Week 10

This week we began our study of Fourier Analysis and Fourier Series. Our system of study for this discussion will be a string on which the wave velocity is $v$ , and which has fixed endpoints at x=0 and x=L where x is a position along the length of the string.

Recall that the wave equation is

$\frac{\partial^2 y(x,t)}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y(x,t)}{\partial t^2}$

and that we can write any solution to this wave equation as a superposition of normal modes:

$y(x,t) = \sum_{n=1}^{\infty} A_n \cos \left[ \omega_n t - \delta_n \right] \sin \left[ \frac{n \pi x}{L} \right]$

where $\omega_n = \frac{n \pi v}{L}$ .

We would like to go from a set of initial conditions (position and velocity of every segment of the string) to the values of each $A_n$ and $\delta_n$ . We were able to do this for systems with a finite number of degrees of freedom, so we should, in principle, be able to do it in the case of systems with a very large number of degrees of freedom.

Let us first consider just a set of positions at time $t =0$. Let’s say we are given these, and they take the form $y(x,t=0) = f(x)$ , where $f(x)$ is any function with $f(0)=f(L)=0$ . These are the boundary conditions obeyed by our segment of string.

If we now apply this to the expression for y in terms of the normal modes, we find

$y(x,t=0) = f(x) = \sum_{n=1}^{\infty} A_n \cos (\delta_n) \sin \left[ \frac{n \pi x}{L} \right]$

Now for the moment, let us combine the coefficients of each $\sin$ into a set of coefficients $B_n = A_n \cos \delta_n$ , in which case we then have

$f(x) = \sum_{n=1}^{\infty} B_n \sin \left[ \frac{n \pi x}{L} \right]$

Let us now try to get the $B$ coefficients from the function $f(x)$ .

To do this, we use a trick involving integrals of sin functions:

$\frac{2}{L} \int_0^L \sin \left[ \frac{n \pi x}{L} \right] \sin \left[ \frac{m \pi x}{L} \right] dx = \left\{ \begin{array}{ll} 1 & n = m \\ 0 & n \ne m \end{array} \right.$

multiplying both sides of the equation for f(x) by $\frac{2}{L} \sin \left[ \frac{m \pi x}{L} \right]$ and integrating over $x$ from $0$ to $L$, we find

$\frac{2}{L} \int_0^L f(x) \sin \left[ \frac{m \pi x}{L} \right] = \sum_{n=1}^{\infty} B_n \cdot \frac{2}{L} \sin\left[ \frac{n \pi x}{L} \right] \sin \left[ \frac{m \pi x}{L} \right] dx = B_m$

This means that if we know the function f(x), and perform some integrals, then we can get the $B_m$ coefficients!

See this weeks Mathematica code to begin to get an idea of how this works:

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/11/Week-10.nb

We can then go a bit further. Again, the goal is to get both the $A_n$ and $\delta_n$ . In order to do this, we need not just initial position information, but also initial velocity information. Right now, we have $B_n = A_n \cos \delta_n$ , so we only have fixed a combination of $A_n$ and $\delta_n$ . Let us say we have that velocity information as a given:

$\left. \frac{\partial y(x,t)}{\partial t} \right|_{t=0} = g(x) = \sum_{n=1}^{\infty} A_n \omega_n \sin \delta_n \sin \left[ \frac{n \pi x}{L} \right]$ .

In this case, we can write $C_n = A_n \omega_n \sin \delta_n$ , and thus we can also express the velocity of each point of the string in terms of a sum over sin functions:

$g(x) = \sum_{n=1}^\infty C_n \sin \left[ \frac{n \pi x}{L} \right]$

We can then use the same trick to get:

$C_n = \frac{2}{L} \int_0^L g(x) \sin \left[ \frac{n \pi x}{L} \right]$

So now you have all of the B’s, and all of the C’s (since we presumed we know both initial position, f(x), and initial velocity, g(x)). We then have

$B_n = A_n \cos \delta_n \text{ and } C_n = A_n \omega_n \sin \delta_n$

From this, we can finally get both $A_n$ and $\delta_n$ :

$A_n = \left[ B_n^2 + \left( \frac{C_n}{\omega_n} \right)^2 \right]^{1/2} \text{ and } \delta_n = \arctan \left[ \frac{ C_n/\omega_n }{B_n} \right]$ .

I will soon be uploading a mathematica notebook that goes through an example of a particular set of initial positions and velocities, and go through how to get the A’s and $\delta$ ‘s for a specific example.

Physics 360

Vibrations and Waves

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