Problem source

$\,$
\begin{center}
\begin{tikzpicture}[scale=1.0]
    % Define coordinates
    \coordinate (A) at (-2, 3);
    \coordinate (B) at (2, 3);
    \coordinate (C) at (6, 3);
    \coordinate (D) at (2, 0);
    
    % Draw lines
    \draw[line width=0.5pt] (D) -- (C);
    \draw[line width=0.5pt] (A) -- (D);
    \draw[line width=0.5pt] (A) -- (-3, 3);
    \draw[line width=0.5pt] (A) -- (C);
    \draw[line width=0.5pt] (C) -- (7, 3);
    
    % Draw the coil/spring decoration
    \draw[line width=0.5pt, domain=0:360*30, samples=2000, variable=\t] 
        plot ({\t/(360*30)*8 -2 +0.1*sin(\t)}, {3-0.25*cos(\t))});
    
    % Add labels
    \node[above left] at (A) {$A$};
    \node[above] at (B) {$B$};
    \node[above right] at (C) {$C$};
    \node[below right] at (D) {$D$};
    
    % Add the filled dot at D
    \fill (D) circle (1.5pt);
    
\end{tikzpicture}
\end{center}
In the above diagram, $ABC$ represents a light spring of natural length $2l$ and modulus of elasticity $\lambda,$ which is coiled round a smooth fixed horizontal rod. $B$ is the midpoint of $AC.$
The two ends of a light inelastic string of length $2l$ are attached to the spring at $A$ and $C$. A particle of mass $m$ is fixed to the string at $D$, the midpoint of the string. The system can be in equilibrium with the angle $CAD$ equal to $\pi/6.$ Show that
\[
mg=\lambda\left(\frac{2}{\sqrt{3}}-1\right).
\]
Write the length $AC$ as $2xl$, obtain an expression for the potential energy of the system as a function of $x$. 
The particle is held at $B$, and the spring is restored to its natural length $2l.$ The particle is then released and falls vertically.
Obtain an equation satisfied by $x$ when the particle next comes to rest. Verify numerically that a possible solution for $x$ is approximately $0.66.$

Solution source

\begin{center}
\begin{tikzpicture}[scale=1.0]
    % Define coordinates
    \coordinate (A) at (-2, 3);
    \coordinate (B) at (2, 3);
    \coordinate (C) at (6, 3);
    \coordinate (D) at (2, 0);
    
    % Draw lines
    \draw[line width=0.5pt] (D) -- (C);
    \draw[line width=0.5pt] (A) -- (D);
    \draw[line width=0.5pt] (A) -- (-3, 3);
    \draw[line width=0.5pt] (A) -- (C);
    \draw[line width=0.5pt] (C) -- (7, 3);
    

    % Add labels
    \node[above left] at (A) {$A$};
    \node[above] at (B) {$B$};
    \node[above right] at (C) {$C$};
    \node[below right] at (D) {$D$};
    
    % Add the filled dot at D
    \fill (D) circle (1.5pt);

    \draw[-latex, blue, ultra thick] (D) -- ($(A)!0.7!(D)$) node[left] {$T_1$};
    \draw[-latex, blue, ultra thick] (D) -- ($(C)!0.7!(D)$) node[right] {$T_1$};
    \draw[-latex, blue, ultra thick] (D) -- ++(0,-1.5) node[left] {$mg$};

    \draw[-latex, red, ultra thick] (A) -- ($(A)!0.3!(D)$) node[left] {$T_1$};
    \draw[-latex, red, ultra thick] (A) -- ($(A)!-0.3!(B)$) node[left] {$T_2$};
    \draw[-latex, red, ultra thick] (A) -- ++(0,0.7) node[left] {$N$};

    \draw[-latex, red, ultra thick] (C) -- ($(C)!0.3!(D)$) node[left] {$T_1$};
    \draw[-latex, red, ultra thick] (C) -- ($(C)!-0.3!(B)$) node[left] {$T_2$};
    \draw[-latex, red, ultra thick] (C) -- ++(0,0.7) node[left] {$N$};

    \pic [draw, angle radius=0.8cm, "$\pi/6$"] {angle = D--A--C};
        
    
\end{tikzpicture}
\end{center}

$|AB| = l \cos \tfrac{\pi}{6} = \frac{\sqrt{3}}{2}l$ therefore $|AC| = \sqrt{3}l$ and the compression is $(2l - \sqrt{3}l)$ and so $T_2 = \frac{\lambda}{2l} (2l - \sqrt{3}l) = \frac12\lambda(2- \sqrt{3})$

\begin{align*}
\text{N2}(\rightarrow, A): && T_1 \cos \tfrac{\pi}{6} - T_2 &= 0 \\
\Rightarrow && T_1 &= \frac12 \frac{2\lambda(2-\sqrt{3})}{\sqrt{3}} \\
&&&=  \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \\
\\
\text{N2}(\uparrow, D): && 2T_1 \cos \frac{\pi}{3} - mg &= 0 \\
\Rightarrow && mg &= \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right)
\end{align*}

Suppose $|AC| = 2xl$, then:

\begin{array}{c|c} 
\text{energy} & \\ \hline
\text{GPE} & -mg \sqrt{l^2 - x^2l^2} \\
\text{EPE} & \frac12 \frac{\lambda (2l - 2lx)^2}{2l} \\
\text{KE} & \frac12 m v^2
\end{array}

Therefore

\[ E = \frac12 mv^2 + \lambda l (1-x)^2-mgl \sqrt{1-x^2}\]

Initially, $E = 0 + 0 + 0 = 0$. When the particle first comes to rest:

\begin{align*}
\text{COE}: && 0 &= E \\
&&&= \lambda l^2 (1-x)^2 - mgl \sqrt{1-x^2} \\
&&&= \lambda l (1-x)^2 - l \lambda \left ( \frac{2}{\sqrt{3}} - 1 \right) \sqrt{1-x^2}  \\
\Rightarrow && (1-x)^2 &= \sqrt{1-x^2} \left ( \frac{2}{\sqrt{3}} - 1 \right) \\
\Rightarrow  && (1-x)^2(1-x^2)^{-1/2} &= \left ( \frac{2}{\sqrt{3}} - 1 \right) \\
\Rightarrow && (1-2x+x^2)(1+\frac12 x^2+\cdots) &= \left ( \frac{2}{\sqrt{3}} - 1 \right) \\
\end{align*}

If $x = \frac23$ then $(1-x)^2(1-x^2)^{-1/2} = \frac19 \cdot \left ( \frac{5}{9} \right)^{-1/2} = \frac{\sqrt{5}}{15}$
If $2\sqrt{3}-3 \approx \frac{\sqrt{5}}5$ we're done.