Answer
$v_{max}=0.458m/s$
Work Step by Step
Recall that the maximum velocity of a simple harmonic oscillator is equal to $$v_{max}=A\omega$$ To find $\omega$, use a formula relating $\omega$ and period, which is $\omega=\frac{2\pi}{T}$. Substituting this expression yields $$v_{max}=\frac{2\pi A}{T}$$ Substituting known values of $A=0.204m$ and $T=2.80s$ yields a maximum speed of $$v_{max}=\frac{2\pi(0.204m)}{2.80s}=0.458m/s$$
