- #1

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Consider a vertical pendulum affected by gravity (See the pdf file i included). Now i can choose two different opposite directions for my unit vectors which give me different equations.

[tex]\downarrow : m\ddot x = mg-kx [/tex]

[tex]\uparrow : m\ddot x = kx-mg[/tex]

Which of course makes perfect sense, changing direction changes the sign. The problem is now if i want to solve them the second case yields a weird result.

so in the first case the (real) solution would be (if we set ##\omega _n^2 = \frac{k}{m}##)

[tex]x = Acos(\omega _n t )+ Bsin(\omega _n t) + \frac{mg}{k}[/tex]

and for the second case

[tex]x = Ae^{\omega _n t} + Be^{-\omega _n t} +\frac{mg}{k}[/tex]

So what I'm wondering why i would get a different solution just by changing the direction of the unit vector and how i can reconcile the approaches or know how i should choose the direction of my unit vectors.

[tex]\downarrow : m\ddot x = mg-kx [/tex]

[tex]\uparrow : m\ddot x = kx-mg[/tex]

Which of course makes perfect sense, changing direction changes the sign. The problem is now if i want to solve them the second case yields a weird result.

so in the first case the (real) solution would be (if we set ##\omega _n^2 = \frac{k}{m}##)

[tex]x = Acos(\omega _n t )+ Bsin(\omega _n t) + \frac{mg}{k}[/tex]

and for the second case

[tex]x = Ae^{\omega _n t} + Be^{-\omega _n t} +\frac{mg}{k}[/tex]

So what I'm wondering why i would get a different solution just by changing the direction of the unit vector and how i can reconcile the approaches or know how i should choose the direction of my unit vectors.