Lagrangian of a mass bewteen two springs with a pendulum hanging down

In summary, the conversation discusses setting up a reference system and determining the generalized coordinates for a system consisting of a mass hanging from a rod attached to another mass that moves horizontally with two springs. The equations for kinetic and potential energy are derived, and there is a discussion about the potential energy of the second mass and why it is equal to kx^2 when the mass is at equilibrium. The conversation also mentions confusion about the potential energy of the second mass and a discrepancy in the position of the second spring when the first spring moves a distance x_2.
  • #1
Davidllerenav
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14
Homework Statement
A particle of mass ##m_1## hangs from a rod of negligible mass and length ##l##, whose support point consists of another particle of mass ##m_2## that moves horizontally subject to two springs of constant ##k## each one. Find the equations of motion for this system.
Relevant Equations
##L=T-V##
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What I first did was setting the reference system on the left corner. Then, I said that the position of the mass ##m_2## is ##x_2##. I also supposed that the pendulum makes an angle ##\theta## with respect to the vertical axis ##y##. So the generalized coordinates of the system would be ##x_2## and ##\theta##. Thus, the coordinates of ##m_1## are:
  • ##x_1=x_2+l\sin\theta##
  • ##y_1=-l\cos\theta##
Then I took the time derivative of both ##x_1## and ##y_1##:
  • ##\dot x_1=\dot x_2+l\dot\theta\cos\theta##
  • ##\dot y_1=l\dot\theta\sin\theta##
The Kinetic energy ##T## would be the sum of the kinetic energy of each mass ##T=T_1+T_2##. Since ##T1=\frac{1}{2}m_2\dotx_2^2## and ##T_2=\frac{1}{2}m_1(\dot x_1^2+\dot y_1^2##, I got:
  • ##T=\frac{1}{2}m_2\dot x_2^2+\frac{1}{2}m_1(\dot x_2^2+l^2\dot\theta^2+2\dot x_2\dot\theta l)##
Now, I have trouble with the potential energy. The potential energy of ##m_1## is easy, it would be just ##V_1=m_1gh##:
  • ##V_1=-m_1gl\cos\theta##
The potential energy of ##m_2## would be the sum of the potential energy that each spring has on the mass. What I did was saying that that the mass ##m_2## moves a distance ##x_2##, thus the lhs spring would have a displacement of ##x_2## and the rhs spring would have a contraction of ##x_2## likewise. So:
  • ##V_2=\frac{1}{2}k x_2^2+\frac{1}{2}k(-x_2)^2=kx_1^2##
But, according to my teacher, if the first spring moves a distance ##x_2##, then the second would contract a distance of ##a-x_2##, where ##a## is the distance between the walls, that is constant. I don't understand why this is the case, can you please explain it better for me to understand? Thanks.
 
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  • #2
Say
[tex]x_2=\frac{a}{2}[/tex] , natual length of spring,
[tex]V_2=0[/tex]
So
[tex]V_2=k(x_2-\frac{a}{2})^2[/tex]
 
  • #3
I don't understand it either. If ##s_1## and ##s_2## are the lengths of both springs then ##a = s_1 + s_2##.

Now, if one spring stretches by ##x## then the other spring must be compressed by the same amount, ## (s_1 + x) + (s_2 - x) = a##

And if ##V_1 = \frac12 k x^2## then ##V_2 = \frac12 k x^2## and so ##V = V_1 + V_2 = k x^2##

Although, I have the origin where ##m_2## is at rest. You say your origin is in the left corner. So the left spring is attached to the origin. Your teacher attaches the the right spring to ##a##.

Now if you move ##m_2## the new location will be at ##x_2## as measured from the origin for the left spring, and at ##a - x_2## which is measured from the right wall for the right spring.
 
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  • #4
Davidllerenav said:
What I first did was setting the reference system on the left corner.

x=0 at the left corner. Do you count V2=0 for x2=0,i.e. the mass is on the left corner ?
 
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  • #5
anuttarasammyak said:
x=0 at the left corner. Do you count V2=0 for x2=0,i.e. the mass is on the left corner ?
No. I count ##V_2=0## when ##x_2=a/2##, i.e. when the mass is halfway between the two walls.
 
  • #6
Marc Rindermann said:
I don't understand it either. If ##s_1## and ##s_2## are the lengths of both springs then ##a = s_1 + s_2##.

