Spring static equilibrium Problem

[Orodruin]

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]

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]

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

 

[Arman777]

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 don't get it this way. I need to proceed . This is painful

 

[Arman777]

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

 

[Orodruin]

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

 

[Arman777]

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
$\dot \alpha_i = \lambda_i \alpha_i$
instead of
$\ddot \alpha_i = \lambda_i \alpha_i$
you wouldn't get an exponential solution if you used the second one