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Derivation

Small Oscillations and Simple Harmonic Motion

D-031 Home PU-101 Threads force · energy · waves Depends on Conservation of Mechanical Energy
Statement

Let a particle of mass m move in one dimension under a smooth conservative potential U(x) possessing a stable equilibrium at x_0, so that U'(x_0)=0 and U''(x_0)>0. Then for sufficiently small displacements q\equiv x-x_0 the motion obeys the harmonic-oscillator equation \(\ddot q+\omega^2 q=0\) with angular frequency \(\omega=\sqrt{U''(x_0)/m}\). Every stable equilibrium of a smooth potential is, to leading order, a simple harmonic oscillator whose frequency is fixed by the curvature of the well and the inertia.

Why it matters

Almost every restoring system in physics — a pendulum, a stretched bond in a diatomic molecule, an atom in a crystal lattice, an LC circuit, a mode of a vibrating string — is governed near its equilibrium by exactly this equation. The harmonic oscillator is therefore not one special system but the universal local behaviour of stability itself.

The result also explains why the mass–spring and the pendulum share the same mathematics despite having nothing physical in common: both are the quadratic minimum of a potential well. Once a problem is linearized, the whole machinery of normal modes, resonance, and phonons follows immediately.

Assumptions
The potential is smooth (at least twice differentiable) near x_0.Without a well-defined U'' there is no leading quadratic term to expand into; a cusp or hard wall such as U\propto|x| gives non-harmonic, amplitude-dependent motion.
The equilibrium is stable: U''(x_0)>0.If U''(x_0)<0 the stationary point is a maximum and \(\omega^2<0\), giving exponential runaway (an inverted pendulum) rather than oscillation.
Motion is conservative — no friction or driving during the derivation.Dissipation adds a \dot q term and turns pure oscillation into damped decay; the frequency then shifts and the amplitude is no longer constant.
The displacement is small enough that cubic and higher Taylor terms are negligible, i.e. \tfrac12 U''(x_0)q^2 \gg \tfrac16 U'''(x_0)q^3.If dropped, anharmonic corrections appear: the frequency becomes amplitude-dependent and effects such as thermal expansion (from an asymmetric well) survive.
The curvature is non-degenerate, U''(x_0)\neq0.If U''(x_0)=0 as well, the leading term is quartic and the restoring law is \propto q^3, which is not simple harmonic at any amplitude.
Derivation
1
\[ \tfrac12 m\dot x^2 + U(x) = E \]
Conservation of mechanical energy for a conservative one-dimensional force; E is a constant of motion (the assumed prior result). A
2
\[ m\dot x\,\ddot x + U'(x)\,\dot x = 0 \;\Longrightarrow\; m\ddot x = -U'(x) \]
Differentiate the energy equation with respect to t (chain rule) and cancel the common factor \dot x; this is Newton's law with F=-U'. A
3
\[ q\equiv x-x_0,\qquad \ddot x=\ddot q \]
Change variable to displacement from equilibrium; since x_0 is constant the acceleration is unchanged. A
4
\[ U(x_0+q)=U(x_0)+U'(x_0)\,q+\tfrac12 U''(x_0)\,q^2+\mathcal{O}(q^3) \]
Taylor-expand the smooth potential about x_0; legal because U is at least C^2 there. B
5
\[ U'(x_0+q)=U'(x_0)+U''(x_0)\,q+\mathcal{O}(q^2)=U''(x_0)\,q+\mathcal{O}(q^2) \]
Differentiate the expansion; the equilibrium condition U'(x_0)=0 removes the constant term. B
6
\[ m\ddot q = -U''(x_0)\,q + \mathcal{O}(q^2) \]
Substitute step 5 into step 2. Dropping the \mathcal{O}(q^2) remainder is justified only for |q| small compared with the length scale U''/U''' over which U' curves. C
7
\[ k\equiv U''(x_0),\qquad \omega^2\equiv \frac{k}{m} \]
Define the effective stiffness and frequency; k>0 by the stability assumption, so \omega is real. A
8
\[ \ddot q+\omega^2 q=0 \]
Divide by m and rearrange — the harmonic-oscillator equation. A
9
\[ q(t)=A\cos(\omega t+\varphi) \]
General solution of a linear, constant-coefficient ODE (characteristic roots \pm i\omega) with two real constants A,\varphi fixed by initial conditions. B
Result
\[ \ddot q=-\omega^2 q,\qquad \omega=\sqrt{\frac{U''(x_0)}{m}},\qquad T=\frac{2\pi}{\omega}=2\pi\sqrt{\frac{m}{U''(x_0)}} \]

