Choosing unit vectors for harmonic motion problems

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 1K views
Incand
Messages
334
Reaction score
47
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.
 

Attachments

Physics news on Phys.org
Incand said:
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.

Whatever your direction for x, the force of the spring is always in the opposite direction to the displacement. So, in the second case, you should have:

[tex]\uparrow : m\ddot x = -kx-mg[/tex]
 
Thank you, I understand now! You won't believe I've been thinking about this for several hours before i wrote this :)