Identical Springs: Compressed vs Uncompressed Mass

AI Thread Summary
The discussion centers on the mass difference between a compressed and an uncompressed spring, both at rest. It is established that the compressed spring has more mass due to the potential energy associated with its compression, which can be related to Einstein's equation E=mc². Participants clarify that while the extra mass is often referred to simply as "energy," it specifically represents potential energy in this context. The conversation also touches on the conservation of total energy in systems involving oscillation, emphasizing that the total energy remains constant despite the interchange between kinetic and potential energy. Ultimately, the compressed spring does weigh slightly more than the uncompressed one due to the added potential energy.
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Two springs (with indentical, invariant, Rest mass) are indentical except that one spring is compressed. Both springs are at rest. Doesn’t the compressed spring have more mass than the uncompressed spring? (With mass equivalent to the energy required to compress the spring?) Assuming the answer is yes, what is the “extra” mass of the compressed spring called?” --it cannot be kE, both springs are at rest.
 
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The extra energy is called potential energy.
 
Thank you, but the question was what is the extra mass called. are you saying the "extra" mass has no other name than potential energy?
 
PCB said:
Thank you, but the question was what is the extra mass called. are you saying the "extra" mass has no other name than potential energy?
Yes, although we often don't bother with the adjective "potential" and just call it "energy".
We'll often (but not quite as often) do the same thing with kinetic energy.

This is becaue in many problems the distinction between the different kinds of energy isn't especially interesting. For example, suppose I take one of your springs, attach a weight to one end, and start it oscillating (google for "simple harmonic oscillator" if what I say next doesn't make sense to you). When the object is at the extremes of its oscillation, there is a moment when it is at rest with kinetic energy zero; at that moment the spring is either maximally compressed or maximally extended and therefore its potential energy is at a maximum. Then as the object continues to oscillate it passes back through the midpoint of its oscillation; at that point its velocity and kinetic energy is at a maximum and the potential energy of the spring is zero. So we see that the total energy is constant, it just keeps swapping back and forth between kinetic and potential energy.

Now, suppose that the entire harmonic oscillator is inside of a black box, so we cannot see the oscillation (this is a very common and very realistic situation). The weight of the box remains constant because the total amount of energy is constant. Sure, at any moment some of that energy is kinetic and some of it is potential, but we on the outside of the box have no way of knowing what the split is; the total energy is the only that matters outside the box.
 
So in my original scenario, the uncompressed spring weights the same as the compressed spring?
 
No, the compressed springs weigh slightly more.
 
PCB said:
So in my original scenario, the uncompressed spring weights the same as the compressed spring?
I'm not sure how you came to that conclusion. The potential energy in the compressed spring adds mass according to Einstein's ##E=mc^2##, just as does kinetic energy.
 
Dear Nuqatory, I didn't come to the conclusion that the two springs weighted the same, indeed, I was expecting Khashishi's answer. I re-asked the question because it seemed to me that your first reply implied they did weight the same. I obviously mis-understood your reply and I apologise for the confusion.
 
I think he meant to say that the mass of the system stayed constant.
 
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