A string is under tension sot that its length uncreased 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. `sqrtn:1`
D. n:1

Answer

To solve the problem of finding the ratio of the fundamental frequency of longitudinal vibrations (FL) to transverse vibrations (FT) of a string under tension, we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a string whose length has increased by \( \frac{1}{n} \) times its original length. This means if the original length is \( L \), the new length \( L' \) is given by:
\[ L' = L + \frac{L}{n} = L \left(1 + \frac{1}{n}\right) = L \frac{n+1}{n} \]

2. **Fundamental Frequencies**:
- The fundamental frequency of a transverse wave on a string is given by:
\[ f_T = \frac{V_T}{\lambda} \]
where \( V_T \) is the velocity of transverse waves and \( \lambda \) is the wavelength.
- The fundamental frequency of a longitudinal wave is given by:
\[ f_L = \frac{V_L}{\lambda} \]
where \( V_L \) is the velocity of longitudinal waves.

3. **Velocity of Waves**:
- For transverse waves, the velocity \( V_T \) is given by:
\[ V_T = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension and \( \mu \) is the linear mass density of the string.
- For longitudinal waves, the velocity \( V_L \) is given by:
\[ V_L = \sqrt{\frac{Y}{\rho}} \]
where \( Y \) is the Young's modulus and \( \rho \) is the density of the material.

4. **Finding the Ratio of Frequencies**:
- The ratio of the frequencies can be expressed as:
\[ \frac{f_L}{f_T} = \frac{V_L}{V_T} \]
- Substituting the expressions for \( V_L \) and \( V_T \):
\[ \frac{f_L}{f_T} = \frac{\sqrt{\frac{Y}{\rho}}}{\sqrt{\frac{T}{\mu}}} \]

5. **Expressing Linear Mass Density**:
- The linear mass density \( \mu \) can be expressed in terms of the volume density \( \rho \) and the cross-sectional area \( A \):
\[ \mu = \rho \cdot A \]

6. **Substituting for \( \mu \)**:
- Now substituting \( \mu \) in the equation:
\[ \frac{f_L}{f_T} = \frac{\sqrt{\frac{Y}{\rho}}}{\sqrt{\frac{T}{\rho \cdot A}}} = \sqrt{\frac{Y \cdot A}{T}} \]

7. **Using Strain**:
- The strain \( \epsilon \) can be expressed as:
\[ \epsilon = \frac{\Delta L}{L} = \frac{1/n}{1} = \frac{1}{n} \]
- Therefore, the Young's modulus \( Y \) can also be expressed in terms of stress and strain:
\[ Y = \frac{Stress}{Strain} = \frac{T/A}{\epsilon} = \frac{T}{A} \cdot n \]

8. **Final Ratio**:
- Substituting back into the frequency ratio:
\[ \frac{f_L}{f_T} = \sqrt{n} \]
- Thus, the final ratio of the fundamental frequency of longitudinal vibrations to transverse vibrations is:
\[ \frac{f_L}{f_T} = 1 : \sqrt{n} \]

### Final Answer:
The ratio of the fundamental frequency of longitudinal vibrations to transverse vibrations is \( 1 : \sqrt{n} \).