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Mass hanging on a spring - find acceleration

  1. Oct 9, 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 0.29 m. When a mass of 0.33 kg is hung on it, the equilibrium length is 0.92 m. At t=0, the mass (which is at the equilibrium point) is given a velocity of 4.80 m/s downward.
    At t=0.70 s, what is the acceleration of the mass? (Positive for upward acceleration, negative for downward)


    2. Relevant equations

    Fspring = -k*(x(t) - xeqb)

    ƩFy,mass = Fspring - m*g = -m*ay
    ƩFy,mass = -k*(x(t) - xeqb) - m*g = -m*a

    x(t) = A*cos(ω*t)

    3. The attempt at a solution
    Not sure how to go about this. I attempted to find the acceleration using the sum of the forces on the mass (which is accelerating downward) but I ran into a roadblock with finding k. I thought the equation for the position of a mass undergoing SHM might help out but I am not sure how to solve for A, because A is not equal to the equilibrium position.
     
  2. jcsd
  3. Oct 9, 2012 #2

    ehild

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    First find the spring constant.


    ehild
     
  4. Oct 9, 2012 #3
    The spring constant, k, is equal to the weight force divided by the change in position.

    k = (m*g)/Δx

    which we are all given.

    Once I've found the spring constant, now what do I do? I am confused on how to find x(t=0.7s) when I have an unknown amplitude.
     
  5. Oct 9, 2012 #4

    ehild

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    The mass will oscillate about the new equilibrium position. It starts SHM from equilibrium point with maximum velocity. Write up the equations for displacement, velocity, acceleration of this SHM.

    ehild
     
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