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Velocity of a particle large and small times

  1. Oct 14, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    The velocity,v, of a particle moving on the x-axis varies with the time,t,according to the differential equation [itex]\frac{dv}{dt}+2v=sint[/itex]. Find v in terms of t given that v=0 when t=0. Deduce that when t is large;the velocity of the particle varies between -0.45 and 0.45. Show that is t is very small,then [itex]v\approx\frac{1}{2}t^2[/itex]


    2. Relevant equations

    [tex]\frac{d}{dx}(\frac{e^{2t}}{5}(2sint-cost))=e^{2t}sint[/tex]

    3. The attempt at a solution

    [tex]\frac{dv}{dt}+2v=sint[/tex]

    [tex]X e^{2t}[/tex]

    [tex]e^{2t}\frac{dv}{dt}+2e^{2t}v=e^{2t}sint[/tex]

    [tex]\int...dt[/tex]

    [tex]ve^{2t}=\frac{e^{2t}}{5}(2sint-cost)+C[/tex]

    using the initial values they gave me [itex]C=\frac{1}{5}[/itex]

    And so
    [tex]v=\frac{1}{5}(2sint-cost)+\frac{e^{-2t}}{5}[/tex]

    As [itex]t\rightarrow\infty(large),\frac{e^{-2t}}{5}\rightarrow0[/itex]

    but [itex]\frac{1}{5}(2sint-cost)[/itex] does not approach any limit as both cost and sint oscillate between 1 and -1.
     
  2. jcsd
  3. Oct 14, 2007 #2
    no not between -1 and 1 that is for individual terms.You have to combine them into 1 cosine term and the new amplitude is ur answer.
    use the below formula:
    A sint + B cost = (A^2+B^2 ) ^(1/2) * Cos(t-arctan(b/A))
     
  4. Oct 14, 2007 #3

    rock.freak667

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    but wouldn't Cos(t-arctan(b/A)) not approach a limit? Or would t tending to infinity mean that t would eventually be equal toarctan(B/A) and so it would be the cos0=1?
     
  5. Oct 14, 2007 #4
    the final limit will include a function that oscillates between some numbers.
    since that function will oscillate it means that the velocity will vary.
    Between [-0.45,0.45] in this case.
    You will have to plug in the numbers that lead to the maximum value (0.45) and the minimum (-0.45). and that will be it for t->infinity.
     
  6. Oct 14, 2007 #5

    rock.freak667

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    Oh....thank you...shall try the next part now
     
  7. Oct 14, 2007 #6
    yes it is just the amplitude as I said [( 2/5)^2 + (1/5)^2 ] ^1/2 = +-0.45
     
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