How does the beat frequency change when the mirror's speed is much smaller than the speed of light?

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SUMMARY

The discussion focuses on the calculation of beat frequency for an electromagnetic wave reflected from a mirror moving at a speed much smaller than the speed of light. The frequency of the reflected wave is given by the formula f = f_{source} * (c + v) / (c - v). When the mirror's speed v is significantly less than the speed of light c, the beat frequency can be approximated as f_{beat} = 2v / λ, where λ is the wavelength. This simplification allows for easier calculations in scenarios involving low-speed reflections.

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thenewbosco
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There are two parts to this but i solved the first part:
in part a) i was to show, for an electomagnetic wave reflected back to its source from a mirror approaching at speed v, that the reflected wave had frequency [tex]f=f_{source}\frac{c+v}{c-v}[/tex] where fsource is the source frequency and c is the speed of light.

now i am asked: when v is much less than c, the beat frequency is much smaller than the transmitted frequency. In this case use the approximation [tex]f + f_{source}\approx 2f_{source}[/tex] and show that the beat frequency can be written:

[tex]f_{beat}=\frac{2v}{\lambda}[/tex]

i don't know how to go about this. I was thinking to solve for fsource and put it into the equation from part a. but this doesn't work...any help
 
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