Answer
The speed is greatest at $x=0$ when potetial energy is minimal and kinetic energy is maximal.
Work Step by Step
If elastic potential energy $U_{el}=\frac{1}{2}kx^{2}$ and total energy is $E_{1}$ then kinetic energy $K=\frac{mv^{2}}{2}=E_{1} - U_{el}=E_{1}-\frac{1}{2}kx^{2}$.
So speed of the particle equals $v=\frac{2E_{1}-kx^{2}}{m}=-\frac{kx^{2}}{m}+\frac{2E_{1}}{m}$, which represents an x-axis inverted parabola with y-interception at $\frac{2E_{1}}{m}$.
If energy is $E_{2}$ we have the same parabola with y-interception at $\frac{2E_{2}}{m}$.
As seen from the graph the speed is greatest at $x=0$ when potetial energy is minimal and kinetic energy is maximal.
