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Another simple harmonic motion problem

  1. Dec 5, 2011 #1
    1. The problem statement, all variables and given/known data
    An object is undergoing SHM with period 0.9 s and amplitude 0.32 m. At t=0, the object is at x=0.32 m and is instantaneously at rest. Calculate the time it takes the object to go from 0.32 m to 0.16 m, and to go from 0.16m to 0 m.


    2. Relevant equations
    x=Acos(wt+θ)
    f=1/T
    w=2∏f



    3. The attempt at a solution

    I completely understood the first part of the problem when going from x= .32 to x= .16. using the formulas to come up with the angular frequency and then just solving for t, in which i got the answer t= 0.15 s.

    I'm confused about the second part because it isn't starting from t=0. I was thinking that the only difference in the formula would be the phase angle, i just don't understand how to come up with that. i thought that the phase angle is the angle in which x is at the position you are starting at, if you start at a position .16 m with an unknown t-value, how do you know what the phase angle is?
     
  2. jcsd
  3. Dec 5, 2011 #2
    I also got 0.15s for the first part.
    To go from 0.32 to 0 is 1/4 of a cycle. I used this to calculate the time to go from 0.16 to 0
     
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