Suppose that a police car on the highway is moving to the right at 26 m/s, while

Question

Suppose that a police car on the highway is moving to the right at 26 m/s, while a speeder is coming up from almost directly behind at a speed of 34 m/s, both speeds being with respect to the ground. The police officer aims a radar gun at the speeder. Assume that the electromagnetic wave emitted by the radar gun has a frequency of 6.00 109 Hz. (a) Find the magnitude of the difference between the frequency of the emitted wave and the wave that returns to the police car after reflecting from the speeder's car.

Explanation / Answer

Let ve = emitter velocity, vr = reflector velocity, +v = receding ve = +26 m/s, vr = -36 m/s f = f0*((c-vr)/(c+ve))*((c-ve)/(c+vr)) = 6800000453.64718 Hz df = f-f0 = 453.64718 Hz (answer) To start let's put the Hz into m/s. One Hertz has a period of 1 interval per second. So 6.8 x 109 Hz is 741.2 Hz or 741.2 intervals per second. Now, the difference between the cars is easy, its 10 m/s. This is static, I assume. One way to imagine the scenario is to give another variable, which is the distance between the cars; we can make this up and say it's 100m. Now after one second the gap between the cars will be 90m, 2sec will be 80m, and so on. We now know that every second there is a 10m change in relative distance to the police car, who is the observer. Alright, so f=(v + vr)/(v + vo)fo, where f is the observed frequency, v is the emitted wave frequency, vr is the velocity of the reciever (cop), vo is the velocity of the observed car, and fo is the original wave (hz). Now, (vr/vo)fo actually works, since v is the same in both cases. So 26/36 = .7222 (* 6.8 x 109 hz) = 535.3111hz. This makes sense since every second the wave from the gun is getting compressed at a rate of 10m per second.

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