Frame of reference in a simple harmonic motion vertical spring

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SUMMARY

The discussion focuses on establishing a frame of reference for simple harmonic motion in a vertical spring system. The user proposes defining the positive direction as upward, leading to the equation ma = kx for upward displacement. However, the response clarifies that while this formulation is mathematically valid, it neglects the influence of gravity, which must be accounted for in the model. The correct approach involves recognizing that gravity alters the dynamics of the system, necessitating a comprehensive understanding of forces acting on the mass.

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  • Understanding of Newton's Second Law of Motion
  • Familiarity with Hooke's Law and spring constants
  • Basic knowledge of simple harmonic motion principles
  • Concept of gravitational force and its effects on motion
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david22
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I have doubts of how can I put my frame of reference in a simple harmonic motion vertical spring. Normally the books choose the origin in the equilibrium position and the positive distance (x>0) downward, and in this conditions Newton´s second law is: ma=-kx; but instead of putting the positive distance downward I want to put it upward, so the negative distance (x<0) is downward and in this conditions Hook´s law is going to be positive(because positive direction is upward) so Newton´s second law is: ma=kx
I want you to tell me if the last expression is correct for the negative distance downward. I would appreciate your help
 
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Phew - for a moment you had me worried you wanted to work out the frame of reference with the coordinate system attached to the mass (i.e. non-inertial).

Off what you actually want to know... I don't think you found the relation you want ... ma=kx says that the acceleration in the +x direction (upwards) is proportional to the displacement upwards ... so the higher the mass gets, the faster it goes. I think you need to to slow down as it goes higher?

Consider:

##\vec{F}=-k\vec{x}## - because the force is always in the opposite direction to the displacement.
Does not matter if you put +ve upwards or downwards.
In this case, it's all 1D so ##\vec{x}=x\hat{\imath}## and we write:

##-kx\hat{\imath} = ma\hat{\imath}##

...and we can divide out the unit vectors and work in magnitudes.

Except that there's still something wrong with this model: there's no gravity!
The presence of gravity is what makes "up" and "down" special, otherwise it's the same as saying "forward" and "back" - gravity is what changes the equation.

If +ve is up, then gravity is negative:
... you should be able to take it from there :)
 

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