Vertical mass-spring system

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  • #1
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does anybody know if the action of a real mas-spring system suspended vertically, support or refute the law of conversation of energy
 

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  • #2
turin
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I doubt that it refutes the law of conservation of energy. Do you have a specific phenomenon in mind?
 
  • #3
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umm a mass hanging on a spring that is vertical
 
  • #4
turin
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umm a mass hanging on a spring that is vertical
Do you think that this refutes conservation of energy? If so, how? And realize that there are many forces and factors envolved, so try to incorporate all of them in your explanation.
 
  • #5
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I'm also doing a piece of work on this at the minute and it would seem that the vertical mass-spring oscillator is going against the conservation of energy principle.
Using the following data:
k (Spring Constant) = 40.875Nm-1
x - Extension of Spring (below equilibrium position) = 0.01m (1cm)
Maximum Height Above equilibrium position = 0.009m (0.9cm)
Mass on the end of the spring = 0.4kg (4*100g slotted masses)

Using Hooke's Law : 0.5*k*x2
= 0.5 * 40.875Nm-1 * (0.01m)2
0.002044Nm = 0.002044 J

Now the gain in Gravitational Potential Energy (GPE) of the mass :
GPE = Mass * Gravity Constant (9.81N kg-1) * Height Change
GPE = 0.4kg * 9.81Nkg-1 * 0.019m = 0.074556Nm = 0.074556 J

I would appreciate it if anyone could validate the mathematical model I have used and either confirm that the data is suggesting that the vertical mass spring oscillator is going against the conservation of energy principle or point out a flaw in my method and how I can fix this?
 
  • #6
turin
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cmj, I didn't understand your point. You have some elastic potential energy stored in the spring and some gravitational potential energy due to elevation of the mass. Do you understand that both of these energies are allowed simultaneously? How do you think conservation of energy is violated here? (BTW, I didn't check your numbers, just your ideas.)
 
  • #7
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The Elastic Potential Energy is what is there when the mass is extended downwards (so I store energy in the spring)

Then the GPE is the maximum height that the mass gets to.

At the start... all the energy is Elastic Potential (which i have calculated above for you)
then at the end... all the energy should be GPE

but there is more GPE than Elastic Potential... meaning ive got more energy than i started with... which shouldnt happen so far as i understand
 
  • #8
turin
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At the start... all the energy is Elastic Potential ... then at the end... all the energy should be GPE
How did you come to these conclusions? So, you hang a mass from the ceiling by a spring. Then, you pull it down to to some elevation and hold it still. Then, you let it go. Is that the situation? I realize that this is just a thought experiment, but try to write a set of instructions for someone to follow so that they could repeat your experiment. This will do two things: 1) help you organize your thoughts and practice being a scientist, and 2) help me see where you are neglecting something.

BTW, do you think that this situation violates Newton's second law?
 
  • #9
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I've already done the experiment.... this is actual data im quoting. I'm doing A2 Physics (like 2nd year at college, UK). The basic setup is what you said... a fixed spring with a mass attached to the end of it. Then i pull the mass down a known extension (in the data I've used a 1cm extension). Then the spring is released (not pushed up or anything like that but just released from the 1cm extension) It then gets pulled upward by the spring to a certain height. What I'm saying is that the energy in the spring (calculated using Hooke's Law) is less than the GPE the mass gains. Therefore... it appears that the situation is giving me a gain in energy with no interference from other sources so far as i can see.
 
  • #10
Doc Al
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Using Hooke's Law : 0.5*k*x2
= 0.5 * 40.875Nm-1 * (0.01m)2
0.002044Nm = 0.002044 J
The spring potential energy should be measured from the unstretched position of the spring, not from the equilibrium position. Measure the change in spring PE from the highest position to the lowest position and then compare that to the change in gravitational PE between those same points.

Now the gain in Gravitational Potential Energy (GPE) of the mass :
GPE = Mass * Gravity Constant (9.81N kg-1) * Height Change
GPE = 0.4kg * 9.81Nkg-1 * 0.019m = 0.074556Nm = 0.074556 J
 
  • #11
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That would appear to just about solve my problem, although there are still a few small issues that I've had. Not to worry anyway as I've found a different way of calculating what I needed which gives me fairly consistent results. These calculations were only being used to calculate the loss in energy (and from there a resistance force seeing as F=E/s). However my other way of doing it seems to work.
Can I double check with you that in a perfect system (ie no resistance/loss in energy) a spring extended 1cm down would result in a 1cm gain above the equilibrium position?
 
  • #12
Doc Al
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Can I double check with you that in a perfect system (ie no resistance/loss in energy) a spring extended 1cm down would result in a 1cm gain above the equilibrium position?
Yes. If you pull the mass a distance X below the equilibrium point and release it, it will rise to a point a distance X above the equilibrium point. (And then continue oscillating between those two points.)
 

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