Lagrange method problem: Multiple Spring-Mass System

[Motorbiker]

Homework Statement

Derive the equations of motion using lagrange method

Relevant Equations

$V_sys=V_(g,sys) +V_(e,sys)$

Lagrangian$L=T_(sys)-V_(sys)$

4

I am working on problem c and I'm not sure if I'm doing it right, please can you help me understand if I am on the right lines? I want to get a better understanding of lagrange method problems
Here is my working:

I have labelled $k_1,k_2,k_3,k_4, k_5$ left to right

Generalised coordinates:
$q_1=x1$
$q_2=x_2$
$q_3=x_3$
Generalised forces:

##Q_1= F_1cos \omega t##

##Q_2F_2 = cos \omega t##

$Q_3 = F_3cos \omega t$

Energies
$$V_e,sys=0.5(k_1x_1^2) + 0.5k_2(x_2-x_1)^2 + 0.5k_3(x_1-x_3)^2 +0.5k_4(-x_3)^2 +0.5k_5(x_3)^2$$

Then you do Lagrangian $$L= T_(sys)-V_(sys)$$

 

[anuttarasammyak]

I have labelled k1,k2,k3,k4,k5 left to right

The figure says k1=k2=k4=k5=k. k3=5k.

Generalised coordinates:

We need wall to wall distance, say L and natural length of the springs which are L/4 and L/2 for thin m bodies though they are not clearly shown in the problem statement.

 

[Motorbiker]

The figure says k1=k2=k4=k5=k. k3=5k.


We need wall to wall distance, say L and natural length of the springs which are L/4 and L/2 for thin m bodies though they are not clearly shown in the problem statement.

We don't need values for the distance or length of the spring to determine the equation of motion.

 

[berkeman]

The figure you posted shows 4 springs with value k and one spring with value 5k.

 

[anuttarasammyak]

We don't need values for the distance or length of the spring to determine the equation of motion.

Your k1 and k5 springs are fixed on the walls. Coordinates of the walls are necessary to calculate their force and potential energy. Say they are 0 and L,
0<x1<x2<x3<L.

Say L-x3-natural length of k5 spring is positive, pull , and is negative, push.

 

[Motorbiker]

Your k1 and k5 springs are fixed on the walls. Coordinates of the walls are necessary to calculate their force and potential energy. Say they are 0 and L,
0<x1<x2<x3<L.

Say L-x3-natural length of k5 spring is positive, pull , and is negative, push.

Okay thank you, how would this impact upon the lagrange equations, please could you show me?

 

[berkeman]

Okay thank you, how would this impact upon the lagrange equations, please could you show me?

Since this is a schoolwork-type thread, you need to do your best to show us your efforts on this. Can you show us your attempt at writing the Lagrangian for this setup based on the hints given so far?

 

[Motorbiker]

The figure you posted shows 4 springs with value k and one spring with value 5k.

Okay so when doing calculations $k_3$ would just be written as $5*k_3$?

Also can you please help correct the line in the OP for the generalised forces? I am new to latex and struggled to fix the line.

 

[Motorbiker]

Since this is a schoolwork-type thread, you need to do your best to show us your efforts on this. Can you show us your attempt at writing the Lagrangian for this setup based on the hints given so far?

I have attached my work on this problem so far.

Let me know if it looks correct. @anuttarasammyak

 

[anuttarasammyak]

Have you incorporated my comment post #2 in your note ?

 

[Motorbiker]

Have you incorporated my comment post #2 in your note ?

Yes I have incorporated your comment. What am I doing wrong?

 

[anuttarasammyak]

Thanks. Be careful of k's. Let me know how you define your x1,x2,x3 in your Lagrangian. From where they are measured ? After correction of k's your Lagrangian will be all right with that definition.

 

[Motorbiker]

Thanks. Be careful of k's. Let me know how you define your x1,x2,x3 in your Lagrangian. From where they are measured ? After correction of k's your Lagrangian will be all right with that definition.

I'm not sure how to define the x1,x2,x3, usually the x is placed above the respective rigid body-so x1 on top of m1 and so on.And the arrow would be pointing to the right matching the arrow on the force.

