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Homework Help: Doppler Effect and sound

  1. Apr 14, 2007 #1
    Hi, I am a little confused about some aspects of sound waves. I was hoping to just check to see if I am on the right track.

    I have a problem where a guy is moving towards the great wall of china, with a air horn, emitting a constant frequency sound of 500Hz. The sounds is going to hit the great wall and bounce back, introducing some interferance. The speed of sound is 350m/s. The persons is running towards the wall at 3/ms, and I have to find the frequency of amplitude beats he hears and the frequency of the sound he hears reflected.

    ok, so f' = (v/(v +/- Vs)*f

    He is moving towards the wall, so Vs is negative? This gives

    f' = (350/(347)) * 500 = 504.3Hz

    Is the frequency of beats then just the difference, ie (504.3-500)Hz for 4.3Hz beat Hz. Am I making any sense?

  2. jcsd
  3. Apr 14, 2007 #2
    Yes I believe thats right.
  4. Apr 14, 2007 #3


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    Science Advisor
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    Gold Member

    No. There is reflection here, so you need to take into account two different Doppler effects. What you found is the frequency of the wave reaching the wall. Now, this wave is reflected and travels back to the guy. You need to do another Doppler effect calculation with this time the guy being the observer so you will need to multiply the frequency you found by another factor (with now the source, which is the wall, being at rest and the observer moving toward the source at 3 m/s).

    Once you get the final frequency, you just subtract 500Hz to get the beat frequency, as you had done.

  5. Apr 14, 2007 #4
    I see, of course :)
  6. Apr 14, 2007 #5
    ok, I am still a little confused but things are a bit clearer :)

    Actually I made a mistake anyway, the air horn is actually stuck into the ground, so it doesn't move, just the guy running towards the "great wall of china". I should learn to read!

    So we have a setup like

    (((( horn )))) man -> [ wall ]

    So he is moving way from the source, and towards the reflected wave I guess from the wall. So there is two doppler effects? One being:

    f' = (350/(353)) * 500 = 495.75Hz ( what the sound behind him sounds like)


    f' = (350/(347)) * 500 = 504.3Hz (what the reflected sound sounds like )

    i am assuming the sound actually reflected is 500 Hz, the frequency of the emitted sounds, as the doppler effect doesn't change the frequency relative to the other standary objects ie the wall, and source are moving.

    So the difference would be then 504.3 - 495.8 = 8.6 Hz?

    Thanks a lot everyone, this stuff is really messing me up :)

  7. Apr 14, 2007 #6

    Doc Al

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    Staff: Mentor

    Looks like you got it just fine.
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