1. Initial Equilibrium

Initially, the bar of mass $M$ rests on the vertical spring.
Compression $x_i = \frac{Mg}{k}$.

2. Motion Analysis

When block $m$ is placed on $M$ and released, the system oscillates.
New Equilibrium Position: The combined mass $(M+m)$ would rest at a compression $x_f = \frac{(M+m)g}{k}$.
Amplitude of Oscillation ($A$): The motion starts from the position $x_i$ with zero velocity. The distance between the starting point and the new equilibrium is the amplitude. $$ A = x_f – x_i = \frac{(M+m)g}{k} – \frac{Mg}{k} = \frac{mg}{k} $$

3. Force Calculation

We need the maximum force the bar $M$ exerts on block $m$ (Normal Force $N$). Considering the Free Body Diagram of block $m$:

• Forces: $N$ (Up), $mg$ (Down).
• Equation: $N – mg = ma \implies N = m(g + a)$.

To maximize $N$, we need maximum upward acceleration ($a$). In Simple Harmonic Motion (SHM), maximum acceleration occurs at the extreme positions. The maximum upward acceleration occurs at the bottom-most point of the oscillation.

Calculating $a_{max}$: $$ a_{max} = \omega^2 A $$ Where $\omega^2 = \frac{k}{M_{total}} = \frac{k}{M+m}$. $$ a_{max} = \left( \frac{k}{M+m} \right) \cdot \left( \frac{mg}{k} \right) = \frac{mg}{M+m} $$

4. Final Result

Substitute $a_{max}$ into the normal force equation:

$$ N_{max} = m \left( g + \frac{mg}{M+m} \right) $$ $$ N_{max} = mg \left( 1 + \frac{m}{M+m} \right) $$ $$ N_{max} = mg \left( \frac{M+m+m}{M+m} \right) $$ $$ N_{max} = mg \left( \frac{M+2m}{M+m} \right) $$