# Lecture 903, Example Problem

For this question, $k=\pi/2$ and $\omega=\pi$. You can get this directly from the wave function. All that is different is the initial phase. Plug in x=1 and t=0, match $y(0,0)=2sin(\pi/2+\phi)$ with the graph. You will see both 3 and 4 are correct in producing $y(0,0)=-2$