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Dynamics: Spring problem (Oscillation)

  1. Nov 26, 2017 #1
    1. The problem statement, all variables and given/known data
    The mass m = 4 kg and the spring constant k = 64 N/m. The spring is unstretched when x = 0.
    At t = 0, x = 0 and the mass has a velocity of 2 m/s down the inclined surface. What is the value
    of x at t = 0.8 s?
    The angle of incline is 20 degrees, and with the mass moving down to the right with a spring attached on its left surface.

    2. Relevant equations
    ΣFx = max
    x = Asin(wt)+Bcos(wt)

    3. The attempt at a solution
    d2x/dt2 + (16 s-2)x = 3.355 m/s2

    I got to this point and I found that for the equation x = Asin(wt)+Bcos(wt) the solution gives x = Asin(wt)+Bcos(wt) + 0.210 m. I get that 0.210 = 3.355/16, but why is it included in this solution for the equation when all other examples of this problem for a vertically or horizontally held spring mass system use just the x = Asin(wt)+Bcos(wt) equation without this addition value? What is this value, and is it just disregarded as zero in the other forms of this problem?
     
  2. jcsd
  3. Nov 27, 2017 #2

    haruspex

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    It is the equilibrium position, i.e. the spring extension required to balance the downslope gravity.
     
  4. Nov 30, 2017 #3
    Oh okay, that makes sense, thanks!
     
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