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Resonance in standing waves

  1. Apr 13, 2005 #1
    so my problem deals with resonance in standing waves

    if you have a standing wave in a long pipe of length 2.90m and that is closed on the left end and open on the right end and the graph of it is as below, with x-axis the position along pipe and y axis vertical air displacement

    http://i2.photobucket.com/albums/y7/twiztidmxcn/plot.gif

    i found the harmonic number to be 5, but that was by guessing at a wavelength

    i now need to find frequency using the speed of sound in air as 344m/s

    i used the harmonic number and the equation L = (2n-1)*wavelength / 4, pluggin in my length of 2.9m and 5 for n, and then divide 344 by that number

    i have gotten a multitude of numbers, including 266.86Hz, but nothing has worked

    any help in the right direction would be much appreciated

    thx-twiztidmxcn
     
  2. jcsd
  3. Apr 13, 2005 #2

    brewnog

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    Are you trying to find the fundamental frequency? If so, then picking a harmonic number of 5 is no good, since this is not the fundamental. Use the most simple waveform you have, instead of that mutant fish thing, and you should have a more sensible answer.
     
  4. Apr 14, 2005 #3
    it just says 'find the pitch (frequency) of the wave using 344m/s as the speed of sound in air'

    i haven;t been taught any other kind of equations to use with this. my wavelength is inaccurately guessed and found to be close to right, im fairly confused
     
  5. Apr 14, 2005 #4

    learningphysics

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    It looks to me like you're on the right track. From your graph it looks like the wavelength is between 2 and 2.5 (one full sinewave).

    This corresponds to n = 3. Using your equation I get the wavelength = 2.32m
    f=344/2.32 = 148.28Hz
     
  6. Apr 14, 2005 #5

    brewnog

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    I still don't understand why you've chosen a harmonic as the wave, rather than the fundamental.

    The frequency that the pipe will resonate at is the fundamental frequency, which happens with the most simple wave, with the largest possible wavelength. At one end of the tube you have a node, the other end; an antinode.
     
  7. Apr 14, 2005 #6

    learningphysics

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    But look at the graph. It is not resonating at the "fundamental" frequency. The wavelength is less than the length of the tube.
     
  8. Apr 14, 2005 #7

    brewnog

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    But the OP said that he'd guessed that.
     
  9. Apr 14, 2005 #8
    my friend, i thank you. this made sense, and i worked around with it and finally got how it worked out.

    thank you.
     
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