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.


1
Expert's answer
2021-01-19T07:09:19-0500

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


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