EASY · BEYOND SMALL ANGLES
PERCENT ERROR AT 30 DEGREES
A simple pendulum of length L = 1.0 m is pulled to θ₀ = 30° (π/6 rad) and released from rest. The small-angle approximation gives T₀ = 2π√(L/g). The true period is larger. By what percentage does the small-angle formula underestimate the true period?
Linked equations:T = 4\sqrt{\frac{L}{g}}\,K\!\left(\sin\frac{\theta_0}{2}\right)T \approx T_0\!\left(1 + \frac{\theta_0^2}{16} + \frac{11\theta_0^4}{3072}\right)
§ 01
Step-by-step solution
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Step 1
Compute the small-angle period T₀ = 2π√(L/g) for L = 1.0 m.
Hint
Use g = 9.80665 m/s². The result is the familiar isochronous period — independent of amplitude.
Step 2
Compute the exact period T = 4√(L/g) · K(sin(θ₀/2)), where K is the complete elliptic integral of the first kind.
Step 3
Find the percentage by which T₀ underestimates T_exact: pct_err = (T_exact − T₀) / T_exact × 100.
Solution walkthrough
At small angles sin θ ≈ θ, turning a nonlinear equation into a simple harmonic oscillator with period T₀ = 2π√(L/g). At 30° the neglected cubic term θ³/6 contributes about 4.5% of the linear term. The exact period, obtained via the complete elliptic integral K, is T ≈ 2.041 s versus T₀ ≈ 2.006 s — a 1.71% underestimate. This matches the leading-order series correction θ₀²/16 × 100 ≈ 1.71%, confirming the series is accurate at moderate angles.
§ 02
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