Lagrangian of a system of several masses and springs

Click For Summary
SUMMARY

The discussion centers on deriving the Lagrangian for a system of three masses (m_1, m_2, m_3) connected by springs (k_1, k_2) in one dimension. The Lagrangian is expressed as L = L_1 + L_2 + L_3, where L_i = T_i - V_i, with T representing kinetic energy and V representing potential energy. The correct formulation of the potential energy for each spring is crucial, leading to the equations of motion: m_1 \ddot x_1 + k_1 [(x_2 - x_1) - l_1] = 0, m_2 \ddot x_2 + k_1 [l_1 + (x_2 - x_1)] - k_2 [l_2 - (x_3 - x_2)] = 0, and m_3 \ddot x_3 + k_2 [l_2 - (x_3 - x_2)] = 0.

PREREQUISITES
  • Understanding of Lagrangian mechanics
  • Familiarity with kinetic and potential energy concepts
  • Knowledge of Euler-Lagrange equations
  • Basic grasp of spring dynamics and Hooke's Law
NEXT STEPS
  • Study the derivation of the Euler-Lagrange equations in detail
  • Explore potential energy calculations in multi-spring systems
  • Learn about the stability analysis of coupled oscillators
  • Investigate numerical methods for solving Lagrangian systems
USEFUL FOR

Students and professionals in physics, particularly those focusing on classical mechanics, mechanical engineering, and systems dynamics. This discussion is beneficial for anyone looking to deepen their understanding of Lagrangian formulations in multi-body systems.

fluidistic
Gold Member
Messages
3,932
Reaction score
283

Homework Statement


I challenged myself with a problem I invented, but I'm stuck.
Consider a 1 dimensional problem consisting of 3 masses, each one separated by a spring. So that from the left to the right of my sketch we have m_1, a spring (k_1 with natural length l_1), m_2, another spring (k_2, l_2) and m_3. Find the equations of motion of the system.


Homework Equations


L=L_1+L_2+L_3 where L_i=T_i-V_i.
After this, Euler-Lagrange equations.

The Attempt at a Solution


For the first mass I reached L_1=\frac{m_1 \dot x ^2}{2}-\frac{k_1 (\Delta x_1 )^2}{2} though this \Delta x_1 really bothers me.
Now to find L_2, V_2 is a real headache. Because this mass is connected to 2 springs, I'm not sure at all how to calculate the potential energy of it. Maybe adding both springs' extensions? I mean V_2=\frac{k_1 (\Delta x_1)^2 + k_2 (\Delta x_2)^2}{2}?
Its kinetic energy would be T_2=\dot x ^2 + 2 \dot x \Delta \dot x_1 + (\Delta x_1)^2. Am I in the right direction?
 
Physics news on Phys.org
fluidistic said:

Homework Statement


I challenged myself with a problem I invented, but I'm stuck.
Consider a 1 dimensional problem consisting of 3 masses, each one separated by a spring. So that from the left to the right of my sketch we have m_1, a spring (k_1 with natural length l_1), m_2, another spring (k_2, l_2) and m_3. Find the equations of motion of the system.


Homework Equations


L=L_1+L_2+L_3 where L_i=T_i-V_i.
After this, Euler-Lagrange equations.

The Attempt at a Solution


For the first mass I reached L_1=\frac{m_1 \dot x ^2}{2}-\frac{k_1 (\Delta x_1 )^2}{2} though this \Delta x_1 really bothers me.
Now to find L_2, V_2 is a real headache. Because this mass is connected to 2 springs, I'm not sure at all how to calculate the potential energy of it. Maybe adding both springs' extensions? I mean V_2=\frac{k_1 (\Delta x_1)^2 + k_2 (\Delta x_2)^2}{2}?
Its kinetic energy would be T_2=\dot x ^2 + 2 \dot x \Delta \dot x_1 + (\Delta x_1)^2. Am I in the right direction?

