Answer
$L+v_{max}\sqrt{\frac{2m}{K}};L-v_{max}\sqrt{\frac{2m}{K}}$
Work Step by Step
According to the law of conservation of energy:
$U_{max}=K.E_{max}$
$\implies \frac{1}{2}Kx^2=\frac{1}{2}(2m)(v_{max})^2$
This simplifies to:
$x=(\sqrt{\frac{2m}{K}})v_{max}$
Now, the maximum/minimum separation is given as:
$The \space maximum \space separation=L+x$
$The \space maximum \space separation=L+v_{max}\sqrt{\frac{2m}{K}}$
$The \space minimum\space separation=L-x$
$The \space minimum\space separation=L-v_{max}\sqrt{\frac{2m}{K}}$