Now, if one spring stretches by ##x## then the other spring must be compressed by the same amount, ## (s_1 + x) + (s_2 - x) = a##

And if ##V_1 = \frac12 k x^2## then ##V_2 = \frac12 k x^2## and so ##V = V_1 + V_2 = k x^2##

Although, I have the origin where ##m_2## is at rest. You say your origin is in the left corner. So the left spring is attached to the origin. Your teacher attaches the the right spring to ##a##.

Now if you move ##m_2## the new location will be at ##x_2## as measured from the origin for the left spring, and at ##a - x_2## which is measured from the right wall for the right spring.
So, the result I got would only be correct if the origin is halfway between the two walls?
 
  • #7
Yes. And for your original coordinate see #2.
 
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  • #8
anuttarasammyak said:
Say
[tex]x_2=\frac{a}{2}[/tex] , natual length of spring,
[tex]V_2=0[/tex]
So
[tex]V_2=k(x_2-\frac{a}{2})^2[/tex]
I'm I bit confused by the use of ##x_2##, so let's say that ##x_2## is the position of the mass at its equilibrium position, i.e. ##x_2=a/2## as you said. And let ##x## be the position of the mass at some time Then if the mass moves to the right to a point ##x##, the left spring would stretch a distance of ##x-x_2=x-a/2##. But, wouldn't the spring on the right be compressed by the same amount?
 
  • #9
That's the reason why ##k## ,not ##\frac{k}{2}## for a single spring, in ##V_2##. Doubled as you did yourself in #1 and #3.
 
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  • #10
Davidllerenav said:
Homework Statement:: A particle of mass ##m_1## hangs from a rod of negligible mass and length ##l##, whose support point consists of another particle of mass ##m_2## that moves horizontally subject to two springs of constant ##k## each one. Find the equations of motion for this system.
Relevant Equations:: ##L=T-V##

But, according to my teacher, if the first spring moves a distance ##x_2## , then the second would contract a distance of ##a-x_2##, where is the distance between the walls, that is constant. I don't understand why this is the case, can you please explain it better for me to understand?
Your generalized coordinates are relative to an origin that is at ##m_2## when the system is at equilibrium, string vertical, springs relaxed. This choice allows you to use the same symbol ##x_2## to denote both the generalized coordinate and the amount by which each spring is compressed or stretched. Thus, if ##m_2## is displaced to the right, the elastic potential energy is as you have written it ##V_{\text{el}}=\frac{1}{2}kx_2^2+\frac{1}{2}kx_2^2=kx_2^2.##

Now suppose you choose to measure generalized coordinate ##x_2## relative to the left wall. If ##m_2## is again displaced from equilibrium to the right, the amount by which the left spring is stretched is ##x_2-\frac{a}{2}## and the amount by which the right spring is compressed is ##\frac{a}{2}-x_2## and the potential energy is ##V_{\text{el}}=\frac{1}{2}k(x_2-\frac{a}{2})^2+\frac{1}{2}k(\frac{a}{2}-x_2)^2=k(x_2-\frac{a}{2})^2.##

In such problems involving springs and Lagrangians one needs to be careful to enunciate the differences, if any, between generalized coordinates and spring displacements.
 
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FAQ: Lagrangian of a mass bewteen two springs with a pendulum hanging down

1. What is the Lagrangian of a mass between two springs with a pendulum hanging down?

The Lagrangian of a mass between two springs with a pendulum hanging down is a mathematical function that describes the energy of the system. It takes into account the potential and kinetic energy of the mass and pendulum, as well as the forces acting on them.

2. How is the Lagrangian calculated for this system?

The Lagrangian is calculated by summing the potential and kinetic energies of the mass and pendulum. The potential energy is determined by the position of the mass and the pendulum in relation to the springs, while the kinetic energy is determined by their velocities. This calculation is based on the principles of classical mechanics.

3. What is the significance of the Lagrangian in this system?

The Lagrangian is significant because it allows us to describe the dynamics of the system using a single mathematical function. This makes it easier to analyze and make predictions about the behavior of the mass and pendulum.

4. Can the Lagrangian be used to solve for the motion of the mass and pendulum?

Yes, the Lagrangian can be used to derive the equations of motion for the mass and pendulum. By taking the derivatives of the Lagrangian with respect to time, we can obtain the equations of motion, which describe how the position and velocity of the mass and pendulum change over time.

5. Are there any limitations to using the Lagrangian for this system?

While the Lagrangian is a powerful tool for analyzing the dynamics of the system, it does have some limitations. It assumes that the springs are ideal and do not have any damping, and that the pendulum is a simple pendulum with no friction. In reality, these assumptions may not hold true, and the Lagrangian may not accurately represent the behavior of the system.

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