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Simple harmonic motion solved for time

  1. May 17, 2016 #1
    1. The problem statement, all variables and given/known data

    A particle of mass 0.50 kg performs simple harmonic motion along the x axis with amplitude 0.55m and period 4.3 seconds. The initial displacement of the particle is -0.30 m and it is travelling in the positive x-direction. The phase constant of the motion (Φ) = -2.15 rad. Find the time at which the particle first reaches x = +0.3m.
    Relevant equations:

    2. Relevant equations

    x(t) = Acos(ωt + Φ)
    f = 1/T = 0.2326 Hz
    ω = 2πf = 1.4612 rad/s

    3. The attempt at a solution

    Rearrange for t:
    t = [arccos(x/A) - Φ]/ω
    t = [arccos(0.6/0.55) + 2.15]/1.4612
    This is giving me a maths error because arccos(x > 1) does't exist, but 0.6m is the total displacement.
    Even when I use 0.3 as my measurement for x I get a ridiculous value for t:

    t = [arccos(0.3/0.55) + 2.15]/1.4612
    t = 40.4423 s

    Not sure where I'm going wrong. Any help would be great!
     
  2. jcsd
  3. May 17, 2016 #2

    ehild

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    Where did the 0.6 for the displacement come from? The initial displacement was given as -0.3 m.
     
  4. May 17, 2016 #3
    If the initial displacement is -0.3m and I am measuring the time to get to 0.3m, should x not be 0.6m?
     
  5. May 17, 2016 #4

    nrqed

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    A) You *have* to use x=0.3 meter. This equation gives the position as a function of time, it does not give a displacement as function of time.

    B) Your calculator must be un radians to get the correct inverse cos.
     
  6. May 17, 2016 #5
    Many thanks, turns out I'm just an idiot who doesn't know how to work a calculator.
     
  7. May 17, 2016 #6

    ehild

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    No. x is displacement with respect to the origin. Better to refer it as 'position'. You know that x=Acos(wt+Φ)=0.3. w, A, and Φ are given, you can find t, but yes, set the calculator to RAD.
     
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