Lagrangian method for equation of motion

[Ed Quanta]

Assume a masssless pulley and a frictionless surfce constraining two equal masses. Let x be the extension of the spring from mits relaxed length. I have to derive the equations of motion by Lagrangian methods, and solve for x as a function of time with the boundary conditions x=0, x'=0, and t=0. Anyone feel like helping for a smile?

 

[Dr Transport]

Draw a picture and we can hep you out...

 

[M98Ranger]

Sure, I would be happy to help! Before we dive into the Lagrangian method, let's review the setup of the problem. We have two equal masses, let's call them m1 and m2, connected by a massless spring with extension x. The spring is also connected to a frictionless surface and passes over a massless pulley. We want to derive the equations of motion for x as a function of time using the Lagrangian method, with the given boundary conditions.

To start, let's define our variables. We will use x1 and x2 to represent the displacements of m1 and m2, respectively. The total kinetic energy of the system can be written as T = 1/2 m1x1'^2 + 1/2 m2x2'^2, where x1' and x2' are the velocities of m1 and m2, respectively. Similarly, the potential energy of the system can be written as V = 1/2 kx^2, where k is the spring constant.

Now, we can write the Lagrangian L = T - V, which gives us L = 1/2 m1x1'^2 + 1/2 m2x2'^2 - 1/2 kx^2. Using the Euler-Lagrange equation, we can obtain the equations of motion for x1 and x2 as dx1/dt = dL/dx1' and dx2/dt = dL/dx2'. These equations can be simplified to mx1'' = -kx and mx2'' = kx, where x1'' and x2'' are the second derivatives of x1 and x2 with respect to time.

Since m1 and m2 are equal masses, we can combine these equations to get mx'' = -kx, where m is the mass of each mass. This is a second-order differential equation, which can be solved by using the general solution x(t) = A cos(ωt) + B sin(ωt), where A and B are constants and ω is the angular frequency given by ω = √(k/m).

Applying the boundary conditions x=0, x'=0, and t=0, we get A = 0 and B = 0. Therefore, the solution for x(t) is x(t) = 0. This means that the extension