Chapter 8: Problem 62
Period of a pendulum A standard pendulum of length \(L\) that swings under the influence of gravity alone (no resistance) has a period of $$ T=\frac{4}{\omega} \int_{0}^{\pi / 2} \frac{d \varphi}{\sqrt{1-k^{2} \sin ^{2} \varphi}} $$ where \(\omega^{2}=g / L, k^{2}=\sin ^{2}\left(\theta_{0} / 2\right), g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(\theta_{0}\) is the initial angle from which the pendulum is released (in radians). Use numerical integration to approximate the period of a pendulum with \(L=1 \mathrm{m}\) that is released from an angle of \(\theta_{0}=\pi / 4\) rad.
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