Answer
$\Delta x'=\sqrt{2}\Delta x$
Work Step by Step
According to the conservation of energy, all energy must be conserved and transitioned to some form of energy. In this case, kinetic energy is converted to elastic potential energy. This means that $$\frac{1}{2}mv^2=\frac{1}{2}k\Delta x^2$$ Solving for $\Delta x$ yields $$\Delta x=\sqrt{\frac{mv^2}{k}}=v\sqrt{\frac{m}{k}}$$ Changing $m$ to $m'=\frac{1}{2}m$ and changing $v$ to $v'=2v$ yields a $\Delta x$ of $$\Delta x'=2v\sqrt{\frac{m}{2k}}=\frac{2}{\sqrt{2}}v\sqrt{\frac{m}{k}}=\sqrt{2}\Delta x$$
