Pendulum Period Calculator
Calculate the period T of a simple pendulum from its length and gravity using T = 2pi * sqrt(L/g). Solve for period, length, frequency, or gravity; pick a planet's gravity in one click; compare the small-angle approximation with the exact elliptic-integral formula; and watch a live SVG pendulum swing at the computed period.
Your ad blocker is preventing us from showing ads
MiniWebtool is free because of ads. If this tool helped you, please support us by going Premium (ad‑free + faster tools), or allowlist MiniWebtool.com and reload.
- Allow ads for MiniWebtool.com, then reload
- Or upgrade to Premium (ad‑free)
About Pendulum Period Calculator
The Pendulum Period Calculator uses the classic simple-pendulum formula \( T = 2\pi\sqrt{L/g} \) to solve for the period \(T\), the length \(L\), the local gravity \(g\), or the natural frequency \(f\). It includes one-click planetary gravity presets, an exact large-angle correction using the elliptic-integral series, a live SVG pendulum that actually swings at the calculated rate, and energy/velocity output when you supply the bob mass.
How to Use This Pendulum Period Calculator
- Choose what to compute: T (period), L (length), g (gravity), or f (frequency). The form reshapes itself to ask only for the quantities it needs.
- Pick a planetary preset — Earth, Moon, Mars, Jupiter, Sun, ISS, and more — or switch to Custom and type your own g.
- Enter the length, period, or any combination the chosen mode requires.
- Optional: enter a swing amplitude (in degrees) and a bob mass. The calculator then reports the exact (non-small-angle) period, the maximum height, the speed at the bottom of the swing, and the peak kinetic / potential energy.
- Press Calculate and review the live SVG swing, the cross-planet comparison table, the step-by-step working, and the cycle counts per minute / hour / day.
What Makes This Calculator Different
The Pendulum Period Formula
For a point-mass bob hanging on a massless rod, swinging through a small angle in a uniform gravitational field:
\[ T \;=\; 2\pi\sqrt{\dfrac{L}{g}} \qquad\Longleftrightarrow\qquad L \;=\; g\left(\dfrac{T}{2\pi}\right)^{\!2} \qquad\Longleftrightarrow\qquad g \;=\; \dfrac{4\pi^{2}L}{T^{2}} \]
Here \(T\) is the period in seconds, \(L\) is the length from the pivot to the centre of mass of the bob (meters), and \(g\) is the local gravitational acceleration (m/s²). The natural frequency is the reciprocal of the period: \( f = 1/T \), and the angular frequency is \( \omega = 2\pi/T = \sqrt{g/L} \).
Why the Mass Doesn't Matter
If you write Newton's second law for a pendulum bob (mass \(m\)) hanging on a rod of length \(L\) at angle \(\theta\), the gravitational restoring torque is \(-m g L \sin\theta\) and the moment of inertia is \(m L^{2}\). Equation of motion:
\[ m L^{2} \ddot{\theta} \;=\; -m g L \sin\theta \quad\Rightarrow\quad \ddot{\theta} \;=\; -\dfrac{g}{L}\sin\theta. \]
The mass cancels. Two pendulums of identical length swing at exactly the same period regardless of how heavy their bobs are. The bob mass does, however, scale the kinetic and potential energy of the swing (and the tension in the rod) linearly.
