Lagrangian mechanics for two springs

In summary, the conversation discusses a problem involving a spring with a negligible mass and spring constant k, supporting two identical masses in a vertical position. The Lagrangian is written using generalized coordinates and the equations of motion are derived. The conversation then discusses the inclusion of gravitational potential energy and how it affects the overall solution. After further consideration, the correct solution is found.
  • #1
wglmb
17
0

Homework Statement



A spring of negligible mass and spring constant k, hanging vertically with one end at a fixed point O, supports a mass m, and beneath it as second, identical spring carrying a second, identical mass.
Using a generalised coordinates the vertical displacements x and y of the masses from their positions with the springs unextended, write down the Lagrangian.
Find the position of equilibrium and the normal modes and frequencies of vertical oscillations.

Homework Equations





The Attempt at a Solution



KE: [tex]\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})[/tex]
PE: [tex]mg(l+x)+mg(2l+y)+\frac{1}{2}kx^{2}+\frac{1}{2}k(y-x)^{2}[/tex] [tex]=mg(3l+x+y)+kx^{2}+\frac{1}{2}ky^{2}-kxy[/tex]

Lagrangian: [tex]L=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mg(3l+x+y)-kx^{2}-\frac{1}{2}ky^{2}+kxy[/tex]

Equations of motion:
[tex]\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}=0 \Rightarrow m\ddot{x}+mg+2kx-ky=0[/tex]

[tex]\frac{d}{dt}(\frac{\partial L}{\partial \dot{y}})-\frac{\partial L}{\partial y}=0 \Rightarrow m\ddot{y}+mg+ky-kx=0[/tex]

Know that [tex]B_{ij}\ddot{q}_{j}+A_{ij}q_{j}=0[/tex]

[tex]\Rightarrow B=\left( \begin{array}{cc}
m & 0 \\
0 & m \end{array} \right), A=\left( \begin{array}{cc}
2k & -k \\
-k & k \end{array} \right)[/tex]

[tex]B^{-1}A=\left( \begin{array}{cc}
\frac{1}{m} & 0 \\
0 & \frac{1}{m} \end{array} \right)\left( \begin{array}{cc}
2k & -k \\
-k & k \end{array} \right)
=\left( \begin{array}{cc}
\frac{2k}{m} & \frac{-k}{m} \\
\frac{-k}{m} & \frac{k}{m} \end{array} \right)[/tex]

Eigenvalues turn out to be:
[tex]\lambda=\frac{k}{2m}(3+\sqrt{5}) and \lambda=\frac{k}{2m}(3-\sqrt{5})[/tex]

This doesn't look very likely to be right.
Then when I work out the eigenvectors, I get (0,0) in both cases...
... where have I gone wrong?
 
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  • #2
When you moved the masses down x and y respectively, the strain energies increased as you have indicated, but the gravitational energies also increase?
 
  • #3
Haven't I got that included in the [tex]mg(3l+x+y)[/tex] part of the PE? This comes from [tex]mg(l+x)+mg(2l+y)[/tex] which is mg times the height of the first particle plus mg times the height of the second.
 
  • #4
The gravitational potential energy is decreasing with increasing values of x and y, so you need some negative signs.
 
  • #5
Ah, I see what you mean - of course! I don't have time just now but I'll work that through later and see how I manage :)

Edit:

Hmm, well I still get the same matrices A and B, since they come from the double-dotted terms and the x&y terms. The constant terms (now -mg for both of the equations of motion) aren't used.
I have a feeling those constant terms shouldn't even be there, since none of the examples we've been given have any...
 
Last edited:
  • #6
It's OK, I've figured it out now. Thanks :)
 

1. What is Lagrangian mechanics for two springs?

Lagrangian mechanics for two springs is a mathematical framework used to model the motion of two masses connected by two springs. It allows us to study the dynamics of the system by considering the potential and kinetic energy of the masses and their corresponding forces.

2. How is Lagrangian mechanics different from Newtonian mechanics?

Lagrangian mechanics differs from Newtonian mechanics in that it uses a different approach to describe the motion of a system. While Newtonian mechanics focuses on forces and acceleration, Lagrangian mechanics uses the concept of energy and constraints to determine the equations of motion.

3. What are the advantages of using Lagrangian mechanics for two springs?

One major advantage of using Lagrangian mechanics for two springs is that it allows for a more elegant and concise formulation of the equations of motion compared to Newtonian mechanics. It also allows us to take into account constraints and generalized coordinates, making it easier to model complex systems.

4. Can Lagrangian mechanics be used for systems with more than two springs?

Yes, Lagrangian mechanics can be used for systems with any number of springs. The principles and equations of motion remain the same, but the complexity of the system may increase as more springs are added.

5. How is the Lagrangian for two springs calculated?

The Lagrangian for two springs is calculated by summing the kinetic and potential energy of the masses and subtracting any constraints. It can be expressed as L = T - V, where T is the total kinetic energy and V is the total potential energy of the system.

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