Problem source

A thin uniform elastic band of mass $m,$ length $l$ and modulus of elasticity $\lambda$ is pushed on to a smooth circular cone of vertex angle $2\alpha,$ in such a way that all elements of the band are the same distance from the vertex. It is then released from rest. Let $x(t)$ be the length of the band at time $t$ after release, and let $t_{0}$ be the time at which the band becomes slack. 
Assuming that a small element of the band which subtends an angle $\delta\theta$ at the axis of the cone experiences a force, due to the tension $T$ in the band, of magnitude $T\delta\theta$ directed
towards the axis, and ignoring the effects of gravity, show that 
\[
\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha=0,\qquad(0< t< t_{0}).
\]
Find the value of $t_{0}.$ 

Solution source

\begin{center}
    \begin{tikzpicture}

        \def\l{5};
        \def\h{3};
        \def\m{2};
        \def\d{0.2};
        
        \coordinate (O) at (0, \l);

        \coordinate (A) at ({\h}, 0);
        \coordinate (B) at ({-\h}, 0);
        \coordinate (D) at (0, 0);

        \coordinate (Ba) at ({(\l-\m+\d/2)/(\l)*\h}, {\m-\d/2}); 
        \coordinate (Bb) at ({-(\l-\m+\d/2)/(\l)*\h}, {\m-\d/2}); 

        \draw (A) -- (O) -- (B);

        \draw[<->] ({(\l-\m+\d/2)/(\l)*\h+0.5}, {\l}) --  ({(\l-\m+\d/2)/(\l)*\h+0.5}, {\m-\d/2});

        \node at ({(\l-\m+\d/2)/(\l)*\h+0.5}, {0.5*(\l+\m-\d/2)}) [right] {$h$};
        
        \pic [draw, angle radius=0.8cm, "$\alpha$"] {angle = D--O--A};

        \draw[dashed] (O) -- (D);
    
        \draw[domain = -90:90, samples=180, variable = \x]  plot ({sin(\x)*(\l-\m)/(\l)*\h}, {\m+cos(\x)*0.15} );
        \draw[domain = 270:90, samples=180, dashed, variable = \x]  plot ( {sin(\x)*(\l-\m)/(\l)*\h}, {\m+cos(\x)*0.15});

        \draw[domain = -90:90, samples=180, variable = \x]  plot ({sin(\x)*(\l-\m+\d)/(\l)*\h},{\m-\d+cos(\x)*0.15});
        \draw[domain = 270:90, samples=180, dashed, variable = \x]  plot ( {sin(\x)*(\l-\m+\d)/(\l)*\h},{\m-\d+cos(\x)*0.15});

        \draw[-latex, blue, ultra thick] (Ba) -- ++({\l/10}, {\h/10}) node[right] {$R$};    
        \draw[-latex, blue, ultra thick] (Ba) -- ++({-0.7}, {0}) node[left] {$T\delta \theta$};    
        
    \end{tikzpicture}
\end{center}

\begin{align*}
\text{N2}(\nwarrow): && T\delta \theta \sin \alpha &= -m\frac{\delta \theta}{2\pi} \ddot{d}
\end{align*}
Notice that $r = d \sin \alpha$ and $x = 2 \pi r$, so $x = 2\pi d \sin \alpha$ and $\ddot{x} = 2\pi \sin \alpha \ddot{d} \Rightarrow \ddot{d} = \ddot{x} \frac{1}{2 \pi \sin \alpha}$

Notice also that $T = \frac{\lambda}{l}(x-l)$ so.

\begin{align*}
&& \frac{m}{4 \pi^2 \sin\alpha} \ddot{x} &= -\frac{\lambda}{l}(x-l)  \sin\alpha \\
\Rightarrow && \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha&=0
\end{align*}

The solution to the differential equation we have is:

\begin{align*}
&& x(t) &= A \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + B \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + l \\
&&&= A \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) +B \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) + l\\
&& \dot{x}(0) =  0 \\
\Rightarrow && B &= 0 \\
&& x(t) &= (x(0)-l) \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right)  + l \\
&& x(t_0) &= l \\
\Rightarrow && t_0 &= \frac{1}{4\sin \alpha} \sqrt{\frac{ml}{\lambda}}
\end{align*}