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Pendulum Period Calculator

Calculate the period of a simple pendulum from length, gravity, and amplitude.
Includes large-angle correction beyond the small-angle limit.
Supports preset gravities.

Pendulum Period

A simple pendulum is a mass swinging from a string or rod, idealized as a point mass on a massless string. Its period (time for one complete swing) depends only on length and local gravity:

T = 2π × √(L / g)

Mass doesn’t appear. A 1-kilogram bob and a 100-gram bob on the same string swing at exactly the same rate. Galileo reportedly noticed this watching a chandelier sway during a church service, using his own pulse as a stopwatch.

Why this calc matters more than the equation looks

The formula is short, but the assumptions hidden in “simple” are real:

  • Point mass on a massless string. Real pendulums have distributed mass; a wooden pole hanging from a pivot is a physical pendulum and uses a different formula (T = 2π √(I/mgd) with I being the moment of inertia about the pivot). The simple-pendulum formula is accurate when the bob is much heavier than the string and small compared to the length.
  • No air resistance, no friction. Real pendulums lose amplitude with each swing. The period is essentially unaffected, but the swing dies out exponentially.
  • Small angle. The formula is exact only in the limit of zero amplitude. Bigger swings take slightly longer.

The large-angle correction

For amplitudes above about 15°, the period is noticeably longer than the textbook formula. The exact period uses an elliptic integral, but a useful series approximation is:

T = T₀ × [1 + (1/16)θ² + (11/3072)θ⁴ + …]

where θ is the half-amplitude in radians and T₀ = 2π√(L/g) is the small-angle period.

Amplitude (half-angle) Correction factor
1.0005 (0.05% slower)
10° 1.0019 (0.2% slower)
15° 1.0043 (0.43% slower)
30° 1.017 (1.7% slower)
45° 1.040 (4% slower)
60° 1.073 (7.3% slower)
90° 1.180 (18% slower)

A grandfather clock with a 1 m pendulum swung at 45° instead of its design amplitude of about 3° would lose almost 4% on time, roughly 1 hour per day. This is why precision pendulum clocks were always designed with very small swing amplitudes.

Gravity matters too

g varies from 9.780 m/s² at the equator to 9.832 m/s² near the poles, a 0.5% difference. The same pendulum keeps slightly different time in Quito versus Stockholm. Off-Earth, the period changes dramatically:

Location g (m/s²) Period of 1-m pendulum
Equator (sea level) 9.780 2.011 s
Earth standard 9.81 2.006 s
London 9.812 2.006 s
Poles 9.832 2.004 s
Moon 1.62 4.937 s
Mars 3.71 3.262 s
Jupiter (cloud top) 24.79 1.262 s

A 1-second pendulum (the kind used in regulator clocks) needs L ≈ 0.994 m on Earth, but only 0.164 m on the Moon. The Moon’s lower gravity makes pendulum clocks tick slower for any given length.

Worked example: tuning a grandfather clock

A pendulum needs T = 2 s exactly to drive a clock at the right speed. On Earth (g = 9.81):

L = g × T² / (4π²) = 9.81 × 4 / 39.48 = 0.9936 m (about 99.4 cm)

That’s why traditional grandfather clocks are around a meter tall: the pendulum needs to be that long for a 2-second tick. If the clock is running fast, the pendulum gets lengthened; if running slow, shortened. A turn of the regulating nut at the bottom usually moves the bob 0.5 mm, changing the period by about 1 part in 4000, roughly 20 seconds per day.

The seismometer connection

A horizontal pendulum (Zöllner suspension) can be made to have an arbitrarily long natural period by adjusting its tilt. This is the trick used in seismometers: a slow, long-period pendulum stays nearly still while the ground vibrates underneath it, so its motion relative to the frame records the seismic wave. The simple T = 2π√(L/g) formula doesn’t quite apply because gravity acts on a tilted geometry, but it’s the same physics extended to a clever configuration.

Foucault pendulum

A pendulum freely swinging in any plane appears to rotate that plane over time, completing one rotation in T_F = 24 hours / sin(latitude). At the poles, the plane completes one full rotation per day. At the equator, it doesn’t rotate at all. At Paris (Foucault’s original 1851 demonstration at the Panthéon, latitude 48.86°N), the rotation period is about 32 hours. This is direct visual proof that Earth rotates beneath you.

What this calculator does

  • Computes T₀ from L and g (the standard formula)
  • Applies the large-angle correction for amplitudes you specify
  • Shows the small-angle vs corrected period side by side
  • Estimates how much a clock at this period would gain or lose per day if it were intended for exactly 2 seconds

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