Pendulums

You’ll be able to

Derive the simple pendulum's SHM using the small-angle approximation
Compute the period of a simple pendulum
Apply the physical-pendulum period formula T = 2π√(I/mgd)

A pendulum is SHM in disguise

For a simple pendulum of length L, gravity provides a restoring torque that gives d²θ/dt² = −(g/L) sin θ. The sine makes this not quite SHM — until the small-angle approximation sin θ ≈ θ (valid for small swings) turns it into d²θ/dt² = −(g/L)θ. That is the SHM equation with ω² = g/L, so the pendulum oscillates sinusoidally for small amplitudes.

Simple pendulum

ω = √(g/L) T = 2π√(L/g)

Valid in the small-angle limit sin θ ≈ θ. Remarkably, the period depends on length and g only — not on the bob's mass or (for small swings) the amplitude.

The physical pendulum

A physical pendulum is any rigid body swinging about a pivot, not just a point mass on a string. Its period is T = 2π√(I/mgd), where I is the moment of inertia about the pivot and d is the distance from the pivot to the center of mass. The simple pendulum is the special case I = mL², d = L, which recovers T = 2π√(L/g).

Physical pendulum

T = 2π√(I / (mgd))

I is the moment of inertia about the pivot; d is the pivot-to-center-of-mass distance. Reduces to the simple-pendulum result when the mass is concentrated at distance L.

Worked example

A simple pendulum has a length of 1 m. Using g = 9.8 m/s², find its period for small oscillations.

1.Use the simple-pendulum formula T = 2π√(L/g).
2.Compute the ratio: L/g = 1/9.8 ≈ 0.102 s².
3.Take the square root: √0.102 ≈ 0.319 s.
4.Multiply by 2π: T = 2π(0.319) ≈ 2.0 s.

Answer: T = 2π√(L/g) = 2π√(1/9.8) ≈ 2.0 s

Checkpoint

The length of a simple pendulum is quadrupled. Its period:

Checkpoint

The simple-pendulum formula T = 2π√(L/g) relies on which assumption?

On the exam

State the small-angle approximation explicitly on free-response: sin θ ≈ θ is what turns d²θ/dt² = −(g/L)sin θ into SHM. Remember the simple-pendulum period is independent of mass, and use the physical-pendulum formula T = 2π√(I/mgd) whenever the swinging object is an extended body.

Answer the 2 checkpoints as you read.