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What is the pattern for resonant frequencies of a tube closed at BOTH ends?

The cases for the string, the open tube and the tube closed at one ends are well known.

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The rule for tubes is that you need a node on each closed end of the tube. This means that the tube closed on both ends is EXACTLY equivalent to a string. The equation that applies will thus be $f_n=n(v/2L)$, where $n=1, 2, 3, ...$.

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  • $\begingroup$ This is not strictly true... a string do not allows flexure and its transverse stiffness comes from a nonlinear coupling; on the other hand, a clamped tube is modelled by beam theory, which defines the flexural stiffness in terms of geometrical parameters of the cross section. $\endgroup$ – nodarkside Jan 9 at 0:23
  • $\begingroup$ @nodarkside The OP is asking about the resonance of air inside the tube, not resonance of the tube itself. The question doesn't make much sense if it was about transverse bending of a beam, because "closed" does not mean the same as "clamped" or "pinned". $\endgroup$ – alephzero Jan 9 at 0:48
  • $\begingroup$ @alephzero I am sorry, I didn't get that. I still not understand how the air resonance can be equivalent to that of a string, but I am not an expert on acoustic vibrations. $\endgroup$ – nodarkside Jan 9 at 0:52

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