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Doppler effect with no given numbers

  1. Aug 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A source of sound mvoes along one straight line between listeners Abigail and Bertha. The wavelength of the sound waves heard by Abigail is about 1/2 the wavelength heard by Bertha. What is the ratio of of the speed of the source to the speed of sound? Assume that the air carrying the sound waves is stationary.


    2. Relevant equations

    [​IMG]

    3. The attempt at a solution
    So I know that the wavelength that Abigail observes is 1/2 the wavelength that Bertha observes. Neither of them are moving and the speed of sound is 343 m/s. I don't know how to tie these thoughts together to solve this problem.
     
  2. jcsd
  3. Aug 10, 2008 #2

    Doc Al

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    That image you linked to is not viewable.

    In any case, you need to apply the Doppler formula (for the case of a moving source) twice: once for Abigal, once for Bertha. Hint: Call the original wavelength emitted by the source "w" and compare the two equations you end up with.
     
  4. Aug 10, 2008 #3
    I've tried that and just ended up confusing myself, so I'm not sure if that's right. I reviewed some examples from class and he solves for fs and then substitutes that in, but the cancellations didn't work out doing it that way.
     
  5. Aug 10, 2008 #4

    tiny-tim

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    Hi rissa_rue13! :smile:

    Show us what equations you have tried.

    (your image file isn't viewable)
     
  6. Aug 10, 2008 #5

    Doc Al

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    Show what you've tried and I'll take a look.

    By setting up the two equations as I suggested, you can eliminate f_s and solve for v_s in terms of the speed of sound.
     
  7. Aug 10, 2008 #6
    So, I set up fb and solve for fs:
    (v+vs)/wb=fs
    right?
    Then, if I substitute that into fa:
    fa=[(v+vs)/w][v/(v-vs)]
    But, I'm trying to get vs/v, so I don't get how that works?
     
  8. Aug 10, 2008 #7

    Doc Al

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    Try this: Write an equation for Abigal's observed frequency (f_a) in terms of f_s and the other variables. Then write a similar equation for Bertha's observed frequency (f_b). Then you can compare them more easily.
     
  9. Aug 10, 2008 #8
    But fa and fb are different because the wavelength Abigail observes is 1/2 that of the wavelength Bertha observes.
     
  10. Aug 10, 2008 #9
    fa=v(1/2w)
    fb=vw
    so wouldn't that make fs different for Abigail and Bertha?
     
  11. Aug 10, 2008 #10

    Doc Al

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    Sure, and we will make use of that key fact. But first write out the two equations. (Hint: The only difference between the equations will be the sign of the source velocity.)
     
  12. Aug 10, 2008 #11
    fa=fs[v/(v+vs)] -> v/(1/2w)=fs[v/(v+vs)]
    fb=fs[v/(v-vs)] -> v/w=fs[v/(v+vs)]
     
  13. Aug 10, 2008 #12

    Doc Al

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    Good. But all you need is the left side version of those formulas. Now take the next step. Hint: What's f_a/f_b?
     
  14. Aug 10, 2008 #13
    So,
    fa/fb=[fs(v/v+vs)]/[fs(v/v-vs)]
    right?
    Then, wouldn't that reduce to:
    fa/fb=[v(v-vs)]/[v(v+vs)] ?
     
  15. Aug 10, 2008 #14

    Doc Al

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    Absolutely. Replace the left hand side with a specific numerical value, based on information given in the problem; simplify the right hand side a bit. Once you've done that you can solve for v_s in terms of v.
     
  16. Aug 10, 2008 #15
    The only number I'm given is that wa=1/2wb. So would I substitute that in by doing:
    (2v/w)/(v/w) which would reduce to 2.
    And then put that in to:
    2=[(v-vs)/(v+vs)] ?
    Wouldn't that make v=vs ?
     
  17. Aug 10, 2008 #16

    Doc Al

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    I just noticed that in your formulas you have f_a and f_b reversed. Abigal is the one who hears the smaller wavelength--and thus larger frequency--sound since the source is moving towards her. So fix that first.
    Good, except as noted above.
    No, it wouldn't. If v = v_s, that right hand side would equal 0. (Again, first correct the error noted above.)
     
  18. Aug 10, 2008 #17
    So fa/fb should still equal 2, right? It was just the other part of the equation that was wrong?
    2 = (v+vs)/(v-vs)
    2v-2vs=v+vs
    v=3vs
    v/3=vs
    assuming v=343 m/s (speed of sound in air):
    vs= 114 m/s
    Therefore,
    vs/v = (114 m/s)/(343 m/s) = 0.333
    right?
     
  19. Aug 10, 2008 #18

    Doc Al

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    All good! (Just a tip: No need to actually calculate the speed. Once you found that v_s = v/3, it immediately follows that v_s/v = 1/3.)
     
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