
For an oscillating simple pendulum,
Acceleration, $a=-\omega^2 x$
and velocity, $v=\omega \sqrt{A^2-x^2}$
$\Rightarrow v^2=\omega^2\left(A^2-x^2\right)$
$\Rightarrow v^2+\omega^2 x^2 =\omega^2 A^2$
$\Rightarrow \quad v^2+\frac{\omega^4 x^2}{\omega^2}=\omega^2 A^2$
$\Rightarrow \quad v^2+\frac{ \alpha ^2}{\omega^2}=\omega^2 A^2$
$\Rightarrow \quad \frac{v^2}{\omega^2 A^2}+\frac{a^2}{\omega^4 A^2}=1$
This is an ellipse $\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\right)$ with centre at origin.
$x = a\,\sin \,\left( {\omega t + \pi /6} \right)$
After the elapse of what fraction of the time period the velocity of the particle will be equal to half of its maximum velocity?