Energy in a vertical spring mass system

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SUMMARY

The discussion centers on demonstrating that energy in a vertical spring mass system remains constant during oscillations. The total mechanical energy is expressed as E_t = (1/2)Mv² + (1/2)ky² + Mgy, where y represents displacement from equilibrium. The user initially miscalculated potential energy at the bottom of the oscillation, leading to confusion about energy conservation. The correct interpretation clarifies that spring potential energy is calculated as (1/2)kx², emphasizing that x is the stretch from the equilibrium position, not y.

PREREQUISITES
  • Understanding of classical mechanics principles
  • Familiarity with Hooke's Law and spring potential energy
  • Knowledge of kinetic energy equations
  • Basic concepts of gravitational potential energy
NEXT STEPS
  • Review the principles of energy conservation in oscillatory systems
  • Study the derivation of the spring potential energy formula, PE = (1/2)kx²
  • Explore the relationship between displacement and energy in harmonic motion
  • Investigate the effects of damping on energy conservation in spring systems
USEFUL FOR

Students in physics, particularly those studying mechanics, as well as educators and anyone interested in understanding energy conservation in oscillatory systems.

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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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