A stretched sonometer wire is in unison with a tuning fork. When the length is increased by 4%, the number of beats heard per second is 6. Find the frequency of the fork.

Answer

To solve the problem, we need to find the frequency of the tuning fork when a stretched sonometer wire is in unison with it. When the length of the wire is increased by 4%, the number of beats heard per second is 6. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Initial Condition**:
- Let the initial length of the sonometer wire be \( L \).
- The frequency of the sonometer wire when it is in unison with the tuning fork is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
- Let the frequency of the tuning fork be \( f \).

2. **Increasing the Length**:
- When the length is increased by 4%, the new length \( L' \) becomes:
\[ L' = L + 0.04L = 1.04L \]

3. **Calculating the New Frequency**:
- The new frequency \( f' \) of the sonometer wire when the length is increased is:
\[ f' = \frac{1}{2L'} \sqrt{\frac{T}{\mu}} = \frac{1}{2(1.04L)} \sqrt{\frac{T}{\mu}} = \frac{f}{1.04} \]

4. **Understanding the Beat Frequency**:
- The beat frequency is the absolute difference between the frequencies of the tuning fork and the sonometer wire:
\[ |f - f'| = 6 \text{ Hz} \]
- Since \( f' < f \) (because increasing length decreases frequency), we have:
\[ f - f' = 6 \]

5. **Substituting for \( f' \)**:
- Substitute \( f' \) in the beat frequency equation:
\[ f - \frac{f}{1.04} = 6 \]

6. **Simplifying the Equation**:
- Factor out \( f \):
\[ f \left(1 - \frac{1}{1.04}\right) = 6 \]
- Calculate \( 1 - \frac{1}{1.04} \):
\[ 1 - \frac{1}{1.04} = \frac{1.04 - 1}{1.04} = \frac{0.04}{1.04} = \frac{1}{26} \]
- Therefore, we have:
\[ f \cdot \frac{1}{26} = 6 \]

7. **Finding the Frequency \( f \)**:
- Multiply both sides by 26:
\[ f = 6 \times 26 = 156 \text{ Hz} \]

### Final Answer:
The frequency of the tuning fork is \( 156 \text{ Hz} \).