Answer
$0.30\frac{m}{s}$
Work Step by Step
We can find the maximum speed as follows:
$\frac{1}{2}mv_{max}^2=\frac{1}{2}KA^2$
This simplifies to:
$v_{max}=A\sqrt{\frac{K}{m}}$
As $\sqrt{\frac{K}{m}}=\frac{2\pi}{T}$
$\implies v_{max}=A(\frac{2\pi}{T})$
We plug in the known values to obtain:
$v_{max}=(0.023)(\frac{2\pi}{0.48})=0.30\frac{m}{s}$
