If we imagine a 2d semisphere "bowl" with a ball resting at the bottom. The ball can be slightly displaced and will roll back and forth crossing the bottom of the bowl an infinite number of times over and over. I believe this motion can be approximated to be simple harmonic for small displacements from equilibrium, but it isn't completely.
What shape of bowl would be needed (it looks like $y=x^2$ doesn't work: Ball Rolling in a Parabolic Bowl) in order to have a ball placed at any point in the bowl to execute simple harmonic motion. Is this possible?
Edit: in reality I'm actually curious about a point mass "ball" with no moment of inertia for simplicity
Best Answer
The shape needed is one 'loop' or 'festoon' of a cycloid (or an upside-down cycloid). Taking the bottom of a loop as origin, the loop's equation may be written as $$x=a(\theta + \sin \theta),\ \ \ \ y=a(1-\cos\theta)$$ for $-\pi<\theta<\pi$, with $\theta=0$ at the bottom of the loop. We see that the horizontal range of the loop is $-\pi a<x<\pi a$. A bead threaded on this loop will perform SHM (not just approximate SHM) from whatever point on the loop it is released. [We can, of course, envisage a particle sliding in a bowl, vertical sections of which, passing through the lowest point, are cycloidal. A rolling ball will also perform translational SHM – at $\sqrt {\frac 57}$ of the frequency derived below, if the rolling ball is homogeneous.]
We can show that the arc length from the origin is $$s=4a \sin \frac{\theta}2$$ and this gives us the neat relation $$s^2 = 8ay$$ If the amplitude of the motion, expressed as a maximum height is $y_0$ (where $y_0<2a$), and as an arc length is $s_0$ (where $s_0<4a$), we can write the energy conservation equation as $$\tfrac12m \left(\frac{ds}{dt}\right)^2+mgy=mgy_0$$ That is $$\left(\frac{ds}{dt}\right)^2+\frac g{4a} s^2=\frac g{4a} s_0^2$$ You'll find that this is satisfied by $$s=s_0 \cos (\omega t + \phi)$$ with $\omega=\sqrt {\frac g{4a}}$. No approximations and a finite amplitude!
Addendum [inspired by dialogue of comments on Eli's answer]
As an alternative to the last step, if you differentiate the energy equation, $\left(\frac{ds}{dt}\right)^2+\frac g{4a} s^2=\frac g{4a} s_0^2\ \ $ wrt time, $t$, and tidy up you get the equation of motion $$\frac{d^2 s}{dt^2} + \frac g{4a} s = 0.$$