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Help differentiating energy wrt time.

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a problem where I have a mass suspended in a system of springs. I need to differentiate the equation wrt time so I can can show equivalence with Newton's second law.

    The mass and springs are vertically aligned so the motion is in one dimension. The actual problem has several springs, but for simplicity I am describing a system with just two. The equation below I think shows the total energy of the system.


    2. Relevant equations

    E = 1/2 mv^2 + k(x-l)^2 + 2k(l-x)^2 -mgx

    where m=mass, v= velocity, k= stiffness, x=current position and l=spring's natural length.

    3. The attempt at a solution

    I think the way to approach it is to substitute dx/dt in place of the velocity, however I can't see what to do with the spring parts.

    I seem to have some kind of mental block on this, and it's very frustrating. Any assistance on how to approach it would be gratefully received!
     
  2. jcsd
  3. Jan 28, 2013 #2
    E = 1/2 mv^2 + k(x-l)^2 + 2k(l-x)^2 -mgx - don't think this is quite right

    I am guessing that we have one spring above the mass and one below?

    The terms for elastic spring energy are based on 1/2ke^2. Where e is spring extension.

    Isn't the extension of the springs l+x and l-x?

    Are both springs of the same stiffness constant k? Perhaps there is a typo here?

    Make these corrections and multiply out brackets before differentiating.

    Remember
    [tex]\frac{d}{dt}v^2=2v\frac{dv}{dt}=2\frac{dx}{dt}a[/tex]
    Where a is acceleration. So the kinetic energy term after differentiating will have ma in it, which is starting to look like N2L. It will also have dx/dt in it. All other terms of your equation will either go to zero because they are independent of t or be a multiple of dx/dt. So dx/dt will cancel throughout. This will leave terms that are all forces.
     
  4. Jan 28, 2013 #3

    haruspex

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    I don't see how it could be l+x. l-x and x-l make sense if the springs are in series with the endpoints fixed 2l apart, and x being the position of the join. But then, the expression simplifies to 3k(l-x)2.
     
  5. Jan 28, 2013 #4
    Thanks for your suggestions. I think I have it now.

    The lengths were the correct way round, but I didn't make it quite clear in the question I asked. The distance between the two fixtures is known, so the deformation can be expressed in relation to that.

    The answer I got did end up as m.a equated with the spring forces according to Hooke's law and gravity which is what I needed.

    It's amazing the difference a nights sleep makes.
     
  6. Jan 29, 2013 #5
    I was thinking L was the extension of the springs when at equilibrium and x was the displacement from equilibrium. Anyhow it does not matter much since the constants drop out during differentiation.
     
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