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Spring static equilibrium Problem

  • Thread starter Arman777
  • Start date
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1. The problem statement, all variables and given/known data
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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.
2. Relevant equations


3. The attempt at a solution

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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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?
 
i think because you defined y1 and y2 as displacement from equilibrium you dont 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
 
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Is it
##m_2\ddot y = -k_2y_2## and
##m_1\ddot y = k_2y_2-k_1y_1##

I dont use ##mg## cause as @timetraveller123 said its useless ?
 
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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
 
I dont 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

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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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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 couldnt reply earlier cause I had another HM's and exam.
 
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Umm How can I collect the same terms one equation.. ?
 
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Solve the first equation for y1 and substitute it into the 2nd equation.
How can I solve theres ## 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##
 
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How can I solve theres ## 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

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@Arman777 Are you familiar with matrix algebra?
 
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$$y_1=\frac{m_2}{k_2}\ddot y_2+y_2$$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 couldnt get it...
 
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##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 couldnt get it...
From the equation I wrote, what is ##\ddot y_1## equal to?
 

Orodruin

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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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##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)## ?
 
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##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.
 

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