Apparently, a developing theme of this blog is that I don’t like using Newtonian methods for continuum mechanics problems.
Let’s now derive the string equation. Wikipedia, say, does it with vector diagrams (this source uses the same derivation, too). We’ll do it with a method that’s closer to what we’ve been using for other problems.
Here’s the model:
We divide the string into a bunch of springs.
The tension in the string is represented by the springs. The mass of the string we’ll treat as being point masses between the springs.
(Note: There is a very high probability that I will mix up the words “spring” and “string” in this post)
The (linear mass) density of the string is rho, so the total mass of the string is rho times L, so the mass in one differential bit is rho-L / n.
We assume the string is tensioned even if it is completely flat between the two ends. This means the string is not at its natural length, but is delta longer. By Hooke’s law, we can find the (equilibrium) tension in the string:
We’ll say that k is the spring constant for the entire string, so the spring constant for the little springs is k-n. Delta is the deviation of the entire string from its natural length, so delta/n is the share for each individual spring.
We find the additional distortion by using the diagram and the Pythagorean theorem.
This means that the potential energy in a spring (when the string is vibrating) is
The kinetic energy of the point masses is easier to find:
The total Lagrangian (we’ll call it F since we’ve already used L to mean the length of the string) is the sum of all kinetic energy terms minus the sum of all potential energy terms.
Then the Lagrange equation for the point mass at yi is
When we plug the Lagrangian into that, the kinetic energy contributes one term due to the partial-with-respect-to-ydot term. However, the potential energy contributes two terms to the partial-with-respect-to-y term. This is because each point has two springs connected to it, one on the left and one on the right.
So when we plug in, we get
This looks fairly complicated, but we can re-write it a little. First, L/n is dx. We can treat the squareroot term as dL (the differential length). Divide the whole thing by dx and we get
One thing we did when we rewrote that was assume that both dL’s (between i and i-1, and i and i+1) were equal. This is, of course, not necessarily true. However, we again (similar to when we derived the Euler-Bernoulli beam equation) are assuming that the angles here are small. In other words, more like
In most situations with vibrating strings that we’d care about (eg., a vibrating guitar string), this is probably the case.
If angles are small, then dy will generally be small compared to dx, and dL will be approximately dx. dx is a constant (L/n), so this justifies us considering both dL’s to be equal. We can substitute all dL = dx in that equation, giving
kn times Delta/n simply equals k Delta. We earlier showed that this is the tension in the string. I think this is interesting: A standard derivation of the string equation doesn’t even involve the string’s “spring constant”, but we show here that even if you include it, you can replace it with the spring tension instead.
The y terms are a second order difference.
There is one problem with the math so far: There was a sign error. Sign errors seem to be extremely easy to make. I probably “should” go back and edit it out and cover it up, but maybe today I’ll use it as an opportunity to find it.
The error was made after the Euler-Lagrange differential equation was quoted. There is a minus sign on the partial with respect to yi term, but there is ALSO a minus sign on the potential energy term in the Lagrangian anyway, and they cancel out. So the next equation should be rho-y-doubledot plus the derivative of the potential energy with respect to yi.
So what we really have at this point is
Due to the y’s being a second order difference, as n tends to infinity / dx tends to zero, this tends to
This is the string equation (and also a type of wave equation). I suppose we can go the extra distance and solve it, too.
The standard method (and from here on out this will be textbook; we are using no personal tricks here) is to assume that y is a function that is a product of one function that depends only on time, and another that depends only on position.
If we plug that into the string equation and divide by y we get
The left side of the equation depends only on t while the right side depends only on x. That means you could, for example, hold x constant and vary t. The right side cannot change, which means the left side cannot, either, and the conclusion is that both sides must equal a constant.
We don’t know what the constant is (yet), but we’ll cheat a little (by anticipating) and say it’s a negative number.
If you’ve solved enough differential equations, it becomes “obvious” that the solution is
The boundary conditions are that the string must have y=0 at the left and right ends, where it is held down. If we define the coordinate system so x=0 is the left side and x=L is the right, then y(0) = 0 implies B = 0.
y(L) = 0 implies
where m is any integer.
Plugging this back into the last equation with g and we get
As before, we can solve this with a sine function. We have no way of knowing the phase (without an initial condition), but phase has minimal or even no effect on sound. The circular frequency is
So the real frequency is
Of course, usually the loudest frequency produced by a vibrating string is the fundamental, m = 1, unless you force it to vibrate at a higher harmonic.
There is one loose end here. We solved the string equation by assuming y = f(x)g(t). What if this isn’t true? It turns out the string equation is linear, so any solutions to it can be added and produce another solution. So the most general solution is a sum of solutions of the form we have gotten — this is a Fourier series of the general solution. Due to the completeness properties of the Fourier series, any valid solution can be expressed this way.