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. ` 1: n`
B. ` n^(2) : 1`
C. `sqrt(n) : 1`
D. `n : 1`

Explanation

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

### Step-by-step Solution:

1. **Identify the formulas for wave velocities**:
- The velocity of longitudinal waves (\( v_1 \)) in a string is given by:
\[ v_1 = \sqrt{\frac{Y}{\rho}} \]
where \( Y \) is the Young's modulus and \( \rho \) is the density of the material.
- The velocity of transverse waves (\( v_2 \)) in a string 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.

2. **Express mass per unit length**:
- The mass per unit length (\( \mu \)) can be expressed as:
\[ \mu = \rho A \]
where \( A \) is the cross-sectional area of the string.

3. **Substitute \( \mu \) into the equation for \( v_2 \)**:
- Substitute \( \mu \) into the formula for \( v_2 \):
\[ v_2 = \sqrt{\frac{T}{\rho A}} \]

4. **Find the ratio of the velocities**:
- Now, we can find the ratio \( \frac{v_1}{v_2} \):
\[ \frac{v_1}{v_2} = \frac{\sqrt{\frac{Y}{\rho}}}{\sqrt{\frac{T}{\rho A}}} = \sqrt{\frac{Y \cdot A}{T}} \]

5. **Relate Young's modulus to tension and strain**:
- Young's modulus \( Y \) can be expressed in terms of tension and strain:
\[ Y = \frac{T}{\text{strain}} = \frac{T}{\frac{\Delta L}{L}} = \frac{T \cdot L}{\Delta L} \]
- Given that the length is increased by \( \frac{1}{n} \) times its original length, we have:
\[ \Delta L = \frac{L}{n} \]
- Therefore, we can substitute this into the expression for \( Y \):
\[ Y = \frac{T \cdot L}{\frac{L}{n}} = T \cdot n \]

6. **Substitute \( Y \) back into the ratio**:
- Substitute \( Y \) back into the ratio of velocities:
\[ \frac{v_1}{v_2} = \sqrt{\frac{T \cdot n \cdot A}{T}} = \sqrt{n \cdot A} \]

7. **Find the ratio of the fundamental frequencies**:
- The fundamental frequency \( f \) is directly proportional to the wave velocity:
\[ \frac{f_1}{f_2} = \frac{v_1}{v_2} \]
- Therefore, we have:
\[ \frac{f_1}{f_2} = \sqrt{n} \]

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