Resonance, find the lengths

  1. May 2, 2006 #1
    "A tuning fork with a frequency of 440 Hz is held above a resonance tube partially filled with water. Assuming that the speed of sound in air is 342 m/s, for what heights of the air column will resonances occur?"

    f1 = 440 Hz
    v = 342m/s
    I'm not sure if this is refering to an open pipe, or a closed pipe.
    For an open pipe,
    fn = (nv)/(2L) = nf1
    v/(2L) = f1
    L = v/(2f1)
    L = 0.3886 m
    But this is the length for the sound traveling up and down... i think. So the length is 0.194m. However, the answer gives 3 values, 0.194m, 0.583m, 0.972m. Even for a closed pipe, I there is only 1 value...
    I must be doing something wrong. I don't think the speed of sound in water is a factor, because my first answer seems to be correct...
     
  2. jcsd
  3. May 2, 2006 #2

    Andrew Mason

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    This is a closed pipe. There is a node at the bottom and an anti-node at the open end. So resonances occur where the length of the pipe is [itex]L = \lambda/4, 3\lambda/4, 5\lambda/4,... (2n+1)\lambda/4[/itex]

    Using the universal wave equation:

    [tex]\lambda = v/f [/tex]

    Substituting the resonance criterion: [itex]\lambda = 4L/(2n+1) = v/f[/itex]

    resonance occurs at:

    [tex]L = (2n+1)v/4f = (2n+1)*342/4*440 = (2n+1).194 m.[/tex]

    AM
     
  4. May 3, 2006 #3
    That makes sense, except for why f is kept as the initial frequency. The formula I have is:
    L = (mv)/(4fm) where m = 1,3,5,.....
    and
    fm = mf1
    thus
    L = v/(4f1)
    why do I use only the initial frequency here?
     
  5. May 3, 2006 #4

    Andrew Mason

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    In this case, there is a tuning fork which provides only the fundamental frequency - a pure sine wave. There are no higher frequencies to resonate in the air column. So you are dealing with the fundamental frequency only.

    AM
     
  6. May 3, 2006 #5
    Oh, Ok. Thanks!
     
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