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Wavelength (Open Air Column)

  1. May 29, 2008 #1
    Hello there,

    For the following problem, I am arriving at an answer different from the given one so I would appreciate any help or hints.

    Thank you. My work is shown below.


    1. A slightly smaller plastic pipe is inserted inside a second plastic pipe to produce an air column, open at both ends, whose lengths can be varied from 35 cm to 65 cm. A loud speaker, connected to an audio frequency generator, is held over one of the open ends. As the length of the air column is increased, resonance is heard first when the air column is 38 cm long and again when it is 57 cm long.

    a) Calculate the wavelength of the sound produced by the audio frequency generator.


    Since the first resonant length is 38 cm long:

    [itex] L = \frac {\lambda}{2} [/itex]

    [itex] 0.38 m = \frac {\lambda}{2} [/itex]

    [itex] \lambda = 0.76 m [/itex]

    However, my textbook says that it is 0.38 m.
    Last edited: May 29, 2008
  2. jcsd
  3. May 29, 2008 #2


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    This is an air column open on both ends. What are the lengths of the possible standing waves for such a pipe? (A pipe closed at one end has a resonant length which is half of the fundamental or lowest frequency. A fully-open pipe does not...)
  4. May 29, 2008 #3
    Thanks for your reply, dynamicsolo.

    However, I am looking in my textbook and it says that the general formula for length of an open air column (in both ends) in wavelengths is: [itex] L = \frac{\lambda}{2} [/itex].

    What may I be missing?
  5. May 29, 2008 #4


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    That's my mistake: the formula you cite is correct and I flubbed on the half-closed column.

    Here's the issue -- there is a resonance for the source's sound wavelength at a pipe length of 38 cm and another at 57 cm. An open pipe will have standing waves at successive half-wavelengths of the sound wave, since each end of the pipe must be an "anti-node" for the sound wave.

    This means that the pipe has a resonance at some number n of half-wavelengths at the 38 cm length and then at a number (n+1) half-wavelengths when it is extended to 57 cm wavelength. We know this because there was not another resonance at some intermediate length. Therefore, one half-wavelength of the sound wave is 57 - 38 cm = 19 cm. (The first resonant length would be at 19 cm, but the telescoping tube can't become that short; likewise, there would be another resonance at 76 cm, if the tube could be extended that far.)

    My apologies for the goof in post #2.
    Last edited: May 29, 2008
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