Answer
$1.43\frac{m}{s}$
Work Step by Step
We know that
$K_i+U_i=K_f+U_f$
$\implies 0+\frac{1}{2}KA^2=\frac{1}{2}mv^2+\frac{1}{2}Kx^2$
$\implies mv^2=K(A^2-x^2)$
This can be simplified as
$v=\sqrt{\frac{K(A^2-x^2)}{m}}$
We plug in the known values to obtain:
$v=\sqrt{\frac{(13.3)[(0.256)^2-(0.128)^2]}{0.321}}=1.43\frac{m}{s}$
