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

Q: 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

a $X = A \sin (\omega t +\delta)$ $V = A \omega \cos (\omega t +\delta)$ $\frac{ A }{2}= A \sin (\omega t +\delta)$ $\therefore V \text { is tve, } \delta \text { must be }$ $\text { At } t =0$ $\text { in } 1^{\text {st }} \text { quadrant or } 4^{\text {th }}$ $\sin \delta=\frac{1}{2} \Rightarrow \delta=\frac{\pi}{6}, \frac{5 \pi}{6}$ $\text { quadrant }$ $\therefore \text { Common solution is } \delta=\frac{\pi}{6}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$\frac{\pi}{6}$
B$\frac{\pi}{3}$
C$\frac{\pi}{4}$
D$\frac{\pi}{2}$

JEE MAIN 2023, Medium

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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