In an answer, there is this equation:
$$p = \frac{\rho S_D}{2 \pi r} \, a$$
This describes that the sound pressure ($p$) is proportional to the acceleration ($a$) of the cone of a loudspeaker.
($\rho$ is the density of air, $S_D$ is the surface area of the cone, and $r$ is the distance from the cone)
I wonder, where does this equation come from? What's the theory behind it?
Answer
To derive this result, there are various possible starting points. Probably the most rigorous approach is to start from the pressure due to a baffled rigid piston. In this case, at a distance $r$ along the axis of the piston we have
$$p(r,t) = \rho c \left( 1-e^{-ik\phi} \right) v(r,t) \; , $$
where $\rho$ is the density of air, $c$ is the speed of sound,
$$ \phi \doteq \sqrt{r^2+r_0^2} - r \quad\text{and}\quad v(r,t) = v_0 e^{i(\omega t-kr)} \; . $$
Above, $v$ is the (oscillating) piston velocity and $r_0$ is the piston radius. This textbook result can be found, for example, in Section 7.4 of Kinsler (4th Ed). Note that the motion is assumed to be harmonic, such that $\omega$ is the oscillation frequency and $k$ is the wavenumber such that $\omega = ck$. In the far-field, for which $\phi \ll 1$, this reduces to
$$ p(t) = i \rho c k \phi \, v(t) \; . $$
The presence of the $i$ shows that the velocity is out of phase with the pressure. Since the motion is harmonic, we can rewrite this in terms of the acceleration using $a = \partial_t v = i \omega v$. Then, the complex pressure becomes
$$ p(t) = \rho \phi \, a(t) \; . $$
Taking $r \gg r_0$ gives $\phi \sim r_0^2/(2r)$, or
$$p(t) = \frac{\rho r_0^2}{2r} \, a(t) \; .$$
The area of the cone is $S_D=\pi r_0^2$, so we are left with
$$p(t) = \frac{\rho S_D}{2\pi r} \, a(t) \; .$$
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