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Homework Help: Doppler effect: Moving source with reflection

  1. Jun 14, 2005 #1
    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 [itex]f_{L_2} = \frac{v}{v - v_S}f_S[/itex], where v is the speed of sound. The sound reflected from the wall has the same frequency [itex]f_{L_2}[/itex]. L1 is approaching the wall with speed [itex]v_S[/itex], so the relative speed of the wavefronts of sound is [itex]v + v_S[/itex]. The frequency observed by L1 is [itex]f_{L_1} = \frac{v + v_S}{v}f_{L_2} = \frac{v + v_S}{v - v_S}f_{S}[/itex] which is approximately equal to [itex]2f_{L_2}[/itex] for [itex]v_S[/itex] close to [itex]v[/itex]. Does this answer make sense, or is it unnecessarily complicated?
     
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  3. Jun 14, 2005 #2

    James R

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    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.
     
  4. Jun 14, 2005 #3
    The equation does reduce to [itex]2f_{L_2}[/itex] look at the equation again. It seems pretty clear to me that the quesiton is asking about the final effect of the doppler shift on the original frequency [itex]f_{L_2}[/itex]. Could you explain what you mean by this?

    Edit: [itex]2f_S[/itex] should be [itex]2f_{L_2}[/itex]
     
    Last edited: Jun 14, 2005
  5. Jun 14, 2005 #4
    It appears that you misread my original post. The frequency reduces to [itex]2f_{L_2}[/itex], not [itex]2f_S[/itex].
     
  6. Jun 14, 2005 #5

    James R

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    Your equation is:

    [tex]f_{L_1} = \frac{v + v_S}{v - v_S}f_{S}[/tex]

    This is the equation for the frequency [itex]f_{L_1}[/itex], and not the shift, which would be

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

    Simplifying, we get:

    [tex]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}[/tex]

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

    If, on the other hand, [itex]v_S \approx v[/itex] then the shift becomes very large.
     
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