Solve Lagrangian Homework Problem: Tricky Motion of Mass C

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The discussion revolves around solving a Lagrangian mechanics problem involving three identical masses connected by springs, with mass A subjected to a driving force. The user attempts to formulate the Lagrangian by expressing the kinetic and potential energy terms but is unsure about the correct coefficients and how to incorporate the driving force into the potential energy. Feedback suggests correcting the kinetic energy coefficient to 0.5 and redefining the potential energy in terms of the relative positions of the masses without introducing an arbitrary equilibrium position. The user expresses confusion about the dynamic nature of equilibrium positions, questioning the necessity of a constant value for J. The conversation highlights the importance of accurately defining energy terms and understanding the effects of external forces in Lagrangian mechanics.
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Homework Statement


A system consists of 3 identical masses (A,B & C) in a line, connected by 2 springs of spring constant k. Motion is restricted to 1 dimension. at t=0 the masses are at rest. Mass A is the subjected to a driving force given by:

F=F0*cos(omega*t)

Calculate the motion of C

Homework Equations


L=T-V, Euler lagrange equation.


The Attempt at a Solution


I figured that the easiest way to do this is to write a lagrangian and solve for the equation of motion for the masses.

I denoted the initial positions of the masses x1, x2 and x3 respectively.
the kinetic energy term for the lagrangian then becomes

T= 1.5*m*((x1dot)^2 + (x2dot)^2 + (x3dot)^2)

For the potential, i have stated that the distance between the masses at equilibrium is equal to J. Hence the potential due to the springs is:

V= k(x1^2 + 2*x2^2 - 3*x2*x1 - x2*x3 + x1*x3 + J(x1+x3-2*x2)

I am unsure if this is correct and i have no idea how to incorporate the driving force into the equation. Any ideas?
 
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Your kinetic term should have a coefficient of 0.5, not 1.5. If we choose to measure each position from its own equilibrium point, then we can set J=0. Then the potential from the springs should be V = 0.5 k [(x1-x2)^2 + (x2-x3)^2]. You can take the force into account by adding -F(t)x1 to V, because a force is minus the gradient of the potential.
 
i was uncertain whether i could ascribe an equilibrium position to each mass as it is determined by its position relative to the adjacent masses rather than a fixed point in space. Hence it is constantly changing. that is why i thought i had to introduce the value J. Is this wrong?
 
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