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Position of Mass (Spring Mass Sytem)

  1. Oct 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A force of 10 Newtons can stretch a spring by 0.04 m. Suppose a mass of 5 kg is attached to the lower end of the spring. We stretch the mass downward by 0.05 m from its equilibrium position and release it from rest. Determine the position of the mass relative to its equilibrium position at t = 0.5 seconds. Assume no damping.

    2. Relevant equations
    k = F/x
    w= k/m
    x(t) = A cos wt + B sin wt

    3. The attempt at a solution
    k = 10/ 0.04 = 250 N/m
    x(0) = -0.05
    x`(0) = 0
    w= (250/5)^1/2 = 50^1/2
    equation x(t) = -0.05cos( (50^1/2) t))
    substituting t= 0.5, -0.049904837
    but the answer is -0.04617 I do not know where I messed up.
     
  2. jcsd
  3. Oct 20, 2016 #2

    Jonathan Scott

    User Avatar
    Gold Member

    You've taken the cosine of the angle in degrees, but it should be in radians.
     
  4. Oct 20, 2016 #3
    Oooohhh
    Thank you very much!
     
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