Answer
$\phi = 5.6~rad$
Work Step by Step
At $t = 0$, the point at $x = 0$ is moving at a velocity of $-4.0~m/s$, and this speed is $0.8$ of the maximum speed.
We can find two possibilities for $\phi$:
$y(x,t) = (y_m) sin(kx - \omega t +\phi)$
$u(x,t) = -(y_m \omega) cos(kx - \omega t +\phi)$
$u(0,0) = -(y_m \omega) cos(0 - 0 +\phi) = -(0.8)(y_m \omega)$
$cos(\phi) = 0.8$
$\phi = 0.64~rad$ or $\phi = 5.6~rad$
At $t$ increase slightly above zero, the value of $u$ becomes a smaller magnitude and less negative.
$u(x,t) = -(y_m \omega) cos(kx - \omega t +\phi)$
$u(0,t) = -(y_m \omega) cos(-\omega t +\phi)$
Therefore, $\phi = 5.6~rad$
