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Calcimator

Pendulum Calculator

Calculate the period and frequency of a simple pendulum from its length and local gravitational acceleration using T = 2π√(L/g).

About this calculator

A simple pendulum -- a point mass swinging on a massless, inextensible string or rod from a fixed pivot -- has a period that depends on only two things: how long the pendulum is and how strong gravity is where it's swinging. It notably does NOT depend on the mass of the bob or, for small swings, on how far it's displaced -- a heavier bob and a lighter one released from the same small angle swing with the identical period, which is the classical result Galileo is credited with first noticing. The period formula, T = 2π√(L/g), comes from solving the pendulum's equation of motion under the small-angle approximation (sin θ ≈ θ, accurate to within about 0.2% at a 10-degree swing from vertical and about 0.4% at 15 degrees, with error growing further beyond that), which turns the pendulum's true nonlinear motion into simple harmonic motion -- the same mathematical form as a mass on a spring.

Because period scales with the SQUARE ROOT of length, quadrupling the length only doubles the period, not quadruples it; and because gravity sits in the denominator under that same square root, a pendulum on the Moon (g ≈ 1.62 m/s², about a sixth of Earth's) swings noticeably slower -- its period is about 2.46 times longer than the identical pendulum on Earth. This length-and-gravity-only dependence is precisely why pendulum clocks work as reliable timekeepers (a fixed-length pendulum keeps consistent time regardless of how it's wound or how heavy its bob is) and why a pendulum makes a genuine, if crude, gravimeter -- timing a pendulum's period at a location lets you solve the same formula backward to measure local g.

Inputs

ft
m/s²

Results

Period

2.01 s

Frequency

0.5 Hz

Angular Frequency3.13 rad/s
How to Use This Calculator
  1. Enter Pendulum Length and Gravitational Acceleration.
  2. Review Period (s) and Frequency (Hz).
  3. Use Angular Frequency (rad/s) to inform your decision.
  4. Use the chart to visualize the results and explore different scenarios by adjusting inputs.

How the result changes with Gravitational Acceleration

Gravitational AccelerationPeriodFrequency
4.912.84 s0.35 Hz
7.362.32 s0.43 Hz
151.62 s0.62 Hz
251.26 s0.8 Hz

What each input means

Pendulum Length
Length of the pendulum from the pivot to the center of mass.
Gravitational Acceleration
Local gravitational acceleration. Earth surface ≈ 9.81 m/s², Moon ≈ 1.62 m/s².

How this is calculated

Worked example, using the default values

  1. Identify Input Parameters
    Pendulum Length = 1, Gravitational Acceleration = 9.81 = 2 input(s) provided
  2. Calculate Period
    Period = 2π√(L/g)
    2.0061 = 2.0061
  3. Calculate Frequency
    Frequency
    0.4985 = 0.4985
  4. Calculate Angular Frequency
    Angular Frequency
    3.132 = 3.132

Engine last updated . Checked against 1 independently-derived test — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.

Frequently Asked Questions

Why doesn't the pendulum's mass affect its period?

Because the mass appears on both sides of the pendulum's equation of motion and cancels out -- a heavier bob has more inertia resisting acceleration, but gravity also pulls on it with proportionally more force, and those two effects offset exactly. This is the same mass-independence behind objects in free fall accelerating at the same rate regardless of weight. It's why a grandfather clock's pendulum bob can be swapped for a different mass without throwing off the clock's timekeeping, as long as the pendulum's length stays the same.

Why does quadrupling the pendulum length only double its period?

Because period is proportional to the SQUARE ROOT of length, not to length directly -- the formula T = 2π√(L/g) has L inside a square root. Doubling the period requires quadrupling the length (since √4 = 2), and to double the period again would require quadrupling the length again, to 16 times the original. This square-root relationship is why grandfather clocks with roughly 1-meter pendulums (about a 2-second period) are a practical size -- reaching a much longer period would require a dramatically, not proportionally, longer pendulum.

How much slower would this pendulum swing on the Moon than on Earth?

About 2.46 times slower (a longer period), because the Moon's surface gravity (about 1.62 m/s²) is roughly one-sixth of Earth's 9.81 m/s², and period scales with 1/√g. Taking the square root of that roughly sixfold drop in gravity gives the period multiplier: √(9.81/1.62) ≈ 2.46. Enter 1.62 for Gravitational Acceleration with the same pendulum length to see this directly -- it's a common way to illustrate why timekeeping and physical intuition built on Earth don't transfer directly to other worlds.

Does this formula work for a pendulum swung through a wide angle?

No -- T = 2π√(L/g) is a small-angle approximation, accurate to within about 0.2% at a 10-degree swing from vertical and about 0.4% at 15 degrees, and it becomes noticeably wrong for larger swings. A pendulum released from a wide angle (say 90 degrees) takes measurably longer per swing than this formula predicts, because the restoring force is no longer proportional to displacement once the small-angle approximation breaks down. This calculator does not take a swing-angle input and always applies the small-angle result.

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