Wow, interesting stuff and some mechanical engineering that is beyond me. But getting back to speaker drivers and how they affect sound..."A faster driver needs a light cone. Or does it?" is the title of a blog post on the Purifi web site.
It is not a very long post, but if you are really busy, here is a TL;DR summary:
I'll expand a little further on Purifi's blog post, since someone will inevitability ignore the last point above and will insist that acceleration is force divided by mass, and therefore lower mass gives higher acceleration. So how do we find how much acceleration we need?
- Displacement in speaker diaphragm does not generate the pressure wave in air that is the sound we hear. Acceleration does.
- What causes the diaphragm to accelerate (and therefore the air in front of it to pressurize) is force. A lower acceleration only results in lesser sound pressure level (SPL), i.e. less loud, not how quickly it appears, i.e. the lack of speed.
- The speed at which we can modulate "force" is unrelated to the mass of the diaphragm.
The late Siegfried Linkwitz (RIP) gave us a very handy formula to predict the free field SPL generated by a speaker driver, given its size, diaphragm travel, and frequency. [Link, see the box "Theory Behind the Nomographs"] It is:
SPL = 94.3 + 20 log10(x) + 40 log10(f) + 40 log10(d) - 20 log10(r) where: x is the peak-to-peak diaphragm travel in meters, f is frequency in Hz, d is the effective diameter of the diaphragm in meters (d = sqrt(4 * Sd / pi), with Sd = effective area in m^2) r is the listening distance in meters
Now, say we want to generate the same SPL at two different frequencies, f1 and f2, what will the diaphragm travels (x1 and x2) be?
SPL1 = 94.3 + 20 log10(x1) + 40 log10(f1) + 40 log10(d) - 20 log10(r)
and
SPL2 = 94.3 + 20 log10(x2) + 40 log10(f2) + 40 log10(d) - 20 log10(r)
Since we want SPL1 = SPL2 ,and "d" and "r" remain the same, we have:
20 log10(x1) + 40 log10(f1) = 20 log10(x2) + 40 log10(f2)
or
log10(x1) + 2 log10(f1) = log10(x2) + 2 log10(f2)
or
x1 * f1^2 = x2 * f2^2
or
x2 = x1 * f1^2/f2^2
So, the amount of travel the diaphragm needs to produce the same SPL in inversely proportional to the ratio of the frequencies squared (i.e. with the same diaphragm travel, SPL goes up/down by 12 dB/octave).
How about acceleration? Well, given the displacement amplitude x, acceleration = x * (2*pi*frequency)^2. Which means acceleration goes up by frequency squared. Since, for the same SPL:
x2 = x1 * f1^2 / f2^2
and
a2 = x2 * (2*pi * f2)^2 = x1 * (f1^2 / f2^2) * (2*pi * f2)^2 = x1 * (2*pi * f1)^2 = a1
The acceleration amplitudes are the same! And therefore forces. Amazingly we need the same force amplitude to produce the same SPL regardless of frequency. Of course, the rate of fluctuation of force is higher with higher frequencies, but the force magnitude is independent of frequency. We need to wiggle the diaphragm more frequently, but that is completely countered by the fact that we need to wiggle it less far.
There are plenty of other reasons why a woofer is not suitable to produce treble. Mass of the diaphragm ain't one.
How can "speed" be relevant if a driver can reproduce all the frequencies it needs to at the volume needed? Intuitively, you'd think that a lighter tweeter could follow the sound of a cymbal better than one heavier. But if the measurements don't show a different response (in the audible frequency range, of course) there will be no difference.
"Speed" is audiofoolery as far as sound goes if the drivers are appropriate for their required frequency response.