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Does vmax = -wA (SHM)

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A large mass bobs up and down on a heavy spring. Initially it is at the top. It achieves its maximum downward velocity of 94 cm s-1 in 0.25 s from its release.

    What are the period, angular freuqnecy, and amplitude for this motion? Find amplitude .

    2. Relevant equations

    vmax=-wA

    3. The attempt at a solution

    i couldnt figure out the solution so i clicked help and this is what it said, ive never seen this equation used before and was wondering if this is always the case, because to me it doesnt make sense.

    the graph starts at its max at t=0 so i take it as a cos function so i use the equation

    x=Acos(wt)

    derived i get

    v=-wAsin(wt)

    if t=0, sin(wt)=0 and all i get is zero can someone help me understand? A=0


    Edit: OOps i didnt input the correct time for when v = Max when i do this sin = 1 i got it now!

    clarification on the formula would be appreciated.
     
    Last edited: Dec 4, 2011
  2. jcsd
  3. Dec 4, 2011 #2
    You are taking t=0 to be when the object is at max displacement ie x = A .
    This corresponds to Cosωt =1, which means ωt = 0
    This means that ωt = 90 or ∏/2 when the displacement = 0, this is the point of max velocity
    v = ωA is the max velocity
     
  4. Dec 4, 2011 #3
    yes in shm v[itex]_{max}[/itex] = A[itex]\omega[/itex]
     
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