Energy in a vertical spring mass system

AI Thread Summary
The discussion centers on demonstrating that energy in a vertical spring mass system remains constant during oscillations. The user initially struggles with the energy equations, noting discrepancies between the total energy at the top and bottom of the oscillation. They correctly identify that potential energy should not be negative at the bottom, as the displacement is measured from the equilibrium position. A key realization is that the spring potential energy is calculated using the stretch from the equilibrium point, not the total displacement. Ultimately, the user acknowledges the correction and thanks the forum for the assistance.
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Homework Statement


I am doing a lab in which we are to show that the energy in a spring mass system is constant throughout the oscillations.


Homework Equations


I set y initial = 0 to be the point where the spring was in equilibrium when the mass was attached to it. At any given point the I believe the energy should be
E_t=\frac{1}{2}Mv^2 + \frac{1}{2}ky^2 + Mgy
Where y is the displacement from the equilibrium position and y+ isup


The Attempt at a Solution


The problem I'm having is that at the top of its oscillation all 3 of the energies are positive.
On the bottom both kinetic and potential energy should be the same since both the values are square keeping them positive. However the potential energy is negative at the bottom since the displacement in negative.
So I get
E_{top}=\frac{1}{2}Mv^2 + \frac{1}{2}ky^2 + Mgy
E_{bottom}=\frac{1}{2}Mv^2 + \frac{1}{2}ky^2 - Mgy
Which makes it seem that energy is not conserved since E_{top}\ne E_{bottom}

Can anyone point me to what I am missing here?

Thanks
 
Last edited:
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Realize that spring PE = ½kx², where x is the amount of stretch. The stretch is not zero at the equilibrium point, so ½ky² is not the spring energy.
 
Oh I see it now. Thanks for the help :)
 
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