When the displacement of a simple harmonic oscillator is one third...

The total energy of the harmonic oscillator is given by the formula

$E = \dfrac{1}{2}k{A}^{2} . . . \left(1\right)$

where, $k$ is the force constant and $A$ is the amplitude.

The potential energy of the oscillator at the given displacement can be written as

$\begin{cases}U & = & \dfrac{1}{2}k{\left(\dfrac{A}{3}\right)}^{2} \\ & = & \dfrac{k{A}^{2}}{2 \times 9} \\ & = & \dfrac{E}{9}\end{cases}$

Hence, its kinetic energy is given by

$\begin{cases}KE & = & E - \dfrac{E}{9} \\ & = & \dfrac{8E}{9}\end{cases}$

Thus, the required ratio is

$\begin{cases}\dfrac{E}{KE} & = & \dfrac{E}{\dfrac{8E}{9}} \\ & = & \dfrac{9}{8}\end{cases}$

Hence, $x = 9$.