Shankar exercise-2.1.2

Q.

Write the Lagrangian and get the equation of motion for the coupled-mass problem.

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Trial.

coupled-mass-spring-system

Lagrangian

$L=T-V$

$T={1 \over 2}m\dot{x}_1^2+{1 \over 2}m\dot{x}_2^2$

$V={1 \over 2}k\left[x_1^2+(x_2-x_1)^2+x_2^2\right]$

$L={1 \over 2}m\dot{x}_1^2+{1 \over 2}m\dot{x}_2^2-{1 \over 2}k\left[x_1^2+(x_1-x_2)^2+x_2^2\right]\tag{1}$
We have two coordinate systems which origins are $x_1=0$ and $x_2=0$ respectively.

Lagrange’s equation for $x_1, \ \dot{x_1}$

${\partial L \over \partial x_1}=-k\left[x_1+(x_1-x_2)\right]=-k\left[ 2x_1-x_2\right]$

${\partial L \over \partial \dot{x_1}}=m\dot{x_1}$

${d \over dt}\left( m\dot{x_1}\right)=-k\left[x_1+(x_1-x_2)\right]=-k\left[ 2x_1-x_2\right]$
Therefore, Equation of motion is

$m\ddot{x_1}=-k\left[2x_1-x_2\right]$

Lagrange’s equation for $x_2, \ \dot{x_2}$

${\partial L \over \partial x_2}=-k\left[x_2-(x_1-x_2)\right]=-k\left[2x_2-x_1\right]$

${\partial L \over \partial \dot{x_2}}=m\dot{x_2}$

${d \over dt}\left( m\dot{x_2}\right)=-k\left[2x_2-x_1\right]$
Therefore, Equation of motion is

$m\ddot{x_2}=-k\left[2x_2-x_1\right]$