# Rayleigh quotient problem

1. Apr 12, 2009

### blalien

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}) = $$\frac{v^{T} K v}{v^{T} M v}$$

3. The attempt at a solution
M = the 4x4 identity
K = $$\left[ \begin{array}{cccc} 4 & -4 & 0 & 0 \\ -4 & 5 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 1 \end{array} \right]$$

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 $$\sqrt{R}$$ is the second mode frequency. Or at least a reasonable estimate.

Am I on track here? Thanks!

Last edited: Apr 12, 2009