In this question, I'll refer to standard cube rotation notation, seen here: J perm website - cube notation.
The move U, can be performed 4 times and return a solved cube (or any cube) to its original state.
This isn't surprising, it's making use of the 4 sides of a square that is built into the cube.
The set of moves (U R U' L' U R' U' L), which is a standard operation in white cross methods, will return to the original position after 3 repetitions. But this makes use of a triangle that is built into the face of a cube. Obviously, you can cut a square across it's diagonal and end up with a triangle, and this is how I understand that period-3 repetition. (Try it for yourself to see why this is a sensible interpretation).
However I was playing around with the operation (R U R' U) and discovered that after performing the moves 5 times, it returns the cube to it's initial position!
I find this really surprising. It doesn't "feel" right that this 3-dimensional, 6-sided, hyper-4-gon should ever have periodicity 5.
Can anyone explain why this is the case in a geometric or intuitive way?
I think @Stinking Bishop put it best. If you look at the side pieces or corners that move under the operations, they form a 5-cycle. Although it is surprising that this awkward pentagon exists on a cube, if you follow each piece independently, that's really all there is to it. Of course that must be the case because it is the same operation each time. There is no decision tree or pause in movement. The pieces are just on a 5-cycle conveyor belt.
I agree with whoever said it that once you know what is going on, the surprising thing ceases to be "why" it exists and more that it can be generated from something so simple. But I've been around maths for long enough that I've seen that many times before. Conway's game of life anyone?
There is no real good geometric explanation as far as I know (edit: Stinking Bishop's explanation is probably the closest you can get). The explanation is pretty simple if you know some group theory. If I lose you throughout this answer, skip to the end.
You can look at it like this. All the permutations (or rather the algorithms to get to those permutations) of a Rubik's Cube form a group with $43 252 003 274 489 856 000$ elements (where the "product" of 2 algorithms is just doing one after the other).
Your algorithm (R U R' U) is one element of that group (let's call it $g$ for simplicity) and doing it $n$ times after each other is the element $g^n$.
It is a pretty basic fact in group theory (for finite groups) that for every element $h$ there exists a number $k$ that is a divisor of the order (=size) of the group such that $h^k$ = 1. It follows directly from Lagrange's theorem, if you are interested.
Even more: There is Cauchy's theorem which states that for every prime divisor $p$ of that (finite) group, there exists an element $h$ such that $h^p = 1$ and where $p$ is the smallest number, bigger than $0$, for which this is true.
What does this mean for the Rubik's cube?
It means that for every prime number $p$ that divides $43252003274489856000$, there exists an algorithm that you would have to $p$ times to get back to the original state. $5$ is such a number. In the same way there should be an algorithm that you would have to repeat exactly $11$ times to get back to the original state (since $11$ divides $43252003274489856000$).