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Mass-spring system

  1. Dec 19, 2009 #1
    Vibration Free

    Please, are correct?

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

    Where frequency is

    [tex]w = \sqrt{\frac{k}{m}}[/tex]

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

    The characteristic equation is:

    [tex]r^2 + w^2 = 0[/tex]
    [tex]r = +or- iw[/tex] where [tex]i^2 = -1[/tex]


    [tex]x(t) = C_1e^{iwt} + C_2e^{-iwt}[/tex]

    Calculating I can get
    [tex]x(t) = a1cos(wt) + a2sin(wt)[/tex]

    Now, I need to do to get the following equation. how do I find?
    [tex]x(t) = Acos(wt - \delta)[/tex] (I think this is the equation we need to get the free vibration)
  2. jcsd
  3. Dec 19, 2009 #2
    Ya, it's correct, and

    x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)

    if you define

    A \equiv a_1^2 + a_2^2
    \delta \equiv arctan(a_1/a_2)

    This is the equation of free vibration of an undamped single degree of freedom dynamic system with linear elastic stiffness.
  4. Dec 19, 2009 #3
    Why if [tex]A \equiv a_1^2 + a_2^2[/tex] and [tex]\delta \equiv arctan(a_1/a_2)[/tex] we have:
    [tex]x(t) = a_1cos(wt) + a_2sin(wt) = A cos(wt - \delta)[/tex]

    What the calculations involved?
  5. Dec 19, 2009 #4
    Use the multiple angle formula for cos, cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
  6. Dec 20, 2009 #5
    [tex]x(t) = a_1cos(wt) + a_2sin(wt)[/tex]

    I do not understand how to find [tex]A cos(wt - \delta)[/tex]
  7. Dec 20, 2009 #6


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    Science Advisor

    Have you tried rather than just staring at the formulas?

    "cos(a-b)= cos(a)cos(b)+ sin(a)sin(b)" with "[itex]\omega t- \delta[/itex]" instead of a- b gives you [itex]Acos(\omega t-\delta)= Acos(\omega t)cos(\delta)+ A sin(\omega t)sin(\delta)[/itex].

    In order to have that equal to [itex]a_1cos(\omega t)+ a_2sin(\omega t)[/itex], you must have [itex]a_1= A cos(\delta)[/itex] and [itex]a_2= A sin(\delta)[/itex].

    Dividing the first equation by the second gives
    [tex]\frac{A sin(\delta)}{A cos(\delta)}= tan(\delta)= \frac{a_2}{a_1}[/tex]
    so [itex]\delta= tan^{-1}(a_2/a-1)[/itex].

    Squaring and summing the two equations gives
    [tex]a_1^2+ a_2^2= A^2 cos^2(\delta)= A^2[/tex]
    so [itex]A= \sqrt{a_1^2+ a_2^2}[/itex].

    BobbyBear didn't quite have those equations in his post.
    Last edited by a moderator: Dec 20, 2009
  8. Dec 20, 2009 #7
    Yes, thank you.

    The general solution is
    [tex]x(t) = Acos(wt - \delta)[/tex]

    I find A with the initial conditions ?

    [tex]x(0) = Acos(- \delta) = X_o[/tex]

    [tex]x'(0) = -Awsin(- \delta) = V_o[/tex]

    I'll be A in the function [tex] \delta [/tex] ???
  9. Dec 20, 2009 #8
    Oopsie, I guess I tried to do it too quickly. HallsofIvy's relationships are the correct ones :)

    And yes, on applying the initial conditions you obtain a 2 by 2 system (the one you wrote out), which you solve for A (the amplitude) and the angle [tex] \delta [/tex]
  10. Dec 20, 2009 #9
    I could not find A

    [tex]X(0) = Acos( - \delta) = X_o[/tex]
    [tex]X'(0) = -Awsin(- \delta) = V_o[/tex]

    [tex]\delta = tan^{-1} \frac{V_o}{X_o w}[/tex]

    And A ???
  11. Dec 20, 2009 #10
    Well, once you have [tex]

    you can go to any of the two equations and get A, eg

    [tex] A= X_0/cos(\delta) = \frac{X_0}{cos(arctan(V_0/X_0w))} = \frac{X_0}{X_0w/\sqrt{X_0^2w^2+V_0^2}} = \frac{\sqrt{X_0^2w^2+V_0^2}}{w}
  12. Dec 20, 2009 #11
    Sorry, I do not understand your calculation.

    Acos(- \delta ) = X_o
  13. Dec 20, 2009 #12
    [tex] A cos(\delta) = X_0 \rightarrow A = X_0 / cos(\delta) [/tex]


    [tex] \delta = arctan(V_0/X_0w)



    [tex] cos(\delta) = cos(arctan(V_0/X_0w))



    (By the way, cos(-a) = cos(a) that's why I didn't bother with the minus sign)
  14. Dec 20, 2009 #13
    Yes. Thank you
  15. Dec 20, 2009 #14
    I got a doubt in the beginning.
    After obtaining the characteristic equation
    [tex]r = \pm iw[/tex] with [tex]i^2 = -1[/tex]

    How do we know [tex]x(t) = C_1e^{iwt} + C_2e^{-iwt}[/tex] ???
  16. Dec 20, 2009 #15
    Because the theory of differential equations tells us that the general solution of a homogeneous linear ordinary differential equation with constant coefficients is given by a linear combination of exponentials of the form: [tex]

    with r being the roots of the characteristic polynomial, which in this case are [tex]\pm iw[/tex]
  17. Dec 20, 2009 #16
    Thank you very much. You know about damped in my other topic ?
  18. Dec 21, 2009 #17
    Someone has a graphic example of the system to vibration free ?
  19. Dec 21, 2009 #18
    To plot the graph using Maple 12, which values I use for [tex]k[/tex], [tex]x_o[/tex] and [tex]v_o[/tex]?
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