Small Angle vs Exact Period
The familiar \( T = 2\pi\sqrt{L/g} \) is only the leading term of a series. The exact period is
\[ T_{exact} \;=\; 2\pi\sqrt{\dfrac{L}{g}}\;\left(1 + \tfrac{1}{16}\theta_0^{\,2} + \tfrac{11}{3072}\theta_0^{\,4} + \tfrac{173}{737280}\theta_0^{\,6} + \dots\right) \]
where \(\theta_0\) is the half-amplitude in radians. The small-angle approximation under-predicts the period by:
| Amplitude θ₀ | Excess period vs small-angle |
|---|---|
| 1° | ~ 0.002 % |
| 5° | ~ 0.048 % |
| 10° | ~ 0.19 % |
| 20° | ~ 0.77 % |
| 30° | ~ 1.74 % |
| 45° | ~ 3.99 % |
| 60° | ~ 7.32 % |
| 90° | ~ 18.0 % |
The Seconds Pendulum
Setting \(T = 2\) s (so each half-swing is one second) and \(g = 9.80665\) m/s² gives the famous "seconds pendulum" length:
\[ L \;=\; \dfrac{g\,T^{2}}{4\pi^{2}} \;=\; \dfrac{9.80665 \cdot 4}{4\pi^{2}} \;\approx\; 0.9936 \text{ m}. \]
This is the design length of every grandfather clock and was once proposed as the international metre. Because a pendulum's period depends on the local \(g\), a seconds pendulum calibrated in London ticks differently at the equator — historically this is how geodesists mapped the shape of the Earth.
Worked Example: 1 m Pendulum on Earth
- Length \(L = 1.00\) m, gravity \(g = 9.80665\) m/s².
- \( T = 2\pi\sqrt{1 / 9.80665} = 2.0064\) s (small-angle).
- Frequency \( f = 1/T \approx 0.4984 \) Hz; angular frequency \( \omega \approx 3.132 \) rad/s.
- At an amplitude of 20° the exact period is about 2.022 s — 0.77% longer.
- If the bob mass is 0.5 kg and θ₀ = 20°, the max height is \( h = L(1 - \cos 20°) \approx 0.060\) m, peak KE = peak PE \(\approx 0.295\) J, and peak speed \( v = \sqrt{2gh} \approx 1.087\) m/s.
Frequently Asked Questions
What is the formula for the period of a simple pendulum?
For small swings, \( T = 2\pi\sqrt{L/g} \). The period depends only on the length and the local gravity — not on the mass of the bob or the amplitude (as long as the amplitude is small).
Does the bob mass affect the period?
No. The mass cancels from the equation of motion. A 1 kg bob and a 100 g bob on the same string swing at the same rate. Mass does scale the kinetic energy, potential energy, and rope tension, however.
How does the planet affect the pendulum period?
Period scales as \(1/\sqrt{g}\). A 1 m pendulum that swings every 2.01 s on Earth would swing every 4.93 s on the Moon (\(g \approx 1.62\)), and every 1.26 s on Jupiter (\(g \approx 24.79\)). The cross-planet table in the result section makes this concrete.
Why does the period grow with large swing amplitudes?
The small-angle formula \( T = 2\pi\sqrt{L/g} \) comes from replacing \(\sin\theta\) with \(\theta\). For larger angles the restoring "force" is weaker than the linear approximation suggests, so the bob spends more time near the turning points and the period grows. The exact result involves the complete elliptic integral of the first kind.
How long should a pendulum be to swing once per second?
If by "once per second" you mean \(T = 1\) s, you need \(L = g (T/2\pi)^2 \approx 0.0248\) m, i.e. about 25 mm — quite short! The 1 m "seconds pendulum" actually has a 2 s period because the historical "second" referred to each tick or tock individually.
How can a pendulum measure gravity?
Switch the mode to Solve for g. Enter the precisely measured length and period — the calculator returns \( g = 4\pi^2 L / T^2 \). This is the basis of the classical pendulum gravimeter (and Galileo's original experiments).
What is the difference between a simple and a physical pendulum?
A simple pendulum is an idealised point mass on a massless string. A physical (compound) pendulum is any real rigid body that swings about a pivot. Its period is \( T = 2\pi\sqrt{I/(mgd)} \) where \(I\) is the moment of inertia about the pivot and \(d\) is the distance from pivot to centre of mass. The simple-pendulum formula is the limit when all mass is concentrated at one point.
Reference this content, page, or tool as:
"Pendulum Period Calculator" at https://MiniWebtool.com// from MiniWebtool, https://MiniWebtool.com/
by miniwebtool team. Updated: 2026-05-15