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Question

The period of a pendulum is given by T=4g0π/2dθ1k2sin2θ=4gF(k)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, g9.8m/s2g \approx 9.8 \mathrm { m } / \mathrm { s } ^ { 2 } is the acceleration due to gravity, k=sin (θ0/2)\left( \theta _ { 0 } / 2 \right), and θ0\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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Answered 3 months ago
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Given is the following function for period:

T=4lg0π/2dθ1k2sin2θ=4lgF(k)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(θ0/2)k = \sin (\theta_0 / 2)

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