# 3.6.1 Vibration modes of a circular drum

The idealised model of a stretched membrane is closely related to the one we saw in section 3.1.1 for a string. The membrane is allowed to vibrate transversely with a (small) displacement from equilibrium given by $w(x,y,t)$. Suppose it has tension per unit length $T$, and mass per unit area $m$. To obtain the equation of motion, we can consider a small rectangular element of the membrane between positions $x$ and $x+ \delta x$ and $y$ and $y+ \delta y$. (Note that the displacement is exaggerated in the plot, for clarity.)

Using exactly the same argument as in section 3.1.1, Newton’s law for this small element requires

$$m~\delta x~\delta y \frac{\partial^2 w}{\partial t^2} \approx T~\delta y \left[ \left( \frac{\partial w}{\partial x} \right) _{x + \delta x} – \left( \frac{\partial w}{\partial x} \right) _{x} \right]$$

$$+ T~\delta x \left[ \left( \frac{\partial w}{\partial y} \right) _{y + \delta y} – \left( \frac{\partial w}{\partial y} \right) _{y} \right] .\tag{1}$$

Now as $\delta x \rightarrow 0$ and $\delta y \rightarrow 0$ the two terms in square brackets tend towards second derivatives multiplied by $\delta x$ and $\delta y$ respectively. Cancelling a factor $\delta x ~ \delta y$, the equation of motion becomes

$$m \frac{\partial^2 w}{\partial t^2} – T \left[ \dfrac{\partial^2 w}{\partial x^2} + \dfrac{\partial^2 w}{\partial y^2} \right] =0 .\tag{2}$$

We want to use this equation to find the vibration modes and natural frequencies of a circular drum. Obviously, we do not want to analyse that problem using Cartesian coordinates: we first want to transform eq. (2) into polar coordinates $(r, \theta)$. From the usual relationships $x=r \cos \theta$ and $y = r \sin \theta$ we can deduce

$$\frac{\partial r}{\partial x} = \cos \theta, \frac{\partial r}{\partial y} = \sin \theta \tag{3}$$

and

$$\frac{\partial \theta}{\partial x} = – \frac{\sin \theta}{r}, \frac{\partial \theta}{\partial y} = \frac{\cos \theta}{r} . \tag{4}$$

Now we use the chain rule: for example,

$$\frac{\partial w}{\partial x} =\frac{\partial w}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial w}{\partial \theta} \frac{\partial \theta}{\partial x}=\frac{\partial w}{\partial r} \cos \theta – \frac{\partial w}{\partial \theta} \frac{\sin \theta}{r} . \tag{5}$$

Using this approach repeatedly and then gathering up all the terms, we find after a little algebra that

$$\dfrac{\partial^2 w}{\partial x^2} + \dfrac{\partial^2 w}{\partial y^2}=\dfrac{\partial^2 w}{\partial r^2} + \dfrac{1}{r} \dfrac{\partial w}{\partial r}+ \dfrac{1}{r^2} \dfrac{\partial^2 w}{\partial \theta^2} \tag{6}$$

so that the equation of motion (2) becomes

$$m \frac{\partial^2 w}{\partial t^2} – T \left[ \dfrac{\partial^2 w}{\partial r^2} + \dfrac{1}{r} \dfrac{\partial w}{\partial r}+ \dfrac{1}{r^2} \dfrac{\partial^2 w}{\partial \theta^2} \right] =0 . \tag{7}$$

We can solve this for the vibration modes of a circular drum with a fixed edge all the way round a circle of radius $a$, by the method of separation of variables. We try to find a solution of the form $w(r,\theta,t) = f(r) g(\theta) e^{i \omega t}$. Substituting into eq. (7), we then require

$$T \left[ f^{\prime \prime} g +\frac{1}{r} f^{\prime} g+\frac{1}{r^2} f g^{\prime \prime} \right] = -m \omega^2 f g \tag{8}$$

where the prime symbol denotes differentiation. We can rearrange the terms in eq. (8) into the form

$$\frac{ r^2 f^{\prime \prime}}{f} + \frac{ r f^{\prime}}{f} + \frac{m \omega^2}{T}r^2 = – \frac{g^{\prime \prime}}{g} . \tag{9}$$

The significance of this rearrangement may not be immediately apparent, but the key feature is that everything on the left-hand side is a function of $r$ only, while the right-hand side is a function of $\theta$ only. The only way that a function of $r$ could be equal to a function of $\theta$ (for all values of $r$ and $\theta$) is if both expressions are in fact NOT functions of anything, but simply constant. So we put each side separately equal to a constant, and for reasons that will rapidly become apparent, it is convenient to call this constant $n^2$. Look first at the equation for $\theta$: the function $g$ must satisfy

$$g^{\prime \prime} = – n^2 g. \tag{10}$$

The general solution is $g=A \cos n \theta+B \sin n \theta$ with constants $A$ and $B$. Bearing in mind that $\theta$ is the polar angle, this solution only makes physical sense if $n$ is a whole number so that the solution “joins up” when you do one complete circle around the origin. So the allowed solutions have $n=0, 1, 2, 3,…$.

Now we can go back to eq. (9) and look at the corresponding equation for the radial variation $f(r)$. This must satisfy

$$r^2 f^{\prime \prime} + r f^{\prime} + \left(\frac{m \omega^2}{T} r^2 – n^2 \right) f = 0 . \tag{11}$$

This equation is called Bessel’s equation, and it arises in many problems with circular geometry. Being a second-order ordinary differential equation, it has two independent solutions (for a given value of $n$). These solutions are collectively called Bessel functions. One of these solutions always has a singularity (i.e. infinite value) at $r = 0$, and so is not relevant for describing the vibration of a complete circular drum. The second solution is the one we want. It is usually denoted $J_n(kr)$, where $k= \omega \sqrt{m/T}$. Its mathematical properties have been extensively explored, but we need not go into any details here. Bessel functions can be readily computed: they are available as library functions in scientific computation systems such as Matlab, Mathematica, Python and so on. The first few of these Bessel functions are plotted in Fig. 2.

To find vibration modes of a membrane we must now enforce the boundary condition, which is simply

$$w(a,\theta,t) =0, \mathrm{~~or~~} f(a)=0 . \tag{12}$$

The natural frequencies are thus governed by the condition

$$J_n(ka)=0 . \tag{13}$$

It is clear from the graphs above that for any value of $n$ there will be a series of possible values of $ka$ at the successive zeros of $J_n$. Each will lead to a value of natural frequency $\omega=k \sqrt{T/m}$. The vibration mode shapes are then

$$u(r,\theta)=J_n(kr) \cos n \theta \mathrm{~~~or~~~}J_n(kr) \sin n \theta \tag{14}$$

They can be arranged in a two-dimensional family, according to the value of $n$ (which determines how many nodal diameters there are) and the number of nodal circles, governed by which zero of the Bessel function is chosen. The result is sketched in Fig. 3, and the corresponding value of $ka$ is shown by each mode sketch. The frequencies can be calculated from the values of $ka$.