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Critical Damped motion

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the general solution of the damped SHM equation(5.9) for the special case of critical damping , that is , when K = [tex]\Omega[/tex]. Show that , if the particle is initially relaeased from rest at x= a , then the subsequent motion is given by

    x=a*(e^-([tex]\Omega[/tex]*t))*(1+[tex]\Omega[/tex]*t)


    2. Relevant equations

    x''+2Kx'+([tex]\Omega[/tex])2*x=0
    x=A*cos([tex]\Omega[/tex]*t) + B*sin([tex]\Omega[/tex]*t)


    3. The attempt at a solution

    x=e[tex]\lambda[/tex]*t

    x'= [tex]\lambda[/tex]*e[tex]\lambda[/tex]*t

    x''=[tex]\lambda[/tex]2*e[tex]\lambda[/tex]*t

    x'= -A*[tex]\Omega[/tex]*sin([tex]\Omega[/tex]*t) + B*[tex]\Omega[/tex]*cos([tex]\Omega[/tex]*t)

    x''= -A*[tex]\Omega[/tex]2*cos([tex]\Omega[/tex]*t) - B*[tex]\Omega[/tex]2*sin([tex]\Omega[/tex]*t)

    Not sure how to proceed...
     
  2. jcsd
  3. Oct 11, 2008 #2

    gabbagabbahey

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    Take a look at the characteristic polynomial of the DE:

    [tex]x''(t)+2 \Omega x'(t) + \Omega ^2 x(t)=0 \Rightarrow \lambda ^2 +2 \Omega \lambda + \Omega ^2 =0[/tex]

    where [tex] \lambda ^n x(t) \equiv \frac{d^n x(t)}{dt^n}[/tex].

    What are the solutions for [itex]\lambda[/itex]? Are they linearly independent? What does that tell you about the general solution for [itex]x(t)[/itex]?
     
    Last edited: Oct 11, 2008
  4. Oct 11, 2008 #3
    That [tex]\lambda[/tex]=-[tex]\Omega[/tex]; its a double root

    I remember from my ODE class that if a polynomial had more than two root, then x(t)=c1*e-r1*t+c2-r2*t; Otherwise , if the polynomial has only one solution, then x(t)=c1*e-r1*t+c2*t*e-r1*t

    How would I go about showing that x(t)=c1*e-r1*t+c2*t*e-r1*t
    rather than loosely remembering what I learned about the conditions for ODE equations certain solutions
     
    Last edited: Oct 11, 2008
  5. Oct 11, 2008 #4

    gabbagabbahey

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    I assume you mean [itex]x(t)=c_1 e^{\lambda_1 t}+c_2te^{\lambda_1 t}=c_1 e^{-\Omega t}+c_2te^{-\Omega t}[/itex]?

    If so, you can show this by substituting this general solution into the ODE and showing that it satisfies it.

    Then, just use the fact that x(0)=a to show that the specific solution is the one given in the problem.
     
  6. Oct 11, 2008 #5
    Should I start off by assuming that the original ODE for a double root is [itex]x(t)=c_1 e^{\lambda_1 t}+c_2te^{\lambda_1 t}=c_1 e^{-\Omega t}+c_2te^{-\Omega t}[/itex]? or do I need to go about proving that I arrived at this solution ?
     
  7. Oct 11, 2008 #6

    gabbagabbahey

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    I don't think you'll need to prove that it's the general solution, the solutions to constant coefficient ODE's are well known and can be found in almost any text on DE's; but you may want to check with your prof to make sure.
     
  8. Oct 11, 2008 #7
    x(0)=a

    I think the initial conditions for v is zero as well , since the damped motion is undriven.

    x' = -[tex]\Omega[/tex]*a*e-[tex]\Omega[/tex]*t + c2*(e-[tex]\Omega[/tex]*t-x(0)=a

    I think the initial conditions for v is zero as well , since the damped motion is undriven.

    x' = -[tex]\Omega[/tex]*a*e-[tex]\Omega[/tex]*t + c2*(e-[tex]\Omega[/tex]*t-[tex]\Omega[/tex]*t*ex(0)=a

    I think the initial conditions for v is zero as well , since the damped motion is undriven.

    x' = -[tex]\Omega[/tex]*a*e-[tex]\Omega[/tex]*t + c2*(e-[tex]\Omega[/tex]*t-[tex]\Omega[/tex]*t*e*t*[tex]\Omega[/tex]

    How would I go about finding c2?
     
  9. Oct 11, 2008 #8

    gabbagabbahey

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    Your last post didn't make much sense.

    You can determine [itex]c_1[/itex] by using x(0)=a, and [itex]c_2[/itex] by using x'(0)=0 (since the particle is released from rest at t=0)
     
  10. Oct 11, 2008 #9
    sorry for the confusion , but I was essentially saying what you said in your last post: Should I assume for an undriven damped oscillator , that the initial conditions for x' and x are zero at t=0.
     
  11. Oct 11, 2008 #10

    gabbagabbahey

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    There's no need to assume anything here; you are given the initial conditions:

    that means x(t=0)=a and x'(t=0)=0.
     
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