Small Oscillations and Normal Modes

Pluck one string of a guitar and, if it is slightly out of tune with its neighbour, you will hear the sound wander back and forth between the two strings — energy sloshing from one to the other and back. A washing machine on spin can set a whole floor humming. A tall building sways in a wind at particular frequencies. All of these are the same phenomenon: a system of coupled oscillators settling into its normal modes — special patterns of motion in which every part oscillates at one shared frequency. This page shows how to find them, and the answer turns out to be a beautiful marriage of mechanics and linear algebra: finding normal modes is solving an eigenvalue problem.

The plan: expand any Lagrangian near a stable equilibrium and discover it always looks like coupled springs; write the coupled equations as a single matrix equation; then diagonalise to split them into independent oscillators — the modes — each with its own frequency.

Every stable equilibrium is a spring

Take any system with generalised coordinates q_i and a potential V(q) with a stable equilibrium at q_i = q_i^0. Write small displacements \eta_i = q_i - q_i^0 and Taylor-expand the potential. The constant term is irrelevant, the linear term vanishes (equilibrium is where the slope is zero), and the first thing that survives is the quadratic term:

V \approx V_0 + \tfrac12 \sum_{i,j} K_{ij}\,\eta_i \eta_j, \qquad K_{ij} = \left.\frac{\partial^2 V}{\partial q_i \partial q_j}\right|_{0}.

The kinetic energy, likewise, is quadratic in the velocities, and near equilibrium we may freeze its coefficients at their equilibrium values:

T \approx \tfrac12 \sum_{i,j} M_{ij}\,\dot\eta_i \dot\eta_j.

So \mathbf{K} (the stiffness matrix) and \mathbf{M} (the mass matrix) are constant symmetric matrices. Feed L = T - V into the Euler–Lagrange equations and you get one clean matrix equation of motion:

The lesson is universal: close enough to any stable equilibrium, everything is a harmonic oscillator. A pendulum near the bottom, atoms in a crystal, a molecule's bonds, the strings of an instrument — all obey this equation. That is why the harmonic oscillator is the most important system in physics.

The eigenvalue problem for the modes

Guess that the whole system oscillates at a single frequency \omega, with every coordinate rising and falling together: \boldsymbol\eta(t) = \mathbf{a}\,\cos(\omega t) for some fixed shape vector \mathbf{a}. Substituting, \ddot{\boldsymbol\eta} = -\omega^2\boldsymbol\eta, so the equation of motion becomes

-\omega^2 \mathbf{M}\,\mathbf{a} = -\mathbf{K}\,\mathbf{a} \quad\Longrightarrow\quad \mathbf{K}\,\mathbf{a} = \omega^2\,\mathbf{M}\,\mathbf{a}.

This is a generalised eigenvalue problem. The eigenvalues \omega^2 are the squared normal frequencies; the eigenvectors \mathbf{a} are the normal modes — the special shapes that oscillate rigidly at a single frequency. Non-trivial shapes exist only when the matrix (\mathbf{K} - \omega^2\mathbf{M}) is singular, so the frequencies come from the characteristic equation

\det\!\left(\mathbf{K} - \omega^2\mathbf{M}\right) = 0.

A system with n degrees of freedom has exactly n normal modes. Because \mathbf{M} and \mathbf{K} are symmetric, the modes are "orthogonal" (in the metric set by \mathbf{M}) and form a complete basis — so any small motion, however complicated, is a superposition of normal modes. Diagonalising \mathbf{M}^{-1}\mathbf{K} is exactly the change of coordinates that decouples the mess into n independent one-dimensional oscillators, the normal coordinates.

Worked example: two masses, three springs

The classic. Two equal masses m in a line, each tied to a wall by a spring k, and to each other by a middle spring k'. Let \eta_1, \eta_2 be their displacements. The matrices are

\mathbf{M} = \begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}, \qquad \mathbf{K} = \begin{pmatrix} k + k' & -k' \\ -k' & k + k' \end{pmatrix}.

Solving \det(\mathbf{K} - \omega^2\mathbf{M}) = 0 gives two modes. Drag the sliders below: pick a mode and step time to watch the pattern.

If the two masses started with a lopsided push — say only mass 1 moving — that is a superposition of both modes, and because the modes have different frequencies the energy appears to slosh from mass 1 to mass 2 and back. That sloshing is the two-string beat you hear on a guitar, and it is nothing but two normal modes drifting in and out of step.

When you solve \mathbf{K}\mathbf{a} = \omega^2 \mathbf{M}\mathbf{a}, the number the eigenvalue routine hands you is \lambda = \omega^2, not \omega itself. The physical frequency is \omega = \sqrt{\lambda}. Forgetting the square root is the single most common slip — you report a "frequency" that is really its square.

This also gives a neat stability test. At a genuine minimum of the potential all the eigenvalues \omega^2 are positive, so every \omega is real and the modes oscillate. But if the equilibrium is a maximum or a saddle, some \omega^2 is negative: then \omega is imaginary, \cos(\omega t) turns into a growing \cosh, and that "mode" runs away exponentially. A negative eigenvalue is the mathematics telling you the equilibrium is unstable in that direction.

Because a structure absorbs energy most greedily at its normal-mode frequencies — that is resonance. Push a system at one of its \omega's, even gently, and the amplitude grows and grows. Soldiers famously break step when crossing a footbridge so their marching rhythm cannot happen to match a bridge mode. The Millennium Bridge in London wobbled alarmingly on its opening day in 2000 because pedestrians unconsciously synchronised their steps with a sideways sway mode. Engineers compute the normal modes of every large structure precisely so they can keep the frequencies of wind, traffic and earthquakes away from them — or add damping to the modes that matter. The eigenvalue problem on this page is, quite literally, load-bearing.