How Does Speaker Motion Affect Pressure in Open and Closed Spaces?

[Annirak]

I've been studying the behaviour of speakers and headphones for my own interest. I was able to derive the differential equation governing the motion of the speaker itself, what I've had trouble doing is deriving the pressure created by the speaker.

First the equation I've derived for movement of headphones

ILB=A(P_F-P_B)+kx+\mu\dot{x}+m\ddot{x}

where I is the current through the speaker coil
L is the length of the wire in the speaker
B is the field strength of the magnet
A is the area of the speaker cone
P_F is the pressure at the front of the speaker cone
P_B is the pressure at the back of the speaker cone
k is the spring constant of the speaker cone mounting
x is the displacement of the speaker cone
\mu is the coefficient of friction of the speaker cone mounting
m is the mass of the speaker cone assembly

In a closed space (eg closed-back headphones), \lambda > a where a is the minimum dimension of the cavity, so the pressure can be modeled via the ideal gas law:

PV=nRT \Rightarrow P=\frac{nRT}{V}

Correlating this to displacement via V=A(x_0+x),

P=\frac{nRT}{A(x_0+x)}

P_F and P_B each share this model with different values of A and x_0, so that the speaker is essentially a diaphragm mounted part way down a sealed cavity.

Now for an open speaker, I'm not so sure. How is the pressure developed in front of a moving speaker related to the motion of the speaker, assuming that the speaker is not at one end of a closed cavity smaller than the minimum wavelength?

 

[Annirak]

Thinking over it some more, I believe that the pressure in front of a non-enclosed moving membrane must be a velocity term, related to the diffusion equation.

The net result should be:

ILB=m\ddot{x}+\mu\dot{x}+\frac{2nRTx}{A(x^2-x_0^2)} for closed front, closed back,
ILB=m\ddot{x}+(\mu+K_{diff})\dot{x}-\frac{nRT}{A(x_0-x)} for open front, closed back,
ILB=m\ddot{x}+(\mu+2K_{diff})\dot{x} for open front, open back

Does anyone know how to calculate K_{diff}? Or if the above is right?

 

[Studiot]

The differential equation(s) governing loudspeaker action is not a single equation but a coupled set of simultaneous equations, electrical and mechanical.

I made a start in this thread backalong.

https://www.physicsforums.com/showthread.php?t=421136&highlight=coupled+differential+equations

 

[Annirak]

Thanks Studiot,
I'm comfortable with the electrical side of the equations (and I'm content to leave that as ILB). I'm more concerned with the developed pressure; which I don't know how to tackle properly.

As far as I can tell, a differential change in displacement \partial x creates a corresponding differential change in pressure \partial P.

This change in pressure brings a corresponding change in diffusion rates:
N=-D\frac{\partial P}{\partial x}

The diffusion rate causes a corresponding time-varying drop in pressure:
\frac{\partial P}{\partial t}=D\frac{\partial^2 P}{\partial x^2}

I think this is what I'm looking for, but now I can't directly see how to relate this to the ideal gas law, which it must.

The problem is in relating \partial x and \partial P. I'm having trouble seeing how the "volume" changes in order for the pressure to increase. I'm sure that this is modeled by a fairly simple differential equation, but I'm having trouble deriving it.

[Edit:] To be clear, I'm not looking for the derivative of the ideal gas law. That deals strictly with closed containers, which doesn't help here.

 

[Studiot]

You will find the derivation you seek in

The Physics of Vibrations and Waves by H J Pain

pages 144 - 152, pages 146 - 7 in particular.