Back and Forth

Again I will post a long, detailed solution, showing all steps. It can be boring, but I like to detail these things :)

Let $\alpha$ be the angle the pendulum makes with the veritical at any given time. Also, let $h$ be the difference in height bewteen the pendulum when its angle is $\alpha$ and when its angle is $\theta$ (the initial value). Also, let $v$ be its initial linear velocity and let $m$ be its mass.

The change in kinetic energy must be the same as the change in potential energy. That is:

$\large \displaystyle mgh = \frac{mv^2}{2}$

$\large \displaystyle v = \sqrt{2gh}$

The linear velocity is equal to $l$ times the angular velocity:

$\large \displaystyle l \frac{d \alpha}{dt} = \sqrt{2gh}$

Since the vertical distance from the pendulum to its fixed suspension point is $l \cos(\alpha)$ for any angle $\alpha$ it makes with the vertical, the change in height will be equal to $l \cos(\alpha) - l \cos(\theta)$ :

$\large \displaystyle \frac{d \alpha}{dt} = \sqrt{ \frac{2g}{l} [ \cos(\alpha) - \cos(\theta) ] }$

Or:

$\large \displaystyle \frac{dt}{d \alpha} = \sqrt{ \frac{l}{2g} \cdot \frac{1}{ \cos(\alpha) - \cos(\theta) } }$

$\large \displaystyle \int dt = \int \sqrt{ \frac{l}{2g} \cdot \frac{1}{ \cos(\alpha) - \cos(\theta) } } d \alpha$

Since its movement is $\theta \rightarrow 0 \rightarrow -\theta \rightarrow 0 \rightarrow \theta$ again, we can simply conisder $4$ times the integral between $0$ and $\theta$ :

$\large \displaystyle T = 4\ \sqrt{ \frac{l}{2g} } \int_0^{\theta} \frac{1}{ \sqrt{ \cos(\alpha) - \cos(\theta) } } d \alpha$

If we make

$\large \displaystyle u = \arcsin \left ( \frac{ \sin \left ( \frac{\alpha}{2} \right ) } { \sin \left ( \frac{\theta}{2} \right ) } \right )$

Then:

$\large \displaystyle du = \frac{1}{\sqrt{1 - \left ( \frac{ \sin \left ( \frac{\alpha}{2} \right ) } { \sin \left ( \frac{\theta}{2} \right ) } \right )^2 }} \cdot \frac{ \cos \left ( \frac{\alpha}{2} \right ) } { \sin \left ( \frac{\theta}{2} \right ) } \cdot \frac12 d \alpha$

$\large \displaystyle du = \frac{\cos\left( \frac{\alpha}{2} \right)}{ \sqrt{2 \sin \left ( \frac{\theta}{2} \right )^2 - 2 \sin \left ( \frac{\alpha}{2} \right )^2 } } \frac{1}{\sqrt{2}} d \alpha$

Since $\cos(2x) = 1 - 2\sin^2(x)$ :

$\large \displaystyle du = \frac{\sqrt{ 1 - \sin^2 \left( \frac{\alpha}{2} \right)}}{\sqrt{\cos(\alpha) - \cos(\theta)}} \frac{1}{\sqrt{2}} d \alpha$

$\large \displaystyle du = \frac{\sqrt{ 1 - \sin^2 \left( \frac{\theta}{2} \right) \sin^2 (u) }}{\sqrt{\cos(\alpha) - \cos(\theta)}} \frac{1}{\sqrt{2}} d \alpha$

$\large \displaystyle d \alpha = \sqrt{2} \frac{\sqrt{\cos(\alpha) - \cos(\theta)}}{\sqrt{ 1 - \sin^2 \left( \frac{\theta}{2} \right) \sin^2 (u) }} du$

So, the period becomes:

$\large \displaystyle T = 4\ \sqrt{ \frac{l}{g} } \int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{1 - \sin^2(\frac{\theta}{2}) \sin^2(u) }} du$

This is a Complete elliptic integral of the first kind with $k = \sin(\frac{\theta}{2})$ . Using Legendre polynomials until the term of order $4$ (since on the limit the highest term has order $4$ ):

$\large \displaystyle T = 2 \pi \sqrt{\frac{l}{g}} \left [ 1 + \frac14 \sin ^2\left ( \frac{\theta}{2} \right ) + \frac{9}{64} \sin ^4\left ( \frac{\theta}{2} \right ) + \mathcal{O} \left ( \sin ^6\left ( \frac{\theta}{2} \right ) \right ) \right ]$

