Answer
$\frac{\pi}{2} \lt \phi \lt \pi$
Work Step by Step
$a = -a_m~cos(\omega~t+\phi)$
When $x = 0,$ then $a = 0$
If $x \lt 0$, then the acceleration is in the positive direction.
When $x = -x_m$, then $a$ is at its most positive value.
Then: $~~-1 \lt cos(\omega~t+\phi) \lt 0$
Since $t = 0$ in the snapshot, then $\frac{\pi}{2} \lt \phi \lt \pi$
