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 \( f \), length \( L \), and diameter \( D \) of a uniform wire under tension \( T \), we can follow these steps:

### Step 1: Understand the formula for fundamental frequency
The fundamental frequency \( f \) of a stretched wire can be expressed as:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( \mu \) is the linear mass density of the wire.

### Step 2: Express linear mass density \( \mu \)
The linear mass density \( \mu \) can be defined as:
\[ \mu = \frac{m}{L} \]
where \( m \) is the mass of the wire.

### Step 3: Relate mass to density and volume
The mass \( m \) can be expressed in terms of the density \( \rho \) and the volume \( V \):
\[ m = \rho V \]
The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area of the wire.

### Step 4: Calculate the cross-sectional area
For a wire with diameter \( D \), the cross-sectional area \( A \) is given by:
\[ A = \pi r^2 = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \]

### Step 5: Substitute the area into the mass equation
Substituting the area into the volume equation, we have:
\[ V = \frac{\pi D^2}{4} \cdot L \]
Thus, the mass \( m \) becomes:
\[ m = \rho \cdot \frac{\pi D^2}{4} \cdot L \]

### Step 6: Substitute mass back into linear density
Now substituting \( m \) back into the equation for \( \mu \):
\[ \mu = \frac{m}{L} = \frac{\rho \cdot \frac{\pi D^2}{4} \cdot L}{L} = \frac{\rho \pi D^2}{4} \]

### Step 7: Substitute \( \mu \) back into the frequency formula
Now, substituting \( \mu \) back into the fundamental frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\frac{\rho \pi D^2}{4}}} \]
This simplifies to:
\[ f = \frac{1}{2L} \cdot \sqrt{\frac{4T}{\rho \pi D^2}} = \frac{2}{L D} \sqrt{\frac{T}{\rho \pi}} \]

### Step 8: Analyze the relationship
From the final expression, we can see that:
\[ f \propto \frac{1}{L D} \]
This means that the fundamental frequency \( f \) is inversely proportional to both the length \( L \) and the diameter \( D \).

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