Choosing unit vectors for harmonic motion problems

Click For Summary
SUMMARY

The discussion centers on the selection of unit vectors in harmonic motion problems, specifically for a vertical pendulum influenced by gravity. Two opposing directions for unit vectors yield different equations: downward results in \( m\ddot{x} = mg - kx \) while upward leads to \( m\ddot{x} = kx - mg \). The first case produces a real solution of \( x = Acos(\omega_n t) + Bsin(\omega_n t) + \frac{mg}{k} \), whereas the second yields an exponential solution, indicating instability. The key takeaway is that the direction of the unit vector significantly impacts the resulting equations and solutions.

PREREQUISITES
  • Understanding of harmonic motion principles
  • Familiarity with differential equations
  • Knowledge of pendulum dynamics
  • Basic grasp of unit vector selection in physics
NEXT STEPS
  • Study the implications of unit vector direction in harmonic motion
  • Explore solutions to differential equations in physics
  • Investigate the stability of solutions in mechanical systems
  • Learn about the role of forces in pendulum motion
USEFUL FOR

Physics students, mechanical engineers, and anyone studying harmonic motion and pendulum dynamics will benefit from this discussion.

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 :)
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
990
  • · Replies 11 ·
Replies
11
Views
2K