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Energy Problem involving simple harmonic motion

  1. Dec 11, 2007 #1
    1. The problem statement, all variables and given/known data
    The motion of a particle is described by x = 10 sin (piet +pie/2 ). At what time ( in second ) is the potential energy equal to the kinetic energy ?


    2. Relevant equations


    KE = 1/2 mv^2 PE=1/2 Kx^2
    3. The attempt at a solution

    V=pie10cos(piet + pie/3)
    V^2 = pie^2 100cos^2(piet+ pie/3)
    KE = mpie5ocos^2 (piet + pie/3)
    X^2 =100sin^2(piet +pie/2 )
    PE =k50 sin^2(piet +pie/2 )
    KE=PE
    pie^2 100cos^2(piet+ pie/3) =K50 sin^2(piet +pie/2 )
    m/k * pie^2 = tan^2(piet +pie/2 )
    take square root of both sides of equation
    1/pie * pie = tan(piet + pie/3)
    piet + pie/3 =tan-1(1)
    solving for t i get a negative value where iam i going wrong.
     
  2. jcsd
  3. Dec 11, 2007 #2

    Kurdt

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    There appear to be a lot of algebraic errors throughout. Where has the m and k gone from the equation toward the end?

    EDIt: up to the this point you seem to be fine as far as I can tell:

    m/k * pie^2 = tan^2(piet +pie/2 )

    I just want to know how you got rid of the m and k in the next step?
     
  4. Dec 11, 2007 #3
    w(angular frequency) = (k/m)^0.5 so
    1/w=m/k
    siince w = pie
    m/k = 1/pie
     
  5. Dec 11, 2007 #4

    Kurdt

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    I tried a different approach and got the same answer. since the motion is SHM one would think that value +[itex]n2\pi[/itex] should work.
     
  6. Dec 11, 2007 #5
    what is n2PI is it equal to m/k ?
     
  7. Dec 11, 2007 #6

    rl.bhat

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    You can wright PE = 1/2*m*w^2*x^2 and KE = 1/2*m*w^2*(A^2 - x^2)
    When PE = KE, we get 2x^2 = A^2 or x = A/sqrt2. Put this value in the equation of SHM.
    A/sqrt2 = Asin(pi*t + Pi/2 ) or 1/sqrt2 = sin(pi*t + pi/2). Sin(pi/4) = sin(3*pi/4) = 1/sqrt2 To avoide negative time, take sin(3pi/4) = sin(pi*t + pi/2)
    That gives you t = 1/4 s.
     
    Last edited: Dec 11, 2007
  8. Dec 11, 2007 #7
    the anwser is 0.9 s
     
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