Large-Angle Pendulum Period (Exact)
Gives the exact period of a simple pendulum at any amplitude via the complete elliptic integral of the first kind: T = 4·√(L/g)·K(sin(θ₀/2))
The equation
What it solves
Gives the exact period of a simple pendulum at any amplitude via the complete elliptic integral of the first kind: T = 4·√(L/g)·K(sin(θ₀/2)). This reduces to the small-angle formula when θ₀ → 0.
When to use it
When amplitude is large (typically > 30°–40°) and precision matters. The elliptic integral is easily evaluated numerically using the AGM algorithm or its series expansion.
When NOT to use it
Overkill for small angles (θ₀ < 15°) where T = 2π√(L/g) is already accurate to 0.5%. Also assumes a simple (point-mass) pendulum; physical pendulums need the moment-of-inertia form.
Common mistakes
Confusing the modulus k = sin(θ₀/2) with the amplitude θ₀ itself. Using K(θ₀/2) instead of K(sin(θ₀/2)). Forgetting the prefactor 4√(L/g) and writing 2π√(L/g)·K(k) instead.
Topics that use this equation
Problems using this equation
- [easy] A simple pendulum of length L = 1.0 m is pulled to θ₀ = 30° (π/6 rad) and released from rest. The sm…
- [challenge] A pendulum is released from rest at θ₀ = 1.0 rad (≈ 57.3°) and its period is measured to be T = 3.0 …
- [exam] A clock pendulum of length L = 1.0 m is pulled to θ₀ = 1.4 rad (≈ 80.2°) and released. (a) Compute t…
- [hard] A pendulum of length L = 2.0 m is released from rest at θ₀ = 1.2 rad (≈ 68.8°). (a) Find the ellipti…