Pendulum Period Calculator
Calculate the period of a simple pendulum from length, gravity, and amplitude.
Includes a large-angle correction beyond the small-angle limit.
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
How we build and check this calculator
This calculator runs entirely in your browser, so the numbers you enter stay on your device. The math behind it is written by hand and tested against worked examples and standard references before the page goes live.
SuperGlobalCalculator is independently built and maintained. See how we build and verify our calculators.