Answer
$v_m = 11\,\text{m/s}$
Work Step by Step
The displacement $x(t)$ of the equilibrium position is given by
\begin{equation}
x = x_{m} \cos (\omega t+\phi)
\end{equation}
$x_{m}$ is the amplitude. The derivative of the displacement gives us the velocity as next
\begin{align*}
v_s &= \dfrac{d}{dt} [ x_{m} \cos (\omega t+\phi) ] =- \omega x_{m} \sin (\omega t+\phi)
\end{align*}
The term $\omega x_{m} $ represents the maximum velocity $v_{m}$ at $\cos (\omega t+\phi) = 1 $. \\
The maximum velocity is given by
\begin{align*}
v_m &= \omega x_{m} = (2\pi/T) x_m= (2\pi/0.04 \,\text{s}) ( 0.07 \,\text{m}) = \boxed{11\,\text{m/s}}
\end{align*}
