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Homework Help: Simple harmonic motion question please help!

  1. Nov 29, 2011 #1
    1. The problem statement, all variables and given/known data

    An object undergoing SHM with a period 0.9s and amplitude of 0.32 m at t=0 the object is at 0.32 m and is instantaneously at rest. Calculate the time it takes the object to go a) from 0.32 to 0.16 m
    b) from 0.16 m to 0

    i can do a but dont understand why when i do the calculation x is not -1/2 and is 1/2

    2. Relevant equations
    x=Asin(wt)




    3. The attempt at a solution

    so for first part 1/2 = 0.32 cos (2+2/9)*pi * t

    i get 0.15 correct answer and for part b i do the same thing because it is the same distance excpt i shift the graph so t1 is at 0.16 like this 0.16 = a cos ((wt) - pi/4)
     
  2. jcsd
  3. Nov 30, 2011 #2
    If 0.15s is the correct value for the time to reach 0.16m from 0.32m, then to reach 0 from 0.32m is the time for exactly one quarter of a period. (=0.9s/4)
    The time from 0.16m to zero is then just the time for the quarter period minus your first answer.
     
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