Simple Pendulum Calculator
Enter any two of the four pendulum variables - length, period, frequency, or gravitational acceleration - and this calculator instantly solves for the others using the small-angle pendulum formula T = 2pi * sqrt(L/g). Switch between metric and imperial length units, choose a planet gravity preset, or enter a custom gravity value. The step-by-step panel shows the full working so you can check every line of arithmetic.
Formula
Worked example
A pendulum 1.00 m long on Earth (g = 9.807 m/s^2): T = 2pi * sqrt(1.00 / 9.807) = 2pi * 0.3193 = 2.006 s. Frequency f = 1 / 2.006 = 0.4985 Hz. Angular frequency omega = 2pi / 2.006 = 3.133 rad/s.
What is a simple pendulum?
A simple pendulum is an idealised system consisting of a small, dense bob suspended from a fixed pivot by a massless, inextensible string or rod. When displaced to one side and released, it swings back and forth under gravity in a repeating arc. The simple pendulum is one of the most studied systems in classical physics because its motion is approximately simple harmonic for small angles, making it straightforward to analyse with a clean closed-form equation. Real pendulums deviate slightly from the ideal because the string has some mass, the bob has finite size, and air resistance damps the motion, but for most practical and educational purposes the simple model is excellent.
The small-angle formula and when it applies
For small angular displacements - generally accepted as up to about 15 degrees from vertical - the restoring force is proportional to the displacement, and the motion is simple harmonic. The period is T = 2pi * sqrt(L/g), which depends only on the length L and the local gravitational acceleration g, not on the mass of the bob or the amplitude of the swing. This is the isochronous property discovered by Galileo. For larger angles the formula underestimates the true period; the exact solution involves complete elliptic integrals of the first kind, adding roughly 0.5% error at 20 degrees and 1.7% at 30 degrees. A corrected approximation for moderate angles is T_corrected = T_small * (1 + theta^2 / 16), where theta is in radians.
How to measure local gravity with a pendulum
Rearranging the formula gives g = 4pi^2 * L / T^2. This means that by timing the oscillations of a pendulum of known length you can determine local gravitational acceleration. The technique was standard practice in geodesy for centuries: scientists would carry a precision pendulum to different locations and measure how the period changed with latitude and altitude, mapping the variations in g across the Earth. Modern gravimeters use quantum interference of falling atoms, but the pendulum method still demonstrates the principle clearly. To get a good measurement, time at least 50 complete oscillations (rather than one) and divide the total time by the number of swings - this averages out timing errors substantially.
The seconds pendulum and pendulum clocks
A "seconds pendulum" is defined as one whose half-period equals exactly one second, so the full period is T = 2 s. On the surface of the Earth this requires a length of L = g / pi^2, which works out to about 0.993 m or 39.1 inches. This is not a coincidence: the metre was originally proposed in the 1790s as exactly one-tenth of a seconds pendulum length, and the fraction was set to make the arithmetic convenient. Pendulum clocks use the isochronous property to keep time: a one-metre pendulum swings 86,400 times in 24 hours (one tick and one tock per oscillation). Temperature changes alter the rod length and therefore the period, so precision clockmakers used temperature-compensating materials such as invar alloys or compensated pendulum designs.
Pendulum periods for common lengths on Earth (g = 9.807 m/s^2)
| Length (m) | Length (ft) | Period (s) | Frequency (Hz) | Use case |
|---|---|---|---|---|
| 0.10 | 0.33 | 0.634 | 1.577 | Lab / toy pendulum |
| 0.25 | 0.82 | 1.003 | 0.997 | Short classroom demo |
| 0.50 | 1.64 | 1.419 | 0.705 | Metronome range |
| 1.00 | 3.28 | 2.006 | 0.499 | Classic 1-m pendulum |
| 0.993 | 3.26 | 2.000 | 0.500 | Seconds pendulum (T = 2 s) |
| 2.00 | 6.56 | 2.837 | 0.352 | Grandfather clock range |
| 5.00 | 16.4 | 4.488 | 0.223 | Large demonstration rig |
| 10.0 | 32.8 | 6.347 | 0.158 | Foucault-type pendulum |
All values use the small-angle formula T = 2pi * sqrt(L/g) and are accurate for amplitudes below 15 degrees.
Frequently asked questions
Does the mass of the bob affect the period?
No. For an ideal simple pendulum the period depends only on the length and local gravity, not on the mass or material of the bob. This was one of Galileo's key observations. In a real pendulum with air resistance, a heavier bob will lose energy more slowly and the damping rate changes, but the natural frequency itself is not affected by mass.
What happens to the period on the Moon?
The Moon's surface gravity is about 1.62 m/s^2, roughly 1/6 of Earth's. Because g appears under the square root in the denominator, a lower g gives a longer period. A 1-metre pendulum that takes 2.006 s on Earth would take 2pi * sqrt(1 / 1.62) = 4.94 s on the Moon - about 2.5 times longer. Select "Moon" in the gravity preset to see this instantly.
How accurate is the small-angle formula?
For amplitudes up to about 5 degrees the error is less than 0.05%. At 10 degrees it is about 0.19%, at 15 degrees about 0.43%, and at 30 degrees about 1.7%. For most classroom and engineering purposes the formula is excellent up to 15 degrees. For larger swings you need an elliptic integral correction or a numerical solver.
What is the difference between period and frequency?
Period (T) is the time in seconds for one complete back-and-forth oscillation. Frequency (f) is the number of complete oscillations per second, measured in hertz (Hz). They are exact reciprocals: f = 1/T and T = 1/f. Angular frequency (omega) is 2pi times the frequency and gives the rate of change of phase in radians per second.
How do I find the length needed for a specific period?
Rearrange the formula to L = g * (T / (2pi))^2. For example, a period of exactly 1 second on Earth (g = 9.807 m/s^2) requires L = 9.807 * (1 / (2pi))^2 = 0.2483 m (about 24.8 cm). Switch "Solve for" to Length in this calculator, enter your desired period, and the answer is instant.
Can I use this calculator for a physical pendulum?
A physical (or compound) pendulum - where the mass is distributed along the rod rather than concentrated at the tip - has a different effective length called the radius of gyration. This calculator covers only the idealised simple pendulum. For a uniform rod pivoted at one end, the effective length is 2/3 of the rod length, so you can approximate it by entering that reduced length.