# Doppler effect: Moving source with reflection

• jdstokes
In summary, 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.
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?

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

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 $2f_{L_2}$

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

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

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.

## 1. What is the Doppler effect?

The Doppler effect is a phenomenon where the perceived frequency of a wave (such as sound or light) changes when the source of the wave is in motion relative to the observer. This results in either a higher or lower frequency being heard or seen by the observer.

## 2. How does the Doppler effect apply to a moving source with reflection?

In this scenario, the source of the wave is moving and there is a reflective surface nearby. As the wave reflects off of the surface, the perceived frequency of the wave changes for the observer. This is due to the relative motion of the source and the reflective surface, which affects the wavelength and frequency of the wave.

## 3. What factors influence the intensity of the Doppler effect?

The intensity of the Doppler effect is influenced by the relative velocities of the source and the observer, as well as the speed of the wave itself. Additionally, the angle at which the wave reflects off of the surface can also affect the intensity of the effect.

## 4. How is the Doppler effect used in real-world applications?

The Doppler effect is used in various applications, such as radar and sonar systems, to determine the speed and direction of moving objects. It is also used in medical imaging techniques, such as Doppler ultrasound, to measure blood flow and detect abnormalities.

## 5. What are some limitations of the Doppler effect?

The Doppler effect is only applicable when there is relative motion between the source and the observer. It also assumes that the wave is not affected by any other factors, such as interference or absorption. Additionally, the Doppler effect may be difficult to detect for waves with very low frequencies.

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