Spring static equilibrium Problem

[Arman777]

Homework Statement

a) If the distances $s_1$ and $s_2$ are the amounts the springs are streched while in static equilibirum, Write down the static equilibirum equations.
b) When the system is distrubed from its equilibirum state, both mases move vertically. Let $y_1(t)$ and $y_2(t)$ represent the displacement of masses from their equilibrium at time t. Write down the Newtons second law for each mass in terms of 2nd order DE.
c) For $y_1(0)=y_2(0)=\dot y_1(0)= 0, \dot y_2(0) = 10$ find the solution.

Homework Equations

The Attempt at a Solution

[/B]
Since the motion is always in one direction, I ll not use vector notation here.

Part(a) : Since its equi state the net force on bodies will be zero. Hence;
$0 = m_1g+k_2s_2-k_1s_1$
$0 = m_2g_2-k_2s_2$

For part (b)
The net force will be not zero this time. And I thought we can write
$m_1\ddot y(t)=m_1g+k_2y_2(t)-k_1y_1(t)$
$m_1\ddot y(t)=m_1g-k_2y_2(t)$

Are these equation true ? Cause later I need to solve these But I amhow also stuck how can I find the $y_1(t)$

 

[Orodruin]

Are these equation true ?

No. How do the displacements relate to the extnsions of the springs?

 

[Chestermiller]

Your equations for part (b) are not correct. The mg's should not be in there. In addition, the y terms are not correct. In terms of y1 and y2, how much does spring k2 stretch?

 

[timetraveller123]

i think because you defined y1 and y2 as displacement from equilibrium you don't have to consider gravity anymore just like you didn't consider the "constant" forcef due to extension of spring

edit :
and furthermore as @Orodruin pointed out i think your displacements might be wrong

 

[Arman777]

Is it
$m_2\ddot y = -k_2y_2$ and
$m_1\ddot y = k_2y_2-k_1y_1$

I don't use $mg$ cause as @timetraveller123 said its useless ?

 

[Arman777]

How do the displacements relate to the extnsions of the springs?

In terms of y1 and y2, how much does spring k2 stretch?

I am not sure I understand it.. or did I do it right?

 

[Chestermiller]

Is it
$m_2\ddot y = -k_2y_2$ and
$m_1\ddot y = k_2y_2-k_1y_1$

I don't use $mg$ cause as @timetraveller123 said its useless ?

No. The equations should read:

$m_2\ddot y = -k_2(y_2-y_1)$ and
$m_1\ddot y = k_2(y_2-y_1)-k_1y_1$

 

[Arman777]

No. The equations should read:

$m_2\ddot y = -k_2(y_2-y_1)$ and
$m_1\ddot y = k_2(y_2-y_1)-k_1y_1$

I didnt understand why the displacement of $m_2$ is $y_2-y_1$. Both of the masses will move downward right ?

I didnt understand why but I can see how

 

[Chestermiller]

I didnt understand why the displacement of y is y2-y1. Both of the masses will move downward right ?

If one end of a spring moves y1 and the other end moves y2, how much does the length of the spring change?

 

[timetraveller123]

I don't use mgmgmg cause as @timetraveller123 said its useless ?

i didnt say it is useless i am saying a constant force only serves to change the equilibrium position

only the extension of top spring depend only on the displacement y1
the bottom spring extension depend both on y1 an y2

 

[haruspex]

I didnt understand why the displacement of $m_2$ is $y_2-y_1$.

It isn't. The displacement of m₂, in the lab frame is y₂. But simultaneously the displacement of m₁ is y₁, so the change in length of the lower spring is y₂-y₁.

 

[Arman777]

Thanks all of you for your comments. I ll try to understand it and then try to solve the equations and share my solution today. I couldn't reply earlier cause I had another HM's and exam.

 

[Arman777]

Umm How can I collect the same terms one equation.. ?

 

[Chestermiller]

Umm How can I collect the same terms one equation.. ?

Solve the first equation for y1 and substitute it into the 2nd equation.

 

[Arman777]

Solve the first equation for y1 and substitute it into the 2nd equation.

How can I solve there's $y_1$ term in it ?

$m_2\ddot y = -k_2(y_2-y_1)$ and
$m_1\ddot y = k_2(y_2-y_1)-k_1y_1$

Those $\ddot y$ should be $\ddot y_1$ , $\ddot y_2$ right so I have

$m_2\ddot y_2 = -k_2(y_2-y_1)$ and
$m_1\ddot y_1 = k_2(y_2-y_1)-k_1y_1$

 

[Chestermiller]

How can I solve there's $y_1$ term in it ?

$m_2\ddot y = -k_2(y_2-y_1)$ and
$m_1\ddot y = k_2(y_2-y_1)-k_1y_1$

Those $\ddot y$ should be $\ddot y_1$ , $\ddot y_2$ right so I have

$m_2\ddot y_2 = -k_2(y_2-y_1)$ and
$m_1\ddot y_1 = k_2(y_2-y_1)-k_1y_1$

$$y_1=\frac{m_2}{k_2}\ddot y_2+y_2$$Now, substitute that into the 2nd equation to get an equation involving only y2. What do you get?

 

[Orodruin]

@Arman777 Are you familiar with matrix algebra?

 

[Arman777]

$$y_1=\frac{m_2}{k_2}\ddot y_2+y_2$$Now, substitute that into the 2nd equation to get an equation involving only y2. What do you get?

$m_1\ddot y_1+m_2\ddot y_2=-ky_1$

or

$m_1\ddot y_1+m_2\ddot y_2=-k_1 (\frac{m_2}{k_2}\ddot y_2+y_2$)

I am either too tired to see or I couldn't get it...

 

[Arman777]

@Arman777 Are you familiar with matrix algebra?

Well kind of yeah. I can do it I guess

 

[Chestermiller]

$m_1\ddot y_1+m_2\ddot y_2=-ky_1$

or

$m_1\ddot y_1+m_2\ddot y_2=-k_1 (\frac{m_2}{k_2}\ddot y_2+y_2$)

I am either too tired to see or I couldn't get it...

From the equation I wrote, what is $\ddot y_1$ equal to?

 

[Arman777]

From the equation I wrote, what is $\ddot y_1$ equal to?

İsnt that goes like $\dddot y_2$... ? or we will take them as zero ?

 

[Orodruin]

Well kind of yeah. I can do it I guess

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?

 

[Chestermiller]

İsnt that goes like $\dddot y_2$... ? or we will take them as zero ?

Yes, it contains $\ddddot y_2$

 

[Arman777]

$m_1(m_2\ddddot y_2/k_2 ) + m_2\ddot y_2 = -k_1 (m_2\ddot y_2/k_2 + y_2)$ ?

 

[Chestermiller]

$m_1(m_2\ddddot y_2/k_2 ) + m_2\ddot y_2 = -k_1 (m_2\ddot y_2/k_2 + y_2)$ ?

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.

 

[Arman777]

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]

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]

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

 

[Arman777]

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 don't think I can do that either

 

[Arman777]

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

 

[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