Answer:
The equations are as below:
[tex]m\ddot{\theta}(l-l_t+x)+2m\dot{\theta}\dot{x}=mgsin\theta\\\ddot{x}=\dfrac{m\dot{\theta}}{m+M}(l-l_t+x)+\dfrac{k}{m+M}(l_o-x)-\dfrac{mg}{m+M}cos\theta[/tex]
Explanation:
As the complete question is not given, the complete question is attached herewith.
From the given data
The mass of the block is M
The length of the spring at any given time is l
The length of the spring initially is lo
The length of the platform is lr
So the coordinates in y and z direction are given as
[tex]y=(l-(l_t-x))cos\theta[/tex]
[tex]\dot{y}=-\dot{\theta}(l-l_t+x)sin\theta+\dot{x}cos\theta[/tex]
[tex]z=(l-(l_t-x))sin\theta[/tex]
[tex]\dot{z}=\dot{\theta}(l-l_t+x)cos\theta+\dot{x}sin\theta[/tex]
Now the Lagrange is given as
[tex]L=\dfrac{1}{2}m\dot{y}^2+\dfrac{1}{2}m\dot{z}^2+\dfrac{1}{2}M\dot{x}^2-mg(l-(l_t-x))cos\theta-\dfrac{1}{2}k(l_o-x)^2[/tex]
[tex]L=\dfrac{1}{2}m(-\dot{\theta}(l-l_t+x)sin\theta+\dot{x}cos\theta)^2+\dfrac{1}{2}m(\dot{\theta}(l-l_t+x)cos\theta+\dot{x}sin\theta)^2+\dfrac{1}{2}M\dot{x}^2-mg(l-(l_t-x))cos\theta-\dfrac{1}{2}k(l_o-x)^2\\L=\dfrac{1}{2}m(\dot{x}^2+\dot{\theta}^2(l-l_t+x)^2)+\dfrac{1}{2}M\dot{x}^2-mg(l-l_t+x)cos\theta-\dfrac{1}{2}k(l_o-x)^2\\\\[/tex]
Now The value is given as
[tex]\dfrac{dL}{d\dot{\theta}}=m\dot{\theta}(l-l_t+x)^2\\\dfrac{dL}{d\theta}=mg(l-l_t+x)sin\theta[/tex]
So the 1st equation becomes
[tex]\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{\theta}}\right)=\dfrac{\partial L}{\partial \ddot{\theta}}\\\dfrac{d}{dt}\left(m\dot{\theta}(l-l_t+x)^2\right)=mg(l-l_t+x)sin\theta\\m\ddot{\theta}(l-l_t+x)+2m\dot{\theta}\dot{x}=mgsin\theta[/tex]
Now the second equation is as
[tex]\dfrac{d}{dt}\left(\dfrac{\partial L}{\partial \dot{x}}\right)=\dfrac{\partial L}{\partial \ddot{x}}\\\ddot{x}=\dfrac{m\dot{\theta}}{m+M}(l-l_t+x)+\dfrac{k}{m+M}(l_o-x)-\dfrac{mg}{m+M}cos\theta[/tex]
