From discrete oscillators to waves

Two coupled masses gave two normal modes — one where they swing together, one where they swing apart. That was already interesting. But nature does not stop at two masses: a guitar string, a drumhead, the air in a room are all built from an enormous number of coupled bits of matter, each tugging its neighbours. What happens as you keep adding masses and springs in a line? This article follows that question from $N=2$ up to $N\to\infty$ . You will see why a chain of $N$ coupled masses has exactly $N$ normal modes, how the mode shapes fill in between "everything together" and "maximal alternation" as $N$ grows, and — the payoff — why an infinite chain of infinitesimally small, infinitesimally spaced masses stops behaving like a chain of masses at all and becomes a continuous vibrating string obeying the wave equation. It is the conceptual bridge from coupled oscillators to wave motion.

The core insight

A chain of $N$ coupled masses has $N$ normal modes ranging from "all in phase" to "maximal alternation," and letting $N\to\infty$ while the spacing shrinks to zero turns Newton's second law for the chain into the wave equation for a continuous string.

The core idea

From two modes to a comb of modes. For two coupled masses there were two normal coordinates and two frequencies, $\omega_1$ (symmetric, masses in phase) and $\omega_2$ (antisymmetric, masses out of phase). Chain $N$ identical masses $m$ in a row, each linked to its neighbours — and to fixed walls at the two ends — by identical springs of stiffness $k$ . Newton's second law for every mass gives $N$ coupled differential equations. Exactly as for $N=2$ , a change of coordinates decouples these into $N$ independent simple-harmonic equations. So the chain has exactly $N$ normal modes, each with its own frequency $\omega_n$ and its own mode shape: how far each mass moves, and in what phase, relative to its neighbours.

Mode shapes interpolate. The lowest mode has every mass moving broadly together, like the symmetric $N=2$ mode — springs barely stretched, frequency low. The highest mode has neighbouring masses alternating as fast as the discreteness allows, like the antisymmetric $N=2$ mode — every spring maximally stretched, frequency high. Modes in between mix the two: some masses nearly still, others moving, the number of direction reversals along the chain rising steadily from lowest mode to highest. As $N$ grows, more modes squeeze into the same range — two points for $N=2$ become a fine "comb" of $N$ points.

The continuum limit. Let $N\to\infty$ while the spacing $a$ between masses shrinks to zero, holding the total length $L=(N-1)a$ fixed. Let the mass per unit length $\mu = m/a$ and the stiffness-related quantity $ka$ both approach fixed, finite values — you cannot simply take $N$ large while leaving $m$ and $k$ untouched. Under this limit the discrete mass index $n$ becomes a continuous position $x$ , and the displacement $x_n(t)$ of the $n$ -th mass becomes a continuous field $y(x,t)$ : the string's transverse displacement at position $x$ and time $t$ .

\frac{x_{n+1}-2x_n+x_{n-1}}{a^2}\ \xrightarrow[a\to0]{}\ \frac{\partial^2 y}{\partial x^2}\,.

where: $x_{n\pm1}, x_n$ are the displacements of neighbouring masses, $a$ is the spacing between them, and $\partial^2y/\partial x^2$ is the second spatial derivative of the continuous displacement field — the curvature of the string at that point. This finite-difference combination is exactly what appears in each mass's equation of motion (from Hooke's law on the two neighbouring springs); as $a\to0$ it becomes, by definition, a second derivative. That single replacement is what turns a chain of coupled ordinary differential equations into one partial differential equation: the wave equation, $\partial^2y/\partial t^2 = v^2\,\partial^2y/\partial x^2$ (derived properly from a string element in the next chapter, not here). The discrete set of $N$ modes similarly becomes a continuum, labelled by a continuous wavenumber $k$ instead of an integer $n$ .

