# Spring static equilibrium Problem

#### Arman777

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Correct. Do you know how to solve that 4th order ODE with constant coefficients? Start off by combining terms and moving everything to one side of the equation.
I ll try

#### haruspex

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Correct. Do you know how to solve that 4th order ODE with constant coefficients? Start off by combining terms and moving everything to one side of the equation.
I think there's an easier way.
Going back to the simultaneous ODEs in post #15, one can look for a linear combination $y=y_1+\alpha y_2$ such that they produce an ODE in y only. Should get a quadratic in $\alpha$.

#### Arman777

Gold Member
If we know the values of $m_1$ and $m_2$ and $k's$ is that helps ?

#### Arman777

Gold Member
So can you write your differential equations on the form
$$\ddot Y = K Y,$$
where $Y$ is a column matrix with two entries and $K$ is a 2x2 matrix?
I dont think I can do that either

#### Arman777

Gold Member
Do we know $y_1+y_2$ ? maybe in terms of s or smthing like that ?

#### Arman777

Gold Member
I have an idea Lets say $$m_1\ddot y_1 = k_2(y_2-y_1)-k_1y_1$$ and $$m_2\ddot y_2 = -k_2(y_2-y_1)$$ Lets substract both sides

$$m_1\ddot y_1-m_2\ddot y_2=2k_2(y_2-y_1)-k_1y_1$$ we know that $m_1=m_2$ thats given actually...and also k values.

so we have $$Y=y_2-y_1$$

$$d^2Y/dt^2=\frac {1} {m} (2k_2Y- k_1y_1)$$ ?

#### haruspex

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I have an idea Lets say $$m_1\ddot y_1 = k_2(y_2-y_1)-k_1y_1$$ and $$m_2\ddot y_2 = -k_2(y_2-y_1)$$ Lets substract both sides

$$m_1\ddot y_1-m_2\ddot y_2=2k_2(y_2-y_1)-k_1y_1$$ we know that $m_1=m_2$ thats given actually...and also k values.

so we have $$Y=y_2-y_1$$

$$d^2Y/dt^2=\frac {1} {m} (2k_2Y- k_1y_1)$$ ?
You are no better off since you have both Y and y1.
Try my suggestion in post #27.
(I think it is effectively the same as Orodruin's method.)

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#### timetraveller123

I think @ Orodruin method might be the easiest you write the system as
$\begin{pmatrix} \ddot {Y_1}\\ \ddot {Y_2} \end{pmatrix} =k \begin{pmatrix} Y_1\\ Y_2 \end{pmatrix} \\$
Where k is
$\begin{pmatrix} -(k_1 + k_2)/m_1 & k_2/m_1 \\ K_2/m_2 & -k_2/m_2 \end{pmatrix}$
If you expand this out you will get back the original system
So to solve this you can diagonalize it and get two uncoupled first order ode which you can solve

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#### Arman777

Gold Member
I think @ Orodruin method might be the easiest you write the system as
$\begin{pmatrix} \ddot {Y_1}\\ \ddot {Y_2} \end{pmatrix} =k \begin{pmatrix} Y_1\\ Y_2 \end{pmatrix} \\$
Where k is
$\begin{pmatrix} -(k_1 + k_2)/m_1 & k_2/m_1 \\ K_2/m_2 & -k_2/m_2 \end{pmatrix}$
If you expand this out you will get back the original system
So to solve this you can diagonalize it and get two uncoupled first order ode which you can solve
I ll going to try it. But I am not hopefull that I can do it :)
Is these type of problems are common in CM classes ?

#### timetraveller123

i am not sure i havent take those classes
do you know about eigenvectors and stuff if you do then i think this might be the fastest way

#### Arman777

Gold Member
Today in lecture our teacher made a solution to this problem by using the Matrix approach as Orodruin pointed out. He write the matrix form then he found the eigenvalues and eigenvectors. And then well he kind of stopped there.

#### timetraveller123

then maybe you should try to take it from there do you know how to diagonalize a matrix

#### Chestermiller

Mentor
then maybe you should try to take it from there do you know how to diagonalize a matrix
Once he has the eigenvalues and eigenvectors, there is no need to diagonalize anything. All he needs to do is use these to determine the coefficients required to satisfy the initial conditions.

