A uniform wire of density `rho` is stretched by `l_(!)` under its propotional limit whose original slength is L. What is lowest frequency of transverse viberation set up in the wire assuming Young's modulus of the material to be Y?

Options

A. `(1)/(2L)sqrt((Yl_(1))/(rhoL))`
B. `(1)/(4L)sqrt((Yl)/(rhoL_(1)))`
C. `(1)/(2L)sqrt((Yl_(1))/(rhoL_(1)))`
D. `(1)/(L)sqrt((Yl_(1))/(rhoL))`

Answer

To solve the problem of finding the lowest frequency of transverse vibration in a uniform wire stretched under its proportional limit, we can follow these steps:

### Step 1: Understand the relationship for the lowest frequency
The lowest frequency \( f \) of a fixed string (or wire) can be expressed using the formula:
\[ f = \frac{V}{2L} \]
where \( V \) is the wave velocity and \( L \) is the length of the wire.

**Hint:** Remember that the lowest frequency corresponds to the fundamental mode of vibration.

### Step 2: Express wave velocity in terms of tension and mass per unit length
The wave velocity \( V \) in a stretched wire can be given by:
\[ V = \sqrt{\frac{T}{\mu}} \]
where \( T \) is the tension in the wire and \( \mu \) is the mass per unit length of the wire.

**Hint:** The wave velocity depends on the tension and the mass distribution of the wire.

### Step 3: Calculate mass per unit length \( \mu \)
The mass per unit length \( \mu \) can be calculated as:
\[ \mu = \frac{m}{L} = \rho A \]
where \( \rho \) is the density of the material and \( A \) is the cross-sectional area of the wire.

**Hint:** Make sure to consider the cross-sectional area when calculating mass per unit length.

### Step 4: Calculate the tension \( T \) in the wire
Using Hooke's law, the tension \( T \) can be expressed as:
\[ T = Y \cdot \text{strain} = Y \cdot \frac{\Delta L}{L} = Y \cdot \frac{L_1}{L} \]
where \( Y \) is the Young's modulus and \( L_1 \) is the extension of the wire.

**Hint:** The tension is directly related to the Young's modulus and the strain in the wire.

### Step 5: Substitute \( T \) and \( \mu \) into the frequency formula
Now substituting the expressions for \( T \) and \( \mu \) into the frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{Y \cdot \frac{L_1}{L}}{\rho A}} \]
This simplifies to:
\[ f = \frac{1}{2L} \sqrt{\frac{Y L_1}{\rho A L}} \]

**Hint:** Ensure that all units are consistent when substituting values.

### Final Expression
Thus, the lowest frequency of transverse vibration set up in the wire is:
\[ f = \frac{1}{2L} \sqrt{\frac{Y L_1}{\rho A L}} \]