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January 24th, 2015, 03:47 AM   #11
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You should form your equations with only $\ddot x_n$ on the left (divide by $m_n$) to do this. Then form your matrices.
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January 24th, 2015, 03:51 AM   #12
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Can you make the solution? I can not do more. how find to eigen frequencies
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January 24th, 2015, 04:02 AM   #13
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You aren't trying!
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January 24th, 2015, 04:11 AM   #14
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I'm trying to do, but it is not. Show route.
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January 24th, 2015, 05:03 AM   #15
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We have $$ \begin{aligned}&& m_1 \ddot x_1 &= k_1(x_2 - x_1) - k_2x_1 \\ && 2\ddot x_1 &= 2x_2 - 3x_1 \\ && \ddot x_1 &= x_2 - \tfrac32 x_1 \\ &\text{similarly} & \ddot x_2 &= \tfrac12 x_1 - \tfrac34 x_1 \end{aligned}$$
Now we assume that $x_n =a_n \mathrm e^{\mathrm i \omega t}$ so that $\ddot x_n = -\omega ^2 x_n$. You can now eliminate $\ddot x_n$ from the equations and form an eigenvalue matrix problem.
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January 24th, 2015, 09:03 AM   #16
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I could not solve my teacher. I'm trying, but no solution. Can you solve it? Although a little way then you are here.
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January 24th, 2015, 09:52 AM   #17
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$$\begin{aligned} &\text{We have} & \ddot x_1 &= x_2 - \tfrac32 x_1 \qquad \ddot x_1 = -\omega^2 x_1 \\ &\text{so} & -\omega^2 x_1 &= x_2 - \tfrac32 x_1 \implies (\omega^2 - \tfrac32)x_1 + x_2 = 0 \end{aligned}$$
You can do the same for $\ddot x_2$. Then you can make a matrix equation $Ax = 0$ who CH means that the determinant of A is zero.

That gives an equation which you can solve for $\omega$.
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January 24th, 2015, 10:57 AM   #18
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Is the right solution?
1)
2)
3)

Last edited by yakamoz29; January 24th, 2015 at 11:12 AM.
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January 25th, 2015, 10:39 AM   #19
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result w1,2=+-34,9 ??
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