Coupled Oscillations

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I have a burning question,

I was trying to find the solutions for a double mass coupled oscillation. So I found out the eigenvectors and then I arrived at this step

[tex] \left( \begin{array}{c} \ddot{x_1} \\ \ddot{x_2} \end{array} \right)=\lambda \left( \begin{array}{c} x_1 \\ \ x_2 \end{array} \right) [/tex]
(the second matrix is without the accents, I think the latex code will take a while to refresh)

ok so my question is, why is one of the solutions displayed as

[tex] x_{1}+x_{2}=A_{1}\cos{(\omega t+\phi)} [/tex]

when from the first equation, it is evident that

[tex] \ddot{x_1}=\lambda{x_1} [/tex]

so

[tex] x_1=A_1\cos{(\omega t+\phi)} [/tex]

I simply don't understand why the above is not acceptable. Also, I am having trouble in relating the addition of the equations (equation 2) to the solution for the eigenvectors. By the way, I also know the solution for the eigenvectors.
 
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Isn't X_1 + X_2 a mode coordinate?

The 2 mode coordinates being the sum and difference of X_1 and X_2. They are ways of looking at the motion of the system as a whole, not X_1 and X_2 individually.
 
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thank you for your reply,

isn't the matrix constructed from x_1 and x_2 individually?

I also do not understand what is a mode coordinate, could you explain this to me if this is important?
 

tiny-tim

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I was trying to find the solutions for a double mass coupled oscillation.

I simply don't understand why the above is not acceptable.
Hi Oerg! :wink:

It is acceptable, but it's not as simple nor as conceptually deep as using normal modes such as x1 ± x2

From http://en.wikipedia.org/wiki/Coupled_oscillation#Coupled_oscillations
The apparent motions of the individual oscillations typically appears very complicated but a more economic, computationally simpler and conceptually deeper description is given by resolving the motion into normal modes.
See also http://en.wikipedia.org/wiki/Normal_mode :smile:
 
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Hi Oerg! :wink:

It is acceptable, but it's not as simple nor as conceptually deep as using normal modes such as x1 ± x2

From http://en.wikipedia.org/wiki/Coupled_oscillation#Coupled_oscillations


See also http://en.wikipedia.org/wiki/Normal_mode :smile:
thanks for your reply too

there was another solution that is

[tex] x_{1}+x_{2}=A_{2}\cos{(\omega t+\phi _{1})} [/tex]

and then with the first equation in the original post, x1 and x2 is then given as

[tex] x_1=\frac{1}{2}(A_{1}\cos{(\omega _{0}t-\phi)} +A_{2}\cos{(\sqrt{3}\omega _{0}t-\phi _{1})}) [/tex]

by the way,

[tex] \lambda =1[/tex]

and

[tex] \lambda =3[/tex]

are the eigenvalues for the problem. So there seems to be a discrepancy for the equations for x_1 and x_2.Where have I gone wrong :confused:
 
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help im drowning arghhhhhhh
 

tiny-tim

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So there seems to be a discrepancy for the equations for x_1 and x_2.Where have I gone wrong :confused:
i don't follow :confused:

what discrepancy are you referring to?
 
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why is the last equation from my last post different from the correct solution to the de?

Also, how do I obtain the correct solutions from the eigenvector
 
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I think i understand a little now, the eigenvalues that I found was when all the masses displayed the same frequency of oscillation.

But how do I prove that the equations for the positions of the masses are a superposition of normal modes with the eigenvectors?
 

tiny-tim

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I think i understand a little now, the eigenvalues that I found was when all the masses displayed the same frequency of oscillation.

But how do I prove that the equations for the positions of the masses are a superposition of normal modes with the eigenvectors?
Hi Oerg! :smile:

Stop using all these technical words

x1 + x2 = A cosBt, x1 - x2 = C cosDt,

obviously x1 = (AcosBt + CcosDt)/2 … that's year-1 arithmetic! :wink:

We solve it that way round because you only have to look at the formula for x1 on its own to see that it's much more difficult to solve than x1 + x2 :smile:

but there's no magic of "superposition" or "normal modes" to understand
 
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Hi Oerg! :smile:

Stop using all these technical words

x1 + x2 = A cosBt, x1 - x2 = C cosDt,

obviously x1 = (AcosBt + CcosDt)/2 … that's year-1 arithmetic! :wink:

We solve it that way round because you only have to look at the formula for x1 on its own to see that it's much more difficult to solve than x1 + x2 :smile:

but there's no magic of "superposition" or "normal modes" to understand
thanks for your reply

hmm, i know about the equations "x1 + x2 = A cosBt, x1 - x2 = C cosDt" for a two mass system, it is just the addition and subtraction and then the acceleration and the position are common terms.

But what about a system with a higher number of masses and springs? How do i know
x?+x?+x?+..=Acoswt+phi

So I was trying to see how by finding out the eigenvectors for a two mass system for simplicity, that x1+x2=Acoswt+phi. Im still at a loss though.
 

tiny-tim

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Homework Helper
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But what about a system with a higher number of masses and springs? How do i know
x?+x?+x?+..=Acoswt+phi

So I was trying to see how by finding out the eigenvectors for a two mass system for simplicity, that x1+x2=Acoswt+phi. Im still at a loss though.
i'm confused :confused: … i think you're answering your own question …

every matrix has eigenvectors, and each eigenvector is a combination of "basis" vectors, and by definition of eigenvector that combination is going to satisfy the shm equation ∑'' = -w2∑, so ∑ = Acoswt+phi :smile:
 
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so we have a Ax=b and b can be expressed as a linear combination of the eigenvalues multiplied by the respective eigenvectors? This is because eigenvectors are orthogonal. so in this spirit we have the solutions for a 3 mass system as

[tex]\left( \begin{array}{cc} \ddot{x_1} \\ \ddot{x_2} \\ \ddot{x_3} \end{array} \right)=\lambda _{1}v_{1}x+\lambda _{2}v_{2}x+\lambda _{3}v_3x [/tex]

where

[tex] x=\left( \begin{array}{cc} x_1 \\ x_2 \\ x_3 \end{array}\right)[/tex]

is this correct?
 
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ahh i think i understand now, I read up on a chapter on diagonalization and now I understand how it can be applied to solve the system of differential equations. Thanks for your help tiny tim.
 

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