Answer
$x_m=0.400m$
Work Step by Step
By law of conservation of energy, the total energy of the system is the sum of the kinetic and potential energy.
Mathematically
$E=K.E+P.E$
$E=\frac{1}{2}mv^2+\frac{1}{2}Kx^2$
but at turning point $E=\frac{1}{2}Kx_m^2$
so $\frac{1}{2}Kx_m^2=\frac{1}{2}mv^2+\frac{1}{2}Kx^2$
$\frac{1}{2}Kx_m^2=\frac{1}{2}(mv^2+Kx^2)$
$Kx_m^2=mv^2+Kx^2$
$x_m^2=\frac{mv^2}{K}+x^2$
taking square root on both sides, we get
$x_m=\sqrt{\frac{mv^2}{K}+x^2}$
putting the values, we get
$x_m=\sqrt{\frac{(0.325)(-13.6)^2}{400}+(0.100)^2}$
$x_m=0.400m$
