Spring static equilibrium Problem

In summary: 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)
  • #1
Arman777
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Homework Statement


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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)##
 

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  • #2
Arman777 said:
Are these equation true ?
No. How do the displacements relate to the extnsions of the springs?
 
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  • #3
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?
 
  • #4
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
 
  • #5
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 ?
 
  • #6
Orodruin said:
How do the displacements relate to the extnsions of the springs?

Chestermiller said:
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?
 
  • #7
Arman777 said:
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##
 
  • #8
Chestermiller said:
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
 
  • #9
Arman777 said:
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?
 
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  • #10
Arman777 said:
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
 
  • #11
Arman777 said:
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.
 
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  • #12
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.
 
  • #13
Umm How can I collect the same terms one equation.. ?
 
  • #14
Arman777 said:
Umm How can I collect the same terms one equation.. ?
Solve the first equation for y1 and substitute it into the 2nd equation.
 
  • #15
Chestermiller said:
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##
 
  • #16
Arman777 said:
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?
 
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  • #17
@Arman777 Are you familiar with matrix algebra?
 
  • #18
Chestermiller said:
$$y_1=\frac{m_2}{k_2}\ddot y_2+y_2$$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...
 
  • #19
Orodruin said:
@Arman777 Are you familiar with matrix algebra?
Well kind of yeah. I can do it I guess
 
  • #20
Arman777 said:
##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?
 
  • #21
Chestermiller said:
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 ?
 
  • #22
Arman777 said:
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?
 
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  • #23
Arman777 said:
İsnt that goes like ##\dddot y_2##... ? or we will take them as zero ?
Yes, it contains ##\ddddot y_2##
 
  • #24
##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)## ?
 
  • #25
Arman777 said:
##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.
 
  • #26
Chestermiller said:
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
 
  • #27
Chestermiller said:
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##.
 
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  • #28
If we know the values of ##m_1## and ##m_2## and ##k's## is that helps ?
 
  • #29
Orodruin said:
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
 
  • #30
Do we know ##y_1+y_2## ? maybe in terms of s or smthing like that ?
 
  • #31
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 $$Y=y_2-y_1$$

$$d^2Y/dt^2=\frac {1} {m} (2k_2Y- k_1y_1)$$ ?
 
  • #32
Arman777 said:
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.)
 
Last edited:
  • #33
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
 
Last edited:
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  • #34
timetraveller123 said:
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 ?
 
  • #35
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
 
<h2>1. What is Spring static equilibrium?</h2><p>Spring static equilibrium is a state in which a spring is at rest and is neither being compressed nor extended. This means that the forces acting on the spring are balanced, resulting in no net force and no movement.</p><h2>2. How is Spring static equilibrium calculated?</h2><p>To calculate Spring static equilibrium, we use the equation F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position. This equation is based on Hooke's Law.</p><h2>3. What factors affect Spring static equilibrium?</h2><p>The factors that affect Spring static equilibrium include the spring constant, the applied force, and the displacement from the equilibrium position. Changes in any of these factors can result in a change in the state of equilibrium.</p><h2>4. What is the significance of Spring static equilibrium?</h2><p>Spring static equilibrium is important in understanding the behavior of springs and their applications in various fields such as engineering, physics, and biology. It allows us to predict how a spring will behave under different conditions and how it can be used in different systems.</p><h2>5. How is Spring static equilibrium different from dynamic equilibrium?</h2><p>Spring static equilibrium is a state in which the forces on a spring are balanced, resulting in no movement. Dynamic equilibrium, on the other hand, is a state in which the forces on an object are balanced, but the object is in motion at a constant speed. In Spring static equilibrium, the restoring force of the spring is equal to the applied force, while in dynamic equilibrium, the net force is zero.</p>

1. What is Spring static equilibrium?

Spring static equilibrium is a state in which a spring is at rest and is neither being compressed nor extended. This means that the forces acting on the spring are balanced, resulting in no net force and no movement.

2. How is Spring static equilibrium calculated?

To calculate Spring static equilibrium, we use the equation F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement from the equilibrium position. This equation is based on Hooke's Law.

3. What factors affect Spring static equilibrium?

The factors that affect Spring static equilibrium include the spring constant, the applied force, and the displacement from the equilibrium position. Changes in any of these factors can result in a change in the state of equilibrium.

4. What is the significance of Spring static equilibrium?

Spring static equilibrium is important in understanding the behavior of springs and their applications in various fields such as engineering, physics, and biology. It allows us to predict how a spring will behave under different conditions and how it can be used in different systems.

5. How is Spring static equilibrium different from dynamic equilibrium?

Spring static equilibrium is a state in which the forces on a spring are balanced, resulting in no movement. Dynamic equilibrium, on the other hand, is a state in which the forces on an object are balanced, but the object is in motion at a constant speed. In Spring static equilibrium, the restoring force of the spring is equal to the applied force, while in dynamic equilibrium, the net force is zero.

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