Can You Hear the Shape of a Drum? (Part 1)

It’s pretty easy to hear the difference between a big bass drum and a small snare drum. Big drums make low sounds and small drums make high sounds.

Why do bigger drums make lower sounds? Can you tell how big a drum is by what sounds it makes?

Could you possibly even tell the shape of the drum by listening? Can you hear the shape of a drum?


In order to understand vibrating drums, we first need to understand the simpler case of vibrating strings.

Vibrating strings vibrate the air, which is what you perceive as sound. So, in order to understand the sound, we need to understand how strings vibrate.

Let’s pluck a guitar string.


When you pluck a string, it stretches the string out, like this:


When you release, it moves. After a fraction of a second, it might look like this:


The vibrations are controlled by a differential equation, i.e., an equation with derivatives in it. If the height of the string (compared to the base, still position), x away from one end, at time t is given by a function f(x,t), then this differential equation is \dfrac{d^2 f}{dt^2} = \dfrac{d^2 f}{dx^2}.1


At each t, f(x,t) gives the position of the string at each x value. For instance, f(x,0) gives the position of the string at the start, t=0. It represents the initial position of the string.

To explain the equation, \dfrac{d^2 f}{dt^2} is the vertical acceleration of the string. Meanwhile, \dfrac{d^2 f}{dx^2} measures how much the string is curving at one instant of time, as you move from left to right. The more the string is curving, the bigger this is.


In other words, the differential equation tells us that the curvature of the string controls the strings acceleration. In the above example, the string would accelerate downward quickly near the top, where \dfrac{d^2 f}{dx^2} is large (and negative), but wouldn’t accelerate much at all where \dfrac{d^2 f}{dx^2} is small. That seems reasonable.

There are infinitely many solutions to this equation. For instance, a more complicated vibrating string might look like this:2


How could we come up with particularly simple examples?

The simplest kind of vibration would be one where the string keeps its shape, but simply scales that shape up and down. In the simplest case, that would look like this:


For the string to keep its shape, its acceleration needs to be (proportional to) its current height. That way, the string accelerates back to equilibrium faster when it’s higher, and more slowly when it’s already low. In other words, we need \dfrac{d^2 f}{dt^2} = -\nu^2 f, where \nu (Greek letter nu) is some constant.3

Mathematically, if that holds, then it’s easy to check that f(x,t) = f(x,0)\cdot cos(\nu t), where f(x,0) is the position of the string when you first let go. In other words, if our string were in this special self-similar vibration, then the original shape would simply vibrate up and down sinusoidally (i.e., like a (co)sine wave), just like in the video above.

What can these special self-similar vibrations look like?

If we plug \dfrac{d^2 f}{dt^2} = -\nu^2 f back into the equation for a vibrating string, we find that we also need \dfrac{d^2 f}{dx^2}(x,0) = -\nu^2 f(x,0). We need the second derivative of the function to be a multiple of the function itself.

In other words, we need f(x,0) to be an eigenfunction of \dfrac{d^2}{dx^2}!


The first place most students see eigen-stuff, like eigenvectors and eigenvalues, is in a college linear algebra course, but the basic idea is pretty simple.

When you multiply a vector v by a matrix A you end up with a new vector A\cdot v. For instance, \begin{bmatrix}2&1\\0&2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} =  \begin{bmatrix}2\\0\end{bmatrix}.

A good way to think about this is that the matrix A transforms vectors (like v)  into a new vector, that we call A \cdot v. For most vectors, this transformed vector will have nothing much to do with the original.

But eigenvectors are special; they are the vectors that A transforms in the simplest possible way, by simply scaling them.

In terms of an equation, if you have a matrix A and a vector v so that A \cdot v happens to equal \lambda \cdot v, for some number \lambda, then v is an eigenvector of A, with eigenvalue \lambda. The eigenvalue v is the amount that A scales the eigenvector v.

