QJEEM23Ph6384: JEE Main 2023 Physics - Class 11 JEEM23Ph6384 Solution: A particle executes S.H.M. of amplitude A along x-axis. At t =…

The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$ .

If we substitute these initial conditions into the equation for the displacement of the particle:

$x = A \sin(wt + \delta)$

We have:

$\frac{A}{2} = A \sin(\delta)$

Dividing both sides by $A$ gives us:

$\frac{1}{2} = \sin(\delta)$

The angle whose sine is $\frac{1}{2}$ is $\delta = \frac{\pi}{6}$ radians (or 30 degrees).

Therefore, the correct answer is $\delta = \frac{\pi}{6}$ .