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Math Help - Differential equations

  1. #1
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    Differential equations

    Any hint of how to represent a mass spring system with vibration damped?
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  2. #2
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    You could search the net for spring dashpot.
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  3. #3
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    Quote Originally Posted by lvleph View Post
    You could search the net for spring dashpot.


    I need to find the equation for damped system.

    I found without damping:

    y(t) = y.coswt + \frac{v.}{w}sinwt
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  4. #4
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    I find damped system:


    damping force. Fd = -c \frac{dx}{dt}


    For the spring mass system with damping force through a fluid in the opposite direction of motion of the mass

    m \frac{dx^2}{dt^2} = -kx - c \frac{dx}{dt}

    m \frac{dx^2}{dt^2} + c \frac{dx}{dt} + kx = 0


    Auxiliary equation

    mw^2 + cw + k = 0

    w1 = \frac{-c + \sqrt{c^2-4mk}}{2m}

    w2 = \frac{-c - \sqrt{c^2-4mk}}{2m}


    The system overdamped
    c^2 - 4mk > 0

    Then:

    x = C1e^{w1t} + C2e^{w2t}


    How do I determine C1 and C2?
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  5. #5
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    That c is inherent to the system, i.e., it come from experimental data. Therefore, in a mathematics problem it should be given.
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  6. #6
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    Quote Originally Posted by lvleph View Post
    That c is inherent to the system, i.e., it come from experimental data. Therefore, in a mathematics problem it should be given.
    C is inherent to the system.

    I do not understand, C1 and C2 are what?
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  7. #7
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    Oh sorry, those are determined by initial conditions. You should have initial conditions that say something like x(0) = f(x) and \frac{dx(0)}{dt}=g(x). With these initial conditions you can solve for both C_1 and C_2. They are basically constants of integration that define the family of solutions.
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  8. #8
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    Quote Originally Posted by lvleph View Post
    Oh sorry, those are determined by initial conditions. You should have initial conditions that say something like x(0) = f(x) and \frac{dx(0)}{dt}=g(x). With these initial conditions you can solve for both C_1 and C_2. They are basically constants of integration that define the family of solutions.


    I understand what you said. But if x is function of time. x(t)

    I do not know which determine the condition for x(0), ie t = 0
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  9. #9
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    Well, x(0) would be something given. Say you know the spring is extended a distance of L at the initial time then we know x(0) = L and therefore
    x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L.
    Because there are two unknowns we needed the second initial condition x_t(0) = g(x). Say, g(x) = 0 then we can solve for C_1,C_2 using
     x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0
    Thus,
     C_1 = -\frac{w_2}{w_1} C_2
    then substituting this into x(0) = C_1 + C_2 = L
    we have
     C_2 = \dfrac{w_1}{w_1 - w_2} L
    and
     C_1 = \dfrac{w_2}{w_2 - w_1} L
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  10. #10
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    Quote Originally Posted by lvleph View Post
    Well, x(0) would be something given. Say you know the spring is extended a distance of L at the initial time then we know x(0) = L and therefore
    x(0) = C_1e^{w_1 \cdot 0} + C_2e^{w_2\cdot 0} = C_1 + C_2 = L.
    Because there are two unknowns we needed the second initial condition x_t(0) = g(x). Say, g(x) = 0 then we can solve for C_1,C_2 using
     x_t(0) = w_1\cdot C_1 e^{w_1\cdot 0} + w_2 \cdot C_2 e^{w_2\cdot 0} = w_1\cdot C_1 + w_2 \cdot C_2 = 0
    Thus,
     C_1 = -\frac{w_2}{w_1} C_2
    then substituting this into x(0) = C_1 + C_2 = L
    we have
     C_2 = \dfrac{w_1}{w_1 - w_2} L
    and
     C_1 = \dfrac{w_2}{w_2 - w_1} L


    Sorry, it did not understand the second initial condition, which is g (x)?
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  11. #11
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    Just like the initial condition x(0) = f(x) is given x_t(0) = g(x) will be given. They are defined by the problem. What are you trying to do? I will be of more help if I knew specifics.
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  12. #12
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    Quote Originally Posted by lvleph View Post
    Just like the initial condition x(0) = f(x) is given x_t(0) = g(x) will be given. They are defined by the problem. What are you trying to do? I will be of more help if I knew specifics.
    Is a deduction of the vibration model.

    My solution so far


    Damped spring mass system

    mx'' = -kx - cx'
    mx'' + cx' + kx = 0

    x(t) = Acoswt + Bsinwt
    x'(t) = -Awsinwt + Bwcoswt
    x''(t) = -Aw^2coswt - Bw^2sinwt
    c is damping constant


    Substituting I can not find: mw^2 + cw + k =0
    But is the correct equation, or not ?
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  13. #13
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    Okay, the way to solve this differential equation is by finding the roots to the equation
    m w^2 + cw + k = 0
    like you wanted to. This can be done using the quadratice formula, i.e.,
     w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}
    Unfortunately, you do not no if w \in \mathbb{R}, w \in \mathbb{C} or w is a repeated root and each one of these cases leads to a different general solution. Therefore, you would have to solve the differential equation using each of these cases and then add all the solutions together.
    If w are repeated real roots then the solution is of the form
     x(t) = C_1 + C_2 t.
    If w \in \mathbb{R} the solution is of the form
    x(t) = C_3 e^{w_1t} + C_4e^{w_2t}.
    Finally, if w \in \mathbb{C} the solution is of the form
    x(t) = C_5 \cos w t + C_6 \sin w t.

    The constants C_1,C_2,C_3,C_4,C_5,C_6 are all determined by the initial conditions.

    EDIT: I am sorry this keeps getting more and more complicated, but your questions are so general they have to have a general answer.
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  14. #14
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    Quote Originally Posted by lvleph View Post
    Okay, the way to solve this differential equation is by finding the roots to the equation
    m w^2 + cw + k = 0
    like you wanted to. This can be done using the quadratice formula, i.e.,
     w = \dfrac{-c \pm \sqrt{c^2 - 4mk}}{2m}
    Unfortunately, you do not no if w \in \mathbb{R}, w \in \mathbb{C} or w is a repeated root and each one of these cases leads to a different general solution. Therefore, you would have to solve the differential equation using each of these cases and then add all the solutions together.
    If w are repeated real roots then the solution is of the form
     x(t) = C_1 + C_2 t.
    If w \in \mathbb{R} the solution is of the form
    x(t) = C_3 e^{w_1t} + C_4e^{w_2t}.
    Finally, if w \in \mathbb{C} the solution is of the form
    x(t) = C_5 \cos w t + C_6 \sin w t.

    The constants C_1,C_2,C_3,C_4,C_5,C_6 are all determined by the initial conditions.

    EDIT: I am sorry this keeps getting more and more complicated, but your questions are so general they have to have a general answer.

    But I not find the equation m w^2 + cw + k = 0

    m(-Aw^2coswt - Bw^2sinwt) + c(-Awsinwt + Bwcoswt) + k(Acoswt + Bsinwt) = 0
    (mw^2+cw+k)(-Acoswt -Bsinwt -Asinwt + Bcoswt + Acoswt + Bsinwt) = 0
    (mw^2+cw+k)(-Asinwt+Bcoswt) = 0
    This equation not correct !!!
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  15. #15
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    I guess I don't understand what you are asking. If we assume that -A\sin wt + B \cos wt \ne 0 this leads you to conclude the only time this is zero is when mw^2 + cw + k = 0. Therefore, the we must solve for the roots of the characteristic equatioin mw^2 + cw + k = 0. Again, I am not sure what you are having issues with, and so I am sorry I am not of much help.
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