Friday

August 1, 2014

August 1, 2014

Posted by **Liz** on Saturday, August 21, 2010 at 1:57pm.

I know you have to set up two equations (one for before the bounce off the speeding car and one after), but I can't seem to work it to a reasonable number.

The equation is:

fd=(v±vd)/(v±vs)fs

where fd= frequency recieved by a detector

fs= frequency of source

v= velocity of the wave

vd= velocity of the detecor

vs= velocity of the source

I've been working on it for days, but the best thing I've gotten was that the car was traveling at 101 m/s, which doesn't sound possible to me lol. Could anyone look at this and explain to me what I'm doing wrong?

- physics -
**drwls**, Saturday, August 21, 2010 at 8:40pmThe equation you have been given is actually not correct for light and radar guns, but it is close enough. For velocities much less than that of light, there is a frequency shift nthat is more easily written as just

deltaf/f = 2*V/c

The frequency shift is deltaf. The factor of two is the result of the echo doubling the Doppler shift. The signal received by the car is shifted delta f/f and the reflected signal is twice that because the car is moving away from the receiver (radar gun).

The formula you have been told to use should give the same result, to three significant figures or more, if used correctly.

The exact relativisitic Doppler formula for light can be found at

http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

In your case

deltaf/f = 4750/2*10^10 =

= 2.38*10^-7 = 2 V/(3*10^8)

V = 35.6 m/s = 128 km/h

- physics -
**Liz**, Sunday, August 22, 2010 at 1:29amI actually used that formula and got that answer (35.625 m/s) but does that take into account the speed of the police car? (or is that a red herring)

- physics -
**Liz**, Sunday, August 22, 2010 at 1:29amI actually used that formula and got that answer (35.625 m/s) but does that take into account the speed of the police car? (or is that a red herring)

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