Answer
$\phi = \pi$
Work Step by Step
It is given that $y(0,0) = 0$
We can find $\phi$:
$y(x,t) = y_m sin(kx-\omega t+\phi)$
$y(0,0) = (4.0~cm)sin(0-0+\phi) = 0$
$sin~\phi = 0$
$\phi = 0$ or $\phi = \pi$
As $t$ increases slightly above zero, the $y$ versus $t$ graph becomes positive and it increases.
$y(x,t) = y_m sin(kx-\omega t+\phi)$
$y(0,t) = y_m sin(0-\omega t+\phi)$
$y(0,t) = y_m sin(-\omega t+\phi)$
Therefore, $\phi = \pi$
