Let's say you start with a small array, then add N elements, one by one. When you run out of space in your underlying, you allocate a newer array, copy all the elements over, then free the old array.

If your new array is always a constant size bigger than the old one, then the total amount of work you do is O(n^2). If your new array is a constant factor bigger than the old one, the total amount of work you do is O(n).

This is ignoring memory allocator costs, which can change things a bit.