# Electromagnetic wave reflection

1. Jan 6, 2009

1. The problem statement, all variables and given/known data

A speeder is pulling directly away and increasing his distance from a police car that is moving at 29 m/s with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of 8.0x109 Hz. The wave reflects from the speeder\'s car and returns to the police car where its frequency is measured to be 318Hz less than the emitted frequency. Find the speeder's speed with respect to the ground.

3. The attempt at a solution

Should I use here Doppler's Effect?

u - speed of light
Vs - speed of speedster
Vo - speed of observator(police car)

f'=fo*$$\frac{u+Vs}{u-Vo}$$

2. Jan 6, 2009

You will need to use the doppler equation twice, once for the emitted sound wave and again for the return sound wave. If you want me to solve it and show you how you will just need to clarify when you say "8.0x109 Hz" do you really mean 8.0x109 Hz or do you mean 8.0x10^9 (8 times 10 to the exponent of 9).

3. Jan 6, 2009

8*109 meaning 872... Thank you for help. Generally i have no idea how to use Doppler's equation twice in this case...

4. Jan 6, 2009

### Redbelly98

Staff Emeritus
Please note that doing that is against Physics Forums policy:

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For more details, click on "Rules" at the top of this page, then scroll down to the "Homework Help" section.

Hopefully, your good hint of applying the Doppler shift twice will get the OP on the right track.

Regards,

Mark

5. Jan 6, 2009

Anyway, I had some attempts, probably not correct. I need it for tommorow. Its 1 AM here, so in like about 8 hours I will be kicked out by a professor from room, because I dont have this done :yuck:

6. Jan 6, 2009

Here is the formula I use, it is the same thing as yours just simplified.

f'=(1+Vobs/Vsnd)f

First Doppler Shift

f'=(1+29m/s / 343m/s)*872Hz

Second Doppler Shift

f''=f'/(1+Vsource/Vsnd)

872Hz-318Hz=f'(1-n/343m/s)

554Hz=f'(1-n/343m/s)

554Hz/f'=1-n/343m/s)

(554Hz/f')-1=n/343m/s

Cross Multiply

((554Hz/f')-1)*343m/s=n

There you go hopefully you can understand that. 343m/s is the speed of air at a normal temperature. n is what you solve for and f' you find in the first doppler shift. I think I worked everything out right, good luck.

7. Jan 6, 2009

### Redbelly98

Staff Emeritus
No, I don't think you mean that. (872 Hz is not a radio frequency. 8.0 x 109 Hz IS a radio frequency.)

You can start out using it once, and get an expression relating the following:
• the frequency of the wave "received" by the speeding car, and
• the speed of the car relative to the police

8. Jan 6, 2009