Understanding torsional oscillations and the concept of moment of inertia is crucial to solve this problem. When a rod is suspended at its middle and undergoes torsional oscillations, its frequency depends on its moment of inertia about the suspension point and the torsional constant of the wire.

Initial Torsional Oscillation Setup

Initially, we have a rod of mass 'M' and total length '2L' suspended at its center. This rod performs torsional oscillations. The frequency of torsional oscillation is given by the formula:

$$\text{f} = \frac{1}{2\pi} \sqrt{\frac{\kappa}{\text{I}}}$$

Where:

• $\text{f}$ is the frequency of oscillation.
• $\kappa$ (kappa) is the torsional constant of the wire (which remains unchanged throughout the problem).
• $\text{I}$ is the moment of inertia of the oscillating system about the axis of rotation.

Rod's Moment of Inertia

For a uniform rod of mass 'M' and total length 'L_total' rotating about an axis through its center, the moment of inertia is given by:

$$\text{I}_{\text{rod}} = \frac{1}{12} \text{M} (\text{L}_{\text{total}})^2$$

In this problem, the total length of the rod is '2L'. So, substitute $\text{L}_{\text{total}} = 2\text{L}$ into the formula:

$$\text{I}_{1} = \frac{1}{12} \text{M} (2\text{L})^2$$

$$\text{I}_{1} = \frac{1}{12} \text{M} (4\text{L}^2)$$

$$\text{I}_{1} = \frac{1}{3} \text{M} \text{L}^2$$

This $\text{I}_{1}$ represents the initial moment of inertia for the system when only the rod is oscillating.

Initial Oscillation Frequency

The initial frequency of oscillation, let's denote it as $\text{f}_{1}$, is derived by substituting $\text{I}_{1}$ into the frequency formula:

$$\text{f}_{1} = \frac{1}{2\pi} \sqrt{\frac{\kappa}{\text{I}_{1}}}$$

$$\text{f}_{1} = \frac{1}{2\pi} \sqrt{\frac{\kappa}{\frac{1}{3} \text{M} \text{L}^2}}$$

Modified Torsional Oscillation Setup

Next, two point masses, each of mass 'm', are attached to the rod. Each mass is positioned at a distance of 'L/2' from the center of the rod, on opposite sides.

Moment of Inertia of Added Masses

For a point mass 'm' at a distance 'r' from the axis of rotation, its moment of inertia is $\text{mr}^2$. Since there are two such masses, and each is at a distance 'L/2' from the center, their combined moment of inertia is:

$$\text{I}_{\text{masses}} = \text{m} \left( \frac{\text{L}}{2} \right)^2 + \text{m} \left( \frac{\text{L}}{2} \right)^2$$

$$\text{I}_{\text{masses}} = 2 \times \text{m} \left( \frac{\text{L}^2}{4} \right)$$

$$\text{I}_{\text{masses}} = \frac{1}{2} \text{m} \text{L}^2$$

Total Moment of Inertia

The total moment of inertia of the system after adding the masses, denoted as $\text{I}_{2}$, is the sum of the moment of inertia of the rod and the moment of inertia of the two added masses:

$$\text{I}_{2} = \text{I}_{1} + \text{I}_{\text{masses}}$$

$$\text{I}_{2} = \frac{1}{3} \text{M} \text{L}^2 + \frac{1}{2} \text{m} \text{L}^2$$

We can factor out $\text{L}^2$:

$$\text{I}_{2} = \text{L}^2 \left( \frac{1}{3} \text{M} + \frac{1}{2} \text{m} \right)$$

Final Oscillation Frequency

The final frequency of oscillation, $\text{f}_{2}$, is found by using the total moment of inertia $\text{I}_{2}$:

$$\text{f}_{2} = \frac{1}{2\pi} \sqrt{\frac{\kappa}{\text{I}_{2}}}$$

Substitute the value of $\text{I}_{2}$:

$$\text{f}_{2} = \frac{1}{2\pi} \sqrt{\frac{\kappa}{\text{L}^2 \left( \frac{1}{3} \text{M} + \frac{1}{2} \text{m} \right)}}$$

Frequency Reduction and Ratio Calculation

The problem states that attaching the masses reduces the oscillation frequency by 10%. This means the new frequency $\text{f}_{2}$ is 90% of the original frequency $\text{f}_{1}$.

$$\text{f}_{2} = \text{f}_{1} - 0.10 \text{f}_{1}$$

$$\text{f}_{2} = 0.90 \text{f}_{1}$$

Now, substitute the expressions for $\text{f}_{1}$ and $\text{f}_{2}$ into this relationship:

$$\frac{1}{2\pi} \sqrt{\frac{\kappa}{\text{L}^2 \left( \frac{1}{3} \text{M} + \frac{1}{2} \text{m} \right)}} = 0.90 \times \frac{1}{2\pi} \sqrt{\frac{\kappa}{\frac{1}{3} \text{M} \text{L}^2}}$$

To simplify this equation, we can cancel out the common terms $\frac{1}{2\pi}$ and $\sqrt{\kappa}$ from both sides. Then, square both sides to eliminate the square root:

$$\frac{1}{\text{L}^2 \left( \frac{1}{3} \text{M} + \frac{1}{2} \text{m} \right)} = (0.90)^2 \times \frac{1}{\frac{1}{3} \text{M} \text{L}^2}$$

$$\frac{1}{\text{L}^2 \left( \frac{1}{3} \text{M} + \frac{1}{2} \text{m} \right)} = 0.81 \times \frac{1}{\frac{1}{3} \text{M} \text{L}^2}$$

We can further cancel out $\text{L}^2$ from both denominators:

$$\frac{1}{\frac{1}{3} \text{M} + \frac{1}{2} \text{m}} = \frac{0.81}{\frac{1}{3} \text{M}}$$

Now, cross-multiply to solve for the ratio $\frac{\text{M}}{\text{m}}$:

$$\frac{1}{3} \text{M} = 0.81 \left( \frac{1}{3} \text{M} + \frac{1}{2} \text{m} \right)$$

Distribute 0.81 on the right side:

$$\frac{1}{3} \text{M} = 0.81 \times \frac{1}{3} \text{M} + 0.81 \times \frac{1}{2} \text{m}$$

Group terms involving 'M' on one side and 'm' on the other:

$$\frac{1}{3} \text{M} - \frac{0.81}{3} \text{M} = \frac{0.81}{2} \text{m}$$

Factor out 'M' on the left side:

$$\text{M} \left( \frac{1}{3} - \frac{0.81}{3} \right) = \frac{0.81}{2} \text{m}$$

$$\text{M} \left( \frac{1 - 0.81}{3} \right) = \frac{0.81}{2} \text{m}$$

$$\text{M} \left( \frac{0.19}{3} \right) = \frac{0.81}{2} \text{m}$$

Finally, isolate the ratio $\frac{\text{M}}{\text{m}}$:

$$\frac{\text{M}}{\text{m}} = \frac{0.81}{2} \times \frac{3}{0.19}$$

$$\frac{\text{M}}{\text{m}} = \frac{2.43}{0.38}$$

Performing the division:

$$\frac{\text{M}}{\text{m}} \approx 6.3947$$

Rounding this value to one decimal place, we get 6.4.

The calculations confirm that the value of the ratio $\text{M}/\text{m}$ is approximately 6.4.