In the equilibrium state, the hanging mass $M$ is stationary, and the disc $m$ moves in a stable circle of radius $R$.
The tension $T$ in the cord provides the centripetal force for $m$ and balances the weight of $M$. $$ T = m \omega^2 R $$ $$ T = Mg $$ Therefore, the equilibrium condition is: $$ Mg = m \omega^2 R \quad \text{— (1)} $$

When the system oscillates, the radius $r$ changes. Since the tension force passes through the center (the hole), there is no torque on $m$. Thus, its angular momentum $L$ is conserved. $$ L = m v r = m (r\omega_{new}) r = m \omega_{new} r^2 = \text{const} $$ At equilibrium: $L = m \omega R^2$. At any radius $r$: $$ \omega_{new} = \frac{L}{m r^2} = \frac{m \omega R^2}{m r^2} = \frac{\omega R^2}{r^2} $$

Let the mass $M$ be pulled down by a small distance $x$.
– New position of hanging mass: $y = y_0 + x$ (downward).
– New radius of disc: $r = R – x$ (inward).
– Since string length is constant, accelerations are linked: $a_M = \ddot{x}$ and $a_{m, radial} = -\ddot{x}$.

Consider the system (M + m) moving along the line of the string:
The driving force is the difference between the “Centrifugal force” pushing out and Gravity pulling down. $$ F_{net} = m r \omega_{new}^2 – Mg $$ Mass of system = $(M + m)$. $$ (M+m)\ddot{x} = Mg – m r \omega_{new}^2 $$ *(Note: We define x positive downwards for M, which means r decreases. The restoring force acts to return to equilibrium)*.

Substitute $\omega_{new} = \frac{\omega R^2}{r^2}$: $$ (M+m)\ddot{x} = Mg – m r \left( \frac{\omega R^2}{r^2} \right)^2 $$ $$ (M+m)\ddot{x} = Mg – \frac{m \omega^2 R^4}{r^3} $$

Substitute $r = R – x$. Since $x \ll R$, we use the binomial approximation $(1 – \frac{x}{R})^{-3} \approx 1 + \frac{3x}{R}$. $$ \frac{1}{r^3} = \frac{1}{(R-x)^3} = \frac{1}{R^3} \left( 1 – \frac{x}{R} \right)^{-3} \approx \frac{1}{R^3} \left( 1 + \frac{3x}{R} \right) $$ Substitute this back into the force equation: $$ (M+m)\ddot{x} = Mg – m \omega^2 R^4 \left[ \frac{1}{R^3} \left( 1 + \frac{3x}{R} \right) \right] $$ $$ (M+m)\ddot{x} = Mg – m \omega^2 R \left( 1 + \frac{3x}{R} \right) $$ From Eq (1), we know $Mg = m \omega^2 R$. $$ (M+m)\ddot{x} = Mg – Mg \left( 1 + \frac{3x}{R} \right) $$ $$ (M+m)\ddot{x} = Mg – Mg – Mg \left( \frac{3x}{R} \right) $$ $$ (M+m)\ddot{x} = – \left( \frac{3Mg}{R} \right) x $$ Using $Mg = m\omega^2 R$, we can rewrite the stiffness term: $$ \frac{3Mg}{R} = \frac{3(m\omega^2 R)}{R} = 3m\omega^2 $$ So, $$ (M+m)\ddot{x} = – (3m\omega^2) x $$