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) $\dfrac{T}{12}$
Concept. From the mean position $x = A\sin\omega t$.
Step 1. Phase. $\sin\omega t = \dfrac12 \Rightarrow \omega t = \dfrac{\pi}{6}$.
Step 2. Time. $t = \dfrac{\pi/6}{2\pi/T} = \dfrac{T}{12}$.
Answer. $\dfrac{T}{12}$ (Option B).
Solution by Sreeraj P, M.Sc Physics