Lagrangian Equations for three masses. Two of those being hung using a spring and the third at rest on a horizontal plane

Question

How to find equations of system dynamics using Lagrange’s approach?

I was able to write the Kinetic Energy and I was also able to write the potential energy of the two springs but how do I write the potential energy of the three masses?

Answer 1

Fairly quickly, assuming $x_1,x_2$ are deviation from unstretched positions (but the inextensible string joining ceiling via the pulleys to the horizontal mass is taut).

At $x_1,x_2$, the position of the horizontal mass is $2x_1$ to the left, the $2m$ mass hanging is $x_1$ down and the hanging $m$ mass is $x_2$ down from the unstretched position. Hence \begin{align*} T &=\frac12 m (2\dot x_1)^2 + \frac12 (2m)(\dot x_1)^2 + \frac12 m (\dot x_2)^2\\ U &=\frac12 k (2x_1)^2 + \frac12 k (x_2-x_1)^2 - 2mgx_1 - mgx_2. \end{align*}

You should be able to complete the answer from here.

Answer 2

Calling $x_0$ the coordinate to the horizontal moving mass $m$ we have

$$ T = \frac 12\left(m\dot x_0^2+2m\dot x_1^2+m\dot x_2^2\right) $$

$$ V = \frac k2\left(x_0^2+x_1^2+(x_2-x_1)^2\right)+g\left(2m x_1+m x_2\right) $$

now we have to consider the kinematic restriction betwee $x_0$ and $x_1$ which is

$$ \dot x_0 = 2\dot x_1\Rightarrow x_0 = 2x_1 + C $$

now assuming $C=0$ and substituting we have finally

$$ T = \frac 12\left(4m\dot x_1^2+2m\dot x_1^2+m\dot x_2^2\right) = \frac 12\left(6m\dot x_1^2+m\dot x_2^2\right) $$

and

$$ V = \frac k2\left(4x_1^2+x_1^2+(x_2-x_1)^2\right)+g\left(2m x_1+m x_2\right) = \frac k2\left(5x_1^2+(x_2-x_1)^2\right)+g\left(2m x_1+m x_2\right) $$