Vibrations of a circular drum

The vibrations of an idealized circular drum, essentially an elastic membrane of uniform thickness attached to a rigid circular frame, are solutions of the wave equation with zero boundary conditions.

There exists infinitely many ways in which a drum can vibrate, depending on the shape of the drum at some initial time and the rate of change of the shape of the drum at the initial time. Using separation of variables, it is possible to find a collection of "simple" vibration modes, and it can be proved that any arbitrarily complex vibration of a drum can be decomposed as a series of the simpler vibrations (analogous to the Fourier series).

Contents

Motivation

Analyzing the vibrating drum problem explains percussion instruments such as drums and timpani. However, there is also a biological application in the working of the eardrum. From an educational point of view the modes of a two-dimensional object are a convenient way to visually demonstrate the meaning of modes, nodes, antinodes and even quantum numbers. These concepts are important to the understanding of the structure of the atom.

The problem

Consider an open disk \Omega of radius a centered at the origin, which will represent the "still" drum shape. At any time t, the height of the drum shape at a point (x, y) in \Omega measured from the "still" drum shape will be denoted by u(x, y, t), which can take both positive and negative values. Let \partial \Omega denote the boundary of \Omega, that is, the circle of radius a centered at the origin, which represents the rigid frame to which the drum is attached.

The mathematical equation that governs the vibration of the drum is the wave equation with zero boundary conditions,

\frac{\partial^2 u}{\partial t^2} = c^2 \left(\frac{\partial^2 u}{\partial x^2}%2B\frac{\partial^2 u}{\partial y^2}\right) \text{ for }(x, y) \in \Omega \,
u = 0\text{ on }\partial \Omega.\,

Here, c is a positive constant, which gives the "speed" of vibration.

Due to the circular geometry of \Omega, it will be convenient to use cylindrical coordinates, (r, \theta, t). Then, the above equations are written as

\frac{\partial^2 u}{\partial t^2} = c^2 \left(\frac{\partial^2 u}{\partial r^2}%2B\frac {1}{r}\frac{\partial u}{\partial r}%2B\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2}\right) \text{ for } 0 \le r < a, 0 \le \theta \le 2\pi\,
u = 0\text{ for } r=a.\,

The radially symmetric case

We will first study the possible modes of vibration of a circular drum that are radially symmetric. Then, the function u does not depend on the angle \theta, and the wave equation simplifies to

\frac{\partial^2 u}{\partial t^2} = c^2 \left(\frac{\partial^2 u}{\partial r^2}%2B\frac {1}{r}\frac{\partial u}{\partial r}\right) .

We will look for solutions in separated variables, u(r, t) = R(r)T(t). Substituting this in the equation above and dividing both sides by c^2R(r)T(t) yields

\frac{T''(t)}{c^2T(t)} = \frac{1}{R(r)}\left(R''(r) %2B \frac{1}{r}R'(r)\right).

The left-hand side of this equality does not depend on r, and the right-hand side does not depend on t, it follows that both sides must equal to some constant K. We get separate equations for T(t) and R(r):

T''(t) = Kc^2T(t) \,
rR''(r)%2BR'(r)-KrR(r)=0.\,

The equation for T(t) has solutions which exponentially grow or decay for K>0, are linear or constant for K=0, and are periodic for K<0. Physically it is expected that a solution to the problem of a vibrating drum will be oscillatory in time, and this leaves only the third case, K<0, when K=-\lambda^2. Then, T(t) is a linear combination of sine and cosine functions,

T(t)=A\cos c\lambda t %2B B\sin c \lambda t.\,

Turning to the equation for R(r), with the observation that K=-\lambda^2, all solutions of this second-order differential equation are a linear combination of Bessel functions of order 0,

R(r) = c_1 J_0(\lambda r)%2B c_2 Y_0(\lambda r).\,

The Bessel function Y_0 is unbounded for r\to 0, which results in an unphysical solution to the vibrating drum problem, so the constant c_2 must be null. We will also assume c_1=1, as otherwise this constant can be absorbed later into the constants A and B coming from T(t). It follows that

R(r) = J_0(\lambda r).

The requirement that height u be zero on the boundary of the drum results in the condition

R(a) = J_0(\lambda a) = 0.

The Bessel function J_0 has an infinite number of positive roots,

0< \alpha_{01} < \alpha_{02} < \cdots

We get that \lambda a=\alpha_{0n}, for n=1, 2, \dots, so

R(r) = J_0\left(\frac{\alpha_{0n}}{a}r\right).

Therefore, the radially symmetric solutions u of the vibrating drum problem that can be represented in separated variables are

u_{0n}(r, t) = \left(A\cos c\lambda_{0n} t %2B B\sin c\lambda_{0n} t\right)J_0\left(\lambda_{0n} r\right)\text{ for }n=1, 2, \dots, \,

where \lambda_{0n} = \alpha_{0n}/a.

The general case

The general case, when u can also depend on the angle \theta, is treated similarly. We assume a solution in separated variables,

u(r, \theta, t) = R(r)\Theta(\theta)T(t).\,

Substituting this into the wave equation and separating the variables, gives

\frac{T''(t)}{c^2T(t)} = \frac{R''(r)}{R(r)}%2B\frac{R'(r)}{rR(r)} %2B \frac{\Theta''(\theta)}{r^2\Theta(\theta)}=K

where K is a constant. As before, from the equation for T(t) it follows that K=-\lambda^2 with \lambda>0 and

T(t)=A\cos c\lambda t %2B B\sin c \lambda t.\,

From the equation

\frac{R''(r)}{R(r)}%2B\frac{R'(r)}{rR(r)} %2B \frac{\Theta''(\theta)}{r^2\Theta(\theta)}=-\lambda^2

we obtain, by multiplying both sides by r^2 and separating variables, that

\lambda^2r^2%2B\frac{r^2R''(r)}{R(r)}%2B\frac{rR'(r)}{R(r)}=L

and

-\frac{\Theta''(\theta)}{\Theta(\theta)}=L,

for some constant L. Since \Theta(\theta) is periodic, with period 2\pi, \theta being an angular variable, it follows that

\Theta(\theta)=C\cos m\theta %2B D \sin m\theta,\,

where m=0, 1, \dots and C and D are some constants. This also implies -L=m^2.

Going back to the equation for R(r), its solution is a linear combination of Bessel functions J_m and Y_m. With a similar argument as in the previous section, we arrive at

R(r) = J_m(\lambda_{mn}r),\, m=0, 1, \dots, n=1, 2, \dots,

where \lambda_{mn}=\alpha_{mn}/a, with \alpha_{mn} the n-th positive root of J_m.

We showed that all solutions in separated variables of the vibrating drum problem are of the form

u_{mn}(r, \theta, t) = \left(A\cos c\lambda_{mn} t %2B B\sin c\lambda_{mn} t\right)J_m\left(\lambda_{mn} r\right)(C\cos m\theta %2B D \sin m\theta)

for m=0, 1, \dots, n=1, 2, \dots

Animations of several vibration modes

A number of modes are shown below together with their quantum numbers. The analogous wave functions of the hydrogen atom are also indicated as well as the associated angular frequency \omega=\lambda_{mn}.

See also

References