Reading. Near any stable equilibrium the restoring force is proportional to displacement, so the motion is sinusoidal at a single frequency set by the curvature of the well and the inertia. A stiffer well (larger U'') or a lighter mass oscillates faster; crucially, the frequency is independent of amplitude — the hallmark of simple harmonic motion (isochronism).

Units check. U'' has units of energy per length squared, \(\mathrm{J\,m^{-2}=kg\,s^{-2}}\). Then U''/m has units \(\mathrm{s^{-2}}\), so \(\omega=\sqrt{U''/m}\) is in \(\mathrm{rad\,s^{-1}}\) and \(T=2\pi/\omega\) is in seconds. For a spring \(U=\tfrac12 kx^2\), U''=k and \(\omega=\sqrt{k/m}\), recovering the familiar result.

Limiting cases
  • Mass–spring: \(U=\tfrac12 kx^2\) is already exactly quadratic, so U''=k and \(\omega=\sqrt{k/m}\) at all amplitudes — no linearization needed.
  • Simple pendulum: \(U=mgL(1-\cos\theta)\) gives U''(0)=mgL with generalized inertia mL^2, so \(\omega=\sqrt{g/L}\) — the small-angle result.
  • Stiff well (U''\to\infty): \(\omega\to\infty\), T\to0 — a hard, rigid trap oscillates arbitrarily fast.
  • Marginal stability (U''(x_0)\to0^{+}): \(\omega\to0\), T\to\infty — the flat-bottomed well gives critically slow, ultimately anharmonic motion.
Breaks when
  • Large amplitude. Once |q| is comparable to the anharmonic scale U''/U''', the cubic and quartic terms matter: the pendulum period grows as \(T=T_0\left(1+\tfrac{1}{16}\theta_0^2+\cdots\right)\), so the frequency becomes amplitude-dependent and isochronism fails.
  • Unstable or inflection equilibria. At a potential maximum U''<0 gives imaginary \omega and exponential divergence; at an inflection (U''=0) the leading restoring force is cubic, not linear.
  • Non-smooth potentials. Hard walls, kinks, or U\propto|x| have no Taylor quadratic term; the motion is periodic but not sinusoidal and its period depends on amplitude.
  • Strong damping or driving. Adding a term -b\dot q shifts the frequency to \(\sqrt{\omega_0^2-\gamma^2}\) and, past critical damping, removes oscillation entirely.
Failure modes
  • Confusing the force constant with the potential value. The frequency depends on the second derivative U''(x_0), not on U(x_0) or the slope U' (which is zero). Students often plug in the depth of the well.
  • Forgetting the effective inertia in generalized coordinates. For a pendulum in angle \theta, the "mass" is the moment of inertia mL^2, not m; using m gives the wrong frequency.
  • Keeping the constant term U(x_0) and thinking it matters. It shifts energy but drops out on differentiation; it has no effect on the dynamics.
  • Assuming any restoring force gives SHM. A restoring force merely guarantees bounded oscillation; only a force linear in displacement gives amplitude-independent frequency.
  • Using degrees instead of radians. The small-angle result \sin\theta\approx\theta and \(\omega=\sqrt{g/L}\) require \theta in radians.
Discussion

The deep content of this derivation is not the harmonic oscillator itself but the claim of universality: the same quadratic form \tfrac12 U''(x_0)q^2 appears at the bottom of every smooth stable well. This is why the SHM equation is worth mastering above almost any other in physics — it is the generic small-amplitude limit of stability. The frequency encodes only two pieces of information: how sharply the potential curves and how much inertia resists.

