Pendulum Period Calculator
Solve T = 2π√(L/g) for period or length.
Frequency ≈ 0.498 Hz · small-angle approximation
A simple pendulum’s period is T = 2π√(L ÷ g) — it depends only on length and gravity, not the bob’s mass or swing width. A 1 m pendulum on Earth (g = 9.81 m/s²) swings with a period of 2π√(1 ÷ 9.81) = 2.006 s, about two seconds per full swing.
The simple pendulum
For small swings, a pendulum’s period depends only on its length and the strength of gravity: T = 2π√(L/g). Remarkably, the mass of the bob and the width of the swing do not appear at all — that is what makes a pendulum a reliable timekeeper.
T = period (s), L = length (m), g = gravity (≈ 9.81 m/s²)
Worked example
A 1 m pendulum on Earth, with g = 9.81 m/s². Find its period:
- 1 Divide length by gravity. L ÷ g = 1 ÷ 9.81 = 0.10194.
- 2 Take the square root. √0.10194 = 0.31928.
- 3 Multiply by 2π. T = 2π × 0.31928 = 2.006 s — about two seconds per full swing.
Period of a 1 m pendulum by location
Same length, different gravity: weaker gravity means a slower, longer-period swing.
| Location | g (m/s²) | Period T |
|---|---|---|
| Earth | 9.81 | 2.006 s |
| Moon | 1.62 | 4.937 s |
| Mars | 3.72 | 3.258 s |
| Jupiter | 24.79 | 1.262 s |
Assumptions and common mistakes
The small-angle approximation. The formula assumes swings below roughly 15°. For larger amplitudes the real period is slightly longer, and T = 2π√(L/g) underestimates it.
Length grows as the square of the period. Because length sits under a square root, quadrupling the length only doubles the period — so a 4 m pendulum swings with a period of about 4 s, not 8 s.
Common mistake: forgetting that mass and amplitude do not matter. A heavy bob and a light bob of the same length keep the same time.