Question

Starting from the 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. $\frac{1}{6}\,s$
• B. $\frac{1}{4}\,s$
• C. $\frac{1}{3}\,s$
• D. $\frac{1}{12}\,s$

Answer 1

Answer: (A)

$KE$ of a body undergoing $SHM$ is given by
$KE=\frac{1}{2}m\omega^{2}\,A^{2}\,cos^{2}\,\omega t $ and $KE_{max}=\frac{m\omega^{2}A^{2}}{2}$
[symbols represent standard quantities] From given information
$KE=\left( KE_{max}\right)\times\frac{75}{100}$
$\Rightarrow \frac{m\omega^{2}A^{2}}{2} cos^{2}\,\omega t=\frac{m\omega^{2}A^{2}}{2}\times\frac{3}{4}$
$\Rightarrow cos\,\omega t=\pm \frac{\sqrt{3}}{2}$
$\Rightarrow \omega t=\frac{\pi}{6} \Rightarrow \frac{2\pi}{T}\times t=\frac{\pi}{6} \Rightarrow t=\frac{T}{12}=\frac{1}{6}s$