Loudspeaker attached to cylinder - closed or open?

I have a loudspeaker that is facing into a long cylindrical piece of PVC pipe of length L. The loudspeaker is sealed in so that no air can escape around the edges. The pipe is open at the other end.

The loudspeaker is attached to a signal generator and made to vibrate at frequency f.

In this configuration, should the system be modeled as a closed pipe or an open pipe? Or does the choice of model depend on f ?

My intuition is that the loudspeaker represents a displacement antinode (and therefore a pressure node), and so the system should be modeled as an open pipe (pressure nodes on both ends). The system would then resonate at frequencies where $$L=\frac{\lambda}{2}=\frac{c}{2f}$$. Is this a correct assumption?

If one were to place a circular piece of PVC with a small hole in the center across the face of the speaker, so that air is compressed inside a small cavity before being released into the pipe, would this reverse the scenario so that the speaker end becomes a pressure antinode?

I decided to test this using approx 1m long PVC pipe from Bunnings and a small loudspeaker.

I had two pieces of light milk-carton plastic which I placed over the front of the loudspeaker - one with a small hole in the center, and one with a large hole. Then for each of the pieces of plastic, I placed the PVC pipe on top, pressing down the plastic between the loudspeaker and the pipe to seal it.

Using the plastic with a large hole, the lowest loud resonance was at 166.5Hz, which corresponds to a wavelength of 2.06m. This means that the 1m pipe was generating half-wavelength standing waves, and therefore the system should be modeled as an open cylinder with pressure nodes at each end. This contradicts Niels Nielson's answer (sorry).

Using the plastic with a small hole, the lowest loud resonance was one octave lower at 83.25Hz, which corresponds to a wavelength of 4.12m. This means that the 1m pipe was generating quarter-wavelength standing waves, and therefore the system should be modeled as an closed cylinder with a pressure node at the open end and a pressure antinode at the speaker end.

So my intuition was correct!

It is possible that I bungled the experiment so if anyone has anything to add I would love to hear it.

Cheers

You are right, a tube driven at its closed end and open at the other will resonate like a tube with one closed end. Putting a punched disc in front of the speaker will cause the hole in its center to radiate as a point source, which will excite the same resonance in the tube and introduce a helmholtz resonance in which the cone compliance bounces off the mass of air entrained within that hole, complicating the frequency response of the system.

• Thanks Niels, but I am slightly confused. Is a tube that's driven at its' closed end modelled as an open tube at both ends (my answer), or a closed tube at one end (your answer)? – Jeremiah Rose Feb 23 '19 at 0:25
• it's modeled as a one end closed plus other end open system, here is why. if I introduce a sound source into such a tube, the tube doesn't care much exactly where the sound source is located- as the waves bounce around in there, eventually a standing wave will build up, and here's how you test this idea: with a speaker in one end of the tube and the other end open, find the resonance and drive it so it's good and strong. note the frequency. then suddenly cut the signal to the speaker and listen for any pitch shift. – niels nielsen Feb 23 '19 at 0:31