The Physical Pendulum & Torsional Oscillators

You’ll be able to

Derive the period of a physical pendulum from the rotational form of Newton's second law
Apply the small-angle approximation and state where it fails
Compare a physical pendulum with a simple pendulum of the same length

When the mass is not concentrated at a point

A simple pendulum idealizes all the mass as a point at the end of a massless string. A physical pendulum is any rigid body swinging about a pivot that is not its center of mass — a meter stick, a swinging leg, a pendulum clock's rod and bob together. The analysis uses the rotational second law Στ = Iα instead of F = ma, and the restoring torque comes from gravity acting at the center of mass a distance d from the pivot: τ = −Mgd sin θ.

Physical, simple and torsional pendulums

T = 2π√(I/Mgd) simple: T = 2π√(L/g) torsional: T = 2π√(I/κ)

d is the distance from the pivot to the center of mass. The simple pendulum is the special case I = ML² and d = L, which reduces the first formula to the second.

The small-angle approximation is what makes it harmonic

The restoring torque −Mgd sin θ is not proportional to θ, so a pendulum is not truly a simple harmonic oscillator. For small angles, sin θ ≈ θ in radians, and the equation becomes Iα = −Mgdθ — the SHM form, with ω² = Mgd/I. The approximation is excellent for small swings: at 10° the error in the period is about 0.2%, but by 45° it exceeds 4% and the period grows noticeably with amplitude. Simple harmonic motion has an amplitude-independent period; a real pendulum only approximately does, and only for small swings.

Torsional oscillators

A torsional oscillator — a disk hanging from a wire, or the balance wheel of a mechanical watch — twists rather than swings. Its restoring torque is supplied by the wire and obeys the rotational analogue of Hooke's law, τ = −κθ, where κ is the torsion constant. The equation of motion is Iα = −κθ, identical in form to the mass-spring system with I replacing m and κ replacing k, so the period is T = 2π√(I/κ). Notably, gravity does not appear at all: a torsional clock keeps the same time on the moon.

Worked example

A uniform rod of length 1.00 m is pivoted at one end and swings as a physical pendulum. Find its period, and compare it with a simple pendulum of the same 1.00 m length.

1.For a rod about its end, I = ML²/3; its center of mass is at d = L/2.
2.Substitute: T = 2π√(I/Mgd) = 2π√((ML²/3)/(Mg L/2)) = 2π√(2L/3g). The mass cancels.
3.T = 2π√(2(1.00)/(3 × 9.8)) = 2π√(0.0680) = 2π(0.2608) ≈ 1.64 s.
4.A simple pendulum of length 1.00 m gives T = 2π√(1.00/9.8) ≈ 2.01 s, so the rod swings faster — its equivalent simple length is 2L/3 = 0.667 m.

Answer: T ≈ 1.64 s for the rod against 2.01 s for a simple pendulum of the same length, because distributing the mass along the rod brings the effective length down to 2L/3

Checkpoint

The period of a physical pendulum is independent of its mass because —

Watch out

A pendulum is only approximately a simple harmonic oscillator. If a question specifies a large amplitude, or asks whether the period depends on amplitude, the expected answer is that the small-angle approximation fails and the period increases with amplitude. Claiming amplitude independence without qualification is the trap being set.

Checkpoint

A torsional oscillator consisting of a disk on a wire is taken to the moon. Its period —

Answer the 2 checkpoints as you read.