Doppler effect: Moving source with reflection

[jdstokes]

Suppose a sound source S is moving towards a stationary wall and that a listener L1 is moving with the sound source. Describe in a few lines why the reflected sound heard by the listener L1 is Doppler-shifted by about twice the amount that a second listener L2 would hear standing by the wall.

The sound waves in front of the moving source are compressed causing an apparent increase in frequency to $f_{L_2} = \frac{v}{v - v_S}f_S$, where v is the speed of sound. The sound reflected from the wall has the same frequency $f_{L_2}$. L1 is approaching the wall with speed $v_S$, so the relative speed of the wavefronts of sound is $v + v_S$. The frequency observed by L1 is $f_{L_1} = \frac{v + v_S}{v}f_{L_2} = \frac{v + v_S}{v - v_S}f_{S}$ which is approximately equal to $2f_{L_2}$ for $v_S$ close to $v$. Does this answer make sense, or is it unnecessarily complicated?

 

[James R]

It's not quite right. Your expression certainly does NOT reduce to 2 fs for vs close to v.

But you're asked about the shift, not the final frequency.

 

[jdstokes]

The equation does reduce to $2f_{L_2}$ look at the equation again. It seems pretty clear to me that the question is asking about the final effect of the doppler shift on the original frequency $f_{L_2}$. Could you explain what you mean by this?

Edit: $2f_S$ should be $2f_{L_2}$

 

[jdstokes]

It appears that you misread my original post. The frequency reduces to $2f_{L_2}$, not $2f_S$.

 

[James R]

Your equation is:

$$f_{L_1} = \frac{v + v_S}{v - v_S}f_{S}$$

This is the equation for the frequency $f_{L_1}$, and not the shift, which would be

$$f_{L_1} - f_S = \left(\frac{1 + v_S/v}{1 - v_S/v} - 1\right)f_S$$

Simplifying, we get:

$$f_{L_1} - f_S = \frac{1 + v_S/v - (1 - v_S/v)}{1 - v_S/v}f_S = \frac{2 v_S/v}{1 - v_S/v}$$

For $v_S << v$, we can see that the shift is approximately $2 v_S/v$.

If, on the other hand, $v_S \approx v$ then the shift becomes very large.