Equilibrium of a Suspended Mass on a Spring: Understanding Potential Energy

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SUMMARY

The potential energy of a mass suspended from a spring in equilibrium is zero, as the gravitational potential energy and the spring potential energy cancel each other out. When the mass is displaced by a distance x from the equilibrium position, the change in potential energy is calculated using the formula (1/2)kx², where k represents the spring constant. This formula accounts for both gravitational potential energy and spring potential energy when measured from the equilibrium position. For accurate analysis of simple harmonic motion (SHM), it is simpler to use the equilibrium position as the reference point.

PREREQUISITES
  • Understanding of potential energy concepts
  • Familiarity with Hooke's Law and spring constants
  • Knowledge of simple harmonic motion (SHM) principles
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the potential energy formula (1/2)kx² in the context of springs
  • Learn about the relationship between gravitational potential energy and spring potential energy
  • Explore examples of simple harmonic motion involving springs and masses
  • Investigate the effects of varying spring constants on potential energy calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillations, as well as educators looking to clarify concepts related to potential energy in spring systems.

Skomatth
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Say a mass is suspended vertically from a spring and is in equilibrium.

What is its potential energy? I think its zero because while it has gravitational potential that would allow it to do work downwards, it also has potential energy from the spring upwards. So do these cancel out? It seems like they have to for SHM equations to work out.

What is its potnetial energy when it is pulled down m meters? (1/2)km^2

This isn't homework just a a concept I'm confused on.
 
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the m in your equation is relative to the equilibrium position so I think you're probably right. Someone else may want to weigh in on this.
 
With a mass hanging from a spring, it's simpler to measure displacement from the equilibrium point. (Thus, if the mass is at the equilibrium point, the potential energy is zero.) If you do that, then when the mass is moved a distance x from that equilibrium point, the change in potential energy is given by 1/2 k x^2. Note that this includes gravitational PE.

If you wanted to keep measuring things from the original unstretched position of the spring, then you'd have to add the gravitational PE term. But you'd get the same net \Delta {PE} when you displace the mass from equilibrium. (Convince yourself of this by doing the calculation.)

It's of course much easier to use the equilibrium position as the reference for analyzing the resulting SHM.
 
Ok, I think I understand just tell me if this is right...

You only need gravitational PE if you consider equilibrium to be mass on the unstretched string. If you consider equilibrium to be mass on the stretched string(not moving) then gains in gravitational PE are included in the .5kx^2 term. This is probably why I've never seen a problem that tells you the height at which the mass is oscillating. Sounds pretty much like I reiterated what Doc Al said but it helped typing it. I'd do the math but I have a policy of no homework the day before the test :smile:
 

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