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Homework Help: Tricky ODE system

  1. Sep 4, 2009 #1
    1. The problem statement, all variables and given/known data
    I have 3 masses in 1-D connected by two springs. A driving force is exerted on the first mass and i need to derive the equation of motion of the last mass. I have worked out the Lagrangian to determine the equations of motion but cannot solve for z.

    2. Relevant equations
    The equations are
    d^2(x)/dt^2 = A(y + B*cos(omega*t) - x)
    d^2(y)/dt^2 = A(z+x-2y)
    d^2(x)/dt^2 = A(y-z)

    x,y,and z are the positions of the masses in 1-D and A,B are constants

    3. The attempt at a solution
    I tried general solutions of x= Csin(kt) +Dcos(kt) ect and also x= Ce^(ikt) but could not work it through.
    Where am i supposed to start?
  2. jcsd
  3. Sep 5, 2009 #2


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    Hi s.g.g! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    (i assume the last one should start d2z/dt2? :wink:)

    You seem to have selected unhelpful variables …

    variables like (x - y) might make it easier …

    and what is d2(x+y+z)/dt2 ? :smile:
  4. Sep 5, 2009 #3
    Sorry, yeah the last equation should be
    d^2z/dt2 = A(y-z)

    x, y and z are all functions of time.

    d^2(x+y+z)/dt^2 = d^2x/dt^2 + d^2y/dt^2 + d^2z/dt^2 isnt it?
  5. Sep 5, 2009 #4


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    (please use the X2 tag just above the Reply box :wink:)
    Yes! :smile:

    and that equals … ? :wink:
  6. Sep 5, 2009 #5
    Ha I'm starting to think the answer is so easy i cant see it.

    That equals the LHS of the sum of the equations. So therefor

    d2/dt2(x+y+z) = AB*cos (omega*t) after cancelling,

    integrating twice with respect to t gives

    x(t)+y(t)+z(t) = -(AB/omega2)cos(omega*t) + C

    How do i solve this for x(t) y(t) and z(t)?
    I know that the masses are originally at rest in their equilibrium position before mass x is subject to a driving force, so I have the initial conditions that
    x(0) = x0 , dx(0)/dt=0
    y(0) = y0 , dy(0)/dt=0
    z(0) = z0 , dz(0)/dt=0
  7. Sep 6, 2009 #6


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    Hi s.g.g! :smile:

    (have an omega: ω :wink:)
    good so far :smile:

    but you haven't yet tried my other suggestion, of using (x - y) etc.

    Note that (x+y+z) gives you the physical centre of mass, and you would expect that to have an easy equation …

    similarly you would expect the difference of any two of x y and z to have a better physical significance than x y or z on its own, and therefore to have a simpler equation. :wink:
  8. Sep 6, 2009 #7
    I still cant do it, im trying to compute d2(x-y) /dt2
    and letting (x-y) =gamma. But i cannot solve for gamma
  9. Sep 6, 2009 #8


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    You need to show us how far you got. :smile:

    (and did you try other combinations, like (y-z)?)

    (and don't use gamma! … use something easy and sensible, like p or q or u! :rolleyes:)
  10. Sep 6, 2009 #9
    Ha sorry:smile:, Heres what iv done.

    say y-x=p & z-y=q
    d2x/dt2 = d2p/dt2+d2y/dt2,
    d2y/dt2 = d2p/dt2+d2x/dt2 = d2q/dt2 + d2z/dt2 &
    d2z/dt2= d2q/dt2+d2y/dt2

    the original equations then become

    but im then stuck with the same problem. how do i solve for p and q
  11. Sep 6, 2009 #10
    sorry tim, disregard that rubbish last remark. it is completely incorrect.

    i say that x-y=p & y-z=q

    i then find that
    d2p/dt2=Bcos(omega*t) +Aq
    & d2q/dt2=A(p-2q)

    But same as before, i cannot solve for p and q, what am i missing?
  12. Sep 6, 2009 #11


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    Differentiate the second equation twice more:
    [itex]d^4q/dt^4= A(d^2p/dt^2- 2 d^2q/dt^2)[itex].

    Now, from the first equation, [itex]d^2p/dt^2= Bcos(omega t)+ Aq[itex] so you have
    [itex]d^4 q/d^4= A(Bcos(omega t)+ Aq)- 2d^2q/dt^2[/itex] or

    [itex]d^4q/dt^4+ 2d^2q/dt^2- Aq= ABcos(omega t)[/itex]

    Once you have found q, integrate [itex]d^2p/dt^2= Bcos(omegat t) + Aq[/itex] to find p.
  13. Sep 6, 2009 #12


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    Assuming this is right (I haven't checked it) …

    try looking for an eigenvector (ie. a combination so that d2/dt2(Bp + q) is a multiple of Bp +q)
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