Geometrically, linearization replaces the true potential curve by its osculating parabola at the minimum. The approximation is excellent precisely because a smooth minimum looks parabolic under magnification — the error is third order in the displacement. The scale over which it holds is set by the ratio U''/U''', the natural anharmonic length of the well.

The same expansion extends directly to many degrees of freedom: near a stable equilibrium the potential becomes \tfrac12\sum_{ij}K_{ij}q_iq_j with the Hessian matrix K_{ij}=\partial^2 U/\partial x_i\,\partial x_j evaluated at equilibrium. Simultaneously diagonalizing K against the mass matrix yields the normal modes — independent harmonic oscillators — which in a crystal become phonons and in field theory become the quanta of a free field. A subtle corollary: the harmonic term is symmetric in q, so a purely harmonic solid would show no thermal expansion, because \langle q\rangle stays at the minimum at every temperature. Real expansion is entirely an anharmonic effect of the cubic term \tfrac16 U'''(x_0)q^3. Small oscillations are thus the doorway from mechanics to solid-state and quantum field theory.

Common misconceptions. SHM is not defined by "oscillating" or even by "a restoring force" — a ball rolling in a V-shaped valley oscillates with a restoring force but is not simple harmonic. The defining feature is linearity: F=-kq, equivalently a parabolic potential, equivalently an amplitude-independent period. Anything else is only approximately SHM, and only for small displacements.

Worked examples
1
Simple pendulum from its potential: \(U(\theta)=mgL(1-\cos\theta)\).
Gravitational potential of a bob at angle \theta from the vertical, measured from the lowest point. Symbols first, numbers later. A
2
\[ \frac{dU}{d\theta}=mgL\sin\theta,\qquad \frac{d^2U}{d\theta^2}=mgL\cos\theta \]
First derivative vanishes at \theta=0 (equilibrium); the second derivative there is mgL>0 (stable). A
3
\[ \omega=\sqrt{\frac{U''(0)}{I}}=\sqrt{\frac{mgL}{mL^2}}=\sqrt{\frac{g}{L}} \]
The effective inertia in the coordinate \theta is the moment of inertia I=mL^2. B
4
\[ L=1.00\ \mathrm{m},\quad g=9.81\ \mathrm{m\,s^{-2}}\ \Rightarrow\ \omega=\sqrt{9.81}=3.13\ \mathrm{rad\,s^{-1}} \]
Insert numbers only now that the symbolic form is fixed. A
\[ \omega=3.13\ \mathrm{rad\,s^{-1}},\qquad T=\frac{2\pi}{\omega}=2.01\ \mathrm{s} \]

Reading. A one-metre pendulum swings once every ~2 s, independent of the mass and (to leading order) of the amplitude. Units check. \(\sqrt{(\mathrm{m\,s^{-2}})/\mathrm{m}}=\mathrm{s^{-1}}\).

1
Vibration of a molecular bond, Morse potential \(U(r)=D_e\left[1-e^{-a(r-r_0)}\right]^2\).
D_e is the well depth, a its inverse width, r_0 the bond length. B
2
\[ U'(r)=2D_e a\,e^{-a(r-r_0)}\left[1-e^{-a(r-r_0)}\right],\qquad U'(r_0)=0 \]
Confirm r_0 is the equilibrium (linear term vanishes). B
3
\[ U''(r_0)=2D_e a^2,\qquad \omega=\sqrt{\frac{U''(r_0)}{\mu}}=a\sqrt{\frac{2D_e}{\mu}} \]
Curvature at the minimum; use the reduced mass \mu=m_1 m_2/(m_1+m_2) as the two-body inertia. C
4
\[ D_e=7.6\times10^{-19}\ \mathrm{J},\ a=1.9\times10^{10}\ \mathrm{m^{-1}},\ \mu=1.6\times10^{-27}\ \mathrm{kg}\ (\mathrm{{}^1H^{35}Cl}) \]
Representative values for the HCl bond. A
\[ \omega=1.9\times10^{10}\sqrt{\frac{2(7.6\times10^{-19})}{1.6\times10^{-27}}}\approx 5.5\times10^{14}\ \mathrm{rad\,s^{-1}} \]

