Let $l$ be the length of the string. Then
$\begin{align*}x &= - l \cos \theta \\y &= - l \sin \theta\end{align*}$
implies that
$\begin{align*}\dot{x} &= l \dot{\theta} \sin \theta \\\dot{y} &= - l \dot{\theta} \cos \theta\end{align*}$
which implies that
$\begin{align*}\ddot{x} &= l \dot{\theta}^2 \cos \theta + l \ddot{\theta} \sin \theta \\\ddot{y} &= l \dot{\theta}^...$
Thus,
$\begin{align}a^2 &= \ddot{x}^2 + \ddot{y}^2 \label{eqn:3} \\&= l^2 \dot{\theta}^4 + l^2 \ddot{\theta}^2 \label{eqn:4}...$
So, we need to find $\dot{\theta}$ and $\ddot{\theta}$ as functions of $\theta$ . For $\dot{\theta}$ , this can be done by equating the kinetic energy that the mass has when its angle is $\theta$ to the change in the mass's potential energy between its initial position and its position when its angle is $\theta$ :
$\begin{align*}m g l \sin \theta = \frac{1}{2} m l^2 \dot{\theta}^2\end{align*}$
Thus,
$\setcounter{equation}{2}\begin{align}\dot{\theta}^2 = \frac{2g}{l} \sin \theta \label{eqn:1}\end{align}$
In order to find $\ddot{\theta}$ , we use Lagrange's equation
$\begin{align*}\frac{\partial L}{\partial \theta} = \frac{d}{dt} \frac{\partial L}{\partial \dot{\theta}}\end{align*}$
The kinetic energy of the particle is
$\begin{align*}T = \frac{1}{2}m l^2 \dot{\theta}^2\end{align*}$
The potential energy of the particle is
$\begin{align*}V = - l \sin \theta m g\end{align*}$
Therefore the Lagrangian of the particle is
$\begin{align*}L = \frac{1}{2}m l^2 \dot{\theta}^2 + l \sin \theta m g\end{align*}$
Using Langrange's equation, we find that
$\setcounter{equation}{3}\begin{align}\ddot{\theta} = \frac{g}{l} \cos \theta \label{eqn:2}\end{align}$
Plugging equations (3) and (4) into equation (1)-(2) gives
$\begin{align*}a^2 &= g^2 \left(4 \sin \theta^2 + \cos^2 \theta \right) \\&= g^2 \left(3 \sin \theta^2 +1 \right) \end...$
or
$\begin{align*}\boxed{a = g \sqrt{3 \sin \theta^2 +1 }} \\\end{align*}$
Therefore, answer (E) is correct.

Solution to 1992 Problem 93