Small Oscillations About Equilibrium

Ask a modern materials-simulation code what a solid will do and its first substantive act is this article's calculation at industrial scale: relax the structure to equilibrium (find where all forces vanish), then compute the Hessian — the matrix of second derivatives of the energy with respect to every atomic coordinate — and diagonalize it into phonon frequencies. That harmonic spectrum decides whether a proposed crystal is dynamically stable (one negative eigenvalue — one unstable direction — and the structure is rejected), what its heat capacity and thermal conductivity are, and how it will show up in infrared and Raman spectra. The 2026 twist is that machine-learned interatomic potentials now stand in for quantum chemistry in these pipelines, and a standard benchmark for whether a learned potential is trustworthy is precisely whether it reproduces the curvatures: get $U''$ wrong and every predicted spectrum and stability call goes wrong with it. The same physics runs today's most precise machines. In a trapped-ion quantum computer, each ion sits at the minimum of an electromagnetic pseudo-potential; its motional modes are small oscillations at a few megahertz, and two-qubit entangling gates work by nudging exactly those modes — gate fidelities live and die on how harmonic the trap really is. Kilometre-scale gravitational-wave interferometers hang their mirrors as pendulums so that, above the sub-hertz $\sqrt{g/L}$ resonance, each mirror swings freely enough to feel a passing ripple of spacetime. From a qubit's quiver to a hanging 40 kg mirror, the design rule is the one formula: find the minimum, take two derivatives, divide by the inertia.

Picture any potential-energy landscape as hilly terrain and drop a marble into one of its valleys. Now zoom in on the valley floor with a microscope. However lumpy and asymmetric the landscape looks from far away, the closer you look at the bottom, the more it looks like one specific shape: a parabola. Not approximately-sort-of — systematically, with the mismatch shrinking as the square of how far you zoom out from the minimum. That parabola is a harmonic oscillator. The ubiquity of harmonic motion in nature is this one geometric fact wearing a thousand costumes.

Two features of the picture do all the work. First, why the parabola and not a straight line: at the very bottom of a valley the slope is zero — that is what makes it the bottom. A Taylor expansion of the landscape about that point therefore has no linear term to offer; the leading survivor is the quadratic. A force that grows linearly with displacement is exactly Hooke's law, so the marble behaves as if attached to an invisible spring of stiffness $k_{\rm eff}=U''(x_0)$. Second, curvature beats depth. The frequency does not care how deep the well is, how wide it is, or how much energy it would take to escape — only how sharply the floor curves at the very bottom. Two wells of identical depth, one narrow and one broad, ring at different pitches; the narrow one, with larger $U''$, snaps its marble back faster. This is why chemical bonds of similar strength can have very different vibration frequencies: the frequency reads the stiffness of the bond, not its depth. Not every equilibrium is a valley, and the second derivative is the referee:

A minimum ($U''>0$) pushes the marble back and produces a real frequency. A maximum ($U''<0$) pushes it away — the same algebra returns an imaginary $\omega$, which is mathematics' way of writing exponential runaway. And the borderline case $U''=0$, a flat-bottomed well, still traps the marble but refuses to give it a fixed pitch: its period depends on how hard you hit it. Everything else in this article is the quantitative version of this one figure.

The entire theory is one Taylor expansion, so let us write it with care. Take a smooth potential $U(x)$ and a point $x_0$ where the force vanishes, and expand about it. Nothing is assumed about the global shape of $U$ — only smoothness at $x_0$:

where: $U', U''$ — first and second derivatives of $U$ evaluated at $x_0$; the dots hide cubic and higher terms. Physically the three written terms are: an energy offset you can calibrate away, a slope term that is exactly zero because $x_0$ is an equilibrium ($F=-U'(x_0)=0$), and the curvature term — the first one that actually pushes. That silent disappearance of the linear term is the entire secret of universal harmonicity. Differentiating equation (1) turns the statement about energy into a statement about force. With the shorthand $\xi\equiv x-x_0$ for the displacement from equilibrium, the force on the particle is

where: $U''(x_0)$ — a positive constant for a stable equilibrium, playing exactly the role of the spring constant $k$ in Hooke's law $F=-k\xi$. No spring is present; the "spring" is the local curvature of whatever potential the particle actually lives in. The sign matters: positive curvature means the force opposes the displacement, i.e. restores.

