If displacement of particle undergoing SHM is given by,

$x=A\sin \omega t$, then velocity of the particle will be given by, $v=A\omega \cos \omega t$.

From the graph, we can see, at point $A$ and $B$ velocity of the particle will be zero. Clearly, time taken is half of time period, which is $\frac{T}{2}$.

Therefore, 

$$\begin{gathered} \frac{T}{2}=0.5 \\ \Rightarrow T=1 \mathrm{s} \end{gathered}$$

Now, angular frequency will be, $\omega =\frac{2\pi }{T}=2\pi \mathrm{rad} {\mathrm{s}}^{-1}$