Sound standing waves paradox....

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Kelvin273
Messages
1
Reaction score
0
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?
 
Physics news on Phys.org