#### Answer

(a) The fraction of energy that is potential energy is 0.25.
The fraction of energy that is kinetic energy is 0.75
(b) $x = 0.707~A$

#### Work Step by Step

(a) We can find an expression for the total energy in the system when all the energy is in the form of elastic potential energy.
$E = \frac{1}{2}kA^2$
We can find an expression for the elastic potential energy when $x = \frac{1}{2}A$
$U_s = \frac{1}{2}kx^2$
$U_s = \frac{1}{2}k~(\frac{1}{2}A)^2$
$U_s = \frac{1}{2}k~(\frac{1}{4}A^2)$
We can find the fraction of energy that is potential energy.
$\frac{U_s}{E} = \frac{\frac{1}{2}k~(\frac{1}{4}A^2)}{\frac{1}{2}kA^2}$
$\frac{U_s}{E} = \frac{1}{4} = 0.25$
The fraction of energy that is potential energy is 0.25. Therefore, the fraction of energy that is kinetic energy is 0.75
(b) We can find $x$ when half the total energy is potential energy.
$U_s = \frac{E}{2}$
$\frac{1}{2}kx^2 = \frac{\frac{1}{2}kA^2}{2}$
$x^2 = \frac{A^2}{2}$
$x = \sqrt{\frac{1}{2}}~A$
$x = 0.707~A$