# Yo-yoing over the harmonic oscillator

1. Dec 8, 2012

### DiracPool

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?

2. Dec 8, 2012

### Physics2.0

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''+=+-x

3. Dec 8, 2012

### grzz

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.