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SHM - Simple Harmonic Motion

  1. Jan 30, 2008 #1
    SHM - Simple Harmonic Motion Question

    1. The problem statement, all variables and given/known data

    Two particles oscillate in SHM along a common straight-line segment of length A. Each particle has a period of 1.5s, but they differ in phase by π/6 rad.
    a. How far apart are they 9in terms of A) 0.50s after the lagging particle leaves one end of the path?
    b. Are they then moving in the same direction, towards each other, or away from each other?

    2. Relevant equations

    X = A cos(ωt + θ)
    T = 2π/ω

    3. The attempt at a solution

    T = 2π/ω Tω = 2π ω = 2π/T
    X = A cos(ωt + θ) = A cos [(2π/T)t + θ] =
    A cos [(2π/1.5)t + 0] and A cos [(2π/1.5)t + π/6]
    A cos [(2π/1.5)t + 0] - A cos [(2π/1.5)t + π/6], t equals .5s
    (-.50)A - (-.87)A = .37A, but the answer in the book is .18A.
    What did I do wrong?
    Last edited: Jan 30, 2008
  2. jcsd
  3. Jan 30, 2008 #2
    Bumping up so someone could tell me what the think.
  4. Jan 31, 2008 #3
  5. Sep 9, 2011 #4
    Just to give the correct answer. Your passages were all good except for the first one. The equations describing the motion of the particles are: (A/2)cos(wt + H) and (A/2)cos(wt) and not just A, because the segment is long A and so (knowing that the cos goes from -1 to + 1) a partile moves from A/2 to -A/2. Substituting A/2 the computation should be right!
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