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

Given: $$x = A\sin(\omega t + \delta)$$, at $$t = 0$$, $$x = A/2$$ and particle moves in positive x-direction.

At $$t = 0$$:

$$A/2 = A\sin\delta$$

$$\sin\delta = 1/2$$

$$\delta = \pi/6$$ or $$\delta = 5\pi/6$$

The velocity is: $$v = A\omega\cos(\omega t + \delta)$$

At $$t = 0$$: $$v = A\omega\cos\delta > 0$$ (moving in positive direction)

For $$\delta = \pi/6$$: $$\cos(\pi/6) = \sqrt{3}/2 > 0$$ ✓

For $$\delta = 5\pi/6$$: $$\cos(5\pi/6) = -\sqrt{3}/2 < 0$$ ✗

Therefore $$\delta = \pi/6$$.

The correct answer is Option 2.