2D Vibrations of a Membrane

The dispersion relation (6.3) can also be obtained by using separation of variables.

Example 6. .1Consider the vibrations of a square membrane inside the region $0 \le x \le 1$ and $0 \le y \le 1$. The wave equation is

$${\partial^{2} z\over \partial t^{2}} = C^{2}\left ({\partia... ...\partial x^{2}} + {\partial^{2}z\over \partial y^{2}}\right ),$$

describing the displacement, $z(x,y,t)$, of the membrane. Suppose the boundary conditions are taken as $z=0$ on the sides of the square $x=0$, $x=1$ and $y=0$,$y=1$. Try a solution of the form

$$z = X(x)Y(y)T(t),$$

so that the wave equation may be expressed in the separable form as

$${1\over C^{2}}{1\over T}{d^{2}T\over dt^{2}} = {1\over X}{d^{2}X\over dx^{2}} + {1\over Y}{d^{2}Y\over dy^{2}}.$$

Since the left hand side of the equation is only a function of time and the right hand side depends only on spatial variations, they must both be equal to a constant. This separation process can be continued to show that both terms on the right hand side are constants. Hence,

$${1\over X}{d^{2}X\over dx^{2}} = - a^{2}, \quad \Rightarrow... ...}} + a^{2}X = 0, \quad \Rightarrow \quad X(x) = \sin m\pi x,$$

where the separation constant is chsoen as

$$a = m\pi,$$

in order to satisfy the boundary conditions in $x$. Here $m$ is an integer. Similarly,

$${1\over Y}{d^{2}Y\over dy^{2}} = - b^{2}, \quad \Rightarrow... ...}} + b^{2}Y = 0, \quad \Rightarrow \quad Y(y) = \sin n\pi y,$$

where

$$b = n\pi,$$

and $n$ is an integer so that the boundary conditions in $y$ are satisfied. Combining these results we obtain the equation for $T$ as

$${1\over T}{d^{2}T\over dt^{2}} = - C^{2}(a^{2} + b^{2}), \q... ...tarrow \quad {d^{2}T\over dt^{2}} + C^{2}(a^{2} + b^{2})T = 0.$$

Thus, we have

$$T(t) = A\cos \omega t + B\sin \omega t,$$

where the frequency $\omega$ satisfies the dispersion relation

$$\omega^{2} = \omega^{2}_{mn} = C^{2}(m^{2} + n^{2}) \pi^{2}.$$

Thus, the general solution is obtained by adding together all the possible solution to give

$$z = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\sin (m\pi x)\sin... ...ft [A_{mn}\cos \omega_{mn}t + B_{mn}\sin \omega_{mn}t\right ].$$

The constants $A_{mn}$ and $B_{mn}$ are determined by initial conditions on $z$ and $\partial z/\partial t$. Each value of $m$ and $n$ corresponds to a different possible mode of oscillations. The fundamental mode is given by $m=1$ and $n=1$ with

$$z = \sin (\pi x) \sin (\pi y)\left (A_{11}\cos \omega_{11}t + B_{11}\sin \omega_{11}t\right ),$$

and has

$$\omega_{11} = \sqrt{2}\pi C.$$

It corresponds to the whole membrane oscillating up and down. The mode $m=2$ and $n=1$ has a spatial form given by

$$\sin (2\pi x) \sin (\pi y),$$

and has

$$\omega_{21} = \sqrt{5}\pi C.$$

These two normal modes are illustrated in Figure 6.1

Figure 6.1: The fundamental mode and the harmonics in the $x$-direction.

Note that we could have obtained exactly the same result by assuming the solution could be expressed, in complex form, as

$$e^{i(kx + ly - \omega t)}.$$

This is, of course, a separable solution since

$$e^{i(kx + ly - \omega t)} = e^{ikx}e^{ily}e^{-i\omega t}.$$

Expressing this in terms of real and imaginary parts gives

$$(\cos kx + i \sin kx)(\cos ly + i \sin ly)(\cos \omega t - i\sin \omega t),$$

and so one solution is

$$z = A \sin kx \sin ly \cos \omega t,$$

where

$$\omega^{2} = C^{2}(k^{2} + l^{2}),$$

$A$ is a constant and $k = m\pi$ and $l=n\pi$ in order to satisfy the boundary conditions in $x$ and $y$.