Physics · Classical Mechanics · Oscillations & Waves
Pendulum Period Calculator
Calculates the period of a simple pendulum given its length and local gravitational acceleration.
Calculator
Formula
T is the period of oscillation (seconds) — the time for one complete back-and-forth swing. L is the length of the pendulum (meters), measured from the pivot point to the center of mass of the bob. g is the local gravitational acceleration (m/s²), approximately 9.81 m/s² on Earth's surface. This formula is valid for small-angle oscillations (amplitudes below roughly 15°), under which sin(θ) ≈ θ in radians.
Source: Halliday, Resnick & Walker — Fundamentals of Physics, 10th Edition, Chapter 15.
How it works
A simple pendulum consists of a point mass (the bob) suspended from a fixed pivot by a massless, inextensible string or rod. When displaced from its equilibrium position and released, gravity provides a restoring force that causes the bob to oscillate. For small angular displacements — typically less than about 15° from vertical — the restoring force is approximately proportional to the displacement, making the motion simple harmonic. This small-angle approximation is what allows the elegant closed-form solution for the period.
The period is given by T = 2π√(L/g), where T is the time in seconds for one complete oscillation (one full swing out and back), L is the effective length of the pendulum in meters from the pivot to the center of mass of the bob, and g is the local gravitational acceleration in m/s². From this formula, the frequency f = 1/T (oscillations per second, measured in Hertz) and the angular frequency ω = √(g/L) (radians per second) are immediately derivable. Notably, neither the mass of the bob nor the amplitude of swing (for small angles) appears in the formula — only length and gravity determine the period.
Pendulum dynamics have enormous practical relevance. Historically, pendulum clocks were the most accurate timekeepers for centuries, exploiting the isochronous (constant-period) nature of small oscillations. Today, the formula is used to measure local gravitational acceleration with high precision, to calibrate seismometers, to design Foucault pendulums that demonstrate Earth's rotation, and to study resonance phenomena in structures and mechanical systems. The simple pendulum also serves as a foundational model for understanding more complex oscillators in engineering and physics.
Worked example
Suppose you want to find the period of a pendulum with a length of L = 2.5 m located on Earth's surface where g = 9.81 m/s².
Step 1 — Compute L/g: 2.5 ÷ 9.81 = 0.2549 s²
Step 2 — Take the square root: √0.2549 = 0.5049 s
Step 3 — Multiply by 2π: 2π × 0.5049 = 3.1727 s
The pendulum completes one full oscillation every 3.17 seconds. Its frequency is 1 ÷ 3.1727 ≈ 0.3151 Hz, and its angular frequency is √(9.81 ÷ 2.5) = √3.924 ≈ 1.9809 rad/s.
As a quick sanity check, a 1-meter pendulum on Earth has a well-known period of approximately 2.006 s. Scaling by √2.5 ≈ 1.581 gives 2.006 × 1.581 ≈ 3.17 s, which matches our result perfectly.
Limitations & notes
The formula T = 2π√(L/g) relies on the small-angle approximation (sin θ ≈ θ), which is accurate to within 1% only for amplitudes below roughly 7° and within about 4% up to 15°. For larger swings, the true period is longer and requires an elliptic integral correction: T = 2π√(L/g) × [1 + (1/16)θ₀² + ...], where θ₀ is the initial angle in radians. The model also assumes a point mass bob and a massless, rigid rod; real pendulums with distributed mass require the physical pendulum formula using the moment of inertia. Air resistance and pivot friction are neglected, both of which cause amplitude to decay over time (damped oscillations) and slightly shift the effective period. The formula further assumes a uniform, constant gravitational field — it does not account for the slight variation in g with altitude, latitude, or subsurface density anomalies, though these effects are small for most practical purposes. Finally, for pendulums with very long lengths (approaching planetary scales) or operating in non-inertial reference frames, relativistic or rotational corrections may be necessary.
Frequently asked questions
Does the mass of the pendulum bob affect its period?
No. In the simple pendulum formula T = 2π√(L/g), mass does not appear. This is because both the gravitational force on the bob and its inertia are proportional to mass, so the two effects cancel exactly. This result was famously demonstrated by Galileo and later formalized by Newton.
How does pendulum length affect the period?
The period is proportional to the square root of the length. Doubling the length increases the period by a factor of √2 ≈ 1.414. To double the period, you must quadruple the length. This square-root relationship means that very long pendulums are needed to achieve significantly longer periods.
What value of g should I use for my location?
The standard value of g = 9.80665 m/s² is the internationally defined standard gravity. In practice, g varies from about 9.76 m/s² near the equator to 9.83 m/s² near the poles due to Earth's rotation and oblateness. For most calculations, 9.81 m/s² is a good approximation, but for precision measurements you should use a locally measured value.
What is the period of a 1-meter pendulum on Earth?
A pendulum of exactly 1 meter length on Earth (g = 9.81 m/s²) has a period of T = 2π√(1/9.81) ≈ 2.006 seconds. This is a useful benchmark: a so-called 'seconds pendulum' that ticks once per second (half-period = 1 s) has a length of approximately 0.994 meters, very close to 1 meter.
How would a pendulum behave on the Moon?
The Moon's surface gravity is about 1.62 m/s², roughly 1/6 of Earth's. A 1-meter pendulum on the Moon would have a period of T = 2π√(1/1.62) ≈ 4.94 seconds — about 2.46 times longer than on Earth. This illustrates how pendulums can be used to measure gravitational acceleration on other bodies.
Last updated: 2025-01-15 · Formula verified against primary sources.