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Yet another FODE question

  1. Mar 4, 2006 #1
    Hey.
    I've been doing more mechanics recently - further kinematics in M3. I've come across another question i'm confused by.

    A particle moves on the positive x-axis. When its displacement from O is x meters, is acceleration is of magnitude x^-3 m/s^2 and directed towards O. Given that initially, when t=0, the particle is at rest with x=1, find the time the particle takes to reach x=1/4.

    First i integrate with respect to x to find:

    V^2 = x^-2 + C where C = -1
    v^2 = x^-2 -1

    V = x^-1 - 1

    Now i integrate again to find velocity as (dx/dt):

    dx/dt = x^-1 - 1

    Which is a FODE, however i cant solve it.

    Thank you,
    Bob
     
  2. jcsd
  3. Mar 4, 2006 #2

    LeonhardEuler

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    That isn't quite right. The problem says the accleration is given by [itex]x^{-3}[/itex] towards the origin, so
    [tex]\frac{d^2x}{dt^2}=-x^{-3}[/tex]
    Multiplying by x3dt2 gives:
    [tex]x^3d^2x =-dt^2[/tex]
    Now integrate twice.
     
  4. Mar 4, 2006 #3

    HallsofIvy

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    No, you can't treat a second derivative like that, "differentials", dx and dy separately, are only defined for the first derivative: we could separate [itex]\frac{dv}{dt}= x^{-3}[/itex] into [itex]cv= x^{-3}dt[/itex] but then you can't integrate x-3 with respect to t.

    You can do this by "quadrature" (looks to me like this is what mr bob did)- since t does not appear explicitly, let [itex]v= \frac{dx}{dt}[/itex]. Then [itex]\frac{d^2x}{dt^2}= \frac{dv}{dt}[/itex] and, by the chain rule, [itex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/itex]

    That makes the equation [itex]v\frac{dv}{dx}= x^{-3}[/itex] which we can separate: [itex]v dv= x^{-3}dx[/itex] and, integrating, [itex]\frac{1}{2}v^2= -\frac{1}{2}x^{-2}+ C[/itex] or [itex]v^2= -x^{-2}+ C[/itex] almost what mr bob said (he is missing the "-" as x increases the particle slows). Taking x= 1, v= 0 when t= 0, we get 0= -1+ C so C= 1 and [itex]v^2= -x^{-2}+ C[/itex] again almost as mr bob said.

    However, (are you listening mr bob?) taking the square root does NOT give v= 1- x-1!!! [itex]\sqrt{a^2+ b^2} \ne a+ b[/itex]!!!

    Instead [itex]v= \frac{dx}{dt}= \sqrt{1- x^{-1}}= \sqrt{\frac{x^2- 1}{x^2}}= \frac{\sqrt{x^2-1}}{x}[/itex]. That separates as [itex]\frac{xdx}{x^2- 1}= dt[/itex] which should be easy to integrate.
     
    Last edited: Mar 5, 2006
  5. Mar 4, 2006 #4

    HallsofIvy

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    No, you can't treat a second derivative like that: we could separate [itex]\frac{dv}{dt}= x^{-3}[/itex] into [itex]cv= x^{-3}dt[/itex] but then you can't integrat x-3 with respect to t.

    You can do this by "quadrature"- since t does not appear explicitly, let [itex]v= \frac{dx}{dt}. Then [itex]\frac{d^2x}{dt^2}= \frac{dv}{dt}[/itex] and, by the chain rule, [itex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}= v\frac{dv}{dx}[/itex]

    That makes the equation [itex]v\frac{dv}{dx}= x^{-3}[/itex] which we can separate: [itex]v dv= x^{-3}dx[/itex] and, integrating, [itex]\frac{1}{2}v^2= -\frac{1}{2}x^{-2}+ C[/itex] or [itex]v^2= -x^{-2}+ C[/itex] almost what mr bob said (he is missing the "-" as x increases the particle slows). Taking x= 1, v= 0 when t= 0, we get 0= -1+ C so C= 1 and [itex]v^2= 1 -x^{-2}[/itex] again almost as mr bob said.

    However, (are you listening mr bob?) taking the square root does NOT give v= 1- x-1!!! [itex]\sqrt{a^2+ b^2} \ne a+ b[/itex]!!!

    Instead [itex]v= \frac{dx}{dt}= \sqrt{1- x^{-1}}= \sqrt{\frac{x^2- 1}{x^2}}= \frac{\sqrt{x^2-1}}{x}[/itex]. That separates as [itex]\frac{xdx}{\sqrt{x^2- 1}}= dt[/itex] which should be easy to integrate.
     
  6. Mar 5, 2006 #5
    Thank you very much. I really appreciate the help. Some of these mechanics questions are a bit hard. Need to improve my integration skills abit.
     
  7. Mar 5, 2006 #6
    I've integrated to [itex]t = \sqrt(x^2 - 1)[/itex] but cant get the answer correct. Find t when x = 1/4, which gives the root of a negative number, not real. But the book's answer is [itex]\sqrt(15)[/itex] which i notice is [itex]\sqrt(x^-2 - 1)[/itex]. How is this possible?

    Thank you,
    Bob
     
    Last edited: Mar 5, 2006
  8. Mar 5, 2006 #7

    HallsofIvy

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    The acceleration is of magnitude x-3 and directed toward O!!

    I missed that myself: the differential equation is not
    [tex]\frac{dv}{dt}= x^{-3}[/tex]
    it is
    [tex]\frac{dv}{dt}= -x^{-3}[/tex]
    !!!

    That's why you had what I (mistakenly) thought was a sign error!
     
    Last edited: Mar 5, 2006
  9. Mar 5, 2006 #8
    When i integrate i get [itex]t = 1 - (1 - x^2)[/itex].
    I'm getting confused by this now, i think my brain's just died, ah well. Sorry for keeping asking questions, i can see how annoying it would be if the answer was so obvious yet i dont grasp it.

    Thanks,
    Bob
     
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