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.
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 |
|---|---|
| 5° | 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