CHALLENGE · THE SIMPLE PENDULUM

LARGE ANGLE PERIOD CORRECTION

A pendulum of length L = 1.0 m is released from rest at θ₀ = 0.6 rad (~34°). This amplitude lies outside the comfortable small-angle zone. Compute both the small-angle approximation period T_small and the exact period T_exact (using the elliptic integral formula T = 4√(L/g) K(sin(θ₀/2))). Then find the fractional error δ introduced by the approximation.

§ 01

Step-by-step solution

Work through one named subgoal at a time. Each step is checked deterministically against the canonical solver — no AI required to verify correctness. Get an AI explanation when you're stuck.

Step 1

Compute T_small = 2π√(L/g). This is what the small-angle approximation predicts. What is T_small in seconds?

Hint

This is the standard formula. The result does not depend on θ₀ — that is the approximation's key assumption.

Step 2

Compute the elliptic modulus k = sin(θ₀ / 2). What is k?

Step 3

Compute T_exact = 4√(L/g) × K. What is the exact period in seconds?

Step 4

Compute the fractional error δ = (T_exact − T_small) / T_small. Express as a decimal (not percentage).

Solution walkthrough
The simple pendulum's exact equation of motion is θ'' = −(g/L) sin θ. It has no closed-form solution in terms of elementary functions. The small-angle trick replaces sin θ ≈ θ, turning it into a harmonic oscillator and giving T = 2π√(L/g) — but this introduces error that grows with amplitude. Step 1 — T_small: apply the standard formula. T_small = 2π√(1.0/9.80665) ≈ 2.006 s. Step 2 — elliptic modulus: the exact period follows from separating variables in the energy equation. After integration, the quarter-period is (1/√(g/L)) × K(sin(θ₀/2)). So k = sin(θ₀/2) = sin(0.3) ≈ 0.2955. Step 3 — K(k): use the AGM. Starting with a = 1, b = √(1−0.2955²) = 0.9553. Iterate: a₁ = 0.9776, b₁ = √(0.9553) = 0.9774; convergence is rapid. The limiting value gives K ≈ 1.6069. Step 4 — T_exact: T_exact = 4√(L/g) × K = 4 × √(1/9.80665) × 1.6069 = 4 × 0.31934 × 1.6069 ≈ 2.053 s. Step 5 — fractional error: δ = (2.053 − 2.006)/2.006 ≈ 0.0230, i.e. about 2.3%. The approximation overestimates the frequency (gives a period that is too short) because it underestimates the restoring force at large angles (sin θ < θ for θ > 0). For scientific clocks, even 0.1% matters — which is why Huygens added cycloidal cheeks.
§ 02

Try it with AI

Continue the conversation with the Physics tutor — the problem context is pre-loaded.

Open in Physics.Ask