Answer
0.5 m
Work Step by Step
$E=\frac{1}{2}mv^{2}+\frac{1}{2}kx^{2}=\frac{1}{2}kx_{m}^{2}$
$\implies mv^{2}+kx^{2}=kx_{m}^{2}$
or $ \frac{mv^{2}}{k}+x^{2}=x_{m}^{2}$
or $x_{m}= \sqrt {\frac{mv^{2}}{k}+x^{2}}= \sqrt {\frac{2.00kg(3.415\,m/s)^{2}}{100\,N/m}+(0.129\,m)^{2}}=0.5\,m$
