PARTIAL DIFFERENTIAL EQUATIONS
Imagine that a stretched string is vibrating.
The wave equation says that, at any position on the string, acceleration in the direction perpendicular to the string is proportional to the curvature of the string.
First, we make a wild assumption: suppose that u is a product of a function of x and a function of t:
Then the wave equation becomes
The only way for this equation to be true for allx and for all t is for both sides to be constant; that is, T’’(t)/a2T(t) = and X”(x)/X(x) =.
If < 0,T(t)and X(x) are both trig functions.*
*If = 0, they’re linear, and if > 0, they’re exponential. But let’s not worry about that now. (Explanation on next slide.)
There are three types of solutions to the equation
in slide 8.
In a stringed instrument, each end of the string is fixed; if the string has length L, then, for all t,
Since T(t) ≠ 0 for all t, X(0) = X(L) = 0; thus, X(x) could be a sum of sine terms with zeros at x = L:
Now we know two things:
This means that
Now substitute into the equation for T:
And find the solution:
The minus sign is correct. The original slide was wrong.
We now have a family of solutions for the wave equation (one for every n):
Suppose we choose an x0 and look at how the displacement varies at this point as t changes.
This is the equation of a sinusoid with frequency =
where n = 1, 2, 3…
FrettingTake it higher
So there are three ways to increase the frequency of a sound, producing a “higher” note:
With wind instruments, it’s not always true that if L is the length of the tube, u(0, t) = u(L, t) = 0.
This is true with the flute (because the pressure doesn’t change at the ends).
However, what happens with some instruments (like the sax) is that ux(0, t) = ux(L, t) = 0, which means that the X(x) functions are cosine terms rather than sines.
Still more bizarre is the behavior of the clarinet. It has boundary conditions
Go back and substitute in these boundary conditions to see what happens!
The harmonics we hear on a stringed instrument and the overtones on a wind instrument are actually produced by standing waves, which are solutions of the form
Sketch some standing waves (see demo on the web).
How are the standing waves for a clarinet different from those for a flute?
with boundary conditions u(0, t) = u(L, t) = 0.