So would this method be suitable in this problem?I understand that all k's except for k3 are equal however they are still denoted with the relevant subscript right?

 

[anuttarasammyak]

So would this method be suitable in this problem?I understand that all k's except for k3 are equal however they are still denoted with the relevant subscript right?

Why don't you use values of spring constants instead of their names?

I'm not sure how to define the x1,x2,x3, usually the x is placed above the respective rigid body-so x1 on top of m1 and so on.And the arrow would be pointing to the right matching the arrow on the force.

Tell me your guess, when the system is in equillibrium, no vibration, how much are $x_1,x_2,x_3$ ?

 

[Motorbiker]

Why don't you use values of spring constants instead of their names?

View attachment 347170

Tell me your guess, when the system is in equillibrium, no vibration, how much are $x_1,x_2,x_3$ ?

I tend to use their names because that's way it's done in my notes however I could just use the values of the spring constants if that makes the problem simpler or clearer.

When the system is in equilibrium there are 3 x's for 3 rigid bodies? The x is simply a frame of reference for the body right?

I understand that the force and rotation must be zero at equilibrium.

 

[anuttarasammyak]

In short, in equilibrium how much are $x1,x2,x3$ , you guess? You have got formula of potential energy. What $x1,x2,x3$ give zero potential energy?

 

[Motorbiker]

In short, in equilibrium how much are $x1,x2,x3$ , you guess? You have got formula of potential energy. What $x1,x2,x3$ give zero potential energy?

I think zero x1,x2 and x3 give zero potential energy.

 

[anuttarasammyak]

I think zero x1,x2 and x3 give zero potential energy.

Right. x1 =0 corresponds to the case that m1 body is in its equilibrium position. x2=0 corresponds to the case that m2 body is in its equilibrium position.x3=0 corresponds to the case that m3 body is in its equilibrium position.
In other words, x1 is displacement of m1 body from its equilibrium position. x2 is displacement of m2 body from its equilibrium position. x3 is displacement of m3 body from its equilibrium position.

 

[Motorbiker]

Right. x1 =0 corresponds to the case that m1 body is in its equilibrium position. x2=0 corresponds to the case that m2 body is in its equilibrium position.x3=0 corresponds to the case that m3 body is in its equilibrium position.
In other words, x1 is displacement of m1 body from its equilibrium position. x2 is displacement of m2 body from its equilibrium position. x3 is displacement of m3 body from its equilibrium position.

Okay thanks a lot for that, I understand it better and it looks like my labelling for the x's is correct for this case.

 

[Motorbiker]

Right. x1 =0 corresponds to the case that m1 body is in its equilibrium position. x2=0 corresponds to the case that m2 body is in its equilibrium position.x3=0 corresponds to the case that m3 body is in its equilibrium position.
In other words, x1 is displacement of m1 body from its equilibrium position. x2 is displacement of m2 body from its equilibrium position. x3 is displacement of m3 body from its equilibrium position.

I’ve attached my finished solution to this problem, can you just check that it looks okay? I just want to make sure I have a good understanding.

 

[anuttarasammyak]

I have no idea of your
$$F_1 \cos,\ F_2 \cos,\ F_3\ \cos$$
Why you introduce these terms added to zeros of Lagrange equation ? You can check your result by yourself whether you get usual equation of motion F=ma where F comes from only spring, I assume.

 

[Motorbiker]

I have no idea of your
$$F_1 \cos,\ F_2 \cos,\ F_3\ \cos$$
Why you introduce these terms added to zeros of Lagrange equation ? You can check your result by yourself whether you get usual equation of motion F=ma where F comes from only spring, I assume.

I added them because there are three forces acting on the system, F1, F2 and F3 right?

 

[anuttarasammyak]

I added them because there are three forces acting on the system, F1, F2 and F3 right?

When you are dealing with not free oscillation but forced oscillation by time dependent forces, you have to add a new term to potential energy in Lagrangian,
$$U=-x_1F_1(t)-x_2F_2(t)-x_3F_3(t)$$.
by which you do not have to think about other things than Lagrangian.