$$Y = A\sin(\omega t + \phi_0)$$ is the time-displacement equation of a SHM. At $$t = 0$$ the displacement of the particle is $$Y = \frac{A}{2}$$ and it is moving along negative $$x$$-direction. Then the initial phase angle $$\phi_0$$ will be:

The displacement equation is $$Y = A\sin(\omega t + \phi_0)$$. At $$t = 0$$, we are given $$Y = \frac{A}{2}$$ and the particle is moving in the negative $$x$$-direction (i.e., velocity is negative).

Substituting $$t = 0$$: $$\frac{A}{2} = A\sin(\phi_0)$$, which gives $$\sin(\phi_0) = \frac{1}{2}$$. This means $$\phi_0 = \frac{\pi}{6}$$ or $$\phi_0 = \frac{5\pi}{6}$$.

The velocity is $$\frac{dY}{dt} = A\omega\cos(\omega t + \phi_0)$$. At $$t = 0$$, the velocity is $$A\omega\cos(\phi_0)$$. Since the particle moves in the negative direction, we need $$\cos(\phi_0) < 0$$.

For $$\phi_0 = \frac{\pi}{6}$$: $$\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} > 0$$, so velocity is positive. This does not satisfy the condition.

For $$\phi_0 = \frac{5\pi}{6}$$: $$\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} < 0$$, so velocity is negative. This satisfies the condition.

Therefore, the initial phase angle is $$\phi_0 = \frac{5\pi}{6}$$.