Periodic Motion & the SHM Condition

Periodic motion repeats: there is a shortest time T, the period, for which the state at time t + T matches the state at t. From it come the frequency f = 1/T in hertz, and the angular frequency ω = 2πf in rad s⁻¹, which counts the same repetition in radians of phase rather than in whole cycles. The amplitude A is the largest magnitude the displacement reaches, measured from the equilibrium position — so you cannot quote an amplitude until you have identified where equilibrium is. A violin string sounding A4 repeats 440 times a second: T = 1/440 = 2.273 ms and ω = 2π × 440 = 2.765 × 10³ rad s⁻¹. Every one of those words applies just as well to a sawtooth voltage, a bouncing ball or a heartbeat, none of which is simple harmonic. Periodicity is cheap; the condition below is not.

Simple harmonic motion is the statement a = −ω²x, or d²x/dt² + ω²x = 0. Three separate claims are packed into it. The acceleration is proportional to the displacement, not to its square or its square root. The minus sign makes it a restoring acceleration, always directed at the equilibrium point. And the constant is the same at every displacement, which is the claim data can falsify. Check the dimensions: [a/x] = (L T⁻²)/L = T⁻², so the constant is a reciprocal time squared and ω is a rate — 2π/ω comes out as a time, as a period must. The positivity of ω² matters as much as its value. If the force pushed outward instead, the constant in d²x/dt² = +λ²x would be positive on the right, and the solutions would be growing and decaying exponentials rather than oscillations. That algebraic sign is the whole difference between a stable equilibrium and an unstable one.

Newton's second law gives a = F(x)/m, so a = −ω²x requires F(x) = −mω²x: a force exactly proportional to displacement. No real force is. What saves the model is a Taylor expansion about a point of stable equilibrium, taken as x = 0: F(x) = F(0) + F′(0)x + ½F″(0)x² + … . The first term vanishes because equilibrium means no net force there. The second has F′(0) negative — write it as −k, with k > 0 in N m⁻¹ — because stability means the force opposes any small push. So for displacements small enough that kx dominates every later term, F ≈ −kx and ω² = k/m, whatever the system physically is: a stretched bond, a floating hull, an atom in a crystal, a beam in a bridge deck. The model is not a special mechanism but the leading behaviour of almost any stable one.

Suppose x(t) satisfies d²x/dt² = −ω²x, and let c be any constant. Differentiation is linear and the equation has no constant term, so the scaled motion c x(t) satisfies it too. The motion released from 5.0 cm is therefore the motion released from 1.0 cm multiplied by 5 at every instant: same zero crossings, same turning points, same period. The distance covered per cycle is five times larger, but so is the acceleration at every corresponding instant — at ω = 12.0 rad s⁻¹, aₘₐₓ = ω²A rises from 1.44 m s⁻² to 7.20 m s⁻² — and the two effects cancel exactly. A dimensional restatement makes the cancellation look inevitable: the only constant in the equation is ω, of dimension T⁻¹, and an amplitude is a length. With no second length to compare it against, no dimensionless function of A can enter the period. Isochronism is a consequence of linearity, not a property of oscillation in general.

Now add the first correction. If the system looks the same under x → −x, the next allowed term is cubic: Fₓ = −kx − βx³, with β in N m⁻³. Two constants of different dimensions now exist, and k/β has dimension of length squared, so √(k/β) is a length the system builds out of itself. The amplitude finally has something to be measured against, and the dimensionless group is βA²/k. First-order perturbation of the motion gives ω ≈ ω₀[1 + 3βA²/(8k)]. A stiffening force, β > 0, raises ω and shortens the period; a softening force lengthens it. The simple pendulum is the softening case, since sin θ < θ makes the restoring torque fall below linear: there β/k = −1/6 and the correction becomes T ≈ T₀(1 + θ₀²/16), worth 0.19% at θ₀ = 10° and 1.7% at 30°. So simple harmonic is always a claim about a range of displacement, never about a system.

Two experiments settle it. The direct one measures acceleration against displacement and plots the pair: the condition predicts a straight line through the origin, whose slope is −ω², and a cubic term shows up as curvature at the ends. It is the weaker test, because a departure of a per cent or so from straightness disappears into the scatter of an accelerometer trace. The sharper test times the period at several amplitudes and plots T against A²: the condition predicts a horizontal line, and any straight slope measures −3T₀β/(8k) directly. It wins on resolution. Timing fifty cycles of a 0.500 s oscillator, with the stopwatch good to ±0.05 s on the 25.0 s total, fixes the period to ±0.001 s — 0.2% — so a period that drifts by 0.2% between two amplitudes is already visible. Quote ω together with the displacement range over which you checked it, and the claim stays honest.