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Doppler Shift and Beats

  1. Apr 27, 2008 #1
    [SOLVED] Doppler Shift and Beats

    1. The problem statement, all variables and given/known data
    A car horn sounds a pure tone as the car approaches a wall. A stationary listener behind the car hears an average frequency of 250Hz, which pulsates (beats) at 12 Hz. What is the speed of the car?

    The speed of sound is assumed to be 331 m/s

    2. Relevant equations

    f_beat = f1 - f2
    f[tex]_{'}[/tex]=f [1 / (1+ VE/V)] Doppler Shift for Stationary Receiver, Receding Emitter.

    3. The attempt at a solution
    There must be two noises overlapping if the stationary listener is hearing a pulsating beat. If the average frequency heard is 250Hz, then (f1+f2) = 500. Plugging this into f_beat=f1-f2=12Hz, you find that the frequencies must be 244Hz and 256Hz. One of these frequencies is the result of the doppler shift as the listener receives the noise from the receding car. The other I assumed to be the noise that the listener received after it bounced off the wall (which the car is approaching). As far as I can tell, there must then be two equations in regards to the doppler shift for each sound, which share two common variables: f, the tone emitted by the car as if it wasn't moving, and VE, the speed of the car. One of these equations should be 244Hz = f[1 / (1 +VE/ 331m/s)], but I'm not sure how to go about calculating the other equation, or whether or not I'm on the right track in solving this.

    Any help would be greatly appreciated.
     
  2. jcsd
  3. Apr 27, 2008 #2
    One equation is stationary receiver, receding emitter, and the other is for a approaching emitter (should just me a minus sign difference).
    Which frequency is the one corresponding to the emitter going away from you?
    Which frequency is the one corresponding to the emitter approaching the wall (which is then relayed to you by the wall)?
    This gives you the 2 equations with 2 variables.
     
  4. Apr 27, 2008 #3
    So the equation for the frequency heard after bouncing off the wall is just the same as if the wall was a stationary receiver with the approaching emitter? That would give the 2 equations

    f1 = f( 1 / (1 + VE/V) receding
    and
    f2 = f(1 / (1 - VE/V) approaching

    Approaching emitters generate a higher frequency, so you would have

    256 = f( 1 / (1 - VE/V), so f= 256 - 256VE/V, which makes the other equation f1=244= (256-256VE/V) * (1/ (1 + VE/V)). Which makes (244 + 244VE/V) = (256-256VE/V), thus 500VE/331 m/s = 12, making VE = 7.94 m/s

    Assuming my algebra skills are more or less in tact. Thanks, that cleared up a lot. I guess I was trying to make the frequency of the echo heard by the listener different than the frequency "heard" by the wall, even though the formula in front of me said it shouldn't be because both were stationary.
     
  5. Apr 28, 2008 #4
    That looks completely right.
    It is a little weird, but the wall does essentially perfectly convey to you what it "hears."
     
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