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Rayleigh quotient problem

  1. Apr 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Given a four degree of freedom system that consists of four carts. The four carts each have mass m=1, and they are connected by three springs of constant k=4, 1, 1 respectively.
    Let x, y, z, and w be the displacement from equilibrium of the four carts, relative to the floor.
    The first normal mode has x = y = z = w = 1 and a frequency of 0.
    Guess the second mode shape, that is orthogonal to the first mode, and then use Rayleigh's quotient to estimate the second mode frequency.

    2. Relevant equations
    R({v}) = [tex]\frac{v^{T} K v}{v^{T} M v}[/tex]

    3. The attempt at a solution
    I've already computed
    M = the 4x4 identity
    K = [tex]\left[ \begin{array}{cccc}
    4 & -4 & 0 & 0 \\
    -4 & 5 & -1 & 0 \\
    0 & -1 & 2 & -1 \\
    0 & 0 & -1 & 1 \end{array} \right][/tex]

    The first mode shape is [1 1 1 1], so I'm guessing the second mode shape is [3 -1 -1 -1]. I took the liberty of computing the second mode shape, and it's very weird, so I'm going to keep my guess simple.

    What follows is how I think you solve the problem.

    Let v = a * [1 1 1 1] + b * [3 -1 -1 -1]
    Compute R in terms of a and b.
    If v is a normal mode, then R has a local extrema at v. Additionally, R({v}) is the corresponding mode frequency squared.
    So we find the values of a and b such that dR/da = 0 and dR/db = 0. Then v is the second normal mode and [tex]\sqrt{R}[/tex] is the second mode frequency. Or at least a reasonable estimate.

    Am I on track here? Thanks!
     
    Last edited: Apr 12, 2009
  2. jcsd
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