The fundatmental frequency of a sonometer wire increases by `6 Hz` if its tension is increased by `44%` keeping the length constant. Find the change in the fundamental frequency of the sonometer when the length of the wire is increased by `20%` keeping the original tension in the wire.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between frequency, tension, and length
The fundamental frequency \( f \) of a sonometer wire is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) = length of the wire
- \( T \) = tension in the wire
- \( \mu \) = mass per unit length of the wire

### Step 2: Set up the initial conditions
Let the initial frequency be \( f \). According to the problem, when the tension is increased by \( 44\% \), the frequency increases by \( 6 \, \text{Hz} \):
\[ f' = f + 6 \, \text{Hz} \]
The new tension \( T' \) can be expressed as:
\[ T' = 1.44T \]

### Step 3: Write the equations for the two frequencies
Using the frequency formula for the initial and new conditions:
1. For the initial frequency:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \quad \text{(Equation 1)} \]
2. For the new frequency after increasing tension:
\[ f' = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{1.44T}{\mu}} \quad \text{(Equation 2)} \]

### Step 4: Relate the two frequencies
Dividing Equation 2 by Equation 1 gives:
\[ \frac{f + 6}{f} = \frac{\sqrt{1.44}}{1} \]
This simplifies to:
\[ \frac{f + 6}{f} = 1.2 \]
Cross-multiplying gives:
\[ f + 6 = 1.2f \]
Rearranging this equation:
\[ 6 = 1.2f - f \]
\[ 6 = 0.2f \]
Thus, we can solve for \( f \):
\[ f = \frac{6}{0.2} = 30 \, \text{Hz} \]

### Step 5: Calculate the new frequency when length is increased
Now, we need to find the new frequency \( f'' \) when the length is increased by \( 20\% \) while keeping the tension constant:
\[ L'' = 1.2L \]
Using the frequency formula again:
\[ f'' = \frac{1}{2L''} \sqrt{\frac{T}{\mu}} = \frac{1}{2 \times 1.2L} \sqrt{\frac{T}{\mu}} = \frac{1}{1.2} \cdot \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{f}{1.2} \]
Substituting \( f = 30 \, \text{Hz} \):
\[ f'' = \frac{30}{1.2} = 25 \, \text{Hz} \]

### Step 6: Find the change in frequency
The change in frequency \( \Delta f \) is:
\[ \Delta f = f - f'' = 30 \, \text{Hz} - 25 \, \text{Hz} = 5 \, \text{Hz} \]

### Final Answer
The change in the fundamental frequency of the sonometer when the length of the wire is increased by \( 20\% \) is \( 5 \, \text{Hz} \).

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