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Mass-spring system damped

  1. Dec 19, 2009 #1
    Damped vibration

    [tex] m \frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + kx = 0 [/tex]

    Characteristic equation is

    [tex]mr^2 + \gamma r + k = 0[/tex]

    [tex]r_1 = \frac{- \gamma + \sqrt{( \gamma )^2 - 4mk}}{2m}[/tex]
    [tex]r_2 = \frac{- \gamma - \sqrt{( \gamma )^2 - 4mk}}{2m}[/tex]

    In overdamped
    [tex]( \gamma )^2 - 4mk > 0[/tex]

    What I need to calculate to find the general solution:
    [tex] x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}[/tex] ???
     
  2. jcsd
  3. Dec 20, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi viciado123! Welcome to PF! :wink:

    You need two "initial" conditions, to find the two constants C1 and C2.

    Usually, they'll be the values of x at two particular times, or the value of x and dx/dt, or of dx/dt and d2x/dt2, at one particular time. :smile:
     
  4. Dec 20, 2009 #3
    Re: Welcome to PF!

    I do not know how to find the equation [tex]x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}[/tex]

    In the books we have to assume [tex]x(t) = e^{rt}[/tex] Why assume this ?

    [tex]x_1(t) = C_1e^{r_1t}[/tex] and [tex]x_2(t) = C_2e^{r_2t}[/tex]
     
  5. Dec 20, 2009 #4

    tiny-tim

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    It's because you can write the characteristic equation as (D - r1)(D - r2)x = 0 (where D is short for d/dt),

    so the solutions are the solutions to (D - r1)x = 0 and (D - r2)x = 0,

    which are the same as dx/dt = r1x and dx/dt = r2x,

    or x = C1er1t and x = C2er2t. :smile:
     
  6. Dec 20, 2009 #5
    Thanks. Is:
    [tex]\int \frac{dx}{dt} = \int r_1xdt[/tex]
    [tex]ln(x) = r_1t[/tex]
    [tex]x(t) = e^{r_1t}[/tex]

    I correct ?
     
  7. Dec 20, 2009 #6

    HallsofIvy

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    Re: Welcome to PF!

    Be careful here. Textbooks suggest you assume a solution of that form (they don't say you must) because it leads to correct solutions, not necessarily because the solution "must" be of that form. You should quickly learn that you can also get polynomial solutions and trig functions as well as exponentials as solutions and, in fact, products of those kinds of things.

    Also you need the crucial theoretical fact that the set of all solutions to a nth order linear differential equation forms an nth dimensional "vector space". In particular, that means that if you have two independent solutions, f1 and f2, of a second order linear differential equation, then any solution can be written as a linear combination of the two solutions: y(x)= C1f1(x)+ C2f2(x) for any solution, y(x), of the equation.

    As for your original question, knowing that [itex]e^{r_1t}[/itex] and [itex]e^{r_2t}[/itex] are solutions to the differential equation (and knowing that exponentials with different coefficienta in the exponent are independent) tells you that any solution, y(x), can be written in the form [itex]C_1e^{r_1t}+ C_2e^{r_2t}[/itex].

    It is also helpful to observe that, for complex r1 and r2, as you have here, [itex]e^{a+ bi}= e^ae^{bi}= e^a(cos(b)+ i sin(b))[/itex] so that your "exponential solutions" can be written as a combination of exponential and trigonometric solutions.
     
  8. Dec 20, 2009 #7
    Re: Welcome to PF!

    Thanks. I find [tex]C_1[/tex] and [tex]C_2[/tex] with initial conditions ?
    The initial conditions is:
    [tex]x(0) = X_o[/tex] Initial Position ?
    [tex]x'(0) = V_o[/tex] Initial velocity?
    It is ?
     
  9. Dec 20, 2009 #8

    tiny-tim

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    Hint: x(0) = C1+ C2

    x'(0) = … ? :smile:
    No, in this case, the solutions must be of that form.

    And you can prove it just by putting y = (D - r1)x, and solving for y, then solving for x.

    The "textbook suggestions" (ie, mere intelligent guesswork :rolleyes:) really apply to "particular solutions", but not to "general solutions" such as this or any other "polynomial = 0" equation. :wink:
     
  10. Dec 20, 2009 #9
    Am I correct?

    [tex]x(0) = C_1 + C_2 = x_o[/tex]
    [tex]x'(0) = C_1r_1 + C_2r_2 = v_o[/tex]

    I find:

    [tex] C_1 = \frac{x_or_2 - v_o}{r_2 - r_1}[/tex]

    [tex] C_2 = \frac{v_o - x_o r_1}{r_2 - r_1}[/tex]

    Correct ?
     
  11. Dec 20, 2009 #10

    tiny-tim

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    Yup! :biggrin:
     
  12. Dec 20, 2009 #11
    Thanks.

    In case [tex] \gamma^2 -4mk < 0[/tex] is underdamped

    [tex]r_1 = \frac{- \gamma}{2m} + ri[/tex] and [tex]r_1 = \frac{- \gamma}{2m} - ri[/tex]

    By Euler's formula

    [tex]x(t) = e^{-(\frac{\gamma}{2m})t} (C_1 cos(rt) + C_2 sin(rt))[/tex]

    Are the same initial conditions?
    What is r? How do I calculate r ?
     
  13. Dec 20, 2009 #12

    tiny-tim

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    Do you understand why [tex]r_1 = \frac{- \gamma}{2m} + ri[/tex] and [tex]r_2 = \frac{- \gamma}{2m} - ri[/tex] ?

    If so, the value of r is obvious. :wink:

    (and use x0 and v0 as before)
     
  14. Dec 20, 2009 #13

    r is the imaginary part ?

    [tex]r = \frac{\sqrt{\gamma^2 -4mk}}{2m}[/tex]
     
  15. Dec 20, 2009 #14

    tiny-tim

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    That's it! :smile:
     
  16. Dec 20, 2009 #15
    Ok thank you
     
  17. Dec 21, 2009 #16
    Someone has a graphic example of the system to overdamped and underdamped ?
     
  18. Dec 21, 2009 #17
    I tried to plot in matlab overdamped. But it appeared nothing
    Code (Text):

    plot((.1*((-30-sqrt(30^2-100*(4*1.3)))*e^((-30+sqrt(30^2-100*(4*1.3)))*t/(2*1.3))/(2*1.3)-(-30+sqrt(30^2-100*(4*1.3)))*e^((-30-sqrt(30^2-100*(4*1.3)))*t/(2*1.3))/(2*1.3)))/(-sqrt(30^2-100*(4*1.3))/(1.3)), t = 0 .. 5)
     
     
  19. Dec 21, 2009 #18
    In underdamped:
    [tex]C_1 = x_o[/tex]

    [tex]C_2 = \frac{v_o}{r}[/tex]

    It is correct ?
     
  20. Dec 21, 2009 #19

    tiny-tim

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  21. Dec 21, 2009 #20
    I find

    [tex]x'(0) = \frac{- \gamma}{2m}(C_2r) = v_o[/tex]

    [tex] C_2 = - \frac{2mv_o}{\gamma r}[/tex]

    Where is my mistake?
     
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