The two blocks have the same mass $\backslash ($ m $\backslash )$ and are connected wia a risid masles fleng As a result of the gravitational acceleration $\backslash ($ g $\backslash ),$ block $1$ moves horizontally and block $2$ can only move vertically. There is no friction in this system. Moreover, block $1$ is connected to a wall via a linear spring that has a spring constant $\backslash ($ k $\backslash )$ and a negligible free length. Therefore, the elongation of the spring is the position $\backslash ($ x $\backslash )$ of block $1$ from the wall. For block $2,$ its horizontal distance to the wall is $\backslash ($ l $\backslash )$ and its vertical position is $\backslash ($ y $\backslash )$ as shown in Fig. $3.$ Use Newtonian mechanics to answer the following questions.

$($a$)$ Draw a free$-$body diagram of the two blocks.

$($b$)$ Apply Newton's second law to derive the equations of motion of the two blocks. Eliminate constraint force$($s$)$ from your equations of motion to obtain a nonlinear, differential equation governing only the variable $\backslash (\backslash$theta$($t$)\backslash ),$ where $\backslash (\backslash$theta $\backslash )$ is the angle between the rigid rod and the vertical as shown in Fig. $3.$

$($c$)$ Determine an algebraic equation governing equilibrium positions $\backslash (\backslash$theta$\_\{0\}\backslash )$ of the system. The equation should involve parameters such as $\backslash ($ m g $\backslash )$ and $\backslash ($ k l $\backslash ).$ Show that there is only one possible equilibrium for $\backslash (0<mi>\backslash </mi>$theta$\_\{0\}<mi>\backslash </mi>$frac$\{\backslash$pi$\}\{2\}\backslash ).$

$($d$)$ Derive a linearized equation of motion around the equilibrium position. If the two$-$block system is subjected to disturbance, will the system oscillate around the equilibrium position? Why?

Figure $3$: $\backslash (\backslash$Lambda $\backslash )$ two$-$block system with a linear spring and a rigid rod