Q109 / 150 · JEE Main 2006 · Physics

Starting from the origin, a body oscillates simple harmonically with a period of $2 \mathrm{~s}$. After what time will its kinetic energy be $75 \%$ of the total energy?

  • A
    $\frac{1}{12} \mathrm{~s}$
  • Correct answer
    B
    $\frac{1}{6} \mathrm{~s}$
  • C
    $\frac{1}{4} \mathrm{~s}$
  • D
    $\frac{1}{3} s$

Solution

$\frac{1}{2} m v^2=\frac{3}{4}\left(\frac{1}{2} m v_{\max }^2\right)$ $\mathrm{A}^2 \omega^2 \cos ^2 \omega \mathrm{t} \Rightarrow \frac{3}{4} \mathrm{~A}^2 \omega^2$ $\cos \omega \mathrm{t}=\frac{\sqrt{3}}{2}$ $\omega t=\frac{\pi}{6} \Rightarrow t=\frac{1}{6}$ sec

Hence, the correct answer is option: 'B'.

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