Simple and Physical Pendulums

Draw the bob's free-body diagram and you get two forces, weight and tension, at an awkward angle to each other. Take torques about the pivot instead and the tension vanishes — its line of action passes through the pivot — leaving one term: τ = −MgL sin θ, with θ measured from the vertical and the minus sign saying the torque always turns the bob back towards the bottom. Newton's second law for rotation, τ = I d²θ/dt² with I = ML², gives ML² d²θ/dt² = −MgL sin θ, or d²θ/dt² = −(g/L) sin θ. Notice what has already happened: M sits on both sides and cancels. Gravity supplies the restoring torque in proportion to mass, and the bob resists it in proportion to mass, so a lead bob and a cork bob on identical strings keep identical time — in vacuum. In air, buoyancy and drag single out the light bob, and the last section takes that up.

d²θ/dt² = −(g/L) sin θ is not SHM. The acceleration is proportional to sin θ, not to θ, and the equation has no sinusoidal solution. The series sin θ = θ − θ³/6 + θ⁵/120 − … shows the move: keep the first term only and the equation becomes d²θ/dt² = −(g/L)θ, which is a = −ω²x in angular clothing, with ω = √(g/L) and T = 2π√(L/g). The price is whatever the discarded terms were worth. At θ = 10° (0.1745 rad), sin θ = 0.1736, so writing θ for sin θ overstates the restoring torque by 0.51%; at 30° the overstatement is 4.7%. Keeping the θ³ term is exactly what produces the amplitude correction at the end of this lesson.

Swap the bob on a string for a rigid body hung from a pivot — a swinging leg, a bell, a metre rule on a nail. Nothing in the torque argument changes except the bookkeeping. The weight acts at the centre of mass, a distance d from the pivot, so τ = −Mgd sin θ, and the rotational inertia is I about the pivot rather than ML². The small-angle result is T = 2π√(I/(Mgd)). Get I from the parallel-axis theorem, I = I(cm) + Md². For a uniform rod of length L pivoted at one end, I = ML²/3 and d = L/2, so T = 2π√(2L/(3g)); with L = 1.00 m and g = 9.81 m s⁻² that is 1.64 s, against 2.01 s for a simple pendulum of the same length. Mass still cancels, because for a body of fixed shape I is itself proportional to M.

Write the result as T = 2π√(L(eq)/g) with L(eq) = I/(Md). That is the length of the simple pendulum which would keep the same time, and the point L(eq) below the pivot is the centre of oscillation — strike a bat or a hammer there and the pivot feels no jolt. Substituting I = I(cm) + Md² gives L(eq) = d + k²/d, where k² = I(cm)/M. Both extremes are slow: tiny d means almost no restoring torque, large d means a long pendulum. The minimum sits at d = k, where L(eq) = 2k. A uniform metre rod has k² = L²/12, so its quickest swing comes from a pivot 0.289 m from the centre, giving L(eq) = 0.577 m and T = 1.52 s — shorter than the 1.64 s from the end. Kater's reversible pendulum uses the pair of pivots that share one L(eq) to measure g to five figures.

Take the lowest point as zero. Swinging out to angle θ lifts the centre of mass by d(1 − cos θ), so U = Mgd(1 − cos θ), and with a frictionless pivot E = ½I(dθ/dt)² + Mgd(1 − cos θ) is constant. Expand the cosine: 1 − cos θ = θ²/2 − θ⁴/24 + …, so U ≈ ½Mgdθ². A potential energy quadratic in the displacement is the signature of SHM, and it says the same thing as the linear restoring torque, since τ = −dU/dθ. It also draws the approximation honestly: the true well is slightly wider than that parabola, so a wide swing lingers longer near the ends than SHM predicts. Energy also answers speed questions with no approximation at all — a simple pendulum released from θ₀ passes the bottom at v = √(2gL(1 − cos θ₀)).

Keep the θ³ term and the period picks up an amplitude dependence: T ≈ T₀[1 + ¼sin²(θ₀/2)] ≈ T₀(1 + θ₀²/16), with θ₀ in radians. The numbers show why clockmakers cared. At θ₀ = 5° the period runs 0.048% long — 41 s a day. At 10° it is 0.19%, nearly three minutes a day. At 30° it is 1.7%, and at 90° the exact period is 18% above T₀. Near the top the series stops being a correction at all: as θ₀ → 180° the period diverges, because a pendulum balanced exactly upright never comes down. Other assumptions fail more quietly. A bob of real size makes the system a physical pendulum, with T = 2π√((L + k²/L)/g) — slightly slow; a string with mass pulls the centre of mass up instead, and runs slightly fast. Air drag shrinks the amplitude, nudging the period down towards T₀ as the swing dies, and buoyancy trims the effective g.