# Lagrangian of two mass and spring/pulley system

• Elvis 123456789
In summary, the homework statement is two blocks of equal mass, connected by a light string, hanging below a massless pulley. One block sits on a frictionless horizontal table and is attached to a spring of constant k while the other block hangs below the pulley. The equation of motion is xp = A where A is a constant. A particular solution of the equation of motion is xp = A(1 - cos(wt)) and is added to the solution for the homogeneous equation to show that for the initial condition x = x-dot = 0 at t = 0, the full solution is x = A(1 - cos(wt)) and w is determined.
Elvis 123456789

## Homework Statement

Two blocks of equal mass, m, are connected by a light string that passes over a massless pulley. One block hangs below the pulley, while the other sits on a frictionless horizontal table and is attached to a spring of constant k. Let x=0 be the equilibrium position of the block on the table.

a.) Determine the lagrangian of the system.

b.) Determine the equation of motion.

c.) Show that a particular solution of the equation of motion is xp = A. where A is a constant, and determine A. Add this solution to the solution for the homogeneous equation and show that for the initial condition x = x-dot = 0 at t = 0, the full solution is x = A(1 - cos(wt)) and determine w. Recall that for an inhomogeneous differential equation that the solution is the sum of the homogeneous solution, xh, and the particular solution.

## Homework Equations

L = T - V

L = ∂L/∂x - d/dt(∂L/∂x-dot) = 0

## The Attempt at a Solution

I attempted part a and b, and I am not really sure if what i got so far is correct. I also don't know how to start for part c. My work is in the attachment

#### Attachments

• CM HW.jpg
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I see how you coupled them now. Retraction.

RomegaPRogRess said:
Your partial derivative, for one, is off. You should have two separate equations of motion for each mass. Secondly, you have to figure out how the second block is affected by the spring force. Obviously it oscillates, as well as the first mass, because they are connected and to a spring no less.
I used the fact that the string's length is a constant in order to get one generalized coordinate. Is that not correct to do?

Elvis 123456789 said:
I used the fact that the string's length is a constant in order to get one generalized coordinate. Is that not correct to do?
Okay I got it now check mine out and see where they differ

You were close, I just haven't seen anyone use that constraint on the length of the string opposed to tension but then again you're using the Lagrangian method.

RomegaPRogRess said:
You were close, I just haven't seen anyone use that constraint on the length of the string opposed to tension but then again you're using the Lagrangian method.
oh yes I see, i made a careless mistake when taking the derivative. Do you have any hints about how to start part c?

Yup. Attempt it first though. I'm assuming you haven't or are rusty at solving differential equations.

You can avoid the mgl term by defining x2 suitably.
Check your treatment of the ##\frac d{dt}\frac{\partial\mathcal{L}}{\partial\dot x}## term.

haruspex said:
You can avoid the mgl term by defining x2 suitably.
Check your treatment of the ##\frac d{dt}\frac{\partial\mathcal{L}}{\partial\dot x}## term.
Yes i caught my error in taking the derivative. Just to be clear, introducing the "l" for the length of the rope doesn't really affect the answer since it vanishes with the derivative correct?

Elvis 123456789 said:
Yes i caught my error in taking the derivative. Just to be clear, introducing the "l" for the length of the rope doesn't really affect the answer since it vanishes with the derivative correct?
Yup. Thank you Haruspex-Sama

RomegaPRogRess said:
Yup. Thank you Haruspex-Sama
Ok so the differential equation is d/dt2(x) + (k/2m)x +g/2 = 0

so for the first part of part c, i just plug in A, and A will be a particular solution if it is A = -mg/k ?

ok so rewriting the differential equation as
d/dt2(x) + (k/2m)x= -g/2 plugging in A for the particular solution gives A = -mg/k

the homogeneous solution will look like

xh = Bcos(wt + φ)

so the general solution is

x = Bcos(wt + φ) + A

the velocity will be

x-dot = -B*w*sin(wt + φ)

for t = 0

Bcos(φ) + A = 0 and -B*w*sin(φ) = 0 for this to happen φ must be zero

then B = -A

that means that the solution is x = A[ 1 - cos(wt) ] and w2 = k/2m

does this look correct?

Correct.

RomegaPRogRess said:
Correct.
thank you.

Elvis 123456789 said:
thank you.
Anytime and thank you for your work.

## 1. What is the Lagrangian of a two mass and spring/pulley system?

The Lagrangian of a two mass and spring/pulley system is a mathematical function that describes the system's energy and motion. It takes into account the kinetic and potential energies of both masses and the spring/pulley system.

## 2. How is the Lagrangian of a two mass and spring/pulley system calculated?

The Lagrangian is calculated by taking the difference between the kinetic energy and potential energy of the system. For a two mass and spring/pulley system, it is given by L = T - V, where T is the total kinetic energy and V is the total potential energy.

## 3. What is the significance of the Lagrangian in a two mass and spring/pulley system?

The Lagrangian is significant because it allows us to describe the system's motion and energy in a simple and elegant way. It is also used to derive the equations of motion for the system, making it a useful tool in analyzing and predicting the system's behavior.

## 4. Can the Lagrangian be used to study other physical systems?

Yes, the Lagrangian can be used to study a wide range of physical systems, including mechanical, electrical, and even quantum systems. It is a powerful tool in theoretical physics and has many applications in various fields of study.

## 5. How does the Lagrangian approach differ from other methods of studying mechanical systems?

The Lagrangian approach differs from other methods, such as Newton's laws or the Euler-Lagrange equations, in that it takes into account the entire system's energy and motion, rather than just the forces acting on individual objects. It also allows for a more intuitive and elegant understanding of the system's dynamics.

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