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Arbitrary Waveforms Propagating to the Left or Right

The first solution we can discern by noting that the wave equation equates a second derivative in time to a second derivative in space. Suppose we write the solution as $f(u)$ where $u$ is an unknown function of $x$ and $t$ and substitute it into the differential equation and use the chain rule:

\begin{displaymath}
\frac{d^2f}{du^2}(\frac{du}{dt})^2 -
v^2 \frac{d^2f}{du^2}(\frac{du}{dx})^2 = 0
\end{displaymath} (108)

or
\begin{displaymath}
\frac{d^2f}{du^2}\left\{(\frac{du}{dt})^2 -
v^2 (\frac{du}{dx})^2 \right\}= 0
\end{displaymath} (109)


\begin{displaymath}
\frac{du}{dt} = \pm v \frac{du}{dx}
\end{displaymath} (110)

with a simple solution:
\begin{displaymath}
u = x \pm vt
\end{displaymath} (111)

What this tells us is that any function

\begin{displaymath}
y(x,t) = f(x \pm vt)
\end{displaymath} (112)

satisfies the wave equation. Any shape of wave created on the string and propagating to the right or left is a solution to the wave equation, although not all of these waves will vanish at the ends of a string.


next up previous contents
Next: Harmonic Waveforms Propagating to Up: Solutions to the Wave Previous: An Important Property of   Contents
Robert G. Brown 2004-04-12