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Yoyoing over the harmonic oscillator 
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#1
Dec812, 07:06 AM

P: 534

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
Dec812, 07:21 AM

P: 5

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%27%27+%3D+x 


#3
Dec812, 07:24 AM

P: 950

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. 


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