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Question

Consider, The motion of a a damped harmonic oscillator is described by $x^{''}$ + $bx^{'}$ + $kx = 0$, where b is greater than or equal to 0

A) Rewrite as a two dimensional linear system.

$ x^{'} = y$

$y^{'}$ = - $(by + kx)$

B) Sketch the analogue of the trace-determinant plane in the bk-plane. That is, identify the regions in the relevant portion of the bk-plane where the corresponding system has similar phase portraits.

I have cut out most of my answer etc as it seems to of merely confused people, how does one answer this question?

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2 Answers 2

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+50

We can write the matrix for this system as:

$$\tag 1 A = \begin{bmatrix} 0 & 1 \\ -k & -b\end{bmatrix}$$

Using the Trace-Determinant approach, we can find the eigenvalues of $(1)$ by writing:

$$\tag 2 \lambda_{1,2} = \frac{Tr(A) \pm \sqrt{(Tr(A))^2 - 4 \det(A)}}{2}$$

So we have:

  • $Tr(A) = - b$

  • $(Tr(A))^2 = b^2$

  • $\det(A) = k$

Substituting these results into $(2)$ and simplifying, yields the eigenvalues:

$$\tag 3 \lambda_{1,2} = \frac{1}{2} \left(\pm \sqrt{b^2 - 4k} -b \right).$$

Now, we are asked to sketch the analogue of the trace-determinant plane in the bk-plane. That is, identify the regions in the relevant portion of the bk-plane where the corresponding system has similar phase portraits.

You can see an example of at MIT Trace-Determinant plot where it is plotting the determinant as the vertical axis and the trace as the horizontal axis and then identifying the different types of phase portraits you get (what a beautiful generalization and approach).

For this problem, we have a two - parameter family for our eigenvalues, so all it is asking is to use the parameters, $bk$, and plot what happens as those change and overlay this onto the Trace-Determinant graph.

In English, this means to see what happens to the eigenvalues as the values of $b$ and $k$ change as you can get all of the ranges of the types of eigenvalues:

  • Real - both positive, both negative, opposite sign
  • Imaginary - zero real part, positive real part, negative real part.

Each of these admits different phase portraits, so overlay that onto the Trace-Determinant picture and identify what the phase portrait looks like on those regions.

Regards

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I Actually did figure this question out days ago but i awarded the bounty anyway, will post a picture when i get time of what it looks like =) –  Faust7 Mar 11 '13 at 21:53
    
I'd like to see the picure (and there are examples on the web using beautifully colored diagrams - so please post it as it is instructive given that this approach has somewhat fallen out of style, but is very instructive the the general audience. Regards. –  Amzoti Mar 11 '13 at 22:18
    
Excellent! $\Huge +1$ –  amWhy Apr 22 '13 at 0:10
    
@amWhy: thank you - that is a nice approach that is not always taught, but should be! –  Amzoti Apr 22 '13 at 0:14

For A, your rewrite as a two dimensional system is fine. I don't understand where the vectors come from and although you say "eigenvalues are given by" you have only one expression.

For B, I don't know what a Trace-Determinate Parabola picture is.

For C, it depends on the roots of the quadratic $mz^2+bz+k=0$ If the roots are complex, it is underdamped as the exponential of the imaginary part gives oscillation. It is critically damped if the roots are equal and overdamped if they are real and different.

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