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Sound standing waves paradox...

  1. Sep 8, 2015 #1
    If you seal a loudspeaker at the end of a tube and close the other end of the tube you will get standing waves; but what are the boundary conditions at the speaker for the sound pressure wave?
    Pressure =0 or Pressure = MAX? I find no mention of this in the literature.

    To find out I performed a simple experiment with a 622 mm long PVC tube. I placed a small microphone connected to an oscilloscope at one end (flush with the wall and sealed) and a speaker connected to a signal generator at the other end. I swept the signal and I observed the first resonance at a frequency f = 276 Hz; corresponding to a wavelength of lambda = 1244mm = 2L, indicating that the speaker is a hard boundary or a pressure antinode.

    This result implies the air particle displacement at the speaker must be zero as pressure and particle displacement waves have a phase delay of Pi/2. So the air at the speaker does not move....and this sounds like a paradox to me since the speaker is a piston and must be displacing the air adjacent to it.

    Any ideas?
  2. jcsd
  3. Sep 9, 2015 #2


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    The boundary condition at the speaker is not so simple that you can just assume the homogeneous condition. The boundary is undergoing forced motion due to the speaker membrane moving in a particular fashion, this is what creates the sound waves.