If you are modeling a vibrating string (like a guitar or piano) using a finite difference approach as described here:
http://hplgit.github.io/num-methods-for-PDEs/doc/pub/wave/html/._wave001.html#wave:string
How does the number of mesh points in your calculation affect the number of modes that can be reproduced?
Is it a simple matter such that for the first mode, you would need 3 points (the two end points and one in the middle to oscillate)? Then for the first and second mode, you would need 5 points (adding two more points to oscillate at 1/4 way from each end)?
And so on?
If so, what's the simple equation for the number of modes that can be reproduced based on the N number of nodes in a rudimentary finite difference approximation?
Thanks.
My professor covered a similar topic in one of my courses on wave dynamics. To answer your question bluntly, to model $n$ normal modes you will need $n+2$ mesh points. Since the $n$th mode has $n$ peaks, and the two endpoints are fixed, you will definitely need (at least) $n+2$ mesh points. Perhaps this is all you wanted, but I've added a neat discrete model that you can use below for your experimentation purposes.
The following is a discretized model of the continuous wave equation on a finite string. Instead of a vibrating string, we can model the situation as $N$ small masses, each of mass $m$, all connected by springs with spring constant $k$ in a line, as shown in the diagram below.
Apologies for the poor diagram. It was done rather quickly in MSPaint. Let the displacement from equilibrium of the $n$th mass be denoted as $x_n$. It can easily be seen that for the $N-2$ central masses, their equations of motion are $$m\ddot{x}_n=-k(x_n-x_{n-1})+k(x_{n+1}-x_n)=k(x_{n-1}-2x_n+x_{n+1})$$ And for the two end masses, $$mx_1=-kx_1+k(x_2-x_1)=k(-2x_1+x_2)$$ $$mx_N=-k(x_N-x_{N-1})-kx_N=k(x_{N-1}-2x_N)$$ If we let $$\underline{x}=\begin{bmatrix} x_{1}\\ \vdots \\ x_{N} \end{bmatrix}$$ Then we can represent this system as $$\ddot{\underline{x}}=-\frac{k}{m}\mathbf{M}\underline{x}$$ Where $$\mathbf{M}=\begin{bmatrix} 2 & -1 & 0 & \dotsc & 0\\ -1 & 2 & -1 & \dotsc & 0\\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \dotsc & -1 & 2 & -1\\ 0 & \dotsc & 0 & -1 & 2 \end{bmatrix}$$ And is a tridiagonal matrix, which are reasonably well studied. It turns out that the $j$th normal mode of $x_n$, which I'll denote $x_{n,j}$ , is precisely $$x_{n,j}(t)=\sin\left(n\frac{j\pi}{N+1}\right)\cos(\omega_j t+\varphi_j)$$ Here $\varphi_j$ is some arbitrary phase angle that can be computed via initial/boundary conditions, and $\omega_j$ is the $j$th normal frequency, given by $$\omega_j=\frac{2k}{m}\left(1-\cos\left(\frac{j\pi}{N+1}\right)\right)=\frac{4k}{m}\sin^2\left(\frac{j\pi}{2(N+1)}\right)$$ I can provide some references on the above if you wish. Since each $x_n$ can be any linear combination of its normal modes, we can see that $$\underline{x}(t)=\sum_{i=1}^{N}\left(\sum_{j=1}^{\infty}a_j x_{i,j}(t)\right)\hat{\underline{e_i}}$$ The $a_j$s can be computed via Fourier series.