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The Doppler Effect - Finding Velocity of Source

  1. Dec 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A fire truck emitting a 450 Hz signal passes by a stationary detector.
    The difference in frequency measured by the detector is 58 Hz.
    If the speed of sound is 345 m/s, how fast is the fire truck moving?

    2. Relevant equations

    frequency of observer = (velocity of sound + velocity of detector / velocity of sound + velocity of source) f0

    3. The attempt at a solution

    f obs (frequency of observer) = 450 Hz + 58 Hz = 508 Hz
    v sound (velocity of sound) = 345 m/s
    v detector = 0
    f0 (frequency of source) = 450 Hz
    v source = ?

    How would I solve for velocity of source? I tried plugging all of the above variables into the equation and isolating the variable but I got a totally wrong answer and I'm not sure why.
     
  2. jcsd
  3. Dec 5, 2013 #2
    Question 1: is the truck moving towards or away from the detector?
    Question 2: what equation did you use and what was the answer you got?
     
  4. Dec 5, 2013 #3

    Doc Al

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    Staff: Mentor

    Careful with the sign of the source velocity.
     
  5. Dec 5, 2013 #4
    That's the thing... it doesn't say. My teacher told me to just use it as it is going away from the detector.

    I used + for the source going away from the observer, yet I still got it wrong?

    This is what I did:

    508 Hz = (345 m/s + 0 / 345 + v source) 450 Hz
    508 Hz / 450 Hz = 345 m/s / 345 m/s + v source
    1.13 Hz (345 m/s) = 345 m/s / v source

    and... everything went downhill from there...
     
  6. Dec 5, 2013 #5
    It says that the truck is passing by.
    It may be that the 58 Hz is the difference in the frequency between the frequency measured when the truck approaches and the one measured when it is going away.

    Anyway, your algebra is not right. The last line in the above does not look good.
    Check again.
     
  7. Dec 5, 2013 #6

    Doc Al

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    That won't work, since you are multiplying the original frequency by a fraction less than one and setting it equal to a larger frequency.

    I think nasu picked up on the key point: that the truck passes by the observer. So at first it is coming towards the observer (higher frequency observed) then it is going away from the observer (lower frequency observed). Set up equations for both coming and going, taking care of the signs of the source velocity in each segment.
     
  8. Dec 6, 2013 #7
    I still don't get it...
    Or I do, but I'm still getting something wrong with my algebra.
    What do I do in any case, to find the velocity of the source?
     
  9. Dec 6, 2013 #8

    Doc Al

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    Do this:
    (1) Write an expression for the higher frequency observed when the truck is approaching.
    (2) Write an expression for the lower frequency observed when the truck is moving away.
    (3) Compare those two and then you can solve for the source velocity.
     
  10. Dec 7, 2013 #9
    Wait, so... difference meaning it would be half for each?
    i.e. f obs = 450 Hz + (58 Hz / 2); = 450 Hz + 29 Hz = 479 Hz (truck going away)
    f obs = 450 - 29 Hz = 421 Hz (truck approaching)

    ... if the above is correct:
    (1) 421 Hz = (345 m/s + 0 / 345 - v source) 450 Hz; truck approaching
    (2) 479 Hz = (345 m/s + 0 / 345 + v source) 450 Hz; truck going away
    (3) ?
     
  11. Dec 7, 2013 #10
    No, the difference is not necessarily split even
    Why don't you follow the steps suggested by Doc Al?
     
  12. Dec 7, 2013 #11

    Doc Al

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    No. Don't try to guess the frequencies by assuming that half the difference goes to each.
    Also: Which gives the higher observed frequency? Truck approaching the observer or truck going away?

    No. Instead:
    (1) F1 = ...
    (2) F2 = ...

    Don't guess what those frequencies are, but use what you know about how they are related.
     
  13. Dec 7, 2013 #12
    Oh, oops. The truck approaching the observer should be the higher frequency...

    (1) F1 = 450 Hz + 58 Hz = 508 Hz
    (2) F2 = 450 Hz - 58 Hz = 392 Hz
     
  14. Dec 7, 2013 #13

    Doc Al

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    Stop assuming you know the observed frequencies!

    Treat F1 and F2 as unknowns.
     
  15. Dec 7, 2013 #14
    (1) F1 = 450 Hz + x = ?
    (2) F2 = 450 Hz - y = ?

    ...Like this?
     
  16. Dec 7, 2013 #15

    Doc Al

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    No.

    The right hand sides of your equations for F1 and F2 were correct here:
     
  17. Dec 7, 2013 #16
    Oh, I see.

    (1) F1 = (345 m/s + 0 / 345 - v source) 450 Hz; truck approaching
    (2) F2 = (345 m/s + 0 / 345 + v source) 450 Hz; truck going away

    So there are three unknown variables here? The v source, F1, and F2...
     
  18. Dec 7, 2013 #17

    Doc Al

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    Right.

    But there's another fact you can use, from the problem statement. How do F1 and F2 relate?
     
  19. Dec 7, 2013 #18
    F1 - F2 = 58 Hz ?
     
  20. Dec 7, 2013 #19

    Doc Al

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    Staff: Mentor

    Exactly!
     
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