Oscillations and Elastic Potential Energy

[pewpew23]

Homework Statement

A mass of 1.5 kg oscillates vertically at the end of a lightweight spring. The spring has a spring constant of 145 Newtons per meter. The amplitude of the motion is 8.00 cm. From this data, complete the table below.

I have to find velocity, acceleration, elastic potential energy, etc. at given points from the equilibrium point.
These points are 8 cm, 6cm, 4cm, 2 cm, 0 cm (equilibrium),... -8 cm

Homework Equations

Elastic Potential= 1/2kx^2
KE=1/2mv^2
GPE=mgh

The Attempt at a Solution

Someone told me that the X you plug into 1/2kx^2 is not the distance from equilibrium in the table. They said that you use the formula F=-kx.
In this case, F=mg so mg=-kx
you solve for X, and this becomes your new equilibrium point.
Then you go through and adjust the rest of the distances from equilibrium so that the closest ones to equilibrium (2 cm previously) are now 2 cm away from this new X.
And this new X is the one you plug into 1/2kx^2 to solve for elastic potential.
Is this correct?

And what values would I use for h to solve for GPE?
Using the X (distance from equilibrium) values given in the table would result in some negative GPE...
Do I make the lowest point equal to 0 cm, and adjust the rest to their distance from the lowest point?

 

[merryjman]

The nice thing is that you get to choose whichever point you want to be zero GPE. As long as you are consistent you'll get the correct answers. You are right that using the EQ point as zero will give some negative GPE values, but that's okay: that just means that the object is gaining KE and/or Spring PE as it loses GPE. And since the points listed are relative to the EQ point, there's no need to adjust them.

I feel that it really helps to picture this thing bouncing up and down in your head - some of the answers, especially for KE, can come easily this way.

 

[pewpew23]

Okay, so I understand the GPE.

But what about the Elastic Potential?
How do I do that?
Am I supposed to solve for X using mg=-kx?

 

[merryjman]

That will give you the distance the spring stretched already (because of the mass on it), in other words, the equilibrium position, which you already found. Just add the additional distances to that (or subtract, for the negative ones) when plugging into the SPE equation.