- #1

DanielA

- 27

- 2

(note: I'm going to represent the lagrangian as simply L because I don't know how to do script L in latex.)

Two particles of equal masses m are confined to move along the x-axis and are connected by a spring with potential energy ##U = \frac{1}/{2}kx^2## (here x is the extension of the spring, ##x = (x_1-x_2-l)## where l is the unstretched length of the spring. Mass 1 remains to the right of mass 2 at all times

a) Write the lagrangian for the system in terms of the positions and velocities of the individual particles.

b)Rewrite L in terms of the new variables X = ##\frac{1}{2}(x_1 + x_2)## (the center of mass position) and x, the extension of the spring from before.

c) Solve for X(t) and x(t) and describe the motion in terms of these variables

L = T - U

Lagrange Equations:

##\frac{\partial L}{\partial X} = \frac{d}{dt}(\frac{\partial L}{\partial \dot X})##

and

##\frac{\partial L}{\partial x} = \frac{d}{dt}(\frac{\partial L}{\partial \dot x})##

I think I messed up my Lagrangian.

I said T = ##\frac{1}{2}M\dot X^2## because I can't really imagine how a variable for the stretching of a spring would have kinetic energy. After all, no mass is associated with that. Maybe I could represent KE with an "initial potential" variable ##U_i## and say ##T = U_i - U_{spring}##, but that doesn't really change anything since I would still not have a ##\dot x## dependence.

U = ##\frac{1}{2}kx^2## because the CM has no concern for the internal configuration so the CM has no potential energy term.

L = T-U = ##\frac{1}{2}M\dot X^2 - \frac{1}{2}kx^2##

Doing the Legrange equation for the X variable, I get, with M being the total mass of the system, ##M\dot x = constant## or ##x = \frac{C}{M}t## This result simply means that the center of mass will move in the x direction at a constant velocity "C/M" where C is a unknown constant. This is expected and shows that the velocity of the CM is unaffected by the orientation of the masses.

Doing the same for the x variable, I get

-kx = 0

or

x = 0.

This result surprised me, and makes me think I created the lagrangian incorrectly because the masses should be able to oscillate based off of initial conditions. After all, if it starts compressed or stretched, then it'll just oscillate back and forth and the x variable should represent that.

## Homework Statement

Two particles of equal masses m are confined to move along the x-axis and are connected by a spring with potential energy ##U = \frac{1}/{2}kx^2## (here x is the extension of the spring, ##x = (x_1-x_2-l)## where l is the unstretched length of the spring. Mass 1 remains to the right of mass 2 at all times

a) Write the lagrangian for the system in terms of the positions and velocities of the individual particles.

b)Rewrite L in terms of the new variables X = ##\frac{1}{2}(x_1 + x_2)## (the center of mass position) and x, the extension of the spring from before.

c) Solve for X(t) and x(t) and describe the motion in terms of these variables

## Homework Equations

L = T - U

Lagrange Equations:

##\frac{\partial L}{\partial X} = \frac{d}{dt}(\frac{\partial L}{\partial \dot X})##

and

##\frac{\partial L}{\partial x} = \frac{d}{dt}(\frac{\partial L}{\partial \dot x})##

## The Attempt at a Solution

I think I messed up my Lagrangian.

I said T = ##\frac{1}{2}M\dot X^2## because I can't really imagine how a variable for the stretching of a spring would have kinetic energy. After all, no mass is associated with that. Maybe I could represent KE with an "initial potential" variable ##U_i## and say ##T = U_i - U_{spring}##, but that doesn't really change anything since I would still not have a ##\dot x## dependence.

U = ##\frac{1}{2}kx^2## because the CM has no concern for the internal configuration so the CM has no potential energy term.

L = T-U = ##\frac{1}{2}M\dot X^2 - \frac{1}{2}kx^2##

Doing the Legrange equation for the X variable, I get, with M being the total mass of the system, ##M\dot x = constant## or ##x = \frac{C}{M}t## This result simply means that the center of mass will move in the x direction at a constant velocity "C/M" where C is a unknown constant. This is expected and shows that the velocity of the CM is unaffected by the orientation of the masses.

Doing the same for the x variable, I get

-kx = 0

or

x = 0.

This result surprised me, and makes me think I created the lagrangian incorrectly because the masses should be able to oscillate based off of initial conditions. After all, if it starts compressed or stretched, then it'll just oscillate back and forth and the x variable should represent that.

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