Starting from the mean position a body oscillates simple harmonically with a period of $2\,s.$ After what time will its kinetic energy be $75\%$ of the total energy ?
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$K E=\frac{75}{100} E=\frac{75}{100} \times \frac{1}{2} m a^{2} \omega^{2}$

$\frac{1}{2} m a^{2} \omega^{2} \cos ^{2} \omega t=\frac{75}{100} \times \frac{1}{2} m a^{2} \omega^{2}$

$\cos ^{2} \omega t=\frac{3}{4}$

$\cos \omega t=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$

or $\omega t=\frac{\pi}{6} \Rightarrow t=\frac{\pi}{6 \pi}=\frac{1}{6} s$

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