Yo-yoing over the harmonic oscillator

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?

Physics2.0

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

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

[itex]\frac{d^{2}x}{dt^{2}}[/itex] = -(positive constant)x.

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DiracPool	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?

Physics2.0	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

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 [itex]\frac{d^{2}x}{dt^{2}}[/itex] = -(positive constant)x.