Find your rhythm

Pendulum simulator & period calculator

Change the length, gravity and starting angle, press Play and time the swing. For small angles the period of a simple pendulum is T = 2π√(L/g): it depends only on the length and the gravitational acceleration, not on the mass of the bob. A 1 m pendulum on Earth has a period of about 2.01 s.

Printable classroom activity ↗
Pendulum explorerTRY IT YOURSELF
L = 1 mθ = -10°
1.00 s
Period2.006 s
Frequency0.498 Hz
Angle now-10 °
Length1 m
Teal + amber · follow the changing valuesIdealized physics model

What if you changed…

Try a starting point

YOUR NEXT AHA

Make the pendulum twice as slow. How much longer must it be?

See the calculation

T = 2π√(L/g)
  1. T = 2π√(L/g) = 2π√(1/9.81) = 2.01 s.
  2. Frequency = 1/T = 0.498 Hz.
  3. Small-angle motion: θ(t) = 10 cos(2πt/T) degrees.

What this model assumes

Small-angle approximation, massless rod/string, point bob, no damping. Angles are limited to 15°.

Keep an eye on this

Large swings have a longer period than the small-angle approximation predicts.

Understand the physics ↗
Waves

Formulas used in this simulation

Period (small angles)
T = 2π√(L/g)
Accurate below about 15°
Frequency
f = 1/T = (1/2π)√(g/L)
Complete swings per second
Length for a given period
L = gT² / (4π²)
Rearranged period formula
Gravity from a measured period
g = 4π²L / T²
A classic lab measurement of g
Angle over time
θ(t) = θ₀ cos(2πt/T)
Released from rest at θ₀
Formulas explained with a worked example ↗
A little question. A clearer picture.

Pendulum explorer: common questions

Wondering about the why? Start here.

How do you calculate the period of a pendulum?

Use T = 2π√(L/g) with the length L in metres and g in m/s². For L = 0.5 m on Earth, T = 2π√(0.5/9.81) ≈ 1.42 s.

How long is a pendulum with a 1 second period?

L = gT² / (4π²) = 9.81 × 1² / 39.48 ≈ 0.248 m, so about 25 cm. A “seconds pendulum”, which takes 1 s for each one-way swing (a 2 s period), is about 0.994 m long.

Why does the formula only work for small angles?

Its derivation uses sin θ ≈ θ in radians. At 15° the true period is only about 0.4% longer than 2π√(L/g), but at 60° it is about 7% longer.

Does a heavier bob swing faster?

Not in this ideal pendulum model: mass cancels out. Length and gravity set the small-angle period.

Why does amplitude barely change the period?

The formula assumes small angles. The animation follows that approximation; for large angles, amplitude does affect period.