Question #156310

Two blocks of mass m1 and m2 connected by a massless spring of spring constant k are placed on a smooth horizontal table. Determine the equations of motion using lagrangian mechanics.

 write the lagrangian for the system.


Expert's answer

The kinetic energy of the system comes from the motion of the blocks and potential energy from the coupling spring. T=1/2mx˙2+1/2My˙2T=1/2m \dot{x}^{2}+1/2M\dot{y}^{2}

V=1/2k(xy)2V=1/2k(x-y)^{2}

L=1/2mx˙2+1/2My˙21/2k(xy)2L= 1/2m\dot{x}^{2}+1/2M\dot{y}^{2}-1/2k(x-y)^{2}

L/x˙=mx˙,L/x=k(xy)\partial L/ \partial \dot{x}=m\dot{x}, \partial L /\partial x=-k(x-y)

L/y˙=My˙,L/y=k(xy)\partial L / \partial \dot{y}=M \dot{y} , \partial L/ \partial y=k(x-y)

Lagrange's equations are written as

d/dt(L/x˙)L/x=0,d/dt(L/y˙)L/y=0d/dt(\partial L/ \partial \dot{x})-\partial L/ \partial x=0, d/dt(\partial L/ \partial \dot{y})-\partial L/ \partial y=0

The equations of motion can then be written as

mx¨+k(xy)=0m\ddot{x}+k(x-y)=0

my¨+k(yx)=0m\ddot{y}+k(y-x)=0


LATEST TUTORIALS
APPROVED BY CLIENTS