Frame of reference in a simple harmonic motion vertical spring

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In simple harmonic motion for a vertical spring, the standard frame of reference places the origin at the equilibrium position with positive displacement downward. However, changing the frame to have positive displacement upward leads to a different interpretation of Newton's second law, resulting in the equation ma = kx, which implies acceleration increases as the mass rises. This interpretation is incorrect because it does not account for the opposing force of gravity, which must be included in the model. The correct approach involves recognizing that the force due to the spring is always opposite to the displacement, regardless of the chosen direction for positive displacement. Therefore, gravity's influence is crucial in accurately modeling the motion of the spring system.
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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