How Is the Second Mode Frequency Estimated Using Rayleigh's Quotient?

[blalien]

Homework Statement

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.

Homework Equations

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

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!

 

[anathema]

Thank you for your post and for sharing your attempt at a solution. Your approach seems to be on the right track. However, there are a few things that I would like to clarify and suggest for you to consider in your solution.

Firstly, let's define the second mode shape as [x y z w], where x, y, z, and w are the displacements from equilibrium of the four carts. The first mode shape given in the problem is [1 1 1 1], which means all four carts have the same displacement from equilibrium. This implies that all four carts are moving in unison, which results in a frequency of 0.

Now, for the second mode shape, we are looking for a shape that is orthogonal to the first mode shape. This means that the two mode shapes are perpendicular to each other, which also implies that they are linearly independent. Therefore, your guess of [3 -1 -1 -1] for the second mode shape is not orthogonal to the first mode shape [1 1 1 1], as it can be expressed as a linear combination of the first mode shape, i.e. [3 -1 -1 -1] = 3 * [1 1 1 1] - 4 * [1 1 1 1].

To find the second mode shape, we can use the Gram-Schmidt process to orthogonalize the first mode shape. This process involves finding a vector that is perpendicular to the first mode shape, and then normalizing it to obtain the second mode shape. In this case, we can take [1 -1 0 0] as the vector that is perpendicular to [1 1 1 1]. Normalizing this vector gives us the second mode shape [1 -1 0 0].

Now, let's use Rayleigh's quotient to estimate the second mode frequency. As you correctly stated, Rayleigh's quotient is given by R({v}) = \frac{v^{T} K v}{v^{T} M v}. Plugging in the values for M and K, we get R({v}) = \frac{1}{2}(4a^2 + 5b^2 - 4ab + 2a^2 - 2ab + b^2). Now, we need to find the values of a and b that will give us a

 

[nlightner]

Yes, you are on the right track. The Rayleigh quotient is a useful tool for estimating the mode frequencies of a system. Your approach of using the first and second mode shapes to find the values of a and b that minimize the Rayleigh quotient is correct. This will give you an estimate of the second mode frequency, but keep in mind that it is only an approximation and the actual frequency may differ slightly. It is always a good idea to double check your results and make sure they make physical sense. Good luck with your calculations!