Answer
$\frac{KE}{PE}=3$
Work Step by Step
We can calculate the ratio of kinetic energy to potential energy as follows:
$v=\sqrt{\frac{K}{m}(A^2-x^2)}$
Given that $x=\frac{A}{2}$
$\implies v=\sqrt{\frac{K}{m}[A^2-(\frac{A}{2})^2]}$
$v=\frac{\sqrt 3}{2}A\sqrt{\frac{K}{m}}$
Now $\frac{KE}{PE}=\frac{\frac{1}{2}mv^2}{\frac{1}{2}Kx^2}$
$\frac{KE}{PE}=\frac{mv^2}{Kx^2}$
$\frac{KE}{PE}=\frac{m[\frac{3}{4}A^2\frac{K}{m}]}{K(\frac{A^2}{4})}$
$\frac{KE}{PE}=\frac{3A^2K}{A^2K}=3$
