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RE: in a word: acceleration.

"It is not a very long post, but if you are really busy, here is a TL;DR summary:
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.
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?

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)
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)
log10(x1) + 2 log10(f1) = log10(x2) + 2 log10(f2)
x1 * f1^2 = x2 * f2^2
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
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."

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