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"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?

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  • $\begingroup$ I don't think you need a relativistic calculation when the car is only travelling at 30 m/s. Just do a regular Doppler shift calculation. $\endgroup$ – John Rennie Jan 30 '15 at 18:02
  • $\begingroup$ That formula is for transverse doppler effect where the objects are moving perpendicular to the light at relativistic speeds. $\endgroup$ – Rick Feb 2 '15 at 20:37
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You asked:

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?

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.

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Your formula is for using Doppler shift with relativistic speeds. For this problem all of the speeds involved are much smaller then the speed of light, making the problem not relative.

Here's a derivation of a formula:

The radar gun will produce waves with frequency $f_0$ traveling at the speed of light $c$ toward the car.

This results in a wavelength $\lambda_0$ of $\frac{c}{f_0}$

The car will be hit with these $\lambda_0$ waves and the time between peaks will be distance/speed resulting in an observed frequency $f_1$ of $\frac{c+v}{\lambda_0}$

The car will reflect the wave at the same frequency but this time the car will be moving in the same direction as the light so this results in a wavelength $\lambda_1$ of $\frac{c-v}{f_1}$

This wavelength is then observed by the radar gun as a wave with frequency $f_2=\frac{c}{\lambda_1}$

Finally the radar gun measures the beat frequency of $|f_0-f_2|$

Combining all the equations:

$$\lambda_0=\frac{c}{f_0}$$ $$f_1=\frac{f_0(c+v)}{c}$$ $$\lambda_1=\frac{c(c-v)}{f_0(c+v)}$$ $$f_2=f_0\frac{c+v}{c-v}$$

So the final beat frequency is:

$$f_0\left|\frac{c+v}{c-v}-1\right|$$

Rearranging gives:

$$2 f_0\left|\frac{\frac{v}{c}+(\frac{v}{c})^2}{1-(\frac{v}{c})^2}\right|$$

Now the question statement asks for an answer the only considers linear terms of $\frac{v}{c}$ so the square terms can be approximated as zero. This leaves:

$$2 f_0\left|\frac{v}{c}\right|$$

Note that if the radar was pointed at a car moving away, this could be represented as a negative v which in this linearized form doesn't make a difference, so radar can be used in either direction.

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