I think you need three different x positions, one for each mass. That way your answer will be three equations of motion, one for each mass. This way you could also write \Delta x_1 = x_2 - x_1
 
Breaking the Lagrangian up by masses isn't a good idea because the potential energies aren't quantities associated with any single mass, but with each spring. In terms of the xi's, what is the distance between the ends of spring 1? Once you have that, it's easy to write down what the potential energy stored by spring 1 is in terms of your variables.
 
zachzach said:
I think you need three different x positions, one for each mass. That way your answer will be three equations of motion, one for each mass. This way you could also write \Delta x_1 = x_2 - x_1
Good observation. Actually that's what I partly did. Hence my equation for T_2. I took \vec r _2 or your x_2 as x+l_1+\Delta x_1.

vela said:
Breaking the Lagrangian up by masses isn't a good idea because the potential energies aren't quantities associated with any single mass, but with each spring. In terms of the xi's, what is the distance between the ends of spring 1? Once you have that, it's easy to write down what the potential energy stored by spring 1 is in terms of your variables.

Oh... hmm. Looking at zach's reply, x_2-x_1 is the distance between the ends of spring 1. So the potential energy of these 2 masses (or not?) would be \frac{k_1 (l_1 +x_2 - x_1 )^2}{2}? I'm not sure whether it's a "+x_2-x_1" or "-x_2+x_1" and also the second's mass is also attached to the other spring...
 
The masses don't have potential energy; the springs do.

Your expression for the potential energy of spring 1 is almost correct, but not quite.
 
vela said:
The masses don't have potential energy; the springs do.

Your expression for the potential energy of spring 1 is almost correct, but not quite.

Oh you're right.
Correct me if I'm wrong. I assumed 5 Lagrangians. 3 for the masses and 2 for the springs. The springs only carry a potential energy while the masses only a kinetic energy. Is this right?
If so, I reach the following Lagrangian: L=\frac{m_1\dot x_1 ^2}{2}+\frac{k_1}{2}(l_1 - x_2 ^2 + 2 x_1 x_2 - x_1 ^2)+\frac{m_2 \dot x_2 ^2}{2}+\frac{k_2}{2}(l_2-x_3 ^2 + 2 x_2 x_3 - x_2 ^2)+\frac{m_3 \dot x_3 ^2}{2}.

As for the motion equations, I get m_1 \ddot x_1 + k_1 (x_1-x_2)=0.
k_1 (x_1-x_2)+k_2 (x_3-x_2)-m_2 \ddot x_2 =0.
And k_2(x_2-x_3)-m_3 \ddot x_3=0. They look too simple in my opinion... What do you think?
 
The equations of motion look okay. They're just what you'd get if you applied F=ma to each mass, right?

Your Lagrangian, as you typed it, is obviously wrong. The units don't work out in the spring terms. You can't combine l1 with x12 for instance.
 
vela said:
The equations of motion look okay. They're just what you'd get if you applied F=ma to each mass, right?
I guess so. But I applied Euler-Lagrange's equations to my wrong Lagrangian to get them.
Your Lagrangian, as you typed it, is obviously wrong. The units don't work out in the spring terms. You can't combine l1 with x12 for instance.
Ah right... Err... But is the following true:
myself said:
Correct me if I'm wrong. I assumed 5 Lagrangians. 3 for the masses and 2 for the springs. The springs only carry a potential energy while the masses only a kinetic energy. Is this right?
.
 
I see my error. My Lagrangian was L=\frac{m_1 \dot x_1 ^2}{2}+\frac{k_1}{2}[l_1- (x_2 -x_1)]^2 + \frac{m_2 \dot x_2 ^2}{2}+\frac{k_2}{2}[l_2 - (x_3-x_2)]^2+\frac{m_3 \dot x_3 ^2}{2}.
It gives me the equations m_1 \ddot x_1 +k_1 [(x_2 - x_1)-l_1]=0.
m_2 \ddot x_2 + k_1 [l_1 +(x_2 - x_1)]-k_2 [l_2 -(x_3-x_2)]=0 and m_3 \ddot x_3 +k_2 [l_2 - (x_3 - x_2)]=0.
Does this looks nicer?
 
  • #10
Looks good.
 
  • #11
vela said:
Looks good.

Ok thank you. :)
 

Similar threads

Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
5
Views
2K
Replies
26
Views
5K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K