Answer
$\phi = -53^o$
Work Step by Step
The equilibrium position equation is given by
\begin{equation}
x = x_{m} \cos (\omega t+\phi)
\end{equation}
$x_{m}$ is the amplitude and $\phi$ is the phase constant.
The first derivative gives us the velocity by
\begin{align*}
v_s = \frac{d x}{d t} = \dfrac{d}{dt} [ x_{m} \cos (\omega t+\phi) ] =- \omega x_{m} \sin (\omega t+\phi)
\end{align*}
$\omega x_{m} $ represents the maximum velocity $v_{m}$
$v_s$ = 0.04 m/s, $v_m$ = 0.05 m/s and $(t=0)$. Subsitute these values in the above equation to get $\phi $
\begin{gather*}
v_s=v_m \sin (\omega t+\phi) \\
0.04 \,\text{m/s} = (-0.05 \,\text{m/s}) \,\,\sin (\omega (0) + \phi)\\
\sin \phi = - \dfrac{ 0.04 \,\text{m/s} }{ 0.05 \,\text{m/s} } \\
\phi = -53^o
\end{gather*}
