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UnkleRhaukus
 one year ago
\[\begin{vmatrix}2\xi^2&1&0\\1& 2\xi^2&1\\0&1&2\xi^2\end{vmatrix}=0\]
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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UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0FIND THE determinant

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.4Just expand it out and solve for $\xi$ = 0. There's only 2 nonzero terms.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.4BTW, you can use the substitution \[u = 2  \xi^{2}\] \[det A = u^{3}  u\] \[u^{3}  u=0\] to simplify the algebra.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[ \begin{align*} \begin{vmatrix} 2\xi^2&1 &0 \\ 1 & 2\xi^2&1 \\ 0 &1 &2\xi^2 \end{vmatrix} &= 0 \\ \\[1ex] (2\xi^2)\Big[(2\xi^2)^2(1)^2\Big] (1)\Big[(1)(2\xi^2)0\Big] +0 &= 0 \\ (2\xi^2)\Big[(2\xi^2)^21\Big](2\xi^2) &= 0 \\ (2\xi^2)\left(\Big[(2\xi^2)^21\Big]1\right) &= 0 \\ (2\xi^2)\Big((2\xi^2)^22\Big) &= 0 \\ (2\xi^2)\Big((2\xi^2)^22\Big) &= 0 \\ \end{align*}\] oh , where's the 2?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0hmm i guess the usub is a good idea

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0i'll try it again with the \(u\)substitution,

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.4\[det A = u^{3}  2u \] is correct, not \[det A = u^{3}  u \]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align}\begin{vmatrix} u&1 &0 \\ 1 & u&1 \\ 0 &1 &u \end{vmatrix} &= 0 \\ \\[1ex] u\Big[u^2(1)^2\Big] (1)\Big[(1)(u)0\Big] +0 &= 0 \\ u\Big[u^21\Big]u &= 0 \\ u^32u &= 0 \end{align}\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0hmm so u = 0 or u^22 =0

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0that's three solutions for u, and three for xi xi = √(2{√2,0,√2})

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0so the normal node frequencies \(\omega\), where : \(\omega/\omega_0 = \xi\) are....

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.4Should be 6 roots for \(\xi\). 2 for each u.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0*the frequencies must be positive oh wait still in angular frequencies,

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\nu = \omega/2\pi =... \textit{physics}\]

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.4Ok, that makes sense. I don't think I can help too much with the phyics though. Are these coupled springs?

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0\[\omega_0^2 = k/m = 6\,[\text{N/m}]/0.0905\,[\text{kg}]\]so \[\nu_1 = 0.99\,[\text{Hz}],\quad \nu_2 = 1.3\,[\text{Hz}], \qquad \nu_1 =2.34\,[\text{Hz}]\]

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0what a rigmarole, so many variable, so few numbers

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.4I 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.

UnkleRhaukus
 one year ago
Best ResponseYou've already chosen the best response.0one trig function, is better than two
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