A uniform wire of length `L`, diameter `D` and density `rho` is stretched under a tension `T`. The correct relation between its fundamental frequency `f`, the length `L` and the diameter `D` is

Options

A. `fprop(1)/(LD)`
B. `fprop(1)/(Lsqrt(D))`
C. `fprop(1)/(D^(2))`
D. `fprop(1)/(LD^(2))`

Explanation

To find the correct relation between the fundamental frequency \( f \), the length \( L \), and the diameter \( D \) of a uniform wire stretched under tension \( T \), we can follow these steps:

### Step 1: Understand the fundamental frequency formula
The fundamental frequency \( f \) of a vibrating string or wire can be expressed as:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( \mu \) is the mass per unit length of the wire.

### Step 2: Determine the mass per unit length \( \mu \)
The mass per unit length \( \mu \) can be calculated using the density \( \rho \) and the volume of the wire. The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L = \left(\frac{\pi D^2}{4}\right) L \]
where \( A \) is the cross-sectional area of the wire. Thus, the mass \( M \) of the wire is:
\[ M = \rho V = \rho \left(\frac{\pi D^2}{4} L\right) \]
Therefore, the mass per unit length \( \mu \) is:
\[ \mu = \frac{M}{L} = \frac{\rho \left(\frac{\pi D^2}{4} L\right)}{L} = \frac{\pi D^2 \rho}{4} \]

### Step 3: Substitute \( \mu \) into the frequency formula
Now, substituting \( \mu \) back into the frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi D^2 \rho}{4}}} \]
This simplifies to:
\[ f = \frac{1}{2L} \sqrt{\frac{4T}{\pi D^2 \rho}} = \frac{2}{\sqrt{\pi}} \cdot \frac{\sqrt{T}}{L D} \]

### Step 4: Analyze the relationship
From the equation derived, we can see that:
\[ f \propto \frac{1}{L D} \]
This indicates that the frequency \( f \) is inversely proportional to both the length \( L \) and the diameter \( D \).

### Conclusion
Thus, the correct relation between the fundamental frequency \( f \), the length \( L \), and the diameter \( D \) is:
\[ f \propto \frac{1}{L D} \]

### Final Answer
The correct option is:
**Frequency is directly proportional to \( \frac{1}{L D} \)** (Option A).

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