"A radar speed trap operates on a frequency $v_o = 109 Hz$. What is the beat frequency between the transmitted signal and one received after reflection from a car moving at v = 30 m/s toward the radar. Do calculations with accuracy to linear terms in v/c."
I'm not entirely sure how to approach this problem. I have a formula for the Doppler effect with time dilation, $f_{obs} = f_{source} \frac{\sqrt{1-\frac{v^2}{c^2}}}{1+\frac{v}{c}cos\theta}$, where v would be the speed of the source (or the signs could be switched if the observer is moving, right?). I also know I'm going to have to make two calculations – one for when the signal hits the car, and another for when it is reflected towards the source.
To solve this problem, I tried using the formula above with v = 3 m/s to find the frequency that the observer would receive, but then the frequency would not change on the way back since the source is not moving (and is thus not subject to the doppler effect), right? Does that mean that the beat frequency is simply the difference between the initial signal, $v_o = 109 Hz$, and the frequency the observer receives?
Best Answer
You asked:
No, that's not right. The Doppler effect of sound depends on the relative velocitiesof both source and listener, compared to the medium. The Doppler effect for light depends on the relative velocity of the source and listener. Then, upon reflection of the initial signal, the reflector becomes a source and the original source becomes a listener. So there are two frequency shifts for radar gun to car and back to gun.
Aside: Police radar doesn't operate at 109 Hz. Are you sure that it isn't 109 GHz? At the low frequency the amount of shift won't be large enough to reliably and quickly detect the speed of the car.