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The motion of a particle is described by . At what time in s is the potential energy

  1. Jun 28, 2009 #1
    1. The motion of a particle is described by x=10sin(pi*t+pi/3). At what time in s is the potential energy equal to the kinetic energy?

    2. I know these equations come into play
    Kinetic Energy
    K=1/2*m*ω^2*A^2*sin^2(ωt+φ)
    Kmax=1/2 k A^2
    K=1/2 mv^2

    Potential Energy
    U=1/2*k*A^2*cos^2(ωt+φ)
    U=1/2 kx^2

    3. The attempt at a solution

    Well I know A=10, w=pi, T=pi seconds I am really stuck on where to go from here I think I am trying to make it too complicated. Any pointers would be great. Also if someone could explain what the phase constant does to the equation it would be very helpful.
     
  2. jcsd
  3. Jun 28, 2009 #2

    rock.freak667

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    Re: The motion of a particle is described by . At what time in s is the potential ene

    you want where

    [tex]\frac{1}{2}mv^2 = mgx[/tex]


    m will cancel out and you know that v= dx/dt.

    then remember that cos2A+sin2A=1
     
  4. Jun 28, 2009 #3
    Re: The motion of a particle is described by . At what time in s is the potential ene

    from that suggestion I end up with:

    1/2*(10pi*cos(pi*t+pi/3))^2=9.8*10*sin(pi*t+pi/3)

    Simplifying gets

    10pi^2/9.8=sin(pi*t+pi/3)/cos(pi*t+pi/3)^2

    I am still jammed up
     
  5. Jun 28, 2009 #4

    rock.freak667

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    Re: The motion of a particle is described by . At what time in s is the potential ene

    (10pi^2/9.8)cos2(pi*t+pi/3)=sin(pi*t+pi/3)

    Now

    [tex]cos^2(\pi t + \frac{\pi}{3}) + sin^2(\pi t + \frac{\pi}{3})=1[/tex]


    sub for [itex]cos^2(\pi t + \frac{\pi}{3})[/tex]

    then you will have a quadratic in [itex]sin(\pi t + \frac{\pi}{3})[/itex]
     
  6. Jun 28, 2009 #5

    rl.bhat

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    Re: The motion of a particle is described by . At what time in s is the potential ene

    KE = PE
    0.5mv^2 = 0.5kx^2
    v = dx/dt = Aωsin(ωt + φ)
    v^2 = A^2ω^2sin^2(ωt + φ) = ω^2(A^2 - x^2)
    So 0.5mv^2 = 0.5kx^2
    mω^2(A^2 - x^2) = kx^2 = mω^2(x^2)
    or A^2 - x^2 = x^2 or A^2 = 2x^2 or x^2 = A^2/2 or x = A/(2)^1/2
    Substitute this value in the given equation and solve for t. You can see how the phase angle helps to find t.
    In this problem g has no roll at all.
     
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