Question

Starting from the origin, a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy by 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

The correct option is A

$\frac{1}{6}s$

Let the equation of the SHM be $x=Asin\omega t$
As per the question, $\frac{1}{2}m{v}^2=\frac{3}{4}\left(\frac{1}{2}m{v}_{max}^2\right)\Rightarrow v^2=\frac{3}{4}{v}_{max}^2\Rightarrow {\left(\omega Acos\omega t\right)}^2=\frac{3}{4}{\left(\omega A\right)}^2\Rightarrow cos\omega t=\frac{\sqrt{3}}{2}\Rightarrow \omega t=\frac{\pi }{6}\Rightarrow \frac{2\pi }{2}t=\frac{\pi }{6}\therefore t=\frac{1}{6}s$