A string is under tension so that its length is increased by `1//n` times its original length . The ratio of fundamental frequency of longitudinal vibrations and transverse vibrations will be

Options

A. `eta :1`
B. `1:eta`
C. `sqrt eta :1 `
D. `1: sqrt eta`

Explanation

To solve the problem, we need to find the ratio of the fundamental frequencies of longitudinal vibrations (f1) and transverse vibrations (f2) of a string that is under tension and has its length increased by \( \frac{1}{n} \) times its original length.

### Step-by-Step Solution:

1. **Understanding the Frequencies**:
- The fundamental frequency of a wave is given by the formula:
\[ f = \frac{v}{\lambda} \]
- For longitudinal waves, the velocity \( v_1 \) is given by:
\[ v_1 = \sqrt{\frac{Y}{\rho}} \]
where \( Y \) is Young's modulus and \( \rho \) is the density of the material.
- For transverse waves, the velocity \( v_2 \) is given by:
\[ v_2 = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the string and \( \mu \) is the mass per unit length of the string.

2. **Relating Mass per Unit Length**:
- The mass per unit length \( \mu \) can be expressed as:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the string and \( L \) is its original length.
- If the length of the string increases by \( \frac{1}{n} \) times its original length, the new length \( L' \) becomes:
\[ L' = L + \frac{L}{n} = L \left(1 + \frac{1}{n}\right) = \frac{(n+1)L}{n} \]
- The new mass per unit length becomes:
\[ \mu' = \frac{m}{L'} = \frac{m}{\frac{(n+1)L}{n}} = \frac{n \cdot m}{(n+1)L} \]

3. **Finding the Ratio of Velocities**:
- The ratio of the velocities of longitudinal and transverse waves is:
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{Y}{\rho}}}{\sqrt{\frac{T}{\mu}}} \]
- Substituting for \( \mu \):
\[ v_2 = \sqrt{\frac{T \cdot (n+1)L}{n \cdot m}} \]
- Therefore, the ratio becomes:
\[ \frac{v_1}{v_2} = \sqrt{\frac{Y \cdot n \cdot m}{T \cdot (n+1)L}} \]

4. **Finding the Ratio of Frequencies**:
- Since frequency is directly proportional to velocity, we can write:
\[ \frac{f_1}{f_2} = \frac{v_1}{v_2} \]
- Thus, the ratio of the fundamental frequencies becomes:
\[ \frac{f_1}{f_2} = \sqrt{n} \]

### Final Answer:
The ratio of the fundamental frequency of longitudinal vibrations to that of transverse vibrations is:
\[ \frac{f_1}{f_2} = \sqrt{n} \]