Integrating $F=-kx$ gives $V=kx^2/2$ up to a constant. Thus $\boxed{-\hbar^2\phi''/(2m)+m\omega^2x^2\phi/2=E\phi}$, with $\omega=\sqrt{k/m}$. The potential is a symmetric upward parabola and tends to infinity in both directions; normalizable states decay at large $|x|$.


Solution

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S3 / 2 / a / Solution

 Solution (source code)