When the tension of a sonometer wire is increased by 25% the fundamental increases by 10 Hz. If its length is increased by 25 % what will be the new frequency?

Answer

To solve the problem, we need to analyze the effects of changes in tension and length on the fundamental frequency of a sonometer wire.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The fundamental frequency \( f \) of a vibrating string (or sonometer 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, and \( \mu \) is the mass per unit length.

2. **Initial Conditions**:
Let the original tension be \( T \) and the original frequency be \( f \). We know that when the tension is increased by 25%, the new tension becomes:
\[ T' = 1.25T \]
The frequency increases by 10 Hz, so the new frequency is:
\[ f' = f + 10 \]

3. **Setting Up the Equations**:
Using the frequency formula for the new tension:
\[ f + 10 = \frac{1}{2L} \sqrt{\frac{1.25T}{\mu}} \]
We can express the original frequency as:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

4. **Dividing the Equations**:
We can divide the two frequency equations:
\[ \frac{f}{f + 10} = \frac{\frac{1}{2L} \sqrt{\frac{T}{\mu}}}{\frac{1}{2L} \sqrt{\frac{1.25T}{\mu}}} \]
This simplifies to:
\[ \frac{f}{f + 10} = \frac{1}{\sqrt{1.25}} = \frac{1}{1.118} \]

5. **Cross-Multiplying**:
Cross-multiplying gives:
\[ f \cdot 1.118 = (f + 10) \]
Expanding this:
\[ 1.118f = f + 10 \]
Rearranging gives:
\[ 0.118f = 10 \]
Therefore, solving for \( f \):
\[ f = \frac{10}{0.118} \approx 84.75 \text{ Hz} \]

6. **Finding the New Frequency with Increased Length**:
Now, if the length is increased by 25%, the new length \( L' \) becomes:
\[ L' = 1.25L \]
The new frequency \( f'' \) can be expressed as:
\[ f'' = \frac{1}{2L'} \sqrt{\frac{T'}{\mu}} = \frac{1}{2 \cdot 1.25L} \sqrt{\frac{1.25T}{\mu}} = \frac{1}{1.25} \cdot \frac{1}{2L} \sqrt{\frac{1.25T}{\mu}} \]
We can relate this back to the original frequency:
\[ f'' = \frac{1}{1.25} f' \]
Substituting \( f' = f + 10 \):
\[ f'' = \frac{1}{1.25} (84.75 + 10) = \frac{1}{1.25} \cdot 94.75 \approx 75.8 \text{ Hz} \]

### Final Answer:
The new frequency when the length is increased by 25% is approximately **75.8 Hz**.