A wire of density `9gm//cm^(3)` is stretched between two clamps 1.00m apart while subjected to an extension of 0.05cm. The lowest frequency of transverse vibration in the wire is (Assume Yong's modulus `Y=9xx10^(10)N//m^(2)`.

Options

A. 35Hz
B. 45Hz
C. 75Hz
D. 90Hz

Answer

To solve the problem of finding the lowest frequency of transverse vibration in the wire, we can follow these steps:

### Step 1: Understand the Given Data
- Density of the wire, \( \rho = 9 \, \text{g/cm}^3 = 9 \times 10^3 \, \text{kg/m}^3 \)
- Length of the wire, \( L = 1.00 \, \text{m} \)
- Extension of the wire, \( \Delta L = 0.05 \, \text{cm} = 0.0005 \, \text{m} \)
- Young's modulus, \( Y = 9 \times 10^{10} \, \text{N/m}^2 \)

### Step 2: Calculate the Tension in the Wire
Using Young's modulus, we can express the tension \( T \) in the wire. Young's modulus is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta L/L} \]
Where:
- \( A \) is the cross-sectional area of the wire.
- Rearranging gives:
\[ T = Y \cdot A \cdot \frac{\Delta L}{L} \]

### Step 3: Calculate the Mass per Unit Length (Linear Density)
The mass per unit length \( m \) of the wire can be expressed as:
\[ m = \frac{\text{mass}}{\text{length}} = \frac{\rho \cdot A \cdot L}{L} = \rho \cdot A \]

### Step 4: Relate Frequency to Tension and Mass per Unit Length
The frequency \( f \) of the lowest mode of vibration (fundamental frequency) in a wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{m}} \]

### Step 5: Substitute for Tension and Mass per Unit Length
Substituting \( T \) and \( m \) into the frequency equation:
\[ f = \frac{1}{2L} \sqrt{\frac{Y \cdot A \cdot \frac{\Delta L}{L}}{\rho \cdot A}} \]
The area \( A \) cancels out:
\[ f = \frac{1}{2L} \sqrt{\frac{Y \cdot \Delta L}{\rho \cdot L}} \]

### Step 6: Substitute the Values
Now we can substitute the known values into the equation:
\[ f = \frac{1}{2 \cdot 1} \sqrt{\frac{9 \times 10^{10} \cdot 0.0005}{9 \times 10^3}} \]
Calculating the inside of the square root:
\[ = \frac{1}{2} \sqrt{\frac{4.5 \times 10^{7}}{9 \times 10^{3}}} = \frac{1}{2} \sqrt{5 \times 10^{3}} = \frac{1}{2} \cdot 70.71 \approx 35.36 \, \text{Hz} \]

### Step 7: Final Result
Thus, the lowest frequency of transverse vibration in the wire is approximately:
\[ f \approx 35 \, \text{Hz} \]