# Why is it a bad approximation summing accelerations of small loudspeaker-cones for determining sound pressure caused by a vibrating plate?

In a previous answer, it was explained that a loudspeaker creates sound pressure which is proportional to its cone acceleration.

Based on this, I got the idea that calculating sound pressure from a vibrating plate can be approximated well by using a lot of little loudspeakers as the plate. For example, I subdivide the plate into 1000 small elements, measure the acceleration of each of them: $a_i(t)$, where $i$ is the index of the little element, then I do a sum: $$p(t) = \frac{\rho S_D}{2 \pi r} \, \sum_{i=1}^{1000} a_i(t)$$

to determine $p(t)$, the sound pressure at a distant point.

But jcandy told me (the one who answered my previous question) this formula is not suitable for doing such a thing like this.

Why is it?

How can I approximate a the sound pressure from a vibrating plate at a distant point?

• How distant? Close to your plate (a distances small compared to plate dimensions), the sound pressure does not depend on distance. – Pieter May 11 '18 at 19:26
• @Pieter: By distant I meant that the distance is much larger than the plate size ("far field", if that's the correct acoustics term, I've just started learning acoustics) – geza May 11 '18 at 22:54

Actually, the idea of summing acceleration as you suggest is on the right track, but its not directly suitable. Let me outline a general procedure that is conceptually simple and similar to what you propose; namely, the Method of Fundamental Solutions (MFS). The text below is actually a summary of the method presented in this AES article.

The pressure for a radiating system satisfies the Helmholtz equation (assuming a uniform, non-viscous medium at rest in the absence of sound). In the region exterior to all sources, the pressure satisfies the time-dependent wave equation

\begin{equation} \left( \nabla^2 - \frac{1}{c_s^2} \frac{\partial^2}{\partial t^2} \right) p({\bf x},t) = 0 \; , \end{equation}

where the velocity of sound, $c_s$, is given by $c_s = 1/\sqrt{\rho \kappa}$, with $\rho$ the density, and $\kappa$ the adiabatic compressibility of air (see Morse and Ingard). For oscillations at an angular frequency $\omega= 2 \pi f$, the pressure is

\begin{equation} p({\bf x},t) = e^{-i\omega t} p_\omega({\bf x}) \; . \end{equation}

The function $p_\omega$ is then determined by the solution of

\begin{equation} \left( \nabla^2 + k^2 \right) p_\omega({\bf x}) = 0 \; , \label{eq.helmholtz} \end{equation}

where $k = \omega/c_s$ is the wavenumber. In the case of a loudspeaker enclosure, for example, the velocity on each speaker panel is zero, except at the location of radiating elements (woofer, tweeter). On a radiating element, the boundary condition is

\begin{equation} -\rho \frac{\partial \mathbf{u}}{\partial t} = \nabla p \quad\leftrightarrow\quad i \omega v_n({\bf x}) = \frac{\partial p_\omega}{\partial n} \; , \end{equation}

with $v_n$ the velocity dependence of the radiator. More specifically, if the radiator is a rigid flat piston (circular, square, oval, whatever), then the velocity profile is

\begin{equation} v_n = \left\{ \begin{array}{cc} V & \mbox{on piston surface} \; , \\ 0 & \mbox{off piston surface} \; . \end{array} \right. \; \end{equation}

To solve the Helmholtz equation subject to the boundary conditions, we construct a solution by summing Green's functions for the exterior Helmholtz problem

\begin{equation} G_j({\bf x}) = \frac{e^{i k R_j({\bf x})}}{4 \pi R_j({\bf x})} \; , \quad\mbox{with}\quad R_j = \left| {\bf x}-{\bf x}_j \right| \; . \label{eq.greenj} \end{equation}

$G_j$ is equivalently an acoustic point source, and solves the Helmholtz equation everywhere except at ${\bf x} = {\bf x}_j$, where $G_j$ is singular. But as long as we keep the point sources outside the region of interest there is no problem. Now expand the pressure due to $n_s$ point sources as

\begin{equation} \frac{p_\omega({\bf x})}{p_0} = a \sum_{j=1}^{n_s} c_j G_j({\bf x}) \; , \end{equation}

where $p_0$ is the normalizing pressure

\begin{equation} p_0 \doteq i k a \rho c_s V \; . \end{equation}

Above, $a$ is a normalizing length that is taken to be the radius of the radiating piston, and $V$ is a normalizing velocity that is taken to be the uniform piston velocity, $V$. Different normalization can be taken to suit the specifics of the problem.

Thus, instead of summing accelerations, we sum pressures due to point sources. The coefficients $c_j$ in the summation are chosen to satisfy the boundary conditions on all surfaces.

• Thank you very much again, jcandy! This is a lot of material to process (being a newbie to acoustics), I'll be busy for a while to completely understand this. I've never thought that acoustics is that complex... – geza May 11 '18 at 22:41
• I agree that it's probably a lot to digest. Unless you want a very generic far-field, low frequency result, or have a special geometry that is already solved, you'll need an advanced method. The traditional approaches are FEM and BEM -- and these are very complicated to implement. I can recommend Kinsler's textbook for the relevant background material. – jcandy May 12 '18 at 0:10

One reason the summation method might not be valid in this application is as follows:

If it is important for you to know the effect of the driving frequency on the radiated sound pressure, then you cannot approximate (for example) the behavior of a single 18" diameter loudspeaker by considering the behavior of a close-packed array of 1" diameter loudspeakers. This is because their frequency responses are completely different; the 18" will have a peak in its radiation resistance somewhere around 500Hz while the 1" speaker will have a radiation resistance peak at about 12kHz. Their mechanical resonances will also be completely different as well.

There may be other reasons why this approximation technique isn't strictly valid and I invite experts in the field to weigh in.

• Thanks for the answer, actually this is useful for me as well, as I'll look into what causes the phenomenon you mentioned exactly. – geza May 11 '18 at 22:43