1. Given SHM equation

   $y = A\sin(\omega t + \phi_0)$

   At $t=0$:

   $y(0)=A\sin\phi_0$

   It is given that:

   $y(0)=\frac{A}{2}$

   So,

   $A\sin\phi_0=\frac{A}{2}$

   $\sin\phi_0=\frac{1}{2}$

2. Possible values of $\phi_0$

   Since

   $\sin\phi_0=\frac{1}{2}$

   the possible phase angles are:

   $\phi_0=\frac{\pi}{6} \quad \text{or} \quad \frac{5\pi}{6}$

3. Use the direction of motion

   Velocity is:

   $v = \frac{dy}{dt}=A\omega\cos(\omega t+\phi_0)$

   At $t=0$:

   $v(0)=A\omega\cos\phi_0$

   The particle is moving along the negative x-direction, so velocity is negative:

   $v(0)<0$

   Therefore,

   $\cos\phi_0<0$

4. Check the two possible angles

   • For $\phi_0=\frac{\pi}{6}$: $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}>0$ So velocity is positive. Not correct.

   • For $\phi_0=\frac{5\pi}{6}$: $\cos\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}<0$ So velocity is negative. Correct.

5. Final answer

   $\boxed{\phi_0=\frac{5\pi}{6}}$

Hence, the correct option is A.