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 solve the problem of how frequency (f) varies with the radius (r) of the cross-section of a stretched string, we can follow these steps:

### Step 1: Understand the formula for frequency of a stretched string
The frequency (f) of a stretched string can be expressed using the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( T \) is the tension in the string,
- \( \mu \) is the mass per unit length,
- \( L \) is the length of the string.

### Step 2: Relate mass per unit length to the radius
The mass per unit length (\( \mu \)) can be expressed in terms of the density (\( \rho \)) and the cross-sectional area (A) of the string:
\[ \mu = \frac{m}{L} = \frac{\rho \cdot A}{L} \]
For a circular cross-section, the area (A) can be expressed as:
\[ A = \pi r^2 \]
Thus, we can write:
\[ \mu = \frac{\rho \cdot \pi r^2 \cdot L}{L} = \rho \cdot \pi r^2 \]

### Step 3: Substitute \( \mu \) back into the frequency formula
Substituting \( \mu \) into the frequency formula gives:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho \cdot \pi r^2}} \]

### Step 4: Simplify the expression
This can be simplified to:
\[ f = \frac{1}{2L} \cdot \sqrt{\frac{T}{\rho \cdot \pi}} \cdot \frac{1}{r} \]
Here, \( \frac{1}{2L} \cdot \sqrt{\frac{T}{\rho \cdot \pi}} \) is a constant (let's call it \( k \)):
\[ f = \frac{k}{r} \]

### Step 5: Conclude the relationship
From the equation \( f = \frac{k}{r} \), we can see that frequency (f) is inversely proportional to the radius (r) of the cross-section of the string.

### Final Answer
Thus, the variation of frequency (f) with the radius (r) of the cross-section of a stretched string is represented by an inverse relationship, which can be depicted graphically as a hyperbola. Therefore, the correct option is **Option A**.

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