Period equals 2 pi times the square root of length divided by gravity

Solution

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How It Works

A simple pendulum's period (T = 2π√(L/g)) depends only on the string length and local gravity — not on the mass of the bob or the swing angle (for small angles). A longer string swings more slowly; stronger gravity makes it swing faster. The frequency is simply f = 1/T.

Example Problem

A grandfather clock has a pendulum 1 m long. What is its period on Earth (g = 9.81 m/s²)?

  1. T = 2π√(1 / 9.81) = 2π × 0.3193
  2. T ≈ 2.006 s — almost exactly a 2-second period, which is by design.

Frequently Asked Questions

Does the mass of a pendulum affect its period?

No. For a simple pendulum, the period depends only on length and gravity. A heavier bob swings at the same rate as a lighter one (assuming the same string length).

Why does the small-angle approximation matter?

The formula T = 2π√(L/g) is exact only for infinitesimally small swings. For angles up to about 15° the error is less than 0.5%, but for large swings the period increases and the formula becomes less accurate.

Can you use a pendulum to measure gravity?

Yes. Measure the period T and length L, then solve for g = 4π²L/T². This method was historically used to map variations in gravity across the Earth.

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Reference: Lindeburg, Michael R. 1992. Engineer In Training Reference Manual. Professional Publication, Inc. 8th Edition.