EXAM · BEYOND SMALL ANGLES
AMPLITUDE DEPENDENCE OF PERIOD
A clock pendulum of length L = 1.0 m is pulled to θ₀ = 1.4 rad (≈ 80.2°) and released. (a) Compute the small-angle period T₀. (b) Compute the exact period T_exact using the elliptic integral formula. (c) By what percentage does the true period exceed T₀? (d) After how many complete swings does the large-angle pendulum lag behind an identical small-angle pendulum by exactly one full period T₀?
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}}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 T₀ = 2π√(L/g) for L = 1.0 m.
Hint
T₀ ≈ 2.006 s.
Step 2
Compute T_exact = 4√(L/g) · K(sin(θ₀/2)) for θ₀ = 1.4 rad.
Step 3
Compute pct_stretch = (T_exact − T₀) / T₀ × 100.
Step 4
After how many swings does the true pendulum lag the small-angle pendulum by exactly one period T₀? Use n = T₀ / (T_exact − T₀).
Solution walkthrough
At 80.2° the modulus k = sin(40.1°) ≈ 0.644. The elliptic integral K(0.644) ≈ 1.792, well above K(0) = π/2. The exact period is 4√(1/g)·K ≈ 2.284 s, stretching T₀ ≈ 2.006 s by 13.8%. The isochronism Galileo observed at small angles has completely broken down. Every cycle the large-angle pendulum lags the small-angle one by ΔT = 0.278 s. After n ≈ 7.2 swings the accumulated lag equals one full T₀, so the two pendulums are exactly one cycle out of phase — the large-angle one completes only 6.2 swings while the small-angle one does 7.2. This is the physical cost of ignoring sin θ ≈ θ at large amplitudes.
§ 02
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