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Homework Help: What is the maximum potential energy of an oscillating mass?

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data
    A 2-kg mass attached to a spring oscillates in simple harmonic motion and has a speed of 5 m/s at the equilibrium point. What is the maximum potential energy of this oscillating mass?

    2. Relevant equations
    I know that the potential energy is: Ep = 1/2 kx2
    k = mω2
  2. jcsd
  3. Sep 24, 2013 #2


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    In SHM,

    KE+PE = Constant.

    At the equilibrium position, the PE is what value? Once you have that then at the max position of the spring, the KE is what value?
  4. Sep 24, 2013 #3
    So, at the equilibrium position KE = PE, therefore KE = 1/2 mv2 = 25 J. And PE = 25 J. If that's correct. But I don't know what will be the value at the max position
  5. Sep 24, 2013 #4
    Not that KE= PE. It is that KE+PE= a constant. So, you can (for example), find all the different energies at the equilibrium and then find all the energies at your maximum potential. And then you know that they must equal the same constant (in the end you should only have max PE which you don't know).
  6. Sep 24, 2013 #5
    I really don't know how to do it, and I have an exam tomorrow. If you tell me the solution, I might find the logic by looking at it.
  7. Sep 24, 2013 #6


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    KE +PE = Constant

    At the equilibrium position you found that PE=0 so that KE = Constant = 25 J


    KE+PE = 25

    At the maximum positionm what would be the KE? (when the spring is oscillating after it reaches its max position does it keep going or does its velocity change?)
  8. Sep 24, 2013 #7
    Since KE + PE = const, and at the equilibrium position PE = 0, therefore KE = Constant = 25 J, as you rock.freak667 said. And since "they must equal the same constant", as Jufro said, that means that at the maximum position KE = 0 and PE = const = 25 J. Is that correct, or am I mistaken again?
  9. Sep 24, 2013 #8


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    Yes that is correct. So PE=25 J at the max position.

    For SHM KE+PE = constant at any point in the motion
  10. Sep 25, 2013 #9
    Thank you very much :)
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