Answer
$\phi = 1.2~rad$
Work Step by Step
$y(x,t) = y_m~sin (kx-\omega t+ \phi)$
$y(0,0) = y_m~sin (kx-\omega t+ \phi) = 4.5~mm$
$y_m~sin (0-0+ \phi) = 4.5~mm$
$y_m~sin (\phi) = 4.5~mm$
We can take the derivative to find an expression for the transverse velocity $u$:
$u(x,t) = -y_m~\omega~cos (kx-\omega t+ \phi)$
$u(0,0) = -y_m~\omega~cos (0-0+ \phi) = -0.75~m/s$
$y_m~(440~rad/s)~cos (\phi) = 0.75~m/s$
$y_m~cos (\phi) = \frac{0.75~m/s}{440~rad/s}$
$y_m~cos (\phi) = 0.0017~m$
$y_m~cos (\phi) = 1.7~mm$
We can divide the two expressions to find $\phi$:
$tan~\phi = \frac{4.5~mm}{1.7~mm}$
$\phi = tan ^{-1} (\frac{4.5~mm}{1.7~mm})$
$\phi = 1.2~rad$
