The pendulum period formula, explained.
For small angles, the restoring force gives approximately simple harmonic motion. The period grows with the square root of length and decreases with the square root of gravity. Bob mass cancels from the equation.
The relationship to remember
Key formulas
- Period (small angles)
- T = 2π√(L/g)
- Accurate below about 15°
- Frequency
- f = 1/T = (1/2π)√(g/L)
- Complete swings per second
- Length for a given period
- L = gT² / (4π²)
- Rearranged period formula
- Gravity from a measured period
- g = 4π²L / T²
- A classic lab measurement of g
- Angle over time
- θ(t) = θ₀ cos(2πt/T)
- Released from rest at θ₀
Worked example
Problem. A clock pendulum must have a period of exactly 2.00 s on Earth (g = 9.81 m/s²). How long should it be, and what would its period be on the Moon (g = 1.62 m/s²)?
- Rearrange T = 2π√(L/g) to L = gT² / (4π²).
- L = 9.81 × 2.00² / (4π²) = 39.24 / 39.48 = 0.994 m.
- On the Moon: T = 2π√(0.994 / 1.62) = 2π × 0.783 = 4.92 s.
Answer. The pendulum should be about 0.994 m long; on the Moon the same pendulum would take about 4.92 s per swing.
A quick check
At L = 1 m and g = 9.81 m/s², the period is about 2.01 s. Increasing length to 4 m doubles the period.
Inside the default experiment
- T = 2π√(L/g) = 2π√(1/9.81) = 2.01 s.
- Frequency = 1/T = 0.498 Hz.
- Small-angle motion: θ(t) = 10 cos(2πt/T) degrees.
When this model applies
Small-angle approximation, massless rod/string, point bob, no damping. Angles are limited to 15°.
Make a prediction. Then test it.
Make the pendulum twice as slow. How much longer must it be?
Try it: Pendulum simulator & period calculator ↗Frequently asked questions
Wondering about the why? Start here.
Does a heavier bob swing faster?
Not in this ideal pendulum model: mass cancels out. Length and gravity set the small-angle period.
Why does amplitude barely change the period?
The formula assumes small angles. The animation follows that approximation; for large angles, amplitude does affect period.