Figure

A row of five masses on springs on the left evolving into a smooth continuous curve on the right, showing the discrete chain becoming a continuous string as the number of masses grows and the spacing shrinks

Derivation AFrom finite difference to the second derivative

The goal of this derivation is to justify carefully the single replacement that carries the whole article — showing, by a Taylor expansion of the neighbouring displacements about mass $n$ , that the discrete combination $x_{n+1}-2x_n+x_{n-1}$ divided by $a^2$ converges to the second spatial derivative $\partial^2y/\partial x^2$ of the continuous displacement field as the spacing $a$ shrinks to zero, which is exactly the step that turns Newton's second law for a discrete chain into the wave equation for a continuous string.

Step 1. Write each mass's equation of motion. For mass $n$ in the interior of the chain, connected to its neighbours by identical springs of stiffness $k$ , Newton's second law with Hooke's law reads $m\,\ddot x_n = k(x_{n+1}-x_n) - k(x_n - x_{n-1}) = k\big(x_{n+1}-2x_n+x_{n-1}\big)$ . The term in brackets is a finite difference: it measures how much mass $n$ 's displacement deviates from the average of its two neighbours.

Step 2. Relabel the discrete displacements as samples of a continuous field. In the continuum limit, the mass at position $x_n = na$ is replaced by a continuous field $y(x,t)$ , so that $x_n(t) = y(na,t)$ , and the two neighbours sit at $x\pm a$ : $x_{n+1}(t)=y(x+a,t)$ and $x_{n-1}(t)=y(x-a,t)$ .

Step 3. Taylor-expand both neighbours about $x$ . For small spacing $a$ , $y(x+a,t) = y(x,t) + a\,\dfrac{\partial y}{\partial x} + \dfrac{a^2}{2}\dfrac{\partial^2 y}{\partial x^2} + \cdots$ and $y(x-a,t) = y(x,t) - a\,\dfrac{\partial y}{\partial x} + \dfrac{a^2}{2}\dfrac{\partial^2 y}{\partial x^2} - \cdots$ , where the dots are terms of order $a^3$ and higher.

Step 4. Add the two expansions. Adding cancels the odd-order (first-derivative) terms exactly, since one carries $+a\,\partial y/\partial x$ and the other $-a\,\partial y/\partial x$ : $y(x+a,t)+y(x-a,t) = 2y(x,t) + a^2\dfrac{\partial^2 y}{\partial x^2} + O(a^4)$ . Subtracting $2y(x,t)$ from both sides gives $x_{n+1}-2x_n+x_{n-1} = a^2\dfrac{\partial^2 y}{\partial x^2} + O(a^4)$ .

Step 5. Divide by $a^2$ and take the limit. $\dfrac{x_{n+1}-2x_n+x_{n-1}}{a^2} = \dfrac{\partial^2 y}{\partial x^2} + O(a^2) \xrightarrow[a\to0]{} \dfrac{\partial^2 y}{\partial x^2}$ . Substituting this into the equation of motion from Step 1, with $m=\mu a$ (mass per unit length $\mu$ times spacing) and $k a\to$ a finite constant related to the tension, converts $m\ddot x_n = k(x_{n+1}-2x_n+x_{n-1})$ into $\mu\,\dfrac{\partial^2 y}{\partial t^2} = (ka)\,\dfrac{\partial^2 y}{\partial x^2}$ — the wave equation, with wave speed $v=\sqrt{ka/\mu}$ .

Worked example

A steel guitar string of length $L=0.65\ \text{m}$ and total mass $M=3.0\times10^{-3}\ \text{kg}$ is modelled as the continuum limit of $N$ point masses $m=M/N$ joined by springs of natural length $a=L/N$ , under tension $T=73\ \text{N}$ (a typical steel high-E string tension). Find the linear mass density and, using $v=\sqrt{T/\mu}$ , the wave speed on the string.

Step 1. Identify what must stay fixed as $N\to\infty$ . Two ratios must be held finite: the mass per unit length, $\mu=m/a$ , and the stiffness per unit length, related to the tension $T$ . If only $m\to0$ were tracked while $a\to0$ , with $\mu$ left to drift, the limiting object would not correspond to any real string of finite mass — so $\mu$ is computed once, from the string's bulk properties, not mass by mass.

