# Spring static equilibrium Problem

#### Orodruin

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The $\alpha$ are the expansion coefficients that tell you how much of each eigenvector there is in the solution. Generally those coefficients will be time dependent. The idea is that any vector
$$Y = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$$
can be written as a linear combination of the eigenvectors
$$Y = \alpha_1(t) v_1 + \alpha_2(t) v_2$$
where the expansion coefficients generally depend on time. Inserting this into the differential equation gives you separated differential equations for the $\alpha$, i.e., the differential equation for $\alpha_1$ does not depend on $\alpha_2$ and vice versa.

#### Arman777

Gold Member
So we have
For

$\begin{pmatrix} \ddot y_1 \\ \ddot y_2 \\ \end{pmatrix}=-12α_1 \begin{pmatrix} 2 \\ -1 \\ \end{pmatrix} -2α_2 \begin{pmatrix} 1 \\ 2 \\ \end{pmatrix}$

#### Orodruin

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You need to insert the expression for Y in terms of the alphas on the left side as well.

#### Arman777

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You need to insert the expression for Y in terms of the alphas on the left side as well.
Could you write it please..so I can learn it.. I dont get it this way. I need to proceed . This is painful

Last edited:

#### Arman777

Gold Member
$\begin{pmatrix} y_1 \\ y_2 \\ \end{pmatrix}=C_1e^{-12t} \begin{pmatrix} 2 \\ -1 \\ \end{pmatrix}+ C_2e^{-2t} \begin{pmatrix} 1 \\ 2 \\ \end{pmatrix}$

Last edited:

#### Orodruin

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That would be the result if you had a first order derivative and not a second order one in your differential equation.

#### Arman777

Gold Member
You need to insert the expression for Y in terms of the alphas on the left side as well.
You mean it will be $\ddot α_1v_1$ etc

#### timetraveller123

i think what @Orodruin menas is that the aplhas you got in post 56 would be the solution of
$\dot \alpha_i = \lambda_i \alpha_i$
$\ddot \alpha_i = \lambda_i \alpha_i$
you wouldn't get an exponential solution if you used the second one

"Spring static equilibrium Problem"

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