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?
- Identify knowns: L = 1 m, g = 9.81 m/s².
- Compute L / g = 1 / 9.81 ≈ 0.10194 s².
- Take the square root: √0.10194 ≈ 0.31928 s.
- Multiply by 2π: T = 2π × 0.31928 ≈ 2.0060 s.
- 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.
Worked Examples
Horology
What is the period of a 1 m grandfather clock pendulum on Earth?
- T = 2π√(L / g)
- T = 2π√(1 / 9.81)
- T = 2π × 0.31928
- T ≈ 2.006 s — by design, one tick per second.
Real clocks compensate for temperature-driven length changes with a bimetallic or invar rod.
Lunar Physics
How fast does the same 1 m pendulum swing on the Moon?
- T = 2π√(L / g) with g_moon = 1.62 m/s²
- T = 2π√(1 / 1.62)
- T = 2π × 0.7857
- T ≈ 4.94 s — about 2.46× slower than on Earth.
The ratio matches √(g_earth / g_moon) ≈ √6.06 ≈ 2.46.
Playground Physics
How long does it take a child on a 2.5 m swing to complete one back-and-forth?
- T = 2π√(L / g)
- T = 2π√(2.5 / 9.81)
- T = 2π × 0.5048
- T ≈ 3.17 s — about three seconds per swing.
Pumping the legs adds energy at the right phase, so the amplitude grows even though the period stays close to the small-angle value.
Related Calculators
- Pendulum Length Calculator — find the length needed for a target period using L = g(T/2π)²
- Pendulum Hub (Period, Frequency, Physical) — the full pendulum calculator with all solve-for options
- Gravity Calculator — compute local gravitational acceleration at a given altitude
- Hooke's Law Calculator — another oscillator: period of a mass on a spring
- Time Converter — convert period between seconds, minutes, and hours
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