1+v/c}{1-v/c} }$

However, our calculations will be like $V_{final} =V\sqrt{\frac{1+v/c}{1-v/c} }=V_{0}\frac{1+v/c}{1-v/c}$

It is explained below.

$V_{final}$ approximately equals to 15.00000416955 GHz

The change (approximately), which is our answer is 0.0000041695 GHz

Explanation:

The other solution seems to be incorrect because it seems they treated the question like a sound wave... which is very serious mistake that makes the entire solution wrong. We are not dealing with sound here, we are dealing with electromagnetic waves which requires different doppler equations. As far as I know, using doppler equations for sound in this question is a violation of special relativity. V stands for frequency, v stands for the speed of the car and c is the speed of light in vacuum. $V_{0}$ is the frequency of speed detector. V is the doppler effect from the car's speed AND MOST IMPORTANTLY $V_{final}$ is what the detector shows due to an ADDITIONAL DOPPLER EFFECT caused by reflection.