Answer to Question #284315 in Classical Mechanics for Mahii

Question

Two blocks of mass and M connected by a massless spring of spring con stant & are placed on a smooth horizontal table. Determine the equations of

motion using Newtonian mechanics

Answer 1

Answer

The kinetic energy and potential energy from the coupling spring constant k for this system can written as

$T= \frac1{2} m x ˙ 2 + \frac1{2} M y ˙ ​ 2$

And

$V=\frac1{2}k(x−y) ^2$

So lagrangial of this system can witten as

$L=\frac1{2} m x ˙ 2 + \frac1{2} M y ˙ ​ 2-\frac1{2}k(x−y) ^2$

Now calculating as below

$\frac{\partial L}{\partial \dot{x}}=m\dot{x}, \\ \frac{\partial L }{\partial x}=-k(x-y)\\ \frac{∂L}{∂ x} ˙ =m x ˙ ,\\∂L/∂x=−k(x−y)\\ \partial L / \partial \dot{y}=M \dot{y} , \\ \partial L/ \partial y=k(x-y)\\∂L/∂ y ˙ ​ =M$

Now lagrangian equation of motion is given by

$\frac{d}{dt}{\frac{\partial L}{ \partial \dot{x}}}-\frac{\partial L}{\partial x}=0, \\$

$\frac{d}{dt}{\frac{\partial L}{ \partial \dot{y}}}-\frac{\partial L}{\partial y}=0,$

Putting all values then we get

Equation of motion

$m\ddot{x}+k(x-y)=0\\m\ddot{y}+k(y-x)=0$