The solution deserves one sentence of reading. Equation (4) says the particle oscillates sinusoidally about the minimum with amplitude $A$ and phase $\phi$ set by how it was launched, while the frequency is set entirely by the landscape — turn the launch knobs however you like, the pitch does not move. That decoupling of how loud from how fast is the defining fingerprint of harmonic motion, and it survives only as long as the parabolic truncation does.

Real problems rarely hand you a particle on a line. A pendulum is described by an angle, a rolling ball by a contact coordinate, a molecule by a bond length; and the "mass" multiplying $\dot q^2$ in the kinetic energy is then generally not the literal mass. The small-oscillation recipe survives intact, provided both the potential and the kinetic energy are expanded about equilibrium. When the kinetic energy near equilibrium has the form $T=\tfrac12 m_{\rm eff}(q)\,\dot q^2$, the frequency generalizes to

where: $q$ — any generalized coordinate (an angle, an arc length, a bond stretch); $k_{\rm eff}$ — the curvature of the potential in that coordinate; $m_{\rm eff}$ — the coefficient of $\tfrac12\dot q^2$, evaluated at the equilibrium. Note that $k_{\rm eff}$ and $m_{\rm eff}$ separately carry units that depend on the choice of $q$, but their ratio is always $\mathrm{s^{-2}}$ — the physics does not care how you parametrize it.

Which formula answers which question

One more standard result completes the kit. Multiplying the equation of motion by $\dot\xi$ and integrating — or simply adding the kinetic energy to the truncated potential — gives the conserved energy of the small oscillation. It is worth displaying because its constancy is what makes the amplitude a meaningful single number:

where: $E$ — mechanical energy measured from the bottom of the well; the last equality evaluates the conserved sum at a turning point, where all energy is potential. The energy is quadratic in the amplitude — double the swing, four times the energy — and it sloshes between kinetic and potential twice per cycle while the total holds still.

Truncating the Taylor series is a physical decision, and it has a price tag. The honest question is: at what amplitude do the discarded terms start to matter? For a symmetric well the first discarded term is the quartic, and comparing it to the quadratic gives a clean validity criterion.

The reach of this one calculation is easiest to feel through its habitats. A chemical bond is a potential well in the interatomic separation $r$: steep repulsion at short range, a minimum at the bond length $r_0$, a flattening tail toward dissociation. Expand about $r_0$ and the molecule becomes a two-mass spring with $k=U''(r_0)$ vibrating at $\omega=\sqrt{k/\mu}$, where $\mu$ is the reduced mass — this is the machinery behind every infrared spectrum, and the worked example below runs the real numbers for HCl.

A crystal is the same story multiplied: each atom sits in a well built by its neighbours, and the collective quadratic expansion — a matrix of second derivatives, diagonalized — yields the phonon normal modes that carry sound and heat through every solid. In the sky, a small body parked at the stable Lagrange points $L_4$ and $L_5$ of the Sun–Jupiter system feels an effective potential (gravity plus centrifugal) with a genuine minimum; expanding it to quadratic order gives the libration frequencies of the Trojan asteroids, thousands of which are observed doing exactly the small oscillations Lagrange computed on paper in 1772. Different forces, one identical calculation.

1 · A skate bowl from its curvature (confidence builder)

A skateboarder rolls near the bottom of a smooth concrete bowl whose profile near the low point is $y=c\,x^2$ with $c=2.0\ \mathrm{m^{-1}}$. Treating the skater as a sliding point mass, find the small-oscillation frequency and period. Does the skater's 62 kg mass matter?

Step 1. Build the potential in the coordinate $x$. The height above the low point is $y=cx^2$, so $U(x)=mgy=mgc\,x^2$. This is already a parabola — the Taylor expansion is exact here — with $U'(0)=0$ confirming $x_0=0$ as the equilibrium.

Step 2. Differentiate twice for the stiffness: $k_{\rm eff}=U''(0)=2mgc$. Near the bottom the track is nearly flat, so the speed is essentially $\dot x$ and the effective mass is just $m$.

Step 3. Apply equation (5): $\omega=\sqrt{2gc}=\sqrt{2\times 9.81\times 2.0}=\sqrt{39.2}=\boxed{6.26\ \mathrm{rad/s}}$, i.e. a frequency of $f=\omega/2\pi\approx 1.0$ Hz and a period $T=2\pi/\omega=\boxed{1.00\ \mathrm{s}}$.

