
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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January 24th, 2015, 03:47 AM  #11 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
You should form your equations with only $\ddot x_n$ on the left (divide by $m_n$) to do this. Then form your matrices.

January 24th, 2015, 03:51 AM  #12 
Newbie Joined: Jan 2015 From: world Posts: 11 Thanks: 0 
Can you make the solution? I can not do more. how find to eigen frequencies

January 24th, 2015, 04:02 AM  #13 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
You aren't trying!

January 24th, 2015, 04:11 AM  #14 
Newbie Joined: Jan 2015 From: world Posts: 11 Thanks: 0 
I'm trying to do, but it is not. Show route.

January 24th, 2015, 05:03 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
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. 
January 24th, 2015, 09:03 AM  #16 
Newbie Joined: Jan 2015 From: world Posts: 11 Thanks: 0 
I could not solve my teacher. I'm trying, but no solution. Can you solve it? Although a little way then you are here.

January 24th, 2015, 09:52 AM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
$$\begin{aligned} &\text{We have} & \ddot x_1 &= x_2  \tfrac32 x_1 \qquad \ddot x_1 = \omega^2 x_1 \\ &\text{so} & \omega^2 x_1 &= x_2  \tfrac32 x_1 \implies (\omega^2  \tfrac32)x_1 + x_2 = 0 \end{aligned}$$ 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 $\omega$. 
January 24th, 2015, 10:57 AM  #18 
Newbie Joined: Jan 2015 From: world Posts: 11 Thanks: 0 
Is the right solution? 1) 2) 3) Last edited by yakamoz29; January 24th, 2015 at 11:12 AM. 
January 25th, 2015, 10:39 AM  #19 
Newbie Joined: Jan 2015 From: world Posts: 11 Thanks: 0 
result w1,2=+34,9 ??


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