T = 2π√(L/g) — Period of a Simple Pendulum

The period of a simple pendulum — why clock pendulums keep time — with a worked 1-metre example.

T = 2\pi\sqrt{\frac{L}{g}}

Period equals two pi times the square root of length over g.

What it means

A pendulum's period depends only on its length and gravity — not on the bob's mass or (for small swings) the amplitude. Longer pendulums swing slower; stronger gravity swings faster. This mass-independence is why pendulums made reliable clocks for centuries.

Variables

Symbol	Name	Unit	Meaning
$T$	Period	s	Time for one full back-and-forth swing
$L$	Length	m	String length from pivot to bob center
$g$	Gravity	m/s²	Local gravitational acceleration

Assumptions

Small swing angle (below ~15°)
Mass concentrated in the bob (simple pendulum)
Negligible air resistance and pivot friction

Derivation

For small angles the restoring torque is approximately $\tau \approx -mgL\theta$ , giving simple harmonic motion with $\omega^2 = g/L$ . The period $T = 2\pi/\omega$ yields $T = 2\pi\sqrt{L/g}$ .

Worked example

Problem

What is the period of a simple pendulum with a 1.00 m string?

Given:

L = 1.00 m
g = 9.81 m/s²

Write the formula: $T = 2\pi\sqrt{L/g}$
Substitute: $T = 2\pi\sqrt{1.00/9.81}$
Calculate: $\sqrt{0.10194} = 0.31928$
Multiply: $T = 2\pi \times 0.31928 \approx 2.01\ \mathrm{s}$

≈ 2.01 s

A 1-metre pendulum takes about 2 seconds per full swing — the famous 'seconds pendulum' used in grandfather clocks. Remarkably, the bob's mass never enters the formula.