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%TCIDATA{Created=Mon Sep 01 15:10:12 1997}
%TCIDATA{LastRevised=Fri Sep 05 11:42:25 1997}

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\begin{document}

\SetTitle{Transverse waves in one dimension}
\SetAuthor{T. Curtright}
\Setdate{4 September 1997}
\TitlePage{}

In this lecture, we will discuss the physics of a very idealized stretched
rubber band, or string, under tension.

We suppose the string is stretched between supports at $x=0$ and $x=L$, to
be definite, with a straight line static equilibrium configuration along the 
$x$-axis. Now, let us suppose the string is vibrating with each point on the
string displaced perpendicular to the $x$-axis in the $y$ direction, i.e.
the vibrations are ``transverse'' to the equilibrium configuration. Since
the displacement from equilibrium varies from point to point on the string,
we have $y(x)$, a function of $x$. In addition, the transverse displacements
will change with time as the string vibrates, so we also have $y(t)$, a
function of $t$. We may display both $x$ and $t$ dependence simultaneously
by writing $y(x,t)$. This indicates that the function $y$ depends on the two
variables $x$ and $t$.

So, what is the kinetic energy of the string? We assume that it is under a
constant tension $F$ and has a constant mass per unit length $\mu $. Then
the total kinetic energy is the sum of the kinetic energies of each small
(``infinitesimal'') piece of the string of length $dx$. This piece has a
mass $dm=\mu \,dx$, and a velocity $v_{y}=\partial y/\partial t$. Thus the
piece has a kinetic energy $dK=\frac{1}{2}\,dm\;\left( \partial y/\partial
t\right) ^{2}$. Summing the kinetic energies for each such piece of the
string, i.e. integrating $\int dx$ over the length of the string, we obtain
the total kinetic energy. 
\begin{equation}
K=\int_{x=0}^{x=L}\;dK=\int_{x=0}^{x=L}\frac{1}{2}\,\mu \,dx\;\left( \frac{%
\partial y}{\partial t}\right) ^{2}\;.
\end{equation}

The potential energy for the distorted, vibrating string is equal to the
amount of work required to stretch the string beyond it's equilibrium length 
$L$. This is just 
\begin{equation}
U=\int_{x=0}^{x=L}F\,\sqrt{1+\left( \frac{\partial y}{\partial x}\right) ^{2}%
}\,dx\;-FL\;.
\end{equation}
As long as the slope is small, we may approximate the first term as\footnote{%
If you don't believe this, check out the following plot of $\sqrt{1+z}$ and $%
1+z/2$ versus $z$.
\par
\[
\sqrt{1+z},\;1+z/2\FRAME{itbpF}{3in}{2.0003in}{-0.0087in}{}{}{}{%
\special{language "Scientific Word";type "MAPLEPLOT";width 3in;height
2.0003in;depth -0.0087in;display "USEDEF";plot_snapshots TRUE;function
\TEXUX{$\sqrt{1+z},\;1+z/2$};linecolor "black";linestyle 1;linethickness
1;pointstyle "point";xmin "-0.490309";xmax "0.509071";xviewmin
"-0.4903";xviewmax "0.5091";yviewmin "0.7031";yviewmax
"1.266";rangeset"X";phi 45;theta 45;plottype 4;numpoints 49;axesstyle
"normal";xis \TEXUX{z};var1name \TEXUX{$x$};valid_file "T";tempfilename
'C:/SWP25/docs/EG1NET01.wmf';tempfile-properties "XP";}}
\]
For small values of $z$, the two curves are indistinguishable. (Exactly what
you mean by ``small'' depends on how accurate you want the approximation to
be!)}
\begin{equation}
\sqrt{1+\left( \frac{\partial y}{\partial x}\right) ^{2}}\approx 1+\frac{1}{2%
}\left( \frac{\partial y}{\partial x}\right) ^{2}\;.
\end{equation}
So within this ``small slope'' approximation, we have 
\begin{equation}
U=\int_{x=0}^{x=L}\frac{1}{2}\,F\,\left( \frac{\partial y}{\partial x}%
\right) ^{2}\,dx\;.
\end{equation}
The total energy for the string is therefore 
\begin{eqnarray}
E &=&K+U \\
&=&\frac{1}{2}\,\,\int_{x=0}^{x=L}\left( \mu \,\left( \frac{\partial y}{%
\partial t}\right) ^{2}+\,F\,\left( \frac{\partial y}{\partial x}\right)
^{2}\right) dx\;.
\end{eqnarray}

\end{document}