Q 11-13-087JEE MainJEE Main 2023 (10 Apr, Shift 1)Easy
A particle executes S.H.M. of amplitude $A$ along $x$-axis. At $t=0$, the position of the particle is $x=\dfrac A2$ and it moves along positive $x$-axis. The displacement of particle in time $t$ is $x=A\sin(\omega t+\delta)$, then the value of $\delta$ will be
Answer: (B) $\dfrac\pi6$
$\sin\delta=\dfrac12$ with positive velocity ($\cos\delta>0$), so $\delta=\dfrac\pi6$.
Solution by Sreeraj P, M.Sc Physics