Which of the followings represent the variation of frequency (f) with radius of cross section of a stretched string?

Options

A.
B.
C.
D. none of these

Answer

To determine the variation of frequency (f) with the radius of the cross-section (r) of a stretched string, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Fundamental Frequency Formula**:
The fundamental frequency (f) of a stretched string is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{m}} \]
where \( T \) is the tension in the string, \( L \) is the length of the string, and \( m \) is the mass per unit length.

2. **Express Mass per Unit Length (m)**:
The mass per unit length \( m \) can be expressed in terms of the volume and density of the string:
\[ m = \frac{\text{mass}}{\text{length}} = \frac{\text{Volume} \times \text{Density}}{L} \]

3. **Calculate Volume**:
The volume of the string can be represented as:
\[ \text{Volume} = \text{Cross-sectional Area} \times \text{Length} = A \times L \]
For a circular cross-section, the area \( A \) is given by:
\[ A = \pi r^2 \]
Therefore, the volume becomes:
\[ \text{Volume} = \pi r^2 \times L \]

4. **Substitute Volume into Mass per Unit Length**:
Now substituting the volume into the mass per unit length:
\[ m = \frac{\pi r^2 \times L \times \rho}{L} = \pi r^2 \rho \]
where \( \rho \) is the density of the material.

5. **Substitute \( m \) into the Frequency Formula**:
Now, substituting \( m \) back into the frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} \]

6. **Simplify the Frequency Expression**:
Rearranging the formula gives:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\pi \rho}} \cdot \frac{1}{r} \]
This shows that frequency \( f \) is inversely proportional to the radius \( r \) of the cross-section of the string.

### Conclusion:
Thus, the relationship between frequency \( f \) and the radius \( r \) of the cross-section of a stretched string is:
\[ f \propto \frac{1}{r} \]