Suppose I have a vibrating membrane vibrating along some function $u(x,y,t)$ where $t$ is time and $x$ and $y$ are variable indicating a point on the membrane. What sounds are produced by this membrane. I have a vague intuition that each point on the membrane can be viewed as creating its own wave which leads me to believe that it would be an integral u with respect to x and y but I'm not sure this is correct and even if it were, I don't know how interference would affect it.

  • $\begingroup$ In an example of a circular membrane with the initial condition of no motion at the edge in two dimensions, the solution will be symmetric relative to the axis and therefore with no even harmonics. The relative intensity of the odd harmonics will depend on the material and tension of the membrane, as well as on the way you trigger the sound. The decay would be fast and the intensity of higher harmonics low. This is essentially a drum, so you know how it sounds and also differently depending on how you hit it. $\endgroup$
    – safesphere
    Aug 24, 2017 at 4:36

1 Answer 1


You can think of the membrane as an array of decoupled oscillators at points $(a,b)$ with oscillation frequency given by the $u(a,b,t)$. You could fourier transform $u(a,b,t)$ into sines and cosines to read off the frequencies being generated at that point.

But how the air responds to an oscillation is a pretty tricky subject in generality. In your head, you can assume that the sound wave frequencies generated at a point $p$ are exactly the frequencies of the oscillator at $p$, but this is a sloppy model.

Now we also want to assume the air is 'linear medium,' that is to say, any sound made by adjacent parts of the membrane only add linearly. This will make the problem much more tractable, but in reality pressure waves in different mediums can add in complicated ways.

The whole song and dance of knowing the sound at some point comes down to knowing the pressure at that point as a function of time, since sound is a wave in pressure.

A point $\vec{r}$, experiences a net pressure at a time $t$ which is the sum of the contributions from all the oscillators. If sound moves in the medium with speed $c$, then for one oscillator the pressure at $\vec{r}$ is related the motion of the oscillator at a time $c \left|\vec{r}\right|$ before $t$. This is to say that the oscillator makes a pressure wave which propagates at velocity $c$ out radially ro $\vec{r}$.

Considering again the case for a single oscillator, we can assume that the change in pressure near the oscillator due to the oscillation is proportional to the velocity of the oscillator.

$$\dot{\rho} \propto \frac{\partial }{\partial t} u(a,b,t).$$

Again, the medium is going to determine this relationship. For now, I will just assume that I can replace $\propto$ with $=$. This is similar to the assumption that the oscillation modes of the membrane are the same as those generated in the air, which is not true in generality.

Then for ambient pressure $\rho_0$, the pressure at a point $\vec{r}$ due to single oscillator is

$$ \rho(\vec{r}, t) = \rho_0 + \frac{\partial }{\partial t} u(a,b,t)|_{t=-cr}. $$

For many oscillators, the second term becomes an integral over the area of the drum head. We can also make this formula a bit more general by replacing $\frac{\partial }{\partial t} u(a,b,t)$ with $g(u,a,b,t)$, the generalized map which returns the change in pressure near the oscillator in an arbitrary medium depending on its motion.

$$ \rho(\vec{r}, t) = \rho_0 + \int \limits_\text{membrane} g(u(a,b,t))|_{t=-cr}. $$

Now actually extracting the 'sound' at $\vec{r}$ from the pressure is a hard thing to do because the of way the human ear works. But if we suppose the ear canal is a cylindrical column of length $L$, then the frequencies you hear are the fourier modes of the $\rho(\vec{r},t)$ on the interval $t-\frac{L}{c}$ to $t$.

  • $\begingroup$ thanks for the great answer! I, however, am unclear on what exactly g is. $\endgroup$
    – tox123
    Aug 25, 2017 at 0:58

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