Why an open tube at both ends suffers resonance when subjected to a sound that propagates through the air with length of where $L / 2$?

I already know the methodology to calculate the harmonics in an open tube. So, I'm not expecting an answer that is based on the calculations. I want to know why this phenomenon occurs from the molecular point of view of the air, taking into account that air could be considered as an ideal gas whose molecules have no interaction with each other.

  • $\begingroup$ "An ideal gas whose molecules have no interaction with each other" - If there are no interactions between the molecules, where does the pressure in the gas come from? The mean free path of air molecules at atmospheric pressure and temperature is less than 100nm. $\endgroup$
    – alephzero
    Jun 18, 2019 at 16:02
  • $\begingroup$ the source of the wave of this sound gives momentum to some particles and they travel through the space with some density higher than the air around. The pressure in the air is not traveling. What is traveling are particles, that, if they shock with some surface with some regular frequency the pressure will appear. $\endgroup$ Jun 18, 2019 at 16:14
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/321231/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 18, 2019 at 18:47

1 Answer 1


The gas molecules in air do indeed interact with each other, by colliding against each other with surprising speed at room temperature. This means a volume of gas will possess compliance (it can act like a spring). Those gas molecules possess mass too, and hence a volume of air can transmit compression waves through it at a speed that depends on this characteristic compliance and mass, and as a transmissive medium free air in an open space will therefore possess a certain characteristic impedance.

Now the air inside a tube possesses a certain characteristic impedance for waves traveling long its length as well, but this impedance is different from that of free air, which means there is a significant impedance mismatch at the open ends of the tube. When an acoustic wave in a tube meets that mismatch, part of the wave gets reflected off the mismatch (with a 180 degree phase reversal) and travels backwards up the tube, and part of the wave escapes into free space.

The fact that this reflection occurs off the open ends means that even a tube with two open ends will still support a resonance.


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