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?

  • $\begingroup$ Could you please also tell us what $\rho$ and $S_D$ are? $\endgroup$
    – FGSUZ
    May 10, 2018 at 22:18
  • $\begingroup$ @FGSUZ: sure, I've added them $\endgroup$
    – geza
    May 10, 2018 at 22:20

1 Answer 1


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

$$p(t) = \frac{\rho S_D}{2\pi r} \, a(t) \; .$$

  • $\begingroup$ Thank you very much for your precise answer! Does this mean, that if I want to approximate the sound pressure (for audible range) from a vibrating plate (~0.5 m size), at a distance of farther than 5 m, it is a good approximation if I take the acceleration of little plate elements, and integrate them? Like if every plate element would be a little loudspeaker. Or is there a better approximation for this? $\endgroup$
    – geza
    May 11, 2018 at 7:35
  • $\begingroup$ The formula is useful is you want to get the simplest (far-field) approximation for any vibrating element (circular, square, concave, convex). If you want to do better, then you probably want to use a specific, known result. Is your plate circular? Is it baffled? In general you would not sum accelerations; rather, you would solve the problem in such a way as to satisfy normal velocity boundary conditions (some fixed velocity V on the radiator, and V=0 on rigid surfaces). $\endgroup$
    – jcandy
    May 11, 2018 at 15:36
  • $\begingroup$ I'm dealing with guitar top-plate simulation, so in a way, it is baffled (I mean, the air is mostly closed to the body). Currently, I'm using summed accelerations, and it is kinda okay. I'd just use some alternative solution, if it is not much more complicated, so worth the effort. I'm sorry, I don't understand your last sentence (I'm completely new to acoustics). $\endgroup$
    – geza
    May 11, 2018 at 16:27
  • $\begingroup$ Much of the effort in solving the Helmholtz equation is connected with satisfying the boundary conditions on all surfaces. The normal velocity must be zero on fixed surfaces (wall, floor, loudspeaker box) and finite on moving surfaces (string, woofer, tweeter, plate). The formula above in terms of acceleration is really not suitable for this. One way to solve the problem is to expand the pressure in terms of point sources with arbitrary amplitudes, then apply the boundary conditions to determine the unknown amplitudes. To get further clarification you'd need to ask this as a question. $\endgroup$
    – jcandy
    May 11, 2018 at 17:08
  • $\begingroup$ Thanks for the information, you're a great help! I've asked this question. $\endgroup$
    – geza
    May 11, 2018 at 17:38

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