00:01
Okay, there have been a couple problems in our book like this, so we're not going to go into too many details.
00:06
Look towards other doppler shift problems if you're a little bit confused.
00:10
But we have this patrol car and it's sending out a signal and it's hitting some car that's either moving towards him or away from him at some particular speed.
00:22
And then that car is going to reflect the signal back to the highway patrol car.
00:29
Okay, so the trick here is that we're sending out some original frequency because of the doppler shift we're seeing some different frequency get reflect hit the other car that frequency is getting reflected that shifted frequency is getting reflected and then the patrol car is seen back some frequency that's been double shifted okay, so while the normal doppler shift equation looks like this f times one plus u over c plus or minus depending the double doppler shift equation is going to be the original frequency times one plus or minus u over c squared okay and this problem has helpfully told us that one plus or minus u over c squared when x is pretty small when u over c is small, which is the case that we're interested in, this is approximately equal to f times 1 plus or minus 2 times u over c.
01:44
So this is an approximation, does this actually do this all of the time to make numbers easier to work with? okay.
01:52
And so we need to figure out what the speed of the car.
01:55
So we know the change in frequency and we need to solve.
02:00
For you.
02:02
All right, so the change in frequency, which is delta f prime prime is equal to f prime prime minus f.
02:16
That's going to be equal to this whole quantity, f prime prime, minus f.
02:22
So now we have plus or minus two times u over c times f, is just equal to this double to this double to this delta f prime.
02:38
Okay, and we're trying to solve for you...