Energy in a vertical spring mass system

  • Thread starter S.A
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  • #1
S.A
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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
[tex]E_t=\frac{1}{2}Mv^2 + \frac{1}{2}ky^2 + Mgy[/tex]
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
[tex]E_{top}=\frac{1}{2}Mv^2 + \frac{1}{2}ky^2 + Mgy[/tex]
[tex]E_{bottom}=\frac{1}{2}Mv^2 + \frac{1}{2}ky^2 - Mgy[/tex]
Which makes it seem that energy is not conserved since [tex]E_{top}\ne E_{bottom}[/tex]

Can anyone point me to what I am missing here?

Thanks
 
Last edited:

Answers and Replies

  • #2
Doc Al
Mentor
45,209
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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.
 
  • #3
S.A
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Oh I see it now. Thanks for the help :)
 

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