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) 1/6 s (b) 1/4 s (c) 1/3 s (d) 1/12 s

Answer: (a). K = 75% of E at x = A/2, so sin ω t = 1/2 ⇒ ω t = π/6. t = T/12 = 2/12 = 1/6 s So the answer is (a) 1/6 s.

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Q74JEE Main 2006 · AIEEEMedium

Energy in SHM

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

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Answer: (a)

$K = 75\%$ of $E$ at $x = \frac{A}{2}$ , so $\sin\omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{6}$ .
$t = \frac{T}{12} = \frac{2}{12} = \frac{1}{6}$ s
So the answer is (a) $\frac{1}{6}$ s.

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