E8.6A pendulum making 90-degree swings
The exact classical mechanical description of a pendulum is quite complex, and the equations of motion usually only solved in introductory texts for small displacements about equilibrium. In this case, the period, \(T \approx 2\pi\sqrt{L/g}\), and the motion is harmonic.
The general solution requires elliptic integrals, but the special case of a pendulum making \(180^\circ\) swings (i.e. \(\pm 90^\circ\) about its equilibrium position) leads to the following expression for the period:
\[T = 2\sqrt{\frac{2l}{g}}\int_0^{\pi/2} \frac{\mathrm{d}\theta}{\sqrt{\cos\theta}}.\]
The substitution \(x = \sin^2\theta\) transforms this integral into a beta function:
\[\int_0^{\pi/2} \frac{\mathrm{d}\theta}{\sqrt{\cos\theta}} = \textstyle \frac{1}{2}\int_0^1 x^{-1/2}(1-x)^{-3/4}\,\mathrm{d}x = \textstyle\frac{1}{2}B(\frac{1}{2},\frac{1}{4}).\]
Therefore,
\[T = \sqrt{2}B(\textstyle\frac{1}{2},\frac{1}{4})\sqrt{\frac{l}{g}}.\]
To find the period of the pendulum in units of \(\sqrt{l/g}\):
In [x]: import numpy as np
In [x]: from scipy.special import beta
In [x]: np.sqrt(2) * beta(0.5, 0.25)
7.4162987092054875
(Compare with the harmonic approximation, \(2\pi= 6.283185\)).