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Period of a pendulum A standard pendulum of length L
Chapter 5, Problem 48E(choose chapter or problem)
Period of a pendulum A standard pendulum of length L swinging under only the influence of gravity (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 \(\omega^{2}=g / L\), \(k^{2}=\sin ^{2}\left(\theta_{0} / 2\right)\), \(g \approx 9.8 \mathrm{~m} / \mathrm{s}^{2}\) 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 m that is released from an angle of \(\theta_{0}=\pi / 4 \mathrm{rad}\).
Questions & Answers
QUESTION:
Period of a pendulum A standard pendulum of length L swinging under only the influence of gravity (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 \(\omega^{2}=g / L\), \(k^{2}=\sin ^{2}\left(\theta_{0} / 2\right)\), \(g \approx 9.8 \mathrm{~m} / \mathrm{s}^{2}\) 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 m that is released from an angle of \(\theta_{0}=\pi / 4 \mathrm{rad}\).
ANSWER:Problem 48EPeriod of a pendulum A standard pendulum of length L swinging under only the influence of gravity (no resistance) has a period of where , , g 9.8 is the acceleration due to gravity, and 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 m that is released from an angle of (rad)SolutionStep 1In this problem we have to use numerical integration to approximate the period of a pendulum with L = 1 m that is released from an angle of (rad)Given We have , , g 9.8 , L = 1 m, (rad)Thus