1. Concept

When the block \(M\) is free to move, the system oscillates about the common Center of Mass (COM). Since there are no horizontal external forces, the COM remains stationary in the horizontal direction. This is a two-body coupled oscillation problem.

2. Reduced Mass Approach

The kinetic energy of the system relative to the COM can be described using the reduced mass \(\mu\): \[ \mu = \frac{mM}{m+M} \] The effective equation of motion becomes equivalent to a simple pendulum with mass \(\mu\) but the restoring force is still provided by gravity acting on the actual mass \(m\).

3. Derivation

Using Energy method: \[ E = \frac{1}{2}\mu (l\dot{\theta})^2 + \frac{1}{2}mgl\theta^2 \] Differentiating with respect to time: \[ \mu l^2 \ddot{\theta} + mgl\theta = 0 \] \[ \ddot{\theta} + \frac{mg}{\mu l}\theta = 0 \] The angular frequency squared is: \[ \omega^2 = \frac{mg}{\mu l} = \frac{mg(m+M)}{mM l} = \frac{g}{l} \left( \frac{m+M}{M} \right) \]

4. Period Calculation

The time period \(T\) is: \[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{l}{g} \frac{M}{m+M}} \] Given that \(T_0 = 2\pi\sqrt{\frac{l}{g}}\) (period when fixed), we can write: \[ T = T_0 \sqrt{\frac{M}{m+M}} \]

Answer: \( T_0 \sqrt{\frac{M}{m+M}} \)