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Homework Help: Block and Spring (Simple Harmonic Motion Problem)

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data

    At t = 0 a block with mass M = 5 kg moves with a velocity v = 2 m/s at position xo = -.33 m from the equilibrium position of the spring. The block is attached to a massless spring of spring constant k = 61.2 N/m and slides on a frictionless surface. At what time will the block next pass x = 0, the place where the spring is unstretched?

    2. Relevant equations

    phi=phase angle
    angular frequency= sqrt(k/m)=w
    x(t)=Acos(wt+phi)

    3. The attempt at a solution

    After working through the givens, I got my equation for the harmonic motion to be
    x(t) = 0.66 cos (3.4987t - 2pi/3)
    Where the angular frequency is sqrt(61.2/5)=3.4987, I got the amplitude from conservation of energy to be .6607 meters and the phase angle is -2pi/3.

    The next step, if I am not mistaken, is to solve for when the spring is at its equilibrium position, or when x(t1)=0. Here are the steps for what I did...

    0=0.6607 cos (3.4987t - 2pi/3)
    0=cos(3.4987t - 2pi/3)
    cos-1(0)=3.4987t - 2pi/3
    pi/2= 3.4987t -2pi/3
    7pi/6=3.4987t
    t=1.0476 seconds

    which, according to the homework website, is not correct! Can anyone see where I made my mistake? Thank you!!
     
  2. jcsd
  3. Apr 14, 2010 #2
    I think probably since the motion is starting at t=0, then x(t) is better to be written as A cos(wt) , (im not really sure) .. but i think it is something like that ..

    i have a question for you, why did you take phi = -2pi/3
     
  4. Apr 14, 2010 #3

    ehild

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    Homework Helper

    3.4987t - 2pi/3 is sooner -pi/2 than pi/2.

    ehild
     
  5. Apr 14, 2010 #4
    I figured out what I did wrong in this situation... cos-1(0) could be either pi/2 or -pi/2, and because the motion is coming to the end of a complete cycle I should have used -pi/2.

    To answer your question, I determined the phase shift=phi by solving the position equation. I knew at t=0, x(t)= -.33 meters, so you just solve x(t) = A cos (wt + f) for f, and I got -2pi/3.
     
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