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
$KE=\left(KE_{max}\right)\times\frac{75}{100}$
$\Rightarrow \, \quad\frac{m\omega^{2} A^{2}}{2} cos^{2} \, \omega t=\frac{m\omega^{2} A^{2}}{2}\times\frac{3}{4}$
$\Rightarrow \, \quad cos\, \omega t =\pm\frac{\sqrt{3}}{2}$
$\Rightarrow \, \quad\omega t=\frac{\pi}{6} \Rightarrow \frac{2\pi}{T} \times t=\frac{\pi}{6} \Rightarrow t =\frac{T}{12}=\frac{1}{6}s$