Two wires are kept tight between the same pair of supports. The tensions in the wires are in the ratio 2 : 1, the radii are in the ratio 3 : 1 and the densities are in the ratio 1 : 2. Find the ratio of their fundamental frequencies.

Explanation

To find the ratio of the fundamental frequencies of the two wires, we can use the relationship between frequency, tension, density, and area. The fundamental frequency \( f \) of a wire is given by the formula:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the mass per unit length of the wire.

The mass per unit length \( \mu \) can be expressed as:

\[ \mu = \rho A \]

where:
- \( \rho \) is the density of the material,
- \( A \) is the cross-sectional area of the wire.

The cross-sectional area \( A \) for a wire with radius \( r \) is given by:

\[ A = \pi r^2 \]

### Step-by-Step Solution:

1. **Identify the given ratios:**
- Tensions in the wires: \( T_1 : T_2 = 2 : 1 \)
- Radii of the wires: \( r_1 : r_2 = 3 : 1 \)
- Densities of the wires: \( \rho_1 : \rho_2 = 1 : 2 \)

2. **Express the areas in terms of the radii:**
- The area \( A \) is proportional to \( r^2 \).
- Thus, the area ratio \( A_1 : A_2 \) can be calculated as:
\[ A_1 = \pi r_1^2, \quad A_2 = \pi r_2^2 \]
\[ \frac{A_1}{A_2} = \frac{r_1^2}{r_2^2} = \left(\frac{3}{1}\right)^2 = 9 : 1 \]

3. **Calculate the mass per unit length \( \mu \) for both wires:**
- For wire 1:
\[ \mu_1 = \rho_1 A_1 \]
- For wire 2:
\[ \mu_2 = \rho_2 A_2 \]

4. **Substituting the ratios into the fundamental frequency formula:**
- The ratio of frequencies \( \frac{f_1}{f_2} \) is given by:
\[ \frac{f_1}{f_2} = \sqrt{\frac{T_1}{T_2}} \cdot \sqrt{\frac{\mu_2}{\mu_1}} \]

5. **Substituting the known ratios into the equation:**
- From the tension ratio:
\[ \frac{T_1}{T_2} = 2 \]
- From the density ratio:
\[ \frac{\rho_2}{\rho_1} = 2 \quad \Rightarrow \quad \frac{\mu_2}{\mu_1} = \frac{\rho_2 A_2}{\rho_1 A_1} = \frac{2 \cdot \frac{1}{9}}{1} = \frac{2}{9} \]

6. **Putting it all together:**
\[ \frac{f_1}{f_2} = \sqrt{2} \cdot \sqrt{\frac{2}{9}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \]

### Final Answer:
The ratio of the fundamental frequencies of the two wires is:

\[ \frac{f_1}{f_2} = \frac{2}{3} \]