Spring static equilibrium Problem

[Arman777]

I have an idea Let's say $$m_1\ddot y_1 = k_2(y_2-y_1)-k_1y_1$$ and $$m_2\ddot y_2 = -k_2(y_2-y_1)$$ Let's 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 $m_1=m_2$ that's 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]

I have an idea Let's say $$m_1\ddot y_1 = k_2(y_2-y_1)-k_1y_1$$ and $$m_2\ddot y_2 = -k_2(y_2-y_1)$$ Let's 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 $m_1=m_2$ that's 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.)

 

[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

 

[Arman777]

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 haven't take those classes
do you know about eigenvectors and stuff if you do then i think this might be the fastest way

 

[Arman777]

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]

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]

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]

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 learned it that way

 

[Arman777]

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 ?

 

[timetraveller123]

were you given the values
the eigen values seem correct

 

[Arman777]

were you given the values
the eigen values seem correct

Thanks :)

 

[Arman777]

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]

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]

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]

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]

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.

 

[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