Simple Pendulum
Swing a pendulum and compare the exact motion with the small-angle approximation
Drag anywhere on the picture to pull the bob out, then let go.
Angle θ over time
Exact period from T = T₀ / AGM(1, cos(θ₀/2)). The measured value is timed from one peak of the simulated swing to the next.
How a pendulum swings
A pendulum bob of mass m hangs on a light rod of length L. When it is pulled to an angle θ, gravity has a component mg sin θ along the arc, pointing back toward the bottom. Newton's second law along the arc gives θ'' = −(g/L) sin θ. The mass cancels, so a heavy bob and a light bob on the same string swing together.
This equation is nonlinear because of the sin θ. For small angles sin θ ≈ θ (in radians), and the equation becomes θ'' = −(g/L) θ, which is simple harmonic motion with ω₀ = √(g/L). Its period T₀ = 2π√(L/g) does not depend on the amplitude. That is what Galileo noticed, and it made pendulum clocks possible.
At larger angles sin θ is noticeably smaller than θ, so the restoring force is weaker than the linear model predicts and the real pendulum takes longer to come back. The visualizer runs both models side by side: the solid bob follows the exact equation, the faded bob follows the small-angle one. Release them at 10° and they stay together for many swings. Release them at 120° and the exact bob falls behind within one cycle.
Key formulas
Equation of motion
θ'' = −(g/L) sin θ
- θ
- angle from the vertical (rad)
- g
- gravitational field strength (m/s²)
- L
- length from pivot to the bob’s centre (m)
Small-angle period
T₀ = 2π √(L/g), θ(t) ≈ θ₀ cos(ω₀t), ω₀ = √(g/L)
Exact period (any amplitude)
T = T₀ / AGM(1, cos(θ₀/2))
- AGM
- arithmetic-geometric mean: repeatedly replace (a, b) with ((a+b)/2, √(ab))
Series approximation
T ≈ T₀ (1 + θ₀²/16 + 11θ₀⁴/3072 + …)
Where the AGM comes from
The exact period is T = (4/ω₀) K(k) with k = sin(θ₀/2), where K is the complete elliptic integral of the first kind. Gauss showed that K(k) = π / (2 AGM(1, √(1 − k²))), and √(1 − k²) = cos(θ₀/2). The AGM converges so quickly that four or five iterations give full double precision.How much does amplitude matter?
| Release angle θ₀ | T / T₀ | Error of T₀ |
|---|---|---|
| 5° | 1.0005 | 0.05% |
| 10° | 1.0019 | 0.2% |
| 20° | 1.0077 | 0.8% |
| 30° | 1.0174 | 1.7% |
| 45° | 1.0400 | 4.0% |
| 60° | 1.0732 | 7.3% |
| 90° | 1.1803 | 18% |
| 120° | 1.3729 | 37% |
| 150° | 1.7622 | 76% |
| 170° | 2.4394 | 144% |
How to use this visualizer
- 1Drag anywhere on the picture to pull the bob out to a release angle, then let go. Both bobs start from rest at that angle.
- 2Start with a small angle (about 10°). The two bobs and the two graph traces stay on top of each other.
- 3Now try 90° or more. The exact bob lags behind, and the measured period grows well past T₀.
- 4Change L or switch between Earth, Moon, Mars and Jupiter. The period scales as √(L/g), but the ratio T/T₀ depends only on the angle.
- 5Compare the measured period with the exact formula. They agree to about a millisecond.
Practice problems
Problem 1: A pendulum on the Moon
Show solutionHide solution
Earth: T₀ = 2π√(1.0 / 9.81) = 2π × 0.319 = 2.01 s
Moon: T₀ = 2π√(1.0 / 1.62) = 2π × 0.786 = 4.94 s
Same period needs the same L/g: LMoon = 1.0 × 1.62 / 9.81 = 0.165 m
2.01 s on Earth, 4.94 s on the Moon; a 16.5 cm pendulum on the Moon
Problem 2: When the approximation breaks
Show solutionHide solution
T₀ = 2π√(2.0 / 9.81) = 2.837 s
θ₀ = 60° = π/3 = 1.047 rad, so θ₀²/16 = 1.097 / 16 = 0.0685
First correction: T ≈ 2.837 × 1.0685 = 3.031 s
Exact: T = T₀ / AGM(1, cos 30°) = 2.837 / AGM(1, 0.8660) = 2.837 × 1.0732 = 3.045 s
T₀ underestimates by about 7%; the one-term correction is within 0.5% of the exact 3.04 s
Discussion
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