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UnkleRhaukus

  • one year ago

\[\begin{vmatrix}2-\xi^2&-1&0\\-1& 2-\xi^2&-1\\0&-1&2-\xi^2\end{vmatrix}=0\]

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  1. anonymous
    • one year ago
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    (' )-( .) what

  2. anonymous
    • one year ago
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    @THEHELPER123

  3. UnkleRhaukus
    • one year ago
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    FIND THE determinant

  4. UnkleRhaukus
    • one year ago
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    , solve for xi

  5. beginnersmind
    • one year ago
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    Just expand it out and solve for $\xi$ = 0. There's only 2 non-zero terms.

  6. beginnersmind
    • one year ago
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    BTW, you can use the substitution \[u = 2 - \xi^{2}\] \[det A = u^{3} - u\] \[u^{3} - u=0\] to simplify the algebra.

  7. UnkleRhaukus
    • one year ago
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    \[ \begin{align*} \begin{vmatrix} 2-\xi^2&-1 &0 \\ -1 & 2-\xi^2&-1 \\ 0 &-1 &2-\xi^2 \end{vmatrix} &= 0 \\ \\[1ex] (2-\xi^2)\Big[(2-\xi^2)^2-(-1)^2\Big] -(-1)\Big[(-1)(2-\xi^2)-0\Big] +0 &= 0 \\ (2-\xi^2)\Big[(2-\xi^2)^2-1\Big]-(2-\xi^2) &= 0 \\ (2-\xi^2)\left(\Big[(2-\xi^2)^2-1\Big]-1\right) &= 0 \\ (2-\xi^2)\Big((2-\xi^2)^2-2\Big) &= 0 \\ (2-\xi^2)\Big((2-\xi^2)^2-2\Big) &= 0 \\ \end{align*}\] oh , where's the 2?

  8. UnkleRhaukus
    • one year ago
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    hmm i guess the u-sub is a good idea

  9. UnkleRhaukus
    • one year ago
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    i'll try it again with the \(u\)-substitution,

  10. beginnersmind
    • one year ago
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    \[det A = u^{3} - 2u \] is correct, not \[det A = u^{3} - u \]

  11. UnkleRhaukus
    • one year ago
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    \[\begin{align}\begin{vmatrix} u&-1 &0 \\ -1 & u&-1 \\ 0 &-1 &u \end{vmatrix} &= 0 \\ \\[1ex] u\Big[u^2-(-1)^2\Big] -(-1)\Big[(-1)(u)-0\Big] +0 &= 0 \\ u\Big[u^2-1\Big]-u &= 0 \\ u^3-2u &= 0 \end{align}\]

  12. UnkleRhaukus
    • one year ago
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    hmm so u = 0 or u^2-2 =0

  13. UnkleRhaukus
    • one year ago
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    u = 0 or ±√2

  14. UnkleRhaukus
    • one year ago
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    that's three solutions for u, and three for xi xi = √(2-{-√2,0,√2})

  15. UnkleRhaukus
    • one year ago
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    so the normal node frequencies \(\omega\), where : \(\omega/\omega_0 = \xi\) are....

  16. beginnersmind
    • one year ago
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    Should be 6 roots for \(\xi\). 2 for each u.

  17. UnkleRhaukus
    • one year ago
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    *the frequencies must be positive oh wait still in angular frequencies,

  18. UnkleRhaukus
    • one year ago
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    \[\nu = \omega/2\pi =... \textit{physics}\]

  19. beginnersmind
    • one year ago
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    Ok, that makes sense. I don't think I can help too much with the phyics though. Are these coupled springs?

  20. UnkleRhaukus
    • one year ago
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    yes

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  21. UnkleRhaukus
    • one year ago
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    \[\omega_0^2 = k/m = 6\,[\text{N/m}]/0.0905\,[\text{kg}]\]so \[\nu_1 = 0.99\,[\text{Hz}],\quad \nu_2 = 1.3\,[\text{Hz}], \qquad \nu_1 =2.34\,[\text{Hz}]\]

  22. UnkleRhaukus
    • one year ago
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    what a rigmarole, so many variable, so few numbers

  23. beginnersmind
    • one year ago
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    I actually prefered variables to numbers when I studied physics for the most part. I'd much rather write \(\omega^{2}\) than have to square 1.3823 every time. The only difference was when the teacher decided it was a good idea to take a general solution cos(ax+phi1) + sin(ax+phi2) and express it as a single trigonometric function. Except it wasn't ax+phi1 but a more complicated expression of the input variables.

  24. UnkleRhaukus
    • one year ago
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    one trig function, is better than two

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