Simple Pendulum Period Calculator
A simple pendulum consists of a mass (bob) suspended by a string of length L. For small oscillation angles (below about 15 degrees), the period is T = 2π × sqrt(L / g), where g is the local gravitational acceleration. Remarkably, the period is independent of the mass of the bob and of the amplitude (for small angles). This property made pendulums the basis of precise timekeeping for centuries. Enter the pendulum length and gravity to get the period and frequency. You can also reverse-solve: use T = 2π sqrt(L/g) to find the required length for a desired period.
Simple pendulum period formula
T = 2π × √(L / g)
Where T is the period (s), L is the pendulum length (m), and g is gravitational acceleration (9.81 m/s² on Earth's surface). Frequency f = 1 / T (Hz). Valid for small oscillation angles (below about 15 degrees).
Understanding pendulum motion
- The period is proportional to the square root of length: doubling L increases T by a factor of sqrt(2) = 1.414.
- Pendulums swing more slowly at higher altitudes because g decreases slightly with distance from Earth's center.
- A compound pendulum (rigid rod) has period T = 2pi sqrt(I / mgd), where I is moment of inertia about the pivot and d is distance from pivot to center of mass.
- Foucault pendulums demonstrate Earth's rotation: the swing plane rotates at a rate depending on latitude.
- For large angles the exact period requires an elliptic integral and is always longer than the small-angle approximation.
Pendulum period: frequently asked questions
What is the period of a simple pendulum?
The period T is the time for one complete swing (back and forth). For a simple pendulum with small oscillation angles (below about 15 degrees), T = 2pi x sqrt(L/g), where L is the pendulum length and g is gravitational acceleration. The period does not depend on mass or amplitude at small angles.
Why does mass not affect the period?
The restoring force on a pendulum is proportional to mass (F = mg sin theta), but so is the inertia that resists motion (F = ma). The mass cancels in the equation of motion, leaving only length and gravity as determining factors. Galileo demonstrated this experimentally.
At what angle does the small-angle approximation break down?
The formula T = 2pi sqrt(L/g) assumes sin(theta) is approximately equal to theta in radians (valid for angles below about 15 degrees). At 30 degrees the true period is about 1.7% longer than the small-angle formula predicts. At 90 degrees the error exceeds 18%.
How long must a pendulum be to have a 1-second period?
From T = 2pi sqrt(L/g), solving for L: L = g(T/2pi)². For T = 1 s and g = 9.81 m/s²: L = 9.81 x (1/2pi)² = 0.2483 m (about 24.8 cm). A 2-second period (the classic grandfather clock half-swing) requires L = 0.9937 m (about 1 m).
Does a pendulum on the Moon swing at the same rate?
No. The Moon's surface gravity is about 1.62 m/s² compared to Earth's 9.81 m/s². A 1-meter pendulum that has a period of 2.006 s on Earth would have a period of 4.95 s on the Moon, making clocks run slower there.
Official sources
- NIST Reference on Constants, Units, and Uncertainty: Standard acceleration of gravity.
- OpenStax University Physics Volume 1: Pendulums.
Reviewed by the CalculatorHub team, edited by James Graham, 15 June 2026. See our methodology.