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Energy/Harmonic Motion

  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A 2.0 kg mass on a spring is extended 0.3 m from equilibrium position and released. The spring constant is 65 N/m.

    C.Find the speed of the mass when the displacement is 0.2 m

    2. Relevant equations
    Ek=(0.5)(m) (v^2)
    Ep=(0.5)(k) (x^2)
    Ek+Ep=E

    3. The attempt at a solution
    I've already ascertain from the previous parts of this question that the initial potential energy of the spring is 2.925 J, and the maximum speed reached by this mass is 1.71 m/s. According to my textbook the answer is 1.27 m/s but I keep getting...

    (1/2)(m) (v^2)=(1/2)(k) (x^2)
    v^2=(1/2)(65)(0.2^2)
    v=1.14
     
  2. jcsd
  3. Nov 12, 2013 #2
    The spring is initially stretched 0.3m from equilibrium, and then when it is released and is displaced 0.2m it is still stretched but has some velocity. So what is the energy before release, and what is the energy at 0.2m from equilibrium after release?
     
    Last edited: Nov 12, 2013
  4. Nov 12, 2013 #3

    Mentz114

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    I solved this for v and got 1.27. You might have made a mistake.

    (1/2)(m) (v^2)+(1/2)(k) (x^2)=(1/2)(m)(1.71^2)

    [Edit: you should have used 0.3, not 0.2 here

    (1/2)(m) (v^2)+(1/2)(k) (x^2)v^2=(1/2)(65)(0.3^2)
     
    Last edited: Nov 12, 2013
  5. Nov 12, 2013 #4

    haruspex

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    The second term is dimensionally wrong. I guess you meant (1/2)m v2+(1/2)k x2=(1/2)m1.712, though it's probably better to use (1/2)m v2+(1/2)k x12=(1/2)k x02, minimising the growth of rounding errors.
     
  6. Nov 12, 2013 #5

    Mentz114

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    Sorry that v^2 is a typo. Tsk... I have corrected the post, thanks for pointing it out.
     
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