# The Loaded String

## Alexei Gilchrist

### 1 Introduction

Imagine a system of \(N\) masses threaded on a light string with a spacing of \(a\) as pictured below. The masses are free to move up and down vertically only. The string is fixed at each end and we want to know how the system will oscillate. This situation describes \(N\) coupled oscillators, a system which we know how to solve (coupled-oscillators).

### 2 Equations of motion

To solve the system we will assume that the tension is large enough and the motion small enough so that we can treat the tension as a constant. Say \(y_n\) denotes the displacement of the \(n^\mathrm{th}\) mass. Examining that mass and its nearest neighbours (see the diagram above) we can write the total force exerted on it:

Following our usual procedure we substitute in \(N\) trial solutions of the form \(y_n(t)=A_n e^{i\omega t}\) expecting to find solutions where all the masses are moving with the same frequency (i.e. the normal modes). This yields \(N\) equations

For \(N\) not too large, this matrix can be fed into your favourite linear algebra program to obtain the eigenvalues and eigenvectors, but the solution for arbitrary \(N\) may not be so apparent. To solve the equations in general requires other techniques, like the one we will cover in the next section, and the result will be that the normal mode frequencies are

Alternatively, we could write it as a sum of a sine and cosine instead of using a phase

### 3 Deriving the solution

Lets go back to the recursion relation between the amplitudes \eqref{eq:diff-eq}. This is a *linear difference equation with constant coefficients*. In a similar process to solving differential equations, to solve this difference equation we substitute a trial solution, which for this family is of the form \(A_n=\lambda^n\):

Note that the roots will be complex if \(\beta \le 4\), that is if \(\omega \le 2\sqrt{\frac{T}{ma}}\). In this case we can go much further with the solution. The complex roots also have unit modulus in this case so we can write them as \(\lambda_1 = e^{i \theta}\) and \(\lambda_2 = e^{-i \theta}\) and the general solution now reads

The boundary conditions require \(A_0 = A_{N+1}=0\). The first boundary yields:

So we have that the amplitudes must have the form

In the above argument we wrote (for \(\beta \le 4\)) \begin{equation*} \lambda = \frac{1}{2}\left(2-\beta \pm i\sqrt{\beta(4-\beta)}\right) = e^{\pm i\theta}, \end{equation*} the imaginary parts of which give \begin{equation*} \frac{1}{2}\sqrt{\beta(4-\beta)} = \sin{\theta}. \end{equation*} After some simplification and solving for \(\omega\) we get that only specific \(\omega\) values are allowed: \begin{equation*} \omega_j = 2\sqrt{\frac{T}{ma}}\sin\left(\frac{j\pi}{2(N+1)}\right), \end{equation*} and for each \(\omega_j\) the displacements of the masses are given by \begin{equation*} y_x^{(j)}(t) = d_{j} \sin\left(\frac{n j \pi}{N+1}\right)\cos(\omega_j t). \end{equation*} An arbitrary phase could be added to this solution without changing the argument, in general this would be needed to take into account the initial conditions.

Note that once you know what the solution is, it’s possible to find nicer trial functions to substitute in \(\ldots\) but of course this is only in retrospect.

### 4 Example

If we have just seven masses linked together we will have seven normal frequencies (the two end masses represent the fixed points):