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Doppler Effect and car alarm

  1. Jul 11, 2007 #1

    exi

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    1. The problem statement, all variables and given/known data

    A parked car's alarm goes off, producing a sound at 961 Hz.

    As you drive toward, pass, and drive away from this parked car, you observe a frequency change of 98 Hz. At what speed are you driving?

    (Speed of sound is 343 m/s.)

    2. Relevant equations

    [tex]f_o = f_s\left( 1-\frac{v_o}{v} \right)[/tex] (Observer moving away from source)

    And possibly

    [tex]f_o = f_s\left( 1+\frac{v_o}{v} \right)[/tex] (Observer moving towards source)

    3. The attempt at a solution

    Interesting problem. Using the above formula for Doppler shift when an observer is moving away from a source, I was coming up with

    [tex]f_o = 863\left( 1-\frac{v_o}{343 m/s} \right)[/tex]

    which solves to 34.9781 m/s, an incorrect answer.

    Any hints on what I'm doing wrong? I was wondering if two-stepping this and using both equations would get anywhere, but without knowing for sure what the observer's frequency would be on that half of the trip, I'm a bit unsure of things.

    Thanks.
     
    Last edited: Jul 11, 2007
  2. jcsd
  3. Jul 11, 2007 #2

    bel

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    First you derive the expression for the change of frequency as you move towards the parked car, then you add it to the frequency change as you move away. What you have now is the total frequency change, and you have a number to which your expression could be equated, and you have only one unknown since your velocity was assumed to be constant.
     
  4. Jul 11, 2007 #3

    exi

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    Hmm... so:

    [tex]f_o = f_s\left( 1+\frac{v_o}{v} \right) + f_s\left( 1-\frac{v_o}{v} \right)[/tex]

    Distributing the frequency of the source and solving for the observer's velocity results in Vo terms that cancel out. Seems I'm doing something stupid, but... what?
     
  5. Jul 11, 2007 #4

    Dick

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    bel suggested you add the CHANGES in frequency. You can also do this by taking the difference instead of the sum. You want the change of frequency, which is a DIFFERENCE.
     
  6. Jul 11, 2007 #5

    exi

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    Don't I feel silly. I'm coming up with

    [tex]f_o = 2*f_s*v_o[/tex]

    I've tried several different numbers, but I seem to keep getting stupidly small (no greater than 4 m/s) velocities. The frequency change suggests to me that it should be higher, so I'm thinking I'm making a mistake as to just which number is equated to the above equation (in place of f_o).
     
  7. Jul 11, 2007 #6

    Dick

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    What happened to the division by v?? And what are you substituting for the various variables?
     
  8. Jul 11, 2007 #7

    exi

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    I think I just caught another algebra / calculation mistake.

    Well, correcting the above

    [tex]f_o = f_s\left( 1+\frac{v_o}{v} \right) - f_s\left( 1-\frac{v_o}{v} \right)[/tex]

    seems to allow f_s distribution to arrive at

    [tex]f_o = f_s+f_s\frac{v_o}{v} - f_s + f_s\frac{v_o}{v}[/tex]

    and multiplying terms by v to remove that fraction gives

    [tex]f_o(v) = f_s(v) + f_s(v_o) - f_s(v) + f_s(v_o)[/tex]

    the f_s(v) terms cancel, and that boils down to, if I'm correctly understanding Bel,

    [tex]98(343) = 2(961)(v_o)[/tex], or [tex]1922(v_o) = 33614[/tex]

    And 17.4891 m/s.

    Ouais?
     
  9. Jul 11, 2007 #8

    Dick

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    Ouais, certainment. (One of your intermediate eqns is missing some stuff, but let's pass on that).
     
  10. Jul 11, 2007 #9

    exi

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    Any other fundamental errors I'm making here? I'd hate to miss an almost-right question. o:)
     
  11. Jul 11, 2007 #10

    Dick

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    Do you believe it? Does it feel right? Can you think of anyway to check it? I'm not going to be here all the time to check.
     
  12. Jul 11, 2007 #11

    exi

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    Not sure, yes, and not offhand (although I did stumble upon a quadratic form of an equation to do this problem, which gave me a number within .04 of my own answer).

    Et c'était une réponse correcte aussi; merci encore.
     
  13. Jul 11, 2007 #12

    Dick

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    Glad to hear the 'yes'. I believe it. Bon chance! Here's what I've checked. 98Hz is about 10% of 961Hz. That means 5% shift approaching and 5% retreating. 5% of the speed of sound is 15m/sec. And that means you are right. Try doing problems in without a calculator sometimes. Just roughly.
     
    Last edited: Jul 11, 2007
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