For example, \begin{bmatrix}2&1\\0&2\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} = 2 \begin{bmatrix}1\\0\end{bmatrix}, and so \begin{bmatrix}1\\0\end{bmatrix} is an eigenvector of the matrix \begin{bmatrix}2&1\\0&2\end{bmatrix}, with eigenvalue 2.


When/if you take linear algebra in college,  you may not realize how important eigen-stuff is. I know that when I took it, the equation Ax=\lambda x seems like some sort of useless, gimmicky math problem of the sort some teachers love.


But it turns out that eigen-stuff is a key step in understanding any number of different ideas in mathematics and physics.

An eigenfunction is similar to an eigenvector.5 We can take a function (instead of a vector) and transform it in some way. In our string example, the transformation is taking two derivatives of the function. For most functions, this gives some function that doesn’t seem to have much to do with the original, like these two functions:


But eigenfunctions are special. They are the functions that, when you take two derivatives, you end up with the same function, perhaps scaled by a constant:


The scaling constant is the eigenvalue for that eigenfunction.

To bring it back to vibrating strings, if we want a self-similar vibrating string, we need to find eigenfunctions and eigenvalues of the transformation \dfrac{d^2}{dx^2}. Equivalently, we need to find solution functions f(x) for the equation \dfrac{d^2 f}{dx^2}(x) = -\nu^2 f(x). (So, the eigenvalue is \lambda = -\nu^2.)

It turns out the only solutions of this equation are pretty simple to write down. If \lambda is positive, the eigenfunctions are the exponential functions, f(x) = e^{\sqrt{\lambda} x} and f(x) = e^{-\sqrt{\lambda} x}.4 (For these, the \lambda =-\nu^2 we’re using isn’t convenient, but, as you’ll see, we don’t need to worry about these.)

If \lambda is negative, so we can use \lambda = -\nu^2, the eigenfunctions are the sine and cosine functions, f(x) = \sin(\nu x) and f(x) = \cos(\nu x).5

Of course, if we have a vibrating string, we have one more restriction.

Any solution should have the two endpoints fixed. Our vibrating string is clamped down at either end, and so shouldn’t be moving at either end.


That means we can easily throw out the exponential eigenfunctions. They’re never zero, so, while they may be eigenfunctions, they can’t represent a vibrating string.

The boundary condition also restricts what sinusoidal functions we could have.

If we call the left end of our string x=0, that means \cos(\nu x) can’t be a vibrating string. So we’re left with sine.

If the two ends of the string were, say, \pi apart, that means our sine wave has to be zero at x=\pi as well.6 For that to happen, we need \nu to be a natural number, like 1 or 2 or 3.


In our example, \nu is conveniently the number of humps, though that’s because we chose the length of our string to be \pi. More generally, though, the larger the \nu, the more oscillations the solution will have.

These eigenfunctions, like \sin(5x), are supposed to represent initial positions for the string. What do they look like as they begin to move?

We already said that for these special self-similar solutions, if we start at f(x,0), then the position over time is f(x,t) = f(x,0) \cos(\nu t). Thus, since our eigenfunctions \sin(\nu x) are supposed to be the initial position of the string (i..e, f(x,0)), the function f(x,t) = \sin(\nu x) \cos(\nu t) is a solution to the differential equation!

What do these solutions look like?

Here are a couple of them.

Clockwise from upper left: \nu = 1, 2, 3 and 7.

We already talked about how larger \nu mean more bumps. But you might also note that they vibrate faster as well. That’s because the \nu shows up in the \cos(\nu t) factor of the solution as well!

In fact, if t was measured in seconds, the string would vibrate once (so, go from its original position back to its original position) in 2\pi/\nu seconds. Equivalently, the frequency of the oscillation is \dfrac{\nu}{2\pi} hertz.

Remember that the vibrating string vibrates the air, and that’s what you hear. So, if the string is vibrating at, say, 261.6 hertz, you would hear the same frequency–a middle C!

These self-similar solutions are the special solutions that give pure tones.

