alleryA particle performs SHM along x-axis with amplitude $A$ and period $T$. The time taken from $x = 0$ to $x = A/2$ is:
A$T/12$
B$T/8$
C$T/6$
D$T/4$
Answer & Solution
Correct answer: A. $T/12$
$x = A\sin(\omega t)$, starting from mean.
At $x = A/2$: $\sin(\omega t) = 1/2$
⇒ $\omega t = \pi/6$
⇒ $t = \pi/(6\omega) = (\pi/6)/(2\pi/T) = T/12$.
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