Answer
(a) $0.26\frac{m}{s}$
(b) $2.8cm$
Work Step by Step
(a) We know that
$K.E_{max}=U_{max}$
$\implies \frac{1}{2}mv_{max}^2=\frac{1}{2}KA^2$
This can be simplified and rearranged as
$v_{max}=\sqrt{(\frac{26}{0.40})(0.032)^2}=0.26\frac{m}{s}$
(b) $\frac{1}{2}mv_{max}^2+\frac{1}{2}kx^2=\frac{1}{2}KA^2$
This can be rearranged and simplified as
$x=\sqrt{A^2-\frac{mv^2}{K}}$
$\implies x=\sqrt{(0.032)^2-(\frac{(0.40)(0.129)^2}{26})}=2.8cm$
