A particle executes $S.H.M.$ of amplitude A along $x$-axis. At $t =0$, the position of the particle is $x=\frac{A}{2}$ and it moves along positive $x$-axis the displacement of particle in time $t$ is $x=A \sin (\omega t+\delta)$, then the value $\delta$ will be
JEE MAIN 2023, Medium
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$X = A \sin (\omega t +\delta)$ $V = A \omega \cos (\omega t +\delta)$
$\frac{ A }{2}= A \sin (\omega t +\delta)$ $\therefore V \text { is tve, } \delta \text { must be }$
$\text { At } t =0$ $\text { in } 1^{\text {st }} \text { quadrant or } 4^{\text {th }}$
$\sin \delta=\frac{1}{2} \Rightarrow \delta=\frac{\pi}{6}, \frac{5 \pi}{6}$ $\text { quadrant }$
$\therefore \text { Common solution is } \delta=\frac{\pi}{6}$
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