Question

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 ?

• A. $\left(\frac{1}{6}\right)$s
• B. $\left(\frac{1}{4}\right)$s
• C. $\left(\frac{1}{3}\right)$s
• D. $\left(\frac{1}{12}\right)$s

Answer 1

The correct option is A

$\left(\frac{1}{6}\right)$s

we know that $KE=\frac{1}{2}m{\omega }^2{A}^{2}co{s}^2\omega t$ and $TE=\frac{1}{2}m{\omega }^2{A}^2$

Give that $\frac{1}{2}m{\omega }^2{A}^{2}co{s}^2\omega t=0.75\times \frac{1}{2}m{\omega }^2{A}^{2}orcos\omega t=\frac{\sqrt{3}}{2}or\omega t=co{s}^{-1}\frac{\sqrt{3}}{2}=\frac{\pi }{6}$ or

Since T= 2 second,

$\frac{2\pi }{T}t=\frac{\pi }{6}$

$t=\frac{1}{6}$ second