Answer
$-\pi/2$
Work Step by Step
We know that the sinusoidal wave function, at a frozen moment, is given by
$$D_{(x,t)} = A\sin(k x-\omega t + \phi)$$
where $A$ is the amplitude, $\omega=2\pi f$ where $f$ is the frequency, $\phi$ is the phase constant, $k=2\pi/\lambda$.
$$D_{(x,t)} = A\sin\left(\frac{2\pi x}{\lambda}-2\pi ft + \phi\right)$$
From the given snapshot, we can see that the frequency is given by
$$f=\dfrac{v}{\lambda}=\frac{1}{2}=\bf 0.5\;\rm Hz$$
where $\lambda$ is the wavelength.
So, at $t=1$ s, and $x=0$, where $D=A$
$$A= A\sin\left(\frac{2\pi (0)}{\lambda}-2\pi (0.5)(1) + \phi\right) = A\sin\left( - \pi + \phi\right)$$
Thus,
$$\sin\left( - \pi + \phi\right)=1$$
$$-\sin\left( \phi\right)=1$$
$$ \phi=\sin\left( -1\right)^{-1}$$
$$\boxed{\phi=-\dfrac{\pi}{2}}$$
