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DE for motion of a particle

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle of mass m is confined to a horizontal plane. It is elastically bound to its equilibrium position by an isotropic elastic force,

    F=mω^2r
    Where r is the displacement of the particle from equilibrium and ω is a real constant parameter.

    From Newton's law, we obtain the equation of motion:

    r'' + ω^2r = 0.

    In rect. coordinates r(t) = ix(t) + jy(t).

    Obtain expressions for x(t) and y(t).


    2. Relevant equations



    3. The attempt at a solution

    The x-component of force :

    Fx = -x(t)mω^2 = mx''(t)
    x''(t) + x(t)ω^2 = 0
    The solution to this DE is x(t) = c1cos(ωt) + c2sin(ωt)
    Then I'll apply the initial conditions given in the problem to obtain the constants.

    Here's the part I'm unsure about
    When I solve the differential equation for the y component I get
    y(t) = a1cos(ωt) + a2sin(ωt). I'm thinking I have to multiply this by t so that the solutions are linearly independent:

    y(t) = a1*tcos(ωt) + a2*tsin(ωt). Is this correct?
     
  2. jcsd
  3. Apr 1, 2009 #2

    George Jones

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    No. Don't multiply by t.
     
  4. Apr 1, 2009 #3
    I don't understand why. Could you please explain?

    Thanks.
     
  5. Apr 1, 2009 #4

    HallsofIvy

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    You don't understand why? What makes you think you should? It certainly can't be because x depends on those functions- the y-coordinate has nothing to do with the x-coordinate. They are completely indepedent.
     
  6. Apr 1, 2009 #5
    There's no need to multiply by [tex] t [/tex]. Yes, [tex] x(t) [/tex] and [tex] y(t) [/tex] are solutions to the same ODE, but there's no reason for them to be linearly independent; you're not going to use them as a basis in which to express other solutions of the ODE. In fact, you already used such a basis to get [tex] x [/tex] and [tex] y [/tex], i.e., [tex] \cos(\omega t) [/tex] and [tex] \sin(\omega t) [/tex].

    You're probably confusing this with a similar-looking situation encountered with some second-order linear ODEs, which can be converted to a system of coupled linear first-order equations; if the associated matrix of this system is degenerate (that is, has repeated eigenvalues), then a basis for the solution space may include such elements as [tex] t \sin(\omega t) [/tex]. (The reason for this is best understood as a consequence of the fact that the "degenerate" real canonical form of a matrix is the sum of the identity and a nilpotent matrix. Thus, when exponentiated, such a matrix yields a prefactor of [tex] e^t [/tex] times the identity and a sum of higher-order terms in [tex] t [/tex] that eventually terminates.)

    As a final note, it's usually preferable to express solutions like [tex] c_1 \cos(\omega t) + c_2 \sin(\omega t) [/tex] in the form [tex] A \cos(\omega t + \phi) [/tex] for some phase [tex] \phi [/tex].
     
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