Step 4. The mass cancelled: both the stiffness ($2mgc$) and the inertia ($m$) are proportional to $m$, so the 62 kg never enters — the same equivalence-principle cancellation as the pendulum. A heavier skater rocks in the bowl at exactly the same one-second period, just with more energy at the same amplitude by equation (6).

2 · Predicting an infrared spectrum from bond curvature (exam level)

The H–Cl bond potential has curvature $k=U''(r_0)=481\ \mathrm{N/m}$ at its minimum ($r_0=0.127$ nm). Predict the classical vibration frequency of the HCl molecule and the wavenumber of its infrared absorption line, and compare with the measured $2886\ \mathrm{cm^{-1}}$. Then estimate the oscillation amplitude if the vibration holds energy $E=\tfrac12\hbar\omega$ and check the smallness of the oscillation.

Step 1. The oscillating coordinate is the bond stretch $\xi=r-r_0$, and the effective mass for a two-body vibration is the reduced mass. With $m_{\rm H}=1.008\,\mathrm{u}$ and $m_{\rm Cl}=34.97\,\mathrm{u}$: $\mu=\dfrac{m_{\rm H}m_{\rm Cl}}{m_{\rm H}+m_{\rm Cl}}=0.980\,\mathrm{u}=1.63\times10^{-27}\ \mathrm{kg}$ — essentially the hydrogen atom's mass, since chlorine barely moves.

Step 2. Apply equation (5) with $k_{\rm eff}=k$ and $m_{\rm eff}=\mu$: $\omega=\sqrt{\dfrac{481}{1.63\times10^{-27}}}=\boxed{5.44\times10^{14}\ \mathrm{rad/s}}$, i.e. $f=\omega/2\pi=8.65\times10^{13}$ Hz — about ninety trillion stretches per second.

Step 3. Spectroscopists quote the wavenumber $\tilde\nu=f/c=\dfrac{8.65\times10^{13}}{3.00\times10^{10}\ \mathrm{cm/s}}=\boxed{2887\ \mathrm{cm^{-1}}}$ — matching the measured fundamental at $2886\ \mathrm{cm^{-1}}$ to four significant figures. One second derivative of the bond potential predicts where in the infrared this molecule drinks light.

Step 4. Amplitude at $E=\tfrac12\hbar\omega$ (the quantum ground state's energy scale): from equation (6), $\tfrac12 kA^2=\tfrac12\hbar\omega$ gives $A=\sqrt{\hbar\omega/k}=\sqrt{\dfrac{1.055\times10^{-34}\times 5.44\times10^{14}}{481}}=\boxed{1.1\times10^{-11}\ \mathrm{m}}$.

Step 5. Check smallness: $A/r_0=0.011\ \mathrm{nm}/0.127\ \mathrm{nm}\approx 9\%$. The bond stretches by under a tenth of its length, comfortably inside the parabolic region — which is why the harmonic prediction lands on the measured line. (The few-cm$^{-1}$ residual, and the unequal spacing of higher vibrational levels, are precisely the anharmonic corrections of equation (7) at work.)

References. Morin, Introduction to Classical Mechanics, Ch. 5 (Oscillations) · Goldstein, Classical Mechanics 3rd ed., Ch. 6 (Oscillations) · Landau & Lifshitz, Mechanics Vol. 1, §21–23 · Taylor, Classical Mechanics, Ch. 5 Key results on this page: the linear term vanishes at equilibrium and $k=U''(x_0)$; the pendulum reduction gives $\omega=\sqrt{g/L}$ from $U=mgL(1-\cos\theta)$, $U''(0)=mgL$, $m_{\rm eff}=mL^2$; the leading pendulum period correction is $T/T_0\approx 1+\theta_0^2/16$. Worked-example numbers: $\omega=6.26\ \mathrm{rad/s}$, $T=1.00$ s for the bowl with $c=2.0\ \mathrm{m^{-1}}$; $\mu=1.63\times10^{-27}$ kg, $\omega=5.44\times10^{14}$ rad/s, $\tilde\nu=2887\ \mathrm{cm^{-1}}$ vs measured 2886, and $A=1.1\times10^{-11}$ m for HCl with $k=481$ N/m.

Builds on: Simple Harmonic Motion · Energy in Simple Harmonic Motion. Leads to: The Simple Pendulum · The Physical Pendulum · Anharmonic Pendulum and Amplitude Dependence · Complex Exponential Representation of Oscillations · Damped Harmonic Oscillator.