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Mass-spring system in SHM concept question

  1. May 7, 2013 #1
    1. The problem statement, all variables and given/known data
    1raq6x.jpg


    2. Relevant equations



    3. The attempt at a solution

    Well since spring is in SHM, only conservative forces are at play here. So using conservation of energy, the kinetic energy would be the change in potential energy. Which I have set up as

    KE = 1/2k(A^2)2 - 1/2k(A/4)2

    after some algebra, combining like terms with common denominators I am left with,

    KE = (15/32)kA2

    now when I take the fraction of the two, PE/KE of

    (15/32)kA2 / 1/2mv2

    I get (15/16)kA2/mvf2

    So I'm thinking my answer would be 15/16

    I don't know whether to just ignore the variables since there was no value in them or I find a way to relate them since I was given a frequency?
     
  2. jcsd
  3. May 7, 2013 #2

    collinsmark

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    So far so good. :approve:

    Sorry but I'm not following you there. :uhh:

    What's the total energy of the system? Once you find that you can find the ratio of kinetic energy over total energy.

    It might be 15/16. But if so, you haven't quite shown why yet.

    Hint: your next step is to find the total energy of the system.

    [Edit: Another hint: the total energy of the system is conserved, and doesn't change with time. :wink:]
     
    Last edited: May 7, 2013
  4. May 7, 2013 #3

    HallsofIvy

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    The reason this is a "conceptual" problem, as opposed to a "calculation" problem is that you can use the fact that the potential energy of a spring is proportional to the extension or compression of the spring to quickly find the ratio. When the mass is at distance A from the equilibrium all of the energy is potential, when it is distance 0 from equilibrium, none of it is. when it is at distance A/4 from equilbrium (and so 3A/4 from greatest extension) what fraction of the energy is potential energy? What fraction is kinetic?
     
    Last edited: May 7, 2013
  5. May 7, 2013 #4
    Oh I think I've figured it out. The total energy of the system can either be when x is at it's max "A", or when the velocity is at it's max = v_max. So I need an equation for the kinetic energy, which I set up as

    1/2mvf^2 = 1/2kA^2, which gives me the Vfinal I can use to substitute into the kinetic energy equation above which allows me to take the ratio, and everything being cancelled out I am just left with 15/16
     
  6. May 7, 2013 #5

    collinsmark

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    I think you mean the potential energy is proportional to the square of the extension or compression. :smile:

    I am not familiar with whatever you are doing with the "1/2mvf^2." :uhh: Maybe you mean [itex] \frac{1}{2}m(v_{max})^2?[/itex].

    But yes, the potential energy of the system is [itex] \frac{1}{2}kA^2 [/itex] when the extension is equal to A. You can use that (and a hints from HallsofIvy and myself [my previous post]) to determine the total energy of the system. :wink:

    In other words, you could solve for [itex] v_{max} [/itex] and substitute things around. But it's not necessary to even bring [itex] v_{max} [/itex] into the problem. If you know what the potential energy is at maximum extension, then you also know the total system energy.
     
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