1. For a simple harmonic oscillator of amplitude $A$, the total energy is $E = \frac{1}{2}kA^2.$

2. When the displacement is $x = \frac{A}{3}$, the potential energy is $U = \frac{1}{2}kx^2 = \frac{1}{2}k\left(\frac{A}{3}\right)^2 = \frac{1}{18}kA^2.$

3. Hence the kinetic energy is

Taking LCM, $K = \left(\frac{9}{18}-\frac{1}{18}\right)kA^2 = \frac{8}{18}kA^2 = \frac{4}{9}\cdot \frac{1}{2}kA^2.$ So, $K = \frac{4}{9}E.$

4. Therefore, $\frac{E}{K} = \frac{E}{(4/9)E} = \frac{9}{4}.$

5. Given that $\frac{E}{K} = \frac{x}{8},$ we get $\frac{x}{8} = \frac{9}{4}.$ So, $x = 8\cdot \frac{9}{4} = 18.$

Therefore, the required integer is $\boxed{18}.$