Step 2. Compute the mass per unit length. $\mu = \dfrac{M}{L} = \dfrac{3.0\times10^{-3}\ \text{kg}}{0.65\ \text{m}} = 4.6\times10^{-3}\ \text{kg/m}$ .

Step 3. Compute the wave speed. In the continuum limit the discrete spring stiffness per unit length becomes the string's tension $T$ , so the same $v=\sqrt{ka/\mu}$ found in the derivation reduces to the familiar string-wave speed $v=\sqrt{T/\mu}$ . Substituting, $v = \sqrt{\dfrac{73}{4.6\times10^{-3}}} = \sqrt{1.587\times10^{4}} \approx 126\ \text{m/s}$ .

Answer: $\mu \approx 4.6\times10^{-3}\ \text{kg/m}$ and $v \approx 126\ \text{m/s}$ — a wave disturbance on this string, once it behaves as a continuum rather than a chain of discrete masses, travels at about $126\ \text{m/s}$ .

Competitive stretch

An $N=3$ discrete-mass model of the same string (two masses, three springs, two more to the walls) has exactly 3 normal modes: lowest with all masses in phase, highest with neighbours maximally out of phase, middle with the end masses opposing while the centre mass stays nearly still — the picture the continuum limit smooths into a continuum of wave modes labelled by wavenumber $k$ .

Common traps

Assuming "the continuum limit" just means taking $N$ very large without also shrinking $a\to0$ and rescaling $m$ and $k$ so that $\mu=m/a$ and $ka$ stay finite — an uncontrolled limit gives nonsense, not a physical string.
Believing a discrete chain already has infinitely many modes before the continuum limit — $N$ masses give exactly $N$ modes, no more, no less.
Expecting a full quantitative dispersion relation $\omega(k)$ for the discrete chain from matrix or eigenvalue methods — that is out of scope here; only the qualitative mode-shape picture and the Taylor-expansion argument are required.
Confusing this topic with molecular vibrations or lattice dynamics (phonons) — those are solid-state-physics applications, not part of this course.

Ask the tutor: if the Taylor-expansion step from finite difference to second derivative still feels abstract, ask the AI tutor to walk through it slowly with a specific numeric spacing $a$ to see how fast the higher-order terms vanish.

Why does adding more masses to the chain increase the number of normal modes?

Each mass you add contributes one more degree of freedom, and one more coupled equation of motion in Newton's second law. Just as with $N=2$ , these coupled equations decouple into independent simple-harmonic equations in suitable normal coordinates, and there are as many independent equations as there are masses. So $N$ masses always give exactly $N$ normal modes — never more, never fewer, and never infinite unless $N$ itself is infinite, which is precisely the continuum limit this article builds towards.

Why do we need both a to 0 and m, k rescaled, rather than just N to infinity?

If you only send $N\to\infty$ while $a$ , $m$ , and $k$ stay fixed, the chain either grows without bound in physical length (if $a$ stays fixed) or the mass and stiffness per unit length blow up or vanish (if $a\to0$ but $m,k$ are untouched). Neither describes a real string of finite length and finite mass. Holding $\mu=m/a$ and the tension-related $ka$ fixed while $N\to\infty$ , $a\to0$ is what keeps the total length $L=(N-1)a$ and the physical properties of the string finite and sensible — it is a controlled limit, not just "more masses."

Is the wave equation derived properly here, or just motivated?

Motivated, deliberately. This article derives the key mathematical step — the finite difference $x_{n+1}-2x_n+x_{n-1}$ becoming $\partial^2y/\partial x^2$ — carefully, with a Taylor expansion, because that step is exactly what a discrete chain shares with a continuous string. But the full wave equation for a string, $\partial^2y/\partial t^2=v^2\partial^2y/\partial x^2$ , with $v=\sqrt{T/\mu}$ derived from a string element under tension using Newton's second law directly on the continuum, is properly derived in the next chapter on wave motion. Here the goal is conceptual: see why the discrete and continuous pictures must agree.

From discrete oscillators to waves

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads