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Spring static equilibrium Problem

  • Thread starter Arman777
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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.
 
1,480
102
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.
 
1,480
102
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:
1,480
102
##\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.
 
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
 

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