Using the Maclaurin series for sine, and again computing until the term of order $4$ (since on the limit the highest term has order $4$ ):

$\color{#20A900} \boxed{ \large \displaystyle T = T_0 \left [ 1 + \frac{1}{16} \theta^2 + \frac{11}{3072} \theta^4 + \mathcal{O} \left ( \theta^6 \right ) \right ] }$

Now, to the limit:

$\large \displaystyle L = \lim_{\theta \rightarrow 0} \left [ \left ( \frac{2}{\theta} \right ) ^4 \left ( \frac{T}{T_0} - 1 \right ) - \frac{1}{\theta^2} \right ]$

$\large \displaystyle L = \lim_{\theta \rightarrow 0} \left [ \left ( \frac{16}{\theta^4} \right ) \left ( 1 + \frac{1}{16} \theta^2 + \frac{11}{3072} \theta^4 + \mathcal{O} \left ( \theta^6 \right ) - 1 \right ) - \frac{1}{\theta^2} \right ]$

$\large \displaystyle L = \lim_{\theta \rightarrow 0} \left [ \frac{1}{\theta^2} + \frac{11}{192} + \mathcal{O} \left ( \theta^2 \right ) - \frac{1}{\theta^2} \right ]$

$\large \displaystyle L = \lim_{\theta \rightarrow 0} \left [ \frac{11}{192} + \mathcal{O} \left ( \theta^2 \right ) \right ]$

$\color{#3D99F6} \boxed{\large \displaystyle L = \frac{11}{192} }$

If, in general, the pendulum subtends an angle $\alpha$ with the downwards vertical, after starting from rest when $\alpha = \theta$ , then conservation of energy tells us that $\tfrac12m \ell^2 \dot{\alpha}^2 + mg\ell(1 - \cos\alpha) \; = \; mg\ell(1 - \cos\theta)$ so that $\dot{\alpha}^2 \; = \; \tfrac{2g}{\ell}(\cos\alpha - \cos\theta)$ and so the period of oscillation is $T(\theta) \; = \; 4\sqrt{\tfrac{\ell}{2g}} \int_0^\theta \frac{d\alpha}{\sqrt{\cos\alpha - \cos\theta}} \; = \; 2\sqrt{\tfrac{\ell}{g}} \int_0^\theta \frac{d\alpha}{\sqrt{\sin^2\frac12\theta - \sin^2\frac12\alpha}}$ The substitution $\sin\tfrac12\alpha = \sin\tfrac12\theta \sin u$ gives $T(\theta) \; = \; \tfrac{2}{\pi}T_0 \int_0^{\frac12\pi} \frac{du}{\sqrt{1 - \sin^2\frac12\theta \sin^2u}} \; = \; \tfrac{2}{\pi}T_0 K\big(\sin^2\tfrac12\theta\big)$ where $K$ is the complete elliptic integral of the first kind. Now it is a standard result that $K(m) \; = \; \tfrac12\pi\left[1 + \big(\tfrac{1}{2}\big)^2 m + \big(\tfrac{1 \cdot 3}{2 \cdot 4}\big)^2m^2 + \cdots \right] \hspace{2cm} |m| < 1$ so that $\tfrac{2}{\pi}K(m) \; \sim \; 1 + \tfrac14m + \tfrac{9}{64}m^2 + o(m^2) \hspace{2cm} m \to 0$ so that $\begin{aligned} \tfrac{T(\theta)}{T_0} & \sim \; 1 + \tfrac14\sin^2\tfrac12\theta + \tfrac{9}{64}\sin^4\tfrac12\theta + o(\theta^4) \\ & \sim \; 1 + \tfrac14\big(\tfrac12\theta - \tfrac{1}{48}\theta^3\big)^2 + \tfrac{9}{64} \times \tfrac{1}{16}\theta^4 + o(\theta^4) \\ & \sim \; 1 + \tfrac{1}{16}\theta^2 + \tfrac{11}{3072}\theta^4 + o(\theta^4) \end{aligned}$ as $\theta \to 0$ , and hence $\left(\tfrac{2}{\theta}\right)^4\left(\tfrac{T(\theta)}{T_0} - 1\right) - \tfrac{1}{\theta^2} \; = \; \tfrac{11}{192} + o(1) \hspace{2cm} \theta \to 0$ so that $\lim_{\theta \to 0}\left[ \left(\tfrac{2}{\theta}\right)^4\left(\tfrac{T(\theta)}{T_0} - 1\right) - \tfrac{1}{\theta^2}\right] \; = \; \boxed{\tfrac{11}{192}}$