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

[thenewbosco]

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 $$f=f_{source}\frac{c+v}{c-v}$$ 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 $$f + f_{source}\approx 2f_{source}$$ and show that the beat frequency can be written:

$$f_{beat}=\frac{2v}{\lambda}$$

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

 

[thenewbosco]

no help on this...

 

[quicknote]

The beat frequency is defined as the difference between the transmitted and reflected frequencies. In this case, we can use the approximation f + f_source ≈ 2f_source because the speed of the mirror is much smaller than the speed of light. This means that the reflected frequency is very close to the source frequency, and the beat frequency is much smaller than the transmitted frequency.

Using the equation from part a, f = f_source(c+v)/(c-v), we can rearrange it to solve for f_source:

f_source = f(c-v)/(c+v)

Substituting this into the approximation, we get:

f + f_source ≈ 2f_source

f + f(c-v)/(c+v) ≈ 2f(c-v)/(c+v)

Now, we can simplify this expression by multiplying both sides by (c+v)/(c-v):

f(c+v) + f(c-v) ≈ 2f(c-v)

fc + fv + fc - fv ≈ 2fc - 2fv

2fc ≈ 2fc - 2fv

Therefore, we can conclude that:

fv ≈ 0

This means that the beat frequency, which is the difference between the transmitted and reflected frequencies, is approximately equal to 0. In other words, the beat frequency is much smaller than the transmitted frequency.

Now, we can use this information to solve for the beat frequency:

f_beat = f - f_source

f_beat ≈ 0 - 0

f_beat ≈ 0

But we know that the beat frequency cannot be exactly 0, so we can use a small approximation to account for the difference:

f_beat ≈ 0 + ε

where ε is a very small number.

Now, let's rearrange this expression to solve for the beat frequency:

f_beat ≈ ε

But we also know that:

f_beat = (transmitted frequency) - (reflected frequency)

f_beat = (transmitted frequency) - (source frequency)

Substituting this into our previous expression, we get:

(transmitted frequency) - (source frequency) ≈ ε

But we also know that the transmitted frequency is equal to the source frequency plus the beat frequency:

(transmitted frequency) = (source frequency) + (beat frequency)

Substituting this into our previous expression, we get:

(source frequency) + (beat frequency) - (source frequency)