Reading. The wavenumber \(\tilde\nu=\omega/(2\pi c)\approx 2.9\times10^{3}\ \mathrm{cm^{-1}}\) lies in the infrared, close to HCl's measured fundamental near \(2886\ \mathrm{cm^{-1}}\); the small discrepancy is anharmonicity. Units check. \(\mathrm{m^{-1}}\sqrt{\mathrm{J/kg}}=\mathrm{m^{-1}}\sqrt{\mathrm{m^2 s^{-2}}}=\mathrm{s^{-1}}\).

Problems
  1. A particle moves in \(U(x)=\alpha x^4-\beta x^2\) with \alpha,\beta>0. Find the stable equilibria and the small-oscillation frequency about them.
    Solution \(U'(x)=4\alpha x^3-2\beta x=2x(2\alpha x^2-\beta)\). Equilibria at x=0 and \(x=\pm\sqrt{\beta/2\alpha}\). Second derivative \(U''(x)=12\alpha x^2-2\beta\). At x=0, \(U''=-2\beta<0\) (unstable). At \(x=\pm\sqrt{\beta/2\alpha}\), \(U''=12\alpha(\beta/2\alpha)-2\beta=6\beta-2\beta=4\beta>0\) (stable). Frequency \(\omega=\sqrt{4\beta/m}=2\sqrt{\beta/m}\).
  2. A 0.25 kg mass on a spring completes 12 oscillations in 8.0 s. Find the spring constant k.
    Solution Period \(T=8.0/12=0.667\ \mathrm{s}\); \(\omega=2\pi/T=9.42\ \mathrm{rad\,s^{-1}}\). From \(\omega=\sqrt{k/m}\), \(k=m\omega^2=0.25\times(9.42)^2=0.25\times88.8=22.2\ \mathrm{N\,m^{-1}}\).
  3. A bead of mass m rests at the bottom of a frictionless bowl whose surface height near the base is \(y=x^2/(2R)\). Show the small oscillations are SHM and find \omega.
    Solution Potential \(U=mgy=mgx^2/(2R)\), already quadratic, so \(U''=mg/R\). Then \(\omega=\sqrt{U''/m}=\sqrt{g/R}\). This matches a pendulum of length R: the bowl's radius of curvature plays the role of the pendulum length, as expected for the osculating circle of the surface.
  4. For the pendulum, use the leading anharmonic correction to estimate the fractional period increase at amplitude \theta_0=20^\circ.
    Solution The correction gives \(T\approx T_0\left(1+\theta_0^2/16\right)\). With \(\theta_0=20^\circ=0.349\ \mathrm{rad}\), \(\theta_0^2/16=0.122/16=0.0076\). The period is about 0.76% longer than the small-angle value — a small but measurable departure from isochronism.
  5. Two atoms of masses \(m_1=1.0\times10^{-26}\ \mathrm{kg}\) and \(m_2=3.0\times10^{-26}\ \mathrm{kg}\) are joined by a bond of stiffness \(k=500\ \mathrm{N\,m^{-1}}\). Find the vibrational angular frequency.
    Solution Reduced mass \(\mu=m_1 m_2/(m_1+m_2)=\dfrac{(1.0)(3.0)}{1.0+3.0}\times10^{-26}=0.75\times10^{-26}=7.5\times10^{-27}\ \mathrm{kg}\). Then \(\omega=\sqrt{k/\mu}=\sqrt{500/(7.5\times10^{-27})}=\sqrt{6.67\times10^{28}}=2.6\times10^{14}\ \mathrm{rad\,s^{-1}}\). The two-body vibration uses the reduced mass, not either atomic mass alone.