Answer
$a_{max} = 92.2~m/s^2$
Work Step by Step
We can find the angular frequency.
$A~\omega = v_{max}$
$\omega = \frac{v_{max}}{A}$
$\omega = \frac{3.90~m/s}{0.165~m}$
$\omega = 23.64~rad/s$
We can find the maximum magnitude of the acceleration.
$a_{max} = A~\omega^2$
$a_{max} = (0.165~m)(23.64~rad/s)^2$
$a_{max} = 92.2~m/s^2$
