= Solution
Let $x$ be the downward displacement of $M$ from its initial position just beyond the table edge. The smaller <mass> moves towards the edge by the same distance, and the hanging chain length is $x$. All the chain and both endpoint masses move at speed $\dot x$, so their total moving mass is $\mathcal M=M+m+\rho L$. The hanging chain's <gravitational potential energy>, relative to the initial configuration, is $-\rho gx^2/2$. Thus the <Lagrangian> for the <chain sliding over a table edge> is
$$
L=\frac12\mathcal M\dot x^2+Mgx+\frac12\rho gx^2.
$$
The <Euler-Lagrange equation> gives $\mathcal M\ddot x=g(M+\rho x)$. Put $\Omega=\sqrt{\rho g/\mathcal M}$. With $x(0)=\dot x(0)=0$, the solution is
$$
\boxed{x(t)=\frac M\rho[\cosh(\Omega t)-1],\qquad\dot x(t)=\frac{M\Omega}\rho\sinh(\Omega t).}
$$
These are also the smaller mass's displacement and speed towards the edge. The formula applies until $M$ reaches the floor, at $t_f=\Omega^{-1}\operatorname{arcosh}(1+\rho y/M)$. Since $y<L$, the smaller mass has not yet reached the edge during this interval. As $\rho\to0$, the result tends to $x=Mg t^2/[2(M+m)]$, as expected for a massless connector.
Let the spring's <spring constant> be $k$, and write its extension as $e_0+x$, allowing for an initial extension $e_0$. The new <Lagrangian> is
$$
L_s=\frac12\mathcal M\dot x^2+Mgx+\frac12\rho gx^2-\frac12k(e_0+x)^2.
$$
The specified equilibrium is $x_e=y-y_1$, with $k(e_0+x_e)=g(M+\rho x_e)$. Put $\xi=x-x_e$; the equation becomes
$$
\mathcal M\ddot\xi+(k-\rho g)\xi=0,\qquad\boxed{\omega=\sqrt{\frac{k-\rho g}{M+m+\rho L}},\qquad f=\frac{\omega}{2\pi}.}
$$
The <spring-stabilized chain over a table edge> oscillates only when $k>\rho g$. More hanging chain increases its own gravitational pull, reducing the effective restoring stiffness. The equilibrium height alone does not determine $k$ without information about the spring's natural length or preload.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-table-edge-chain-spring.png]
{title=Chain gravity accelerates sliding and subtracts from a spring's restoring stiffness; stable oscillations require k greater than rho g}
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