To solve this problem, we need to understand the relationship between the frequency of a body executing simple harmonic motion (SHM) and the frequency of its potential energy.$\\$In SHM, the displacement of the body can be represented as:$\\ x(t) = A \cos(2 \pi n t) \\$where $A$ is the amplitude and $n$ is the frequency of the motion.$\\$The potential energy $U$ in SHM is given by:$\\ U = \frac{1}{2} k x^2 \\$Substituting the expression for $x(t): \\ U = \frac{1}{2} k (A \cos(2 \pi n t))^2 \\$Simplifying, we get:$\\ U = \frac{1}{2} k A^2 \cos^2(2 \pi n t) \\$Using the trigonometric identity $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2},$ we can rewrite $U$ as:$\\ U = \frac{1}{2} k A^2 \left(\frac{1 + \cos(4 \pi n t)}{2}\right) \\$Simplifying further:$\\ U = \frac{1}{4} k A^2 + \frac{1}{4} k A^2 \cos(4 \pi n t) \\$From this expression, we can see that the potential energy $U$ has a constant term and a cosine term with frequency $2n. \\$Therefore, the frequency of the potential energy is $2n. \\$This corresponds to Option 2.