Question

Starting from origin, a body oscillates simple harmonically with a period of $2\,s$. After what time will its kinetic energy be $75\%$ of the total energy ?

• A. 1/6 s
• B. 1/12 s
• C. 1/3 s
• D. 1/4 s

Answer 1

Answer: (A)

As per,
$KE =\frac{75}{100} E =\frac{75}{100} \times \frac{1}{2} ma ^{2} \omega^{2}$
$\frac{1}{2} ma ^{2} \omega^{2} \cos ^{2} \omega t =\frac{75}{100} \times \frac{1}{2} ma ^{2} \omega^{2}$
$\cos ^{2} \omega t =\frac{3}{4}$
$\cos \omega t =\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6}$
or $\omega t =\frac{\pi}{6}$
$\Rightarrow t =\frac{\pi}{6 \pi}=\frac{1}{6} s$