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Simple Harmonic Motion with Spring

  1. Mar 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A massless spring is hanging vertically. With no load on the spring, it has a length of 0.22 m. When a mass of 0.68 kg is hung on it, the equilibrium length is 0.75 m. At t=0, the mass (which is at the equilibrium point) is given a velocity of 4.88 m/s downward.
    At t=046s, what is the acceleration of the mass? (Positive for upward acceleration, negative for downward)


    2. Relevant equations

    [itex]\Sigma[/itex]Fy=-k(yeq-y0)
    [itex]\Sigma[/itex]Fy=may

    3. The attempt at a solution

    The only two forces acting on the mass are the Tension from the spring and weight. So
    Tsm-(mg)=may
    I am now stuck wondering if that was a good place to start. Is the traditional way to solve this problem to find the spring constant first?
     
  2. jcsd
  3. Mar 6, 2012 #2

    rock.freak667

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    Yes you should find the spring constant first as you will need it later on to get the acceleration at t= 0.46.
     
  4. Mar 6, 2012 #3
    I have k=(mg)/(yeq-y0). I don't think this is right.
     
  5. Mar 7, 2012 #4

    rock.freak667

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    Right, well you are given m, y0 and yeq and you know g. So you can get k.
     
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