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Standing waves and length of tube

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A narrow column of air is found to have standing waves at frequencies of 390 Hz, 520 Hz, and 650 Hz and at no frequencies in between these. The behavior of the tube at frequencies less than 390 Hz or greater than 650 Hz is not known.

    How long is the tube?

    2. Relevant equations

    f = m(v/(2L)) = mf
    m = 1,2,3,4.....

    3. The attempt at a solution

    I keep getting .4358 or .4410 (depending on if I use 344 or 340 for the speed of sound, respectivly)
  2. jcsd
  3. Sep 8, 2007 #2


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    one-end open tube or both ends open tube?
    btw, your answers seem wrong either
    Last edited: Sep 8, 2007
  4. Sep 9, 2007 #3
    open on both ends, basically like a pipe..
  5. Sep 9, 2007 #4


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    did u work out the corresponding m's for your frequencies? indeed, different wave speed will give different L.
  6. Sep 9, 2007 #5

    so each frequency will give a different length? so what length would be correct?
  7. Sep 9, 2007 #6
    well it said 'narrow column of air' so i assumed open on both ends
  8. Sep 9, 2007 #7


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    no, what I meant was, you should work out the value of your m's corresponding to each frequency:
    say [tex]m_{f1} = k, m_{f2}=k+1, m_{f3}=k+2[/tex] where f1, f2 and f3 are the 390, 520, 650 Hz.

    and you would only get different answer if your speed of sound is different.
    it appears that this can only be the open-both-ends case for it to work. (just from the wavelength to m relations)
    Last edited: Sep 9, 2007
  9. Nov 23, 2007 #8
    Um, I'm not sure on this question either. Can someone guide me through it? It's also been giving me problems.
  10. Jan 30, 2009 #9
    Hate to revive an old thread but I can't figure out this one for the life of me....I've tried entering the length for every frequency (with the accompanying mode) and none of them seem to give me the correct length.
  11. Jan 30, 2009 #10


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    Three frequencies 390, 520, and 650 Hz can be wrightten as 3x130, 4x130 and 5x130. The same open tube can resonate in these three modes. Hence fundamental resonant frequency of the tube must be 130 Hz. Using this frequency and the velocity of sound, you can calculate the length of the tube.
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