The Helmholtz resonator is affected by the presence of the speaker, which effectively forms one wall of the enclosure. The stiffness of this "wall" then depends on the speaker Q. This is what David Weems mentioned.

That's absolutely true.

In fact, when we talk about a loudspeaker in a ported box, we need to analyze the complete lumped-element circuit of the whole system. Fortunately, this was done long ago by the titans of acoustics.

R.Small provides the following lumped element circuit.

If we assume we made a perfect box (there is no resistive absorption inside a box, i.e. Rab=0, and there is no small holes for leakage, i.e. Ral=infinity), then we would have an ideal Helmholtz resonator, whose Q factor is described by the formula I cited in my previous post. However, in a real box a Helmholtz resonator and a loudspeaker make up a coupled oscillator system, which inevitably strongly influence each other. I have not solved the coupled equation system for the circuit shown above, but even without solving it, it can be said that the speaker (

__connected to the voltage source__) will have a very strong effect on the overall Q of the whole system due to the high value of the resistor associated with the presence of a strong electromechanical coupling in a loudspeaker (BL^2/[(Rg+Re)Sd^2)]). On the other hand, in a loudspeaker, connected to the current source, this electromechanical coupling is gone because internal resistance of the current source is very high (in limit Rg --> infinity), and overall Q factor of the whole system would be close to that of the perfect Helmholtz resonator.