Answer
(a) $A = 0.10~m$
(b) $v = 0.35~m/s$
Work Step by Step
(a) $\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$
$A = \sqrt{\frac{m}{k}}~v_{max}$
$A = \sqrt{\frac{1.0~kg}{16~N/m}}~(0.40~m/s)$
$A = 0.10~m$
(b) We can use conservation of energy to find the speed.
$\frac{1}{2}mv^2+\frac{1}{2}k(\frac{A}{2})^2=\frac{1}{2}mv_{max}^2$
$v^2=v_{max}^2-(\frac{k}{m})(\frac{A}{2})^2$
$v = \sqrt{v_{max}^2-(\frac{k}{m})(\frac{A}{2})^2}$
$v = \sqrt{(0.40~m/s)^2-(\frac{16~N/m}{1.0~kg})(\frac{0.10~m}{2})^2}$
$v = 0.35~m/s$
