The wave is given by $y(x,t)=A\sin(kx-\omega t+\phi).$

At time $t=0$, the snapshot becomes $y(x,0)=A\sin(kx+\phi).$

To determine $\phi$, we use the graph at $t=0$.

At $x=0$, $y(0,0)=A\sin\phi.$

From the given snapshot, the particle at $x=0$ is at the mean position, so $y(0,0)=0.$ Therefore, $\sin\phi=0 \implies \phi=0,\ \pi,\ 2\pi,\dots$

Now check the slope of the snapshot at $x=0$: $\left.\frac{\partial y}{\partial x}\right|_{t=0}=Ak\cos(kx+\phi).$ So at $x=0$, $\left.\frac{\partial y}{\partial x}\right|_{x=0,t=0}=Ak\cos\phi.$

From the figure, the curve crosses the origin with negative slope. Hence, $\cos\phi<0.$

Among the possible values from step 4, only $\phi=\pi$ gives $\sin\phi=0, \qquad \cos\phi=-1<0.$

Therefore, the phase is $\boxed{\phi=\pi}.$

Checking options:

• A: $\frac{\pi}{2}$ ✗
• B: $\pi$ ✓
• C: $0$ ✗
• D: $-\frac{\pi}{2}$ ✗