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Sound Wave Intensity

The energy density of sound waves is given by:

\begin{displaymath}
\frac{dE}{dV} = \frac{1}{2}\rho \omega^2 s^2
\end{displaymath} (164)

(again, very similar in form to the energy density of a wave on a string). However, this energy per unit volume is propagated in a single direction. It is therefore spread out so that it crosses an area, not a single point. Just how much energy an object receives therefore depends on how much area it intersects in the incoming sound wave, not just on the energy density of the sound wave itself.

For this reason the energy carried by sound waves is best measured by intensity: the energy per unit time per unit area perpendicular to the direction of wave propagation. Imagine a box with sides given by $\Delta A$ (perpendicular to the direction of the wave's propagation) and $v \Delta t$ (in the direction of the wave's propagation. All the energy in this box crosses through $\Delta A$ in time $\Delta t$. That is:

\begin{displaymath}
\Delta E = (\frac{1}{2}\rho \omega^2 s^2) \Delta A v \Delta t
\end{displaymath} (165)

or
\begin{displaymath}
I = \frac{\Delta E}{\Delta A \Delta t} = \frac{1}{2}\rho \omega^2 s^2 v
\end{displaymath} (166)

which looks very much like the power carried by a wave on a string. In the case of a plane wave propagating down a narrow tube, it is very similar - the power of the wave is the intensity times the tube's cross section.

However, consider a spherical wave. For a spherical wave, the intensity looks something like:

\begin{displaymath}
I(r,t) = \frac{1}{2}\rho \omega^2 \frac{s_0^2 \sin^2(kr - \omega
t)}{r^2} v
\end{displaymath} (167)

which can be written as:
\begin{displaymath}
I(r,t) = frac{P}{r^2}
\end{displaymath} (168)

where $P$ is the total power in the wave.

This makes sense from the point of view of energy conservation and symmetry. If a source emits a power $P$, that energy has to cross each successive spherical surface that surrounds the source. Those surfaces have an area that varies like $A = 4\pi r^2$. A surface at $r = 2 r_0$ has 4 times the area of one at $r = r_0$, but the same total power has to go through both surfaces. Consequently, the intensity at the $r = 2 r_0$ surface has to be $1/4$ the intensity at the $r = r_0$ surface.

It is important to remember this argument, simple as it is. Think back to Newton's law of gravitation. Remember that gravitational field diminishes as $1/r^2$ with the distance from the source. Electrostatic field also diminishes as $1/r^2$. There seems to be a shared connection between symmetric propagation and spherical geometry; this will form the basis for Gauss's Law in electrostatics and much beautiful math.


next up previous contents
Next: Doppler Shift Up: Sound Previous: Sound Wave Solutions   Contents
Robert G. Brown 2004-04-12