1. The displacement is given by $x = A\sin(\omega t + \delta).$

2. At $t=0$, the particle is at $x = \frac{A}{2}.$ Substituting in the equation: $\frac{A}{2} = A\sin\delta$ $\sin\delta = \frac{1}{2}.$

3. Therefore, $\delta = \frac{\pi}{6} \quad \text{or} \quad \frac{5\pi}{6}.$

4. Now use the condition that the particle is moving along the positive $x$-axis at $t=0$. Velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t + \delta).$ At $t=0$, $v = A\omega \cos\delta.$

   Since it moves along positive $x$-axis, velocity must be positive: $\cos\delta > 0.$

5. Check the two possible values:

   • If $\delta = \frac{\pi}{6}$, then $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} > 0$ so this is valid.
   • If $\delta = \frac{5\pi}{6}$, then $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2} < 0$ so this is invalid.

6. Hence, $\boxed{\delta = \frac{\pi}{6}}.$

So the correct option is D.