HARD · BEYOND SMALL ANGLES
LARGE ANGLE PERIOD FROM ELLIPTIC INTEGRAL
A pendulum of length L = 2.0 m is released from rest at θ₀ = 1.2 rad (≈ 68.8°). (a) Find the elliptic modulus k = sin(θ₀/2). (b) Compute the small-angle period T₀. (c) Compute the exact period T = 4√(L/g) · K(k). (d) Find the ratio T/T₀. How much longer is the true period compared with the small-angle prediction?
Linked equations:T = 4\sqrt{\frac{L}{g}}\,K\!\left(\sin\frac{\theta_0}{2}\right)K(k) = \int_0^{\pi/2}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}
§ 01
Step-by-step solution
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Step 1
Compute the elliptic modulus k = sin(θ₀/2) for θ₀ = 1.2 rad.
Hint
θ₀/2 = 0.6 rad. sin(0.6) ≈ 0.5646.
Step 2
Compute T₀ = 2π√(L/g) for L = 2.0 m.
Step 3
Compute T_exact = 4√(L/g) · K(k) using the complete elliptic integral K(k).
Solution walkthrough
At 68.8° the elliptic modulus is k = sin(60°) ≈ 0.565. The complete elliptic integral K(0.565) ≈ 1.744, which is significantly above its small-angle value of π/2 ≈ 1.571. The exact period is 4√(L/g)·K(k) ≈ 3.116 s, nearly 10% longer than T₀ ≈ 2.837 s. The two-term series T₀(1 + θ₀²/16 + 11θ₀⁴/3072) ≈ 3.114 s agrees to within 0.07%, showing the series is still a good approximation at this amplitude but is no longer as clean as at smaller angles.
§ 02
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