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Three-Degree-Of-Freedom Spring System

  1. Nov 8, 2013 #1
    Three masses, m1 , m2, m3, on a frictionless, horizontal plane, connected by two springs, both with a spring constant k.

    The system is set in motion by displacing the middle mass, m2, a distance a to the right, whilst holding the end masses, m1 and m2, in equilibrium.

    Also should be noted that m1 = m3 and m2 = βm1.

    Summing the forces on each mass and using Newton's second law I have obtained the following matrix system:

    $$
    \begin{bmatrix}
    βm & 0 & 0\\0 & m & 0\\0& 0& βm\\
    \end{bmatrix}
    \begin{bmatrix}
    \ddot x_{1}\\\ddot x_{2}\\\ddot x_{3}\\
    \end{bmatrix}
    +
    \begin{bmatrix}
    k & -k & 0\\-k & 2k & -k\\0& -k& k\\
    \end{bmatrix}
    \begin{bmatrix}
    x_{1}\\x_{2}\\x_{3}\\
    \end{bmatrix}
    =
    \begin{bmatrix}
    0\\0\\0\\
    \end{bmatrix}
    $$

    which is $$\underline{M}\underline{\ddot X} + \underline{k}\underline{X}=0$$

    After assuming a solution of the form:
    $$ \underline{X}={U}[A_{}cos(ω_{}t)+B_{}sin(ω_{}t)]$$

    it can be shown that
    $$\underline{\ddot X}=-ω^{2}\underline{X}$$
    and then that:
    $$( \underline{k}-ω^{2} \underline{M}) \underline{X}=0$$
    and for there to be a non-trivial solution,
    $$det( \underline{k}-ω^{2} \underline{M})=0$$

    here I have a problem, which might not even be a problem, but the characteristic equation I get only has two positive roots, where I am asked for three. Is there a) something obviously wrong with my method;b) a potential error in my calculations;c) is it possible for there to be only two frequencies?
     
  2. jcsd
  3. Nov 8, 2013 #2
    Your original equation is certainly wrong. You said that ## m_1 = m_3 ## and ## m_2 = \beta m_1 ##. Then your M matrix should be ## \text{diag} \{ m, \beta m, m \} ##.

    Apart from that, it is possible to have roots with multiplicity greater than 1. You should still get different normal mode vectors.
     
  4. Nov 8, 2013 #3

    ehild

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    Homework Helper
    Gold Member

    One of the roots is zero: It corresponds to equal displacement of all the three masses.

    ehild
     
  5. Nov 10, 2013 #4
    Whoops.

    erm, so my roots are:

    $$0, \sqrt\frac{k}m, \sqrt\frac{k(2+β)}{βm}$$

    making 0(?) the fundamental frequency, then the square root of k/m the first overtone and the square root of k(2+β)/m the second overtone, each with a corresponding mode shape vector?
     
  6. Nov 10, 2013 #5

    ehild

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    Homework Helper
    Gold Member

    They are not overtones. They are the angular frequencies of normal modes. All normal modes mean different displacements of the masses with respect to each other. You get these normal mode vectors by plugging back the frequencies into the original equation.

    (One normal mode is pure translation, all bodies have the same displacement. The other one is a symmetric vibration, m1 and m3 move in opposite directions. The third one is asymmetric vibration, m1 and m3 move in the same direction, and m2 moves in the opposite direction)

    The motion of the system with the given initial condition is a linear combination of the three normal modes.

    ehild
     
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