Simple Pendulum Calculator
Calculate the period and frequency of a small-angle simple pendulum given its length and the local gravitational acceleration. The classical T = 2π√(L/g) formula.
Category: Physics
Simple Pendulum Calculator Inputs
Simple Pendulum Calculator Formula
Equation
T = 2π √(\fracL)g
Excel Formula
=T=2PISQRT({L){g}}
Variables
- Length L (m) — Length of the pendulum string (m).
- Gravity g (m/s²) — Local gravitational acceleration (9.81 m/s² on Earth at sea level).
How the Simple Pendulum Calculator Works
A simple pendulum is a point mass on a massless, inextensible string that swings under gravity. For small angular displacements its motion is simple harmonic, with a period that depends only on length and gravity — not on mass or amplitude. This independence is called isochronism, discovered by Galileo.
The core relationship is T = 2\pi \sqrt{\frac{L}{g}}. Typical inputs include Length L, Gravity g.
Enter your values in the simple pendulum calculator above, review the step-by-step solution, and compare against the worked examples below so you can see how each input changes the result. This free online physics tool is built for homework, design checks, and professional verification.
Simple Pendulum Calculator Theory & Explanation
Equation of motion
Let θ be the angular displacement from vertical. The torque from gravity about the pivot is −m g L sin θ, leading to:
\ddotθ + (g)/(L) \sinθ = 0
Small-angle approximation
When θ is small (in radians), sin θ ≈ θ. The equation reduces to the simple harmonic oscillator equation, whose solution has angular frequency ω² = g/L:
\omega = √(\fracg)L \quad\Rightarrow\quad T = (2π)/(\omega) = 2π √(\fracL)g
Period, frequency, angular frequency
Three related quantities characterize the oscillation:
f = (1)/(T) = (1)/(2π)√(\fracg)L, \qquad \omega = 2π f = √(\fracg)L
Large-angle correction (1st order)
For larger amplitudes (θ up to ~30°) the period lengthens. A common 4th-order correction is:
T ≈ 2π√(\fracL)g(1 + (θ_0^2)/(16) + (11\,θ_0^4)/(3072) + ·s)
Validity and limits
The simple-pendulum model assumes a point mass and a massless, rigid string. For a real clock pendulum the rod has finite mass (small correction) and air resistance damps motion over time. The 1-second "seconds pendulum" is roughly 0.2488 m long on Earth.
Why the Period Does Not Depend on Mass
The restoring force on a displaced pendulum bob is mg\sinθ, proportional to mass; the inertia resisting that force is also proportional to mass. The two cancel exactly, leaving a period that depends only on length and gravity.
This is the same cancellation that makes all objects fall at the same rate in vacuum, and it is a direct consequence of gravitational and inertial mass being equal — the equivalence principle. Amplitude drops out too, but only approximately: the small-angle result holds because \sinθ ≈ θ for small θ, and it degrades as the swing widens.
T = 2π√(\fracL)g
The Small-Angle Approximation and Its Error
The exact period involves an elliptic integral, whose series expansion shows how the error grows with amplitude θ_0. At 10° the true period is about 0.19% longer than the small-angle formula predicts; at 30° it is roughly 1.7% longer; at 90° about 18% longer.
For a pendulum clock this matters enormously: a 0.19% error accumulates to nearly three minutes a day. It is why precision clock pendulums swing through only a few degrees, and why the escapement is designed to keep the amplitude constant — a drifting amplitude would mean a drifting rate.
T = 2π√(\fracL)g(1 + (1)/(16)θ_0^2 + (11)/(3072)θ_0^4 + ·s)
Using a Pendulum to Measure g
Rearranging the period formula gives g = 4π^2 L / T^2, one of the oldest and still one of the most accessible precision measurements in physics. Because T is squared, timing dominates the error budget — so the standard technique is to time 50 or 100 complete swings and divide, reducing the effect of reaction-time error by the same factor.
For an honest result, measure L to the centre of mass of the bob rather than to its top, keep the amplitude under about 10° so the small-angle formula holds, and repeat at several lengths: plotting T^2 against L should give a straight line through the origin with slope 4π^2/g. A systematic offset in the line's intercept usually means the length was measured to the wrong point on the bob.
g = (4π^2 L)/(T^2)
Simple Pendulum Calculator Worked Examples
Worked Example
Inputs
- lengthL: 1
- gravityG: 9.81
Result: period: 2.0061 frequency: 0.4985 angularFrequency: 3.1321
Explanation
A 1-meter pendulum on Earth (g = 9.81 m/s²) completes a full swing in T = 2π × √(1/9.81) ≈ 2.006 s. This corresponds to a frequency of about 0.498 Hz and an angular frequency of √(9.81/1) ≈ 3.132 rad/s.
Second Scenario
Inputs
- lengthL: 2.25
- gravityG: 9.81
Result: period: 2.0061 frequency: 0.4985 angularFrequency: 3.1321
Explanation
This scenario uses different inputs (lengthL = 2.25, gravityG = 9.81) to show how changing one variable affects the simple pendulum result. Run the calculator above with these values to get the exact updated output with step-by-step work.
Common Simple Pendulum Calculator Use Cases
- Physics problem sets and labs
- Engineering design checks
- Unit and formula verification
- Simple Pendulum homework and study
- Simple Pendulum design and analysis
Simple Pendulum Calculator FAQs
Does the mass matter?
No. In the small-angle linearized equation, mass cancels. This is unusual — most oscillators have a mass dependence — but it follows directly from the gravitational restoring torque scaling linearly with m while the rotational inertia also scales linearly with m. A heavier pendulum bob swings at the same rate as a lighter one.
Why does the angle get neglected?
sin θ ≈ θ is accurate to within 1% for θ ≤ 14° (≈ 0.245 rad). For amplitudes up to ~30° the error in the period is only a few tenths of a percent, so the simple formula is adequate. Beyond that, higher-order corrections become necessary.
Why do grandfather clocks keep time on the Moon?
They don't — they would run slower. The period scales as 1/√g, so on the Moon (g ≈ 1.62 m/s²) a "seconds pendulum" would have period ≈ 2π × √(L/1.62). A clock designed for Earth would slow by a factor of √(9.81/1.62) ≈ 2.46.
What is a Foucault pendulum?
A Foucault pendulum is a long-period pendulum whose plane of oscillation appears to rotate in the Earth's rotating reference frame — visual evidence of Earth's rotation. The pendulum's period itself is unaffected; what changes is the direction of swing.
Does a heavier bob swing more slowly?
No. Mass cancels out of the period entirely, because the gravitational restoring force and the inertia resisting it are both proportional to mass. A lead bob and a wooden bob on strings of equal length keep the same time in vacuum. In air the lighter bob is damped faster by drag, so its swing dies away sooner, but the period of each swing is unchanged.