3.1 A discrete chain of masses

Imagine a horizontal row of $N$ identical small masses, each of mass $m$, connected to neighbours by identical springs of stiffness $\kappa$. Equilibrium spacing $a$. Let $y_n(t)$ denote the transverse displacement of the $n$-th mass from equilibrium.

The force on mass $n$ from the spring on its left is $\kappa\, (y_{n-1} - y_n)$ (pulling toward $y_{n-1}$); from the spring on its right, $\kappa\, (y_{n+1} - y_n)$. Newton’s second law gives

This is a coupled system of $N$ second-order ODEs. Without the coupling (the term in parentheses), each mass would just oscillate independently at $\omega_0 = \sqrt{\kappa/m}$. With the coupling, the masses talk to each other: a kick on one propagates to its neighbours.

A travelling pulse

Watch what happens if you set the leftmost mass moving and leave the rest at rest. The first mass pulls on the second; the second responds, and starts pulling on the third; the third responds in turn. After some time, the disturbance has reached the far end of the chain. Energy that was originally on the left has been transported, mass by mass, to the right.

The speed of that transport depends on $\kappa$, $m$, and $a$. By dimensional analysis,

(Why? The units of $\kappa/m$ are $1/\text{time}^2$; $a$ has units of length; so $a\sqrt{\kappa/m}$ has units of length/time. There is no other dimensionally consistent combination.) We will derive the exact relation — same form, prefactor $1$ — in the next lesson.

Normal modes

Because the equations are linear and the chain is uniform, the system has $N$ normal modes, each oscillating sinusoidally in time at its own frequency. For a chain with fixed ends, the $k$-th mode has spatial shape

with $k = 1, 2, \ldots, N$. The mode frequencies are

For large $N$ and small $k$, $\sin(k\pi / 2(N+1)) \approx k\pi/(2(N+1))$, so

with $k_\text{wavenumber} = k\pi / L$ for total length $L = (N+1) a$. The low modes of the discrete chain look exactly like the low modes of a continuous string. We will see in the next lesson that this is no accident.

The continuum limit

Now imagine taking $N \to \infty$ and $a \to 0$ while keeping the total length $L = N a$ fixed and the linear mass density $\mu = m/a$ fixed. The chain becomes a continuous string. The discrete second difference $y_{n+1} - 2 y_n + y_{n-1}$ approaches $a^2 \partial^2 y / \partial x^2$. The discrete equation of motion becomes the wave equation — the topic of the next lesson.