MEDIUM · BEYOND SMALL ANGLES
FIRST ORDER PERIOD CORRECTION
A pendulum of length L = 1.5 m is released from θ₀ = 0.8 rad (≈ 45.8°). The period can be expanded as T = T₀(1 + θ₀²/16 + 11θ₀⁴/3072 + …). (a) Find T₀, the small-angle period. (b) Compute the first-order correction Δ T = T₀ · θ₀²/16. (c) Find the first-order corrected period T_corr = T₀ + ΔT. (d) Include the second correction term to get T_series = T₀(1 + θ₀²/16 + 11θ₀⁴/3072).
Linked equations:T \approx T_0\!\left(1 + \frac{\theta_0^2}{16} + \frac{11\theta_0^4}{3072}\right)T = T_0\sum_{n=0}^{\infty}\left[\frac{(2n)!}{2^{2n}(n!)^2}\right]^2\theta_0^{2n}
§ 01
Step-by-step solution
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Step 1
Compute T₀ = 2π√(L/g) for L = 1.5 m, g = 9.80665 m/s².
Hint
√(1.5/9.80665) ≈ 0.3909 s, so T₀ ≈ 2.457 s.
Step 2
Compute the first-order correction ΔT = T₀ · θ₀²/16.
Step 3
Add the correction: T_corr = T₀ + ΔT.
Step 4
Include the second correction: T_series = T₀(1 + θ₀²/16 + 11θ₀⁴/3072).
Solution walkthrough
The Taylor expansion of T in powers of θ₀ comes from expanding the elliptic integral K(sin(θ₀/2)). The leading correction is θ₀²/16, which at 0.8 rad gives a 4% increase over T₀. Adding the next term (11θ₀⁴/3072 ≈ 0.15%) gives T_series ≈ 2.559 s, which matches the exact elliptic-integral value to within 0.006% — well inside experimental precision. Each successive term in the series extends the accuracy to higher amplitudes.
§ 02
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