A particle executes SHM period of 8 seconds  After what

SOLUTION

$\omega =\frac{2\pi }{8}=\frac{\pi }{4}rad{s}^{-1}$
Let 't' be the time after which the energy will be half kinetic and half potential. 
Since motion starts from mean position, $y=A\sin \omega t$ 
$\frac{dy}{dt}=v=A\cos \omega t.\omega$
$K.E=\frac{1}{2}m{v}^2=\frac{1}{2}m.A^2{\omega }^2{\cos }^2\omega t.$ This must be half of the total energy.
$\therefore \frac{1}{2}m{A}^2{\omega }^2{\cos }^2\omega t=\frac{1}{2}.\frac{1}{2}m{\omega }^2{A}^2$
$\therefore \cos \omega t=\frac{1}{\sqrt{2}};\omega t=\frac{\pi }{4};t=\frac{\pi }{4}.\omega =\left(\frac{\pi }{4}\right)\left(\frac{4}{\pi }\right)=1s.$