Answer
$y=2\sin(0.5x).$
Work Step by Step
If a function has form $y=A\sin(\omega x)$, then there $|A|$ stands for the amplitude and $T =\frac{2\pi}{\omega}$ stands for the period.
First we identify $\omega$ and $A$ and then plug them into the form of $y=Asin(\omega x)$.
Hence $A=|2|=2$, $\omega=\frac{2\pi}{T}=\frac{2\pi}{4\pi}=0.5.$
Thus $y=A\sin(\omega x)=2\sin(0.5x).$