#### Orodruin

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Once he has the eigenvalues and eigenvectors, there is no need to diagonalize anything. All he needs to do is use these to determine the coefficients required to satisfy the initial conditions.
Well, the process of finding the eigenvalues and eigenvectors essentially gives you the diagonalisation as well ...

#### timetraveller123

wait even after obtaining eigenvectors and eigenvalues
you still have to change basis from
$y' = p^{-1} y$
and construct a diagonal matrix filled with the eigenvalues right? at least this is what i know
then after solving that revert back to normal basis

#### Orodruin

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wait even after obtaining eigenvectors and eigenvalues
you still have to change basis from
$y' = p^{-1} y$
and construct a diagonal matrix filled with the eigenvalues right? at least this is what i know
You don't really have to do it. You just note that with a complete set of eigenvectors $v_i$, you can expand the solution in terms of them, i.e.,
$$y(t) = \sum_i \alpha_i(t) v_i.$$
Now, inserting into the differential equation would give
$$\ddot y = \sum_i \ddot{\alpha}_i(t) v_i = K \sum_i \alpha_i(t) v_i = \sum_i \lambda_i \alpha_i(t) v_i,$$
where $\lambda_i$ are the eigenvalues. Since the $v_i$ are linearly independent, the coefficients in front of $v_i$ on either side of the equation must be the same and therefore
$$\ddot \alpha_i = \lambda_i \alpha_i.$$
Of course, this is the same thing as you will get if you do the diagonalisation explicitly.

#### timetraveller123

oh wow that's actually rather neat i never learnt it that way

#### Arman777

Gold Member
I find something like

$\begin{pmatrix} \ddot y_1 \\ \ddot y_2 \\ \end{pmatrix} = \begin{pmatrix} -10 & 4 \\ 4 & -4 \\ \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ \end{pmatrix}$

I find the values $λ_1=-12$ and $λ_2=-2$

Correct ?

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#### timetraveller123

were you given the values
the eigen values seem correct

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#### Arman777

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You don't really have to do it. You just note that with a complete set of eigenvectors $v_i$, you can expand the solution in terms of them, i.e.,
$$y(t) = \sum_i \alpha_i(t) v_i.$$
Now, inserting into the differential equation would give
$$\ddot y = \sum_i \ddot{\alpha}_i(t) v_i = K \sum_i \alpha_i(t) v_i = \sum_i \lambda_i \alpha_i(t) v_i,$$
where $\lambda_i$ are the eigenvalues. Since the $v_i$ are linearly independent, the coefficients in front of $v_i$ on either side of the equation must be the same and therefore
$$\ddot \alpha_i = \lambda_i \alpha_i.$$
Of course, this is the same thing as you will get if you do the diagonalisation explicitly.
So $\ddot y_1 = -12y_1$ and $\ddot y_2 = -2y_2$ or which eigenvalue corresponds to which ?

#### Orodruin

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So $\ddot y_1 = -12y_1$ and $\ddot y_2 = -2y_2$ or which eigenvalue corresponds to which ?
No, you need to use the eigenvectors. The equations where the differential equations are not coupled are the ones for the $\alpha$s, not for the $y$s.

#### Arman777

Gold Member
For $λ_1=-12$ I find eigenvector
$\begin{pmatrix} 2 \\ -1 \\ \end{pmatrix}$

and for $λ_1=-2$ I find
$\begin{pmatrix} 1 \\ 2 \\ \end{pmatrix}$

so $\ddot y_1= -12 \begin{pmatrix} 2 \\ -1 \\ \end{pmatrix}$

$\ddot y_2=-2 \begin{pmatrix} 1 \\ 2 \\ \end{pmatrix}$ ?

#### Orodruin

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No. You need to write your differential equations on the form
$$\ddot Y = \begin{pmatrix} \ddot y_1 \\ \ddot y_2 \end{pmatrix} = \ddot \alpha_1 v_1 + \ddot \alpha_2 v_2 = \lambda_1 \alpha_1 v_1 + \lambda_2 \alpha_2 v_2.$$
This will give you differential equations for the $\alpha$s, not for the $y$s.

#### Arman777

Gold Member
What is $α$ ??

I am so confused right now. $ν$ are the eigenvectors okay $λ$ is the eigenvalue.

Its so sad that our teacher never solved a problem like this before. Even once and I guess my algebra sucks.

"Spring static equilibrium Problem"

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