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
$KE=\frac{1}{2}m{\omega }^2{A}^{2}co{s}^2\omega tandK{E}_{max}=\frac{m{\omega }^2{A}^2}{2}$
[symbols represent standard quantities] From given information
$KE=\left(K{E}_{max}\right)\times \frac{75}{100}$
$\Rightarrow \frac{m{\omega }^2{A}^2}{2}co{s}^2\omega t=\frac{m{\omega }^2{A}^2}{2}\times \frac{3}{4}$
$\Rightarrow cos\omega t=\pm \frac{\sqrt{3}}{2}$
$\Rightarrow \omega t=\frac{\pi }{6}\Rightarrow \frac{2\pi }{T}\times t=\frac{\pi }{6}\Rightarrow t=\frac{T}{12}=\frac{1}{6}s$