Q 11-13-016NEETJEE MainMedium
A particle executes SHM with period $T$ starting from the mean position. The minimum time it takes to reach half its amplitude is
Answer: (B) $\dfrac{T}{12}$
$x = A\sin\omega t = \dfrac{A}{2} \Rightarrow \omega t = \dfrac{\pi}{6} \Rightarrow t = \dfrac{\pi/6}{2\pi/T} = \dfrac{T}{12}$.
Solution by Sreeraj P, M.Sc Physics