Most solutions, though, aren’t these special, self similar ones. For instance, we could add together the \nu =1 and \nu=2 ones, and we’d get this vibration:


In this case, we’d hear two tones: the one for \nu =1 and the one for \nu =2.

The complicated example I gave at the beginning of the post was some combination of the solutions for \nu =1 through \nu = 6.


In no way is this obvious, but it turns out that every vibration of the string is really just a (perhaps infinite) summation of bits of the pure tone solutions. For instance, if you wanted the vibration for a string that started out in a triangle, you could add together a bit of each of the solutions up to \nu=20 in the right proportions, like this:


When you listen to one of these combined vibrations, there’s usually a main tone that you hear the loudest. But the secondary, usually higher frequency, tones can also be heard. They are the overtones.

Since any vibration is a combination of bits of the pure tone solutions, when you pluck the string, you hear a bit of each of the pure tones your vibration solution included.

In fact, if you just pluck the string at random, you’d probably get a bit of all the tones the string can produce.

Let’s return to our original question: can you hear the shape of a drum?

Strings, though, don’t really have shapes. As long as the string is made of the same material and is tightened to the same tension, the “shape” of a string is just its length.

So, the equivalent question for a string is: can you hear the length of a string?7


The key is the eigenvalues \lambda = -\nu^2 we found earlier. They control the frequency of the sounds the string can produce.

It’s important “all the tones the string can produce” is not the same as “all the tones.” Since every solution to our differential equation (and thus every vibration) is a combination of our pure tone solutions, the string can only produce tones corresponding to the \nu that we found, or, in other words, corresponding to a given eigenvalue \lambda.8

The spectrum (of \dfrac{d^2}{dx^2}) of the string is the set of all the eigenvalues of \dfrac{d^2}{dx^2}. In this case, that would be \{-1^2, -2^2, -3^2, \cdots\}. To summarize, if you know the spectrum of \dfrac{d^2}{dx^2} (i.e., know its eigenvalues), you know the tones the string can make. And similarly, if you know the tones, you know the spectrum.

So, can you hear the length of a string? More technically, if you can hear all the tones, and thus know the spectrum, can you deduce the length of the string?

The answer is yes, but this post is already so long that an explanation will have to wait till next time!


  1. This is a math blog more than a physics blog, so I won’t derive this equation from the physics. The equation assumes that the string isn’t stretched too much and that there’s no friction/air resistance/etc. Also, usually there would be a constant multiplying one of the sides, but let’s say that the strings we’re using are all the same thickness, material, tightness, etc., so that that constant just happens to be 1. The math isn’t any harder, just a bit more ugly. 
  2. The amount the string vibrates up and down is greatly exaggerated from reality, so that it’s easier to see what’s going on. It also vibrates much, much slower. 
  3. It seems weird I use -\nu^2 instead of just a negative number \lambda. But while it looks confusing here, it will avoid a \sqrt{-\lambda} later, so I think it’s worth it. 
  4. Technically, an eigenfunction is an eigenvector. In fact, it’s often just called an eigenvector. This is because you can think of the space of all functions as a vector space; like vectors, you can add and subtract functions, and scale them by a constant.
  5. I promise, these are the simple ones. Most eigenfunctions are so complicated you can’t write down a formula for them. 
  6. Really, these are the same thing, believe it or not. If you allow imaginary numbers and know how to take e to a complex number, then the eigenfunctions are f(x) = e^{\sqrt{\lambda} x} and f(x) = e^{-\sqrt{\lambda} x} for any complex number \lambda. It’s weird, but exponentials of imaginary things give sine waves. But that’s not important to understand for vibrating strings. 
  7. The distance of \pi was chosen just for simplicity of the coefficients. 
  8.  We aren’t playing a guitar, where we use the fret board to momentarily change the length of the string. We are only plucking it. The length of the string is fixed. 
  9. Sound is quantized! In other words, for a given string with given length, you can only produce a discrete set of tones, and not any tone you want. The quantization of energy in quantum mechanics actually comes about in the same way. The energy levels represent eigenvalues for that more complicated system. 

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s