Mass–spring oscillators

Put a glider of mass m on a frictionless track against a spring and measure x from the position where the spring is neither stretched nor compressed. The only horizontal force is the spring's, Fₓ = −kx, and Newton's second law turns that into m d²x/dt² = −kx, or d²x/dt² = −(k/m)x. That is the simple-harmonic equation with ω² = k/m, so ω = √(k/m) and T = 2π√(m/k). The identification is a dimensional statement as much as an algebraic one: [k] = N m⁻¹ = kg s⁻², so [k/m] = s⁻² and its square root really is an angular frequency. Take k = 24.0 N m⁻¹ and m = 0.150 kg: ω = √160 = 12.6 rad s⁻¹, T = 0.497 s, f = 2.01 Hz. Notice what did not enter. The amplitude, the release point and the initial velocity fix the constants A and φ in the general solution; they cannot touch k/m, which is built from the hardware alone.

Hang the same spring from a support and let y run downwards from the unloaded end. Two forces act now: the weight mg down and the spring force ky up, so m d²y/dt² = mg − ky. This is not the SHM equation — it has a constant term — and the fix is to move the origin rather than to change the physics. At rest the acceleration is zero, so the static extension is Δl = mg/k. Substituting u = y − Δl gives m d²u/dt² = mg − k(u + Δl) = −ku, since kΔl = mg exactly cancels the weight. The equation, and therefore ω, is the horizontal one untouched. With k = 24.0 N m⁻¹ and m = 0.150 kg the sag is Δl = 0.150 × 9.81 ÷ 24.0 = 61.3 mm, and T = 2π√(0.0613/9.81) = 0.497 s, the same figure as on the track. The general rule is worth naming: a constant added force shifts the equilibrium and leaves ω alone, because the equation stays linear with the same coefficient on u.

Side by side, both springs are stretched by the same x, so their forces add: F = (k₁ + k₂)x and k_eff = k₁ + k₂. End to end, the same tension F runs through both, so their extensions add: x = F/k₁ + F/k₂, giving 1/k_eff = 1/k₁ + 1/k₂. Check the limits before using either. Series must be softer than the softer spring, because adding coils can only make a given force stretch the chain further; parallel must be stiffer than the stiffer one. With k₁ = 32.0 N m⁻¹ and k₂ = 48.0 N m⁻¹, series gives 19.2 N m⁻¹ and parallel 80.0 N m⁻¹, and a 0.450 kg block swings with T = 0.962 s or T = 0.471 s. One corollary catches people out: a spring of N identical coils is N identical springs in series, so cutting it in half doubles its stiffness and shortens the period by a factor √2.

So far the spring stored energy but carried none of its own. Give it mass mₛ spread evenly along its natural length L and assume the stretch is uniform, so the element at distance l from the fixed end moves at (l/L)v while the load moves at v. Its kinetic energy is ½(mₛ/L)dl × (lv/L)², and integrating along the spring gives ∫₀^L ½(mₛ v²/L³) l² dl = ½(mₛ/3)v². The spring therefore behaves as an extra mₛ/3 bolted to the load, and ω = √(k/(m + mₛ/3)). The correction is not small in the laboratory: with k = 24.0 N m⁻¹, m = 0.200 kg and a 60 g spring, the naive period is 0.574 s and the corrected one 0.602 s — the corrected period is 4.9% longer than the naive one, enormous against a stopwatch good to a few tenths of a percent. The exact factor is ⅓ only while mₛ ≪ m, because a heavy spring does not stretch uniformly during the motion.

The period came out free of amplitude because F was strictly proportional to x. Real springs are straight only over a range. Push far enough and coils bind in compression; pull far enough and the wire yields, leaving a permanent set and a new k. Between those extremes the curve bends gently, and a first correction Fₓ = −kx − βx³ already breaks isochronism: to leading order ω ≈ ω₀(1 + 3βA²/(8k)). Put numbers on it. With k = 24.0 N m⁻¹ and a hardening β = 5.1 × 10² N m⁻³, an amplitude of 50 mm gives 3βA²/(8k) = 0.0199, so the frequency rises 2.0% and the period of the 0.150 kg glider falls from 0.497 s to 0.487 s — measurable in ten timed swings. A hanging spring adds a harder limit: if A exceeds Δl the coils go slack at the top of the swing, the restoring force stops obeying −ku, and the motion is not simple harmonic at all.

Remove the wall. Two gliders m₁ and m₂ on a frictionless track are joined by a spring of stiffness k and natural length L₀. Let s = x₂ − x₁ − L₀ be the extension. The spring pulls them together, so m₁ d²x₁/dt² = ks and m₂ d²x₂/dt² = −ks. Subtracting gives d²s/dt² = −ks(1/m₁ + 1/m₂) = −(k/μ)s with μ = m₁m₂/(m₁ + m₂), the reduced mass. So the separation oscillates at ω = √(k/μ) while the centre of mass, feeling no external force, drifts at constant velocity. Check it against the case already solved: let m₂ → ∞ and μ → m₁, which is the wall-anchored oscillator, as it must be. With m₁ = 0.250 kg, m₂ = 0.400 kg and k = 24.0 N m⁻¹, μ = 0.154 kg and T = 0.503 s — shorter than either mass alone against a wall, 0.641 s and 0.811 s, because μ is smaller than both.