Question

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

Answer 1

(' )-( .) what

Answer 2

@THEHELPER123

Answer 3

FIND THE determinant

Answer 4

, solve for xi

Answer 5

Just expand it out and solve for $\xi$ = 0. There's only 2 non-zero terms.

Answer 6

BTW, you can use the substitution

\[u = 2 - \xi^{2}\]
\[det A = u^{3} - u\]
\[u^{3} - u=0\]

to simplify the algebra.

Answer 7

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

Answer 8

hmm i guess the u-sub is a good idea

Answer 9

i'll try it again with the \(u\)-substitution,

Answer 10

\[det A = u^{3} - 2u \] is correct, not

\[det A = u^{3} - u \]

Answer 11

\[\begin{align}\begin{vmatrix}
u&-1 &0 \\
-1 & u&-1 \\
0 &-1 &u
\end{vmatrix}
&= 0 \\
\\[1ex]
u\Big[u^2-(-1)^2\Big]
-(-1)\Big[(-1)(u)-0\Big]
+0
&= 0 \\
u\Big[u^2-1\Big]-u
&= 0 \\
u^3-2u
&= 0
\end{align}\]

Answer 12

hmm so u = 0 or u^2-2 =0

Answer 13

u = 0 or ±√2

Answer 14

that's three solutions for u, and three for xi

xi = √(2-{-√2,0,√2})

Answer 15

so the normal node frequencies \(\omega\), where : \(\omega/\omega_0 = \xi\) are....

Answer 16

Should be 6 roots for \(\xi\). 2 for each u.

Answer 17

*the frequencies must be positive

oh wait still in angular frequencies,

Answer 18

\[\nu = \omega/2\pi =... \textit{physics}\]

Answer 19

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

Answer 20

yes

Answer 21

\[\omega_0^2 = k/m = 6\,[\text{N/m}]/0.0905\,[\text{kg}]\]so \[\nu_1 = 0.99\,[\text{Hz}],\quad \nu_2 = 1.3\,[\text{Hz}], \qquad \nu_1 =2.34\,[\text{Hz}]\]

Answer 22

what a rigmarole, so many variable, so few numbers

Answer 23

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.

Answer 24

one trig function, is better than two