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Simple Harmonic Motion Question

  1. May 24, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle moves with SHM in a straight line.
    In the first second after starting from rest,it travels a distance of 1m in a constant direction.
    In the next second, it travels a distance of 2m in the same direction.
    Find its maximum acceleration


    2. Relevant equations
    x=acos(ωt)
    [itex]\dot{x}[/itex]=-ω√(a2-x2)
    [itex]\ddot{x}[/itex]=-ω2x


    3. The attempt at a solution
    Starting from rest ∴ when [itex]\dot{x}[/itex]=0, x=a, [itex]\ddot{x}[/itex]=maximum.
    maximum [itex]\ddot{x}[/itex]=-ω2x
    maximum amplitude = 1.5
    ∴ amplitude [itex]\geq[/itex]1.5

    After many more lines of work, no solution was able to be found.
    Can someone help?
     
  2. jcsd
  3. May 24, 2012 #2
    You have two equations and two unknowns, you don't need to do anything more than plugging values into x=acos(ωt).

    From the given information what values do you suppose you need to plug in?
     
  4. May 24, 2012 #3
    when t=1, x=a-1
    when t=2, x=a-3
     
  5. May 24, 2012 #4
    In the first second after starting from rest,it travels a distance of 1m in a constant direction.
    In the next second, it travels a distance of 2m in the same direction.

    I'd use this, what exactly is this saying?
     
  6. May 24, 2012 #5
    it is saying that after the 1st second the particle has moved 1m from the amplitude position and that after the 2nd second the particle mas moved a total of 3m from the amplitude position.
     
  7. May 24, 2012 #6

    vela

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    You have
    \begin{align*}
    a - a \cos \omega &= 1 \\
    a - a \cos 2\omega &= 3
    \end{align*} Solve for a in the first equation and substitute into the other. Then use a trig identity for ##\cos 2\omega## to write it in terms of ##\cos \omega##.
     
  8. May 25, 2012 #7
    ok. thank you for this. it helps me a lot
     
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