Two sonometer wires A and B are fixed on a sonometer. The ratio of their lengths, diameters, densities and tensions are given below:
`(L_A)/(L_B) = 36/35, (d_A)/(d_B) = 4/1, (T_A)/(T_B) = 8/1, (rho_A)/(rho_B) = 1/2`
If the higher frequency among the two wires is 360 Hz, then what is the best freqency (in Hz) observed when the two wires are sounded together?

Answer

To solve the problem, we need to find the beat frequency when two sonometer wires A and B are sounded together. We will use the given ratios of their lengths, diameters, densities, and tensions to calculate the frequencies of both wires and then find the beat frequency.

### Step-by-Step Solution:

1. **Identify the Given Ratios:**
- Lengths: \( \frac{L_A}{L_B} = \frac{36}{35} \)
- Diameters: \( \frac{d_A}{d_B} = \frac{4}{1} \)
- Tensions: \( \frac{T_A}{T_B} = \frac{8}{1} \)
- Densities: \( \frac{\rho_A}{\rho_B} = \frac{1}{2} \)

2. **Calculate the Mass per Unit Length (μ):**
The mass per unit length \( \mu \) is given by:
\[ \mu = \rho \cdot A \]
where \( A \) is the area of cross-section. The area \( A \) can be expressed in terms of diameter \( d \):
\[ A = \frac{\pi d^2}{4} \]
Therefore, the ratio of mass per unit length for wires A and B can be expressed as:
\[ \frac{\mu_A}{\mu_B} = \frac{\rho_A \cdot A_A}{\rho_B \cdot A_B} = \frac{\rho_A \cdot \frac{\pi d_A^2}{4}}{\rho_B \cdot \frac{\pi d_B^2}{4}} = \frac{\rho_A}{\rho_B} \cdot \left(\frac{d_A}{d_B}\right)^2 \]

3. **Substituting the Ratios:**
Substituting the given ratios:
\[ \frac{\mu_A}{\mu_B} = \frac{1/2}{1} \cdot \left(\frac{4}{1}\right)^2 = \frac{1}{2} \cdot 16 = 8 \]

4. **Calculate the Frequency Ratio:**
The frequency \( f \) of a wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Therefore, the ratio of frequencies \( \frac{f_A}{f_B} \) can be expressed as:
\[ \frac{f_A}{f_B} = \frac{T_A / \mu_A}{T_B / \mu_B} = \frac{T_A}{T_B} \cdot \frac{\mu_B}{\mu_A} \]
Substituting the known ratios:
\[ \frac{f_A}{f_B} = \frac{8}{1} \cdot \frac{1}{8} = 1 \]

5. **Finding the Frequencies:**
Since we know that the higher frequency \( f_B = 360 \, \text{Hz} \), we can find \( f_A \):
\[ f_A = \frac{35}{36} f_B = \frac{35}{36} \cdot 360 \approx 350 \, \text{Hz} \]

6. **Calculate the Beat Frequency:**
The beat frequency \( f_{\text{beat}} \) is given by the difference between the two frequencies:
\[ f_{\text{beat}} = |f_B - f_A| = |360 - 350| = 10 \, \text{Hz} \]

### Final Answer:
The beat frequency observed when the two wires are sounded together is **10 Hz**.