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## Homework Statement

No friction. m

_{3}=5m, m

_{2}=12m, m

_{1}=3m

Prove that the system's acceleration when it's released is a=0.2g. is it the max acceleration?

Show that the equilibrium point is at x

_{0}=0.75L

Show that the kinetic energy of the system is E=mgL

Show that m

_{2}will stop momentarily at ##x_1=\frac{15}{8}L##

## Homework Equations

Harmonic motion, the force is proportional to the displacement: F=kx

$$\ddot x=\omega^2x=0$$

## The Attempt at a Solution

The initial state:

$$\left\{ \begin{array}{l} m_1g-T=(m_1+m_2)a \\ T=m_2a \end{array}\right.\;\rightarrow a=\frac{3}{27}g=0.11g$$

It's wrong.

In the intermediate state:

_{2}accelerates distance x but the rope lengthens by:

$$l_1^2=x^2+\frac{L^2}{\sin^2\alpha}-2x \frac{L}{\sin\alpha}\cos\alpha$$

$$\Delta l=\frac{L}{\sin\alpha}-\sqrt{x^2-\frac{2L}{\tan\alpha}x+\frac{L^2}{\sin^2\alpha}}$$

$$\left\{ \begin{array}{l} m_1g-T_1=m_1\ddot x \\ T_1-T_3\cos\alpha=m_2\ddot\Delta l \\ T_3-m_3g=m_3\ddot\Delta l \end{array} \right.$$

It's complicated, Δl and it's derivative and i am afraid it won't get to a second order differential equation.