Lagrangian of two mass and spring/pulley system

Elvis 123456789

1. 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.

2. Homework Equations
L = T - V

L = ∂L/∂x - d/dt(∂L/∂x-dot) = 0

3. The Attempt at a Solution
I attempted part a and b, and im not really sure if what i got so far is correct. I also dont know how to start for part c. My work is in the attachment

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RomegaPRogRess

I see how you coupled them now. Retraction.

Elvis 123456789

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?

RomegaPRogRess

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

RomegaPRogRess

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.

Elvis 123456789

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?

RomegaPRogRess

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

haruspex

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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.

Elvis 123456789

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 doesnt really affect the answer since it vanishes with the derivative correct?

RomegaPRogRess

Yes i caught my error in taking the derivative. Just to be clear, introducing the "l" for the length of the rope doesnt really affect the answer since it vanishes with the derivative correct?
Yup. Thank you Haruspex-Sama

Elvis 123456789

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 ?

Elvis 123456789

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.