Written by Alexei Gilchrist, updated some time ago
The continuum limit of the loaded string is derived, arriving at the one dimensional wave equation.
Level: 2, Subjects: Coupled Oscillators Waves

1 Introduction

In the article the-loaded-string we examine the motion of \(N\) small masses threaded on a string and free to move in the vertical direction so that the equations of motion for the system are \begin{equation}\label{eq:eqnofmotion} \ddot{y}_n = -\frac{T}{ma}(y_{n-1}-2y_n+y_{n+1}). \end{equation}

The loaded string: \(N\) equal masses threaded on a string and free to oscillate in one dimension.

The normal mode frequencies turn out to be \begin{equation*} \omega_j = 2\sqrt{\frac{T}{ma}}\sin\left(\frac{j \pi}{2(N+1)}\right), \end{equation*} where \(j\in 1,2,\ldots,N\) labels the normal mode. For each normal mode \(j\), the motion of the \(n^\mathrm{th}\) mass is described by \begin{equation*} y_n^{(j)}(t) = d_{j} \sin\left(\frac{n j \pi}{N+1}\right)\cos(\omega_j t + \phi_j). \end{equation*} In general, the position of the \(n^\mathrm{th}\) mass can be described as a superposition of all the normal modes: \begin{equation*} y_n(t) = \sum_{j=1}^{N} d_{j} \sin\left(\frac{n j \pi}{N+1}\right)\cos(\omega_j t + \phi_j), \end{equation*} and the constants \(d_{j}\) and \(\phi_j\) are determined by the initial conditions.

In this article we will look at the continuum limit of this system.

2 Continuum limit

Imagine that we add more and more masses to the string at the same time as we make them lighter and lighter so that the total mass remains constant. Eventually the string will appear to be a continuous object with the mass distributed evenly throughout its length. In other words it will look like, well, a string! To see what out model predicts we start by rewriting the equation of motion \eqref{eq:eqnofmotion} in a more suggestive form: \begin{equation*} \ddot{y}_n(t) = -\frac{T}{m}\left(\frac{y_{n+1}(t)-y_n(t)}{a}-\frac{y_n(t)-y_{n-1}(t)}{a}\right). \end{equation*}

To make the limit to the continuous case we make the following changes:

  1. As we add more and more masses to the string the spacing will become smaller and smaller so replace \(a\) with \(\Delta x\), and we are considering the limit as \(\Delta x\rightarrow 0\).
  2. To keep the total mass \(M\) constant, we can write each little mass in terms of a linear density \(\rho = M/L\) where \(L\) is the total length (\(L=a(N+1)\)), and replace \(m=\rho\Delta x\).
  3. The position index \(n\) will become a continuous variable \(x=an\) and \(n+1\) will become \(x+\Delta x\), \(n-1\) will become \(x-\Delta x\).
  4. The height \(y_n(t)\), which was indexed by \(n\), now becomes a function of both \(x\) and \(t\): \(y(x,t)\).
  5. Differentials become partial differentials as \(y\) is now a function of two variables.
These changes lead to \begin{equation}\label{eq:eqnofmotion3} \frac{\partial^2 y(x,t)}{\partial t^2} = -\frac{T}{\rho\Delta x}\left(\frac{y(x+\Delta x,t)-y(x,t)}{\Delta x}-\frac{y(x,t)-y(x-\Delta x,t)}{\Delta x}\right). \end{equation}

The we can expand the functions \(y(x+\Delta x,t)\) and \(y(x-\Delta x,t)\) to second order in a the Taylor series: \begin{align*} y(x+\Delta x,t)&\approx y(x,t) + \Delta x \frac{\partial y}{\partial x} +\frac{1}{2}(\Delta x)^2\frac{\partial^2 y}{\partial x^2} \\ y(x-\Delta x,t)&\approx y(x,t) - \Delta x \frac{\partial y}{\partial x} +\frac{1}{2}(\Delta x)^2\frac{\partial^2 y}{\partial x^2}. \end{align*} Substituting these into \eqref{eq:eqnofmotion3} leads to a drastic simplification, \begin{equation*} \frac{\partial^2 y(x,t)}{\partial t^2} = c^2\frac{\partial^2 y(x,t)}{\partial x^2}, \end{equation*} where the constant \(c=\sqrt{T/\rho}\) has units of a velocity (\(m/s\)). This is the one dimensional wave equation!

3 Solution

In a similar way we can take the continuum limit of the solution we found for the loaded string. In particular, using \(L=a(N+1)\) and \(m=a\rho\), the normal mode frequencies will transform to \begin{align*} \omega_j &= 2\sqrt{\frac{T}{ma}}\sin\left(\frac{j \pi}{2(N+1)}\right) \\ &\rightarrow \frac{2}{\Delta x}\sqrt{\frac{T}{\rho}}\sin\left(\frac{j \pi \Delta x}{2L}\right) \\ &\approx \frac{2}{\Delta x}\sqrt{\frac{T}{\rho}}\frac{j \pi \Delta x}{2L} \\ &= \frac{j \pi c}{L}. \end{align*} where \(c=\sqrt{T/\rho}\). The sine factor in the normal modes becomes \begin{equation*} \sin\left(\frac{n j \pi}{N+1}\right) = \sin\left(\frac{j \pi}{a(N+1)}na \right) \rightarrow \sin\left(\frac{j \pi}{L}x\right). \end{equation*} So for each normal mode the solution now looks like \begin{equation*} y^{(j)}(x,t) = d_j\sin\left(\frac{j \pi}{L}x\right)\cos\left(\frac{j \pi c}{L} t + \phi_j\right). \end{equation*}