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

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Homework Help Overview

The discussion revolves around the potential energy of a mass suspended from a spring in equilibrium. Participants explore the relationship between gravitational potential energy and spring potential energy, particularly in the context of simple harmonic motion (SHM).

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss whether the potential energies from gravity and the spring cancel each other out at equilibrium. They also consider how to define potential energy when the mass is displaced from the equilibrium position.

Discussion Status

Some participants suggest that measuring potential energy from the equilibrium position simplifies the analysis. There is an acknowledgment of different reference points for potential energy, but no consensus has been reached on the implications of these choices.

Contextual Notes

One participant notes a personal policy against doing homework the day before a test, indicating a potential constraint on their engagement with the problem.

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 [itex]1/2 k x^2[/itex]. 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 [itex]\Delta {PE}[/itex] 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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