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Modelling: Derivation of the wave equation

The wave equation has many physical applications from sound waves in air to magnetic waves in the Sun's atmosphere. However, one of the simplest systems to visualise and describe are waves on a stretched elastic string.

Initially the string is horizontal. Then we distort it by displacing it in the vertical direction and at some time, say $t=0$, we release it and the string starts to oscillate. The aim is to try and determine the vertical displacement of the string, $u(x,t)$, as a function of space, $x$, and time $t$. The typical situation is illustrated in Figure 2.1 at a typical time $t$.

To derive the wave equation we need to make some simplifying assumptions
(i) The density of the string, $\rho $, is constant so that the mass of the string between $P$ and $Q$ is simply $\rho $ times the length of the string between $P$ and $Q$, namely $\Delta s$. Thus,

\begin{displaymath}
\Delta s=\sqrt{(\Delta x)^{2} + (\Delta
u)^{2}} = \Delta x \...
...\approx \Delta x
\sqrt{1+({\partial u \over \partial x})^{2}}.
\end{displaymath}

(ii) The displacement, $u(x,t)$, and its derivatives are assumed small so that

\begin{displaymath}
\Delta s \approx \Delta x
\end{displaymath}

and the mass of the portion of the string is

\begin{displaymath}
\rho \Delta x.
\end{displaymath}

(iii) The only forces acting on this portion of the string are the tensions, $T_{1}$ at $P$ and $T_2$ at $Q$. The gravitational force is neglected.
(iv) The motions are purely vertical. There is no horizontal motion of any portion of the string.

Now we consider the forces acting on the typical string portion shown in Figure 2.1. Tension acts tangential to the string and the gradient of the tangent is, of course, the slope of $u(x,t)$ or simply ${\partial u \over \partial x}$. Now we resolve the forces into their horizontal and vertical components.

Horizontal: At $P$ the tension force is $T_{1} \cos \Theta_{1}$ and it acts to the left, whereas at $Q$ the force is $T_{2} \cos \Theta_{2}$, acting to the right. Since there is no horizontal motion, these forces must balance and so

\begin{displaymath}
T_{1} \cos \Theta_{1} = T_{2} \cos \Theta_{2} = T = {\rm const}.
\end{displaymath} (2.5)

Vertical: From Figure 2.1 it is clear that the force at $Q$ is $T_{2} \sin
\Theta_{2}$ and at $P$ is $-T_{1} \sin \Theta_{1}$. Then Newton's law of motion gives

$\displaystyle {\rm mass \times acceleration = applied \, force}$      

so that
$\displaystyle \rho \Delta x {\partial^{2} u \over \partial t^{2}} = T_{2} \sin \Theta_{2}-
T_{1} \sin \Theta_{1}$      

Figure 2.1: The string at a typical time $t$.
\includegraphics [scale=0.7]{fig1.ps}

Divide by $T$ and use (2.5) to give
$\displaystyle {\rho \Delta x \over T} {\partial^{2} u\over \partial t^{2}} = {T...
...sin \Theta_{1} \over T_{1}
\cos \Theta_{1}} = \tan \Theta_{2} - \tan \Theta_{1}$      

$\tan \Theta_{2}$ is the gradient of the tangent to $u(x,t)$ at $x=x+\Delta x$ and this is just $({\partial u \over \partial x})_{x+ \Delta x}$. In a similar manner,
$\displaystyle \tan \Theta_{1} = ({\partial u \over \partial x})_{x} .$      

Thus, we get
$\displaystyle {\rho \over T} {\partial^{2} u \over \partial t^{2}} = {1 \over \...
...ght )_{x + \Delta x} -
\left ({\partial u \over
\partial x}\right )_{x} \right]$      

Using the definition of the derivative
$\displaystyle \stackrel{\lim}{\Delta x} \rightarrow 0 \left\{ {f(x+\Delta x) - f(x) \over
\Delta x} \right\} ,$      

we let $\Delta x$ tend to zero to obtain
\begin{displaymath}
{1 \over c^{2}} {\partial^{2} u \over \partial t^{2}} = {\partial^{2} u \over
\partial x^{2}} .
\end{displaymath} (2.6)

where $c^{2} = T/ \rho$, the tension divided by the density, is the wave speed squared.


next up previous
Next: D'Alembert's solution of the Up: Partial Differential Equations of Previous: Definitions of Partial Differential
Prof. Alan Hood
2000-10-30