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SOUND WAVES

\epsfig{file=PS/cyl_element.ps,height=1.667in}
$\textstyle \parbox{4.25in}{ Picture a \lq\lq snapshot
Now let $s$ denote the displacement of particles of the medium from their equilibrium positions. This may also differ between one end of the cylindrical element and the other: $s$ on the left vs. $s+ds$ on the right. We assume the displacements to be in the $x$ direction but very small compared to $dx$, which is itself no great shakes.1

The fractional change in volume $dV/V$ of the cylinder due to the difference between the displacements at the two ends is

\begin{displaymath}
{dV \over V} \; = \; {A(s+ds) - As \over Adx}
\; = \; {ds . . . 
 . . . r dx}
\; = \; \left(\partial s \over \partial x \right)_t
\end{displaymath} (1)

where the rightmost expression reminds us explicitly that this description is being constructed around a "snapshot" with $t$ held fixed.

Now, any elastic medium is by definition compressible but "fights back" when compressed ($dV < 0$) by exerting a pressure in the direction of increasing volume. The BULK MODULUS $B$ is a constant characterizing how hard the medium fights back - a sort of 3-dimensional analogue of the spring constant. It is defined by

\begin{displaymath}
P \; = \; - B \, {dV \over V} .
\end{displaymath} (2)

Combining Eqs. (1) and (2) gives
\begin{displaymath}
P \; = \; - B \; \left(\partial s \over \partial x \right)_t
\end{displaymath} (3)

so that the difference in pressure between the two ends is
\begin{displaymath}
dP \; = \; \left(\partial P \over \partial x \right)_t \, d . . . 
 . . .  \; \left(\partial^2 s \over \partial x^2 \right)_t \, dx .
\end{displaymath} (4)

We now use $\sum F_x = m a_x$ on the mass element, giving

\begin{displaymath}
-A dP \; = \; A B \; \left(\partial^2 s \over \partial x^2  . . . 
 . . . rho A dx \, \left(\partial^2 s \over \partial t^2 \right)_x
\end{displaymath} (5)

where we have noted that the acceleration of all the particles in the volume element (assuming $ds \ll s$) is just $a_x \equiv (\partial^2 s / \partial t^2 )_x$.

If we cancel $A dx$ out of Eq. (5), divide through by $B$ and collect terms, we get

\begin{displaymath}
\left(\partial^2 s \over \partial x^2 \right)_t \; - \;
{ . . . 
 . . . \, \left(\partial^2 s \over \partial t^2 \right)_x \; = \; 0
\end{displaymath} (6)

which the acute reader will recognize as the WAVE EQUATION in one dimension ($x$), provided
\begin{displaymath}
c \; = \; \sqrt{B \over \rho}
\end{displaymath} (7)

is the velocity of propagation.

The fact that disturbances in an elastic medium obey the WAVE EQUATION guarantees that such disturbances will propagate as simple waves with phase velocity $c$ given by Eq. (7).





Jess H. Brewer 2009-09-01