Choosing unit vectors for harmonic motion problems

In summary, the conversation discusses a vertical pendulum affected by gravity and how choosing different directions for the unit vectors can lead to different equations. It is noted that changing the direction changes the sign, but the second case yields a strange result. The solution for the first case is provided, and it is explained that in the second case, the force of the spring is always in the opposite direction to the displacement.
  • #1
Incand
334
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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.
 

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  • #2
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]
 
  • #3
Thank you, I understand now! You won't believe I've been thinking about this for several hours before i wrote this :)
 

Related to Choosing unit vectors for harmonic motion problems

1. What are unit vectors and why are they important in harmonic motion problems?

Unit vectors are vectors with a magnitude of 1 and are used to represent direction in a coordinate system. In harmonic motion problems, unit vectors are crucial for determining the direction of motion and the magnitude of displacement.

2. How do I choose the appropriate unit vectors for a specific harmonic motion problem?

The unit vectors chosen for a harmonic motion problem depend on the coordinate system being used. For example, if the problem is in a two-dimensional Cartesian coordinate system, the unit vectors would be the x and y directions. If the problem is in a polar coordinate system, the unit vectors would be the radial and tangential directions.

3. Can I use any unit vectors for a harmonic motion problem?

No, it is important to choose unit vectors that are orthogonal (perpendicular) to each other. This ensures that the motion is only in the specified direction and does not affect the other components.

4. What is the relation between unit vectors and the equations of motion in harmonic motion problems?

In harmonic motion problems, the equations of motion are typically written in terms of the unit vectors. For example, the equations for position, velocity, and acceleration would involve the unit vectors in their respective directions.

5. How can I check if my chosen unit vectors are correct for a harmonic motion problem?

One way to check if the chosen unit vectors are correct is by using the Pythagorean theorem. The sum of the squares of the components of the unit vectors should equal 1, which is the definition of a unit vector. Additionally, the chosen unit vectors should be perpendicular to each other.

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