2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ How distant? Close to your plate (a distances small compared to plate dimensions), the sound pressure does not depend on distance. $\endgroup$
    – user137289
    Commented May 11, 2018 at 19:26
  • $\begingroup$ @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) $\endgroup$
    – geza
    Commented May 11, 2018 at 22:54

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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... $\endgroup$
    – geza
    Commented May 11, 2018 at 22:41
  • $\begingroup$ 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. $\endgroup$
    – jcandy
    Commented May 12, 2018 at 0:10
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer, actually this is useful for me as well, as I'll look into what causes the phenomenon you mentioned exactly. $\endgroup$
    – geza
    Commented May 11, 2018 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.