A tuning fork resonates with a sonometer wire of length 1 m stretched with a tension of 6 N. When the tension in the wire is changed to 54 N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ Hz.

Answer

To solve the problem, we need to find the frequency of the tuning fork based on the information given about the sonometer wire and the beats produced. Here’s the step-by-step solution:

### Step 1: Understand the relationship between frequency, tension, and linear density
The frequency of a vibrating string (or sonometer wire) is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( f \) = frequency of the wire
- \( L \) = length of the wire (1 m in this case)
- \( T \) = tension in the wire
- \( \mu \) = linear density of the wire

### Step 2: Calculate the frequencies for both tensions
Let \( f_1 \) be the frequency of the wire when the tension is 6 N, and \( f_2 \) be the frequency when the tension is 54 N.

Using the formula:
1. For \( T = 6 \, \text{N} \):
\[ f_1 = \frac{1}{2 \cdot 1} \sqrt{\frac{6}{\mu}} = \frac{1}{2} \sqrt{\frac{6}{\mu}} \]

2. For \( T = 54 \, \text{N} \):
\[ f_2 = \frac{1}{2 \cdot 1} \sqrt{\frac{54}{\mu}} = \frac{1}{2} \sqrt{\frac{54}{\mu}} \]

### Step 3: Relate the frequencies using the beats produced
The problem states that when the tension is changed to 54 N, the tuning fork produces 12 beats per second with the wire. The beat frequency is given by:
\[ |f_2 - f_1| = 12 \, \text{Hz} \]

### Step 4: Set up the equation for beats
From the above, we can write:
\[ f_2 - f_1 = 12 \]

Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{1}{2} \sqrt{\frac{54}{\mu}} - \frac{1}{2} \sqrt{\frac{6}{\mu}} = 12 \]

### Step 5: Simplify the equation
Factor out \( \frac{1}{2} \):
\[ \frac{1}{2} \left( \sqrt{\frac{54}{\mu}} - \sqrt{\frac{6}{\mu}} \right) = 12 \]
Multiply both sides by 2:
\[ \sqrt{\frac{54}{\mu}} - \sqrt{\frac{6}{\mu}} = 24 \]

### Step 6: Solve for the square roots
Let \( x = \sqrt{\frac{1}{\mu}} \):
\[ \sqrt{54} x - \sqrt{6} x = 24 \]
\[ (\sqrt{54} - \sqrt{6}) x = 24 \]
Now, calculate \( \sqrt{54} \) and \( \sqrt{6} \):
\[ \sqrt{54} = 3\sqrt{6} \]
Thus:
\[ (3\sqrt{6} - \sqrt{6}) x = 24 \]
\[ (2\sqrt{6}) x = 24 \]
\[ x = \frac{24}{2\sqrt{6}} = \frac{12}{\sqrt{6}} = 2\sqrt{6} \]

### Step 7: Substitute back to find \( \mu \)
Now substituting back to find \( \mu \):
\[ \mu = \frac{1}{x^2} = \frac{1}{(2\sqrt{6})^2} = \frac{1}{24} \]

### Step 8: Find \( f_1 \)
Now substitute \( \mu \) back into the equation for \( f_1 \):
\[ f_1 = \frac{1}{2} \sqrt{\frac{6}{\mu}} = \frac{1}{2} \sqrt{6 \times 24} = \frac{1}{2} \sqrt{144} = \frac{1}{2} \times 12 = 6 \, \text{Hz} \]

### Final Answer
The frequency of the tuning fork is:
\[ \boxed{6 \, \text{Hz}} \]