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Question

# The period of a pendulum is given by $T = 4 \sqrt { \frac { \ell } { g } } \int _ { 0 } ^ { \pi / 2 } \frac { d \theta } { \sqrt { 1 - k ^ { 2 } \sin ^ { 2 } \theta } } = 4 \sqrt { \frac { \ell } { g } } F ( k )$, where $\ell$ is the length of the pendulum, $g \approx 9.8 \mathrm { m } / \mathrm { s } ^ { 2 }$ is the acceleration due to gravity, k=sin $\left( \theta _ { 0 } / 2 \right)$, and $\theta _ { 0 }$ is the initial angular displacement of the pendulum (in radians). The integral in this formula F(k) is called an elliptic integral, and it cannot be evaluated analytically.Approximate F(0.1) by expanding the integrand in a Taylor (binomial) series and integrating term by term.

Solution

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Given is the following function for period:

$T = 4 \sqrt{ \dfrac{l}{g} } \int_0^{\pi/2} \dfrac{ d\theta }{ \sqrt{1-k^2 \sin^2 \theta} } = 4\sqrt{ \dfrac{l}{g} } F(k)$

and $k = \sin (\theta_0 / 2)$

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