I've been looking around and trying to figure it out, but I can't seem to figure out how the cosine function get's into the solution to the HO equation d2x/dt2=-kx/m. I know this is extremely basic, but could someone indulge me?
The second derivative of [itex]-cosθ[/itex] is equal to [itex]cosθ[/itex]. Because [itex][-cosθ]' = -sinθ [/itex] and [itex][-sinθ]' = cosθ [/itex] so [itex][-cosθ]'' = cosθ [/itex]. Maybe this is also helpfull: http://www.wolframalpha.com/input/?i=x''+=+-x
It is not so difficult to use a better notation! Try to see whether x = acos(bt), where a and b are constants, fits with the equation [itex]\frac{d^{2}x}{dt^{2}}[/itex] = -(positive constant)x.