Pendulum Period Calculator

Simple pendulum with length L, mass m, and angle θ
Period equals 2 pi times the square root of length divided by gravity

Solution

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How It Works

The period (T) of a simple pendulum is the time it takes to complete one full swing, and depends only on the string length (L) and the local gravitational acceleration (g). The relationship is T = 2π√(L/g) — note that the mass of the bob does not appear. A longer string gives a longer period; stronger gravity gives a shorter period. The formula assumes small swing angles (below ~15°), where sin(θ) ≈ θ; larger amplitudes require an elliptic-integral correction.

Example Problem

A grandfather clock pendulum is 1 m long. What is its period on Earth, where g = 9.81 m/s²?

  1. Identify knowns: L = 1 m, g = 9.81 m/s².
  2. Compute L / g = 1 / 9.81 ≈ 0.10194 s².
  3. Take the square root: √0.10194 ≈ 0.31928 s.
  4. Multiply by 2π: T = 2π × 0.31928 ≈ 2.0060 s.
  5. The clock ticks once per half-swing, so a 2-second period gives one tick per second — by design.

The 2-second period is why grandfather clocks are built around a roughly 1-metre pendulum: it matches the natural swing of a one-second tick.

Key Concepts

T = 2π√(L/g) is the small-angle simple-pendulum equation. It is independent of bob mass (Galileo's discovery) and independent of swing amplitude as long as the amplitude is small. The frequency is the reciprocal of the period: f = 1/T. For larger angles, the true period exceeds the small-angle value by a few tenths of a percent at 15°, but more dramatically as the amplitude grows.

Applications

  • Designing grandfather clock pendulums to produce a specific tick rate (a 1 m rod gives a 2 s period).
  • Calibrating a pendulum's length so that it swings in resonance with a forcing oscillator.
  • Estimating swing time for a child's swing, a hanging chandelier, or any cable-and-mass system.
  • Validating a measured local gravity value by comparing the predicted period to a stopwatch reading.
  • Teaching simple harmonic motion and the small-angle approximation in introductory physics.

Common Mistakes

  • Including the bob mass in the formula — it cancels out in the derivation; only L and g matter.
  • Measuring length to the bottom of the bob instead of to its center of mass; the effective length is pivot-to-COM.
  • Applying the small-angle formula to swings larger than about 15°; the real period is longer.
  • Forgetting to use SI units — L must be in metres and g in m/s² for T to come out in seconds.
  • Confusing period (one full back-and-forth) with the half-period (one swing direction only).

Frequently Asked Questions

How do you calculate the period of a pendulum?

Use T = 2π√(L/g). Divide the string length L by the gravitational acceleration g, take the square root, then multiply by 2π. The result is the time for one complete swing in seconds.

What is the formula for pendulum period?

T = 2π√(L/g), where L is the string length in metres and g is the local gravitational acceleration in m/s². The bob's mass does not appear in the formula.

Does pendulum mass affect the period?

No. For a simple pendulum at small angles, the period depends only on the string length and gravity — not on how heavy the bob is. A bowling ball and a tennis ball on identical 1 m strings swing at the same rate.

How long is a 1-second pendulum?

Rearranging T = 2π√(L/g) gives L = g(T/2π)². For T = 1 s on Earth, L = 9.81 × (1/(2π))² ≈ 0.248 m, or about 24.8 cm.

Why does the small-angle approximation matter?

T = 2π√(L/g) is exact only for infinitesimally small swings. For 15° amplitudes the error is below 0.5%, but it grows quickly: a 45° amplitude pendulum has a period about 4% longer than the small-angle formula predicts.

How does gravity change the pendulum period?

A pendulum on the Moon (g ≈ 1.62 m/s²) swings about √(9.81/1.62) ≈ 2.46× slower than the same pendulum on Earth. Stronger gravity (a denser planet) speeds up the swing.

Reference:

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

Simple Pendulum Diagram

A simple pendulum is a point mass on a massless string of length L, free to swing through small angles under gravity g. The period only depends on the string length and the gravitational field — not on the bob's mass or, for small angles, the amplitude.

θmLg

L — length from pivot to bob center of mass · m — bob mass (cancels out of T) · θ — swing angle from vertical (must be small for T = 2π√(L/g) to be accurate) · g — local gravitational acceleration.

Worked Examples

Seismology

What is the period of Foucault's 67 m pendulum at the Panthéon in Paris?

In 1851 Léon Foucault hung a 28 kg brass bob from the Panthéon dome on a 67 m wire to prove Earth's rotation. Compute the swing period at Paris's local g ≈ 9.81 m/s².

  • Knowns: L = 67 m, g = 9.81 m/s²
  • T = 2π × √(L / g)
  • T = 2π × √(67 / 9.81)
  • T = 2π × √6.830
  • T = 2π × 2.6135

T ≈ 16.42 s

The slow swing made it easy to mark each plane traversal in chalk on a sand ring; the plane of swing appeared to rotate by about 11° per hour at Paris's latitude (sin 49°). Modern seismometers use much shorter pendulums but the same T = 2π√(L/g) relation.

Planetary Science

How would a 1.5 m pendulum behave inside a Mars habitat?

Imagine a wall clock with a 1.5 m pendulum mounted inside a future Mars surface habitat. Mars has a surface gravity of g ≈ 3.72 m/s² — calculate the natural swing period.

  • Knowns: L = 1.5 m, g_mars = 3.72 m/s²
  • T = 2π × √(L / g)
  • T = 2π × √(1.5 / 3.72)
  • T = 2π × √0.4032
  • T = 2π × 0.6350

T ≈ 3.99 s

An Earth-tuned pendulum clock would run roughly √(9.81/3.72) ≈ 1.62× slower on Mars, losing about 9 hours per Earth day. Habitat clocks would need physically shorter pendulums (≈ 0.57 m for a 1-second tick) or electromechanical timekeeping.

Clockmaking

Verify the historical "seconds pendulum" (T = 2 s, L ≈ 0.994 m)

Before the metre was standardised by Earth-quadrant measurement, French scientists proposed defining it as the length of a pendulum whose half-period equals 1 second. Plug L = 0.994 m and standard gravity g = 9.80665 m/s² into Kater's formula and confirm T = 2 s.

  • Knowns: L = 0.994 m, g₀ = 9.80665 m/s² (standard gravity)
  • T = 2π × √(L / g)
  • T = 2π × √(0.994 / 9.80665)
  • T = 2π × √0.10136
  • T = 2π × 0.31838

T ≈ 2.000 s

Each half-swing is exactly one second — that's why it's called the "seconds pendulum." The proposal was eventually dropped because g varies with latitude and altitude, but high-precision pendulum gravimeters used the same principle to map Earth's gravity field through the 20th century.

Simple Pendulum Period Formula

For small swing angles, the period of a simple pendulum depends only on its length and the local gravitational acceleration — not on the bob's mass and (to first order) not on the amplitude. This makes a pendulum one of the simplest practical ways to measure g.

T = 2π × √(L / g)

Where:

  • T — period — time for one full back-and-forth swing (s)
  • L — length from the pivot to the center of mass of the bob (m)
  • g — local gravitational acceleration (m/s²); about 9.81 on Earth's surface

The formula assumes a massless string, a point-mass bob, and small-angle motion (θ ≲ 15°), where sin θ ≈ θ. At larger amplitudes the period lengthens slightly — the leading correction adds about θ²/16 to T, so a 30° swing is roughly 1.7% slower than the small-angle result. The mass of the bob cancels out entirely; this is why Galileo could conclude that all pendulums of equal length keep time the same way.

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