Answer
$v_m = \dfrac{F_{m}}{b }$
Work Step by Step
To determine the velocity we should take the first derivative of the equilibrium position
\begin{equation}
x = x_{m} \cos (\omega t+\phi)
\end{equation}
Where $x_{m}$ is the amplitude of the displacement. Differentiate to get 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*}
The maginutde of the term $\omega x_m$ represents the maximum velocity $v_m$, so the velocity amplitude $v_m$ is given by \begin{align*} v_{m}=\omega x_{m} =\omega \left( \frac{F_{m}}{b \omega} \right) =\boxed{ \dfrac{F_{m}}{b }} \end{align*}
