Lagrangian of two mass and spring/pulley system

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1. Nov 20, 2016

Elvis 123456789

1. The problem statement, all variables and given/known data
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. Relevant 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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2. Nov 20, 2016

RomegaPRogRess

I see how you coupled them now. Retraction.

3. Nov 20, 2016

Elvis 123456789

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?

4. Nov 20, 2016

RomegaPRogRess

Okay I got it now check mine out and see where they differ

5. Nov 20, 2016

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.

6. Nov 20, 2016

Elvis 123456789

oh yes I see, i made a careless mistake when taking the derivative. Do you have any hints about how to start part c?

7. Nov 20, 2016

RomegaPRogRess

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

8. Nov 20, 2016

haruspex

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.

9. Nov 20, 2016

Elvis 123456789

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?

10. Nov 20, 2016

RomegaPRogRess

Yup. Thank you Haruspex-Sama

11. Nov 20, 2016

Elvis 123456789

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 ?

12. Nov 20, 2016

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?

13. Nov 20, 2016

RomegaPRogRess

Correct.

14. Nov 20, 2016

Elvis 123456789

thank you.

15. Nov 20, 2016

RomegaPRogRess

Anytime and thank you for your work.