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Lagrangian Mechanics for two springs (revisited)

by annamz
Tags: lagrangian, vertical spring
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annamz
#1
Feb22-12, 03:44 PM
P: 3
1. The problem statement, all variables and given/known data
Essentially the problem that I am trying to solve is the same as in this topic except that it is for 3 springs and 3 masses
http://www.physicsforums.com/showthread.php?t=299905


2. Relevant equations
I have found similar equations as in the topic but I face a problem in describing the potential energy due to gravity for the system. The euler-lagrange equations aren't supposed to have any constants but due to the PE I find that I have constants in my euler-lagrange equations.

How do you define the gravitational potential energy for a vertical two spring system?

Thank you all in advance
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fluidistic
#2
Feb22-12, 04:31 PM
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Quote Quote by annamz View Post
1. The problem statement, all variables and given/known data
Essentially the problem that I am trying to solve is the same as in this topic except that it is for 3 springs and 3 masses
http://www.physicsforums.com/showthread.php?t=299905


2. Relevant equations
I have found similar equations as in the topic but I face a problem in describing the potential energy due to gravity for the system. The euler-lagrange equations aren't supposed to have any constants but due to the PE I find that I have constants in my euler-lagrange equations.

How do you define the gravitational potential energy for a vertical two spring system?

Thank you all in advance
Setting the zero of potential energy can be made wherever you want. Personally I'd set it at the top of the system. If I label the x-axis as going along with the springs and pointing downward for increasing x, the potential energy of the upper mass would be [itex]-mgd[/itex] where d is the distance between the top of the system and the first mass. The important thing to keep in mind is "the more you go downward, the less gravitational potential energy there is", I think.
annamz
#3
Feb22-12, 04:36 PM
P: 3
Quote Quote by fluidistic View Post
Setting the zero of potential energy can be made wherever you want. Personally I'd set it at the top of the system. If I label the x-axis as going along with the springs and pointing downward for increasing x, the potential energy of the upper mass would be [itex]-mgd[/itex] where d is the distance between the top of the system and the first mass. The important thing to keep in mind is "the more you go downward, the less gravitational potential energy there is", I think.
Thanks for your reply, that's what I tried to do but I keep getting mixed up between the displacements for example what I found was this:

mg(l+x_1 )+mg(2l+x_2 )+mg(3l+x_3 )

where x_1 is the closest to the ceiling mass and l is the length of each spring at rest. We consider all springs to have the same length. Somehow I feel that the potential I found is not quite right

fluidistic
#4
Feb22-12, 04:59 PM
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Lagrangian Mechanics for two springs (revisited)

Quote Quote by annamz View Post
Thanks for your reply, that's what I tried to do but I keep getting mixed up between the displacements for example what I found was this:

mg(l+x_1 )+mg(2l+x_2 )+mg(3l+x_3 )

where x_1 is the closest to the ceiling mass and l is the length of each spring at rest. We consider all springs to have the same length. Somehow I feel that the potential I found is not quite right
I am not sure but maybe [itex]-[mg(l+x_1)+mg(l+x_1+l+x_2)+mg(l+x_1+l+x_2+l+x_3)]=-mg (6l+3x_1+2x_2+x_3)[/itex] is more appropriated.
Notice that in your Lagrangian must also appear the potential energy of each spring (that you must express in terms of your generalized coordinates x_1, x_2 and x_3).
fluidistic
#5
Feb22-12, 06:59 PM
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Hey guys, so I'm stealing this problem because I study the same course. I put my work into a spoiler so that the OP does not cheat if he doesn't want to. But I need help.
Spoiler
My Lagrangian is [itex]\frac{m}{2}(\dot x_1 ^2 + \dot x_2 ^2 + \dot x _3 ^2)+\frac{k}{2}[2x_2^2+x_1^2+x_3^2-2(x_1x_2+x_2x_3)][/itex].
Using Euler-Lagrange equations, I reach the equations of motion:
(1)[itex]m\ddot x_1 +k (x_2-x_1)=0[/itex].
(2)[itex]m\ddot x_3 +k (x_2-x_3)=0[/itex].
(3)[itex]m\ddot x_2+k (x_3+x_1-2x_2)=0[/itex].
At first glance they look reasonable in my opinion. Now I must find the normal modes and frequencies (and the equilibrium positions too). I do not know how to do any of these, which I know is very important.
Hmm to get the equilibrium position I guess I must solve the system of DE's and find the stationary solutions, but this doesn't look like a piece of cake. No idea about the others. I'd appreciate any help.
annamz
#6
Feb23-12, 10:58 AM
P: 3
Can I ask you something?I find the same Lagrangian but the only difference is that the coefficient of x_1^2 is 3 and instead of the product x_1x_2 I found the product x_3x_1
fluidistic
#7
Feb23-12, 12:53 PM
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Quote Quote by annamz View Post
Can I ask you something?I find the same Lagrangian but the only difference is that the coefficient of x_1^2 is 3 and instead of the product x_1x_2 I found the product x_3x_1
Good. This mean we need another person to correct either both of us or one of us :)


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