## Yo-yoing over the harmonic oscillator

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?
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 The second derivative of $-cosθ$ is equal to $cosθ$. Because $[-cosθ]' = -sinθ$ and $[-sinθ]' = cosθ$ so $[-cosθ]'' = cosθ$. Maybe this is also helpfull: http://www.wolframalpha.com/input/?i=x%27%27+%3D+-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 $\frac{d^{2}x}{dt^{2}}$ = -(positive constant)x.

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