CHALLENGE · BEYOND SMALL ANGLES
LENGTH FROM LARGE ANGLE PERIOD
A pendulum is released from rest at θ₀ = 1.0 rad (≈ 57.3°) and its period is measured to be T = 3.0 s. Using the exact formula T = 4√(L/g) · K(sin(θ₀/2)), find the pendulum length L. Show why the small-angle inversion L = g(T/2π)² gives a significantly wrong answer.
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 = 1.0 rad.
Hint
θ₀/2 = 0.5 rad. sin(0.5) ≈ 0.4794.
Step 2
Evaluate K(k) for k = sin(0.5) ≈ 0.4794 using the arithmetic-geometric mean or table.
Step 3
Invert the exact formula to find L = g · (T / (4 · K))².
Solution walkthrough
Starting from T = 4√(L/g)·K(k), square both sides and isolate L: L = g(T/4K)². At θ₀ = 1.0 rad the modulus is k = sin(0.5) ≈ 0.4794 and K ≈ 1.675. The exact length is L ≈ 1.966 m. The naive small-angle inversion uses K = π/2 ≈ 1.571, giving L_naive ≈ 2.237 m — an overestimate of 13.8%. The verification step 4√(1.966/9.80665)×1.675 = 3.000 s confirms the answer. This problem illustrates how critically the amplitude dependence of K affects precision measurements of g from pendulum timing.
§ 02
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