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Simple Harmonic Motion of an oscillating particle

  1. Nov 9, 2008 #1
    The period of an oscillating particle is 56 s, and its amplitude is 18 cm. At t = 0, it is at its equilibrium position. Find the distances traveled during these intervals.
    (a) t = 0 to t = 14 s
    cm

    (b) t = 14 s to t = 28 s
    cm

    (c) t = 0 to t = 7 s
    cm

    (d) t = 7 s to t = 14 s
    cm

    x = Acos([tex]\omega[/tex] * t * [tex]\delta[/tex])
    [tex]\omega[/tex] = 2[tex]\pi[/tex] / T
    [tex]\omega[/tex] = 2[tex]\pi[/tex] / 56

    Since x = 0 when t = 0
    Then [tex]\delta[/tex] = [tex]\pi[/tex] / 2
    correct?

    for part A)

    x = 18cm * cos[2[tex]\pi[/tex] / 56 * 14s + [tex]\pi[/tex] / 2] - cos[2[tex]\pi[/tex] / 56 * 0s + [tex]\pi[/tex] / 2]

    Is this good?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 14, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi maniacp08,

    The 18cm needs to multiply both cosine functions, not just the first one. Was that just a typo?

    Also you can think about how the time period of 14 seconds is related to the period, to get the answer right away.
     
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