There are two types of problems related to using potential energy. We can be given \(V\relax (x) \) but not at the equilibrium point, or given \(V\relax (x) \) at the equilibrium point. If \(V\relax (x) \) given is not at the equilibrium point, then we first need to find \(x_{0}\) which is the equilibrium point. This is done by solving \(V^{\prime }\relax (x) =0\). Then expand \(V\relax (x) \) near \(x_{0}\) using Taylor series and obtain new \(V\relax (x) \) which is now centered around \(x_{0}\).
The other type of problem, is where we need to find \(V\relax (x) \) at equilibrium, from the physics of the problem. See MC2 as example. For the vertical pendulum problem \(V\relax (x) =\frac {1}{2}kx^{2}-mgx\). This is the potential energy at equilibrium.
We need to convert the above to \(V\relax (y) =\frac {1}{2}ky^{2}+V\relax (0) \) and only now we can write\[ F=-V^{\prime }\relax (y) =-m\omega ^{2}y \] From the above, \(\omega \) can be found.\begin {align*} ky & =m\omega ^{2}y\\ \omega ^{2} & =\frac {k}{m} \end {align*}
Remember, we can only use \(F=-V^{\prime }\relax (y) =-m\omega ^{2}y\) when \(V\relax (y) \) has form \(\frac {1}{2}ky^{2}+V\relax (0) \). Do not use \(\frac {1}{2}kx^{2}-mgx\). There should not be linear term in \(V\left ( x\right ) \).
\(V\relax (y) \) should always be \(0\) at equilibrium. And \(V\left ( y\right ) =\frac {1}{2}m\omega ^{2}y^{2}\) so \(V^{\prime }\relax (y) =m\omega ^{2}y\)