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Homework Help: Simple Harmonic Motion Question

  1. Aug 2, 2010 #1
    1. The problem statement, all variables and given/known data
    A particle undergoes simple harmonic with amplitude of 10cm and frequency 5Hz. If it passes through the equilibrium position (x=0) at time t = 0, calculate the position, velocity and acceleration of the particle at t = 3.5m5 ms.



    3. The attempt at a solution


    Given that:

    A = 0.1m
    f = 5Hz
    therefroe w = 2pif = 10pi
    phase difference = 0.

    So using the standard: x = Acos(wt+PD)

    where PD = phase difference.

    I get x = 0.1m.
    differentiating this I get:
    v = -0.34 m/s
    and again differentiating I get:
    a = -98.0 m/s^2.

    However these answers appear to be wrong. Any advice on where I am messing up?
     
  2. jcsd
  3. Aug 2, 2010 #2

    rock.freak667

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    Homework Helper

    Your attempt at getting 'A' and 'ω' are correct. But the particle starts at x=0, t=0. So your general equation should be sine and not cosine.

    If it said x=A at t=0, then you'd use cosine.

     
  4. Aug 2, 2010 #3

    Thanks - that makes sense. I just read the position from the book without thinking...

    So:

    x = 0.1*sin(10pi*t)
    v = 0.1*10pi*cos(10pi*t)
    a = -0.1*100*pi^2*sin(10pi*t)

    would that be correct? thanks
     
  5. Aug 2, 2010 #4

    rock.freak667

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    Homework Helper

    Yes that should be correct now.
     
  6. Aug 2, 2010 #5
    Thank you very much - i appreciate the hepl.
     
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