A wire of density `9 xx 10^(3) kg//m^(3)` is stretched between two clamps 1 m apart and is subjected to an extension of `4.9 xx 10^(-4) m`. The lowest frequency of transverse vibration in the wire is `(Y = 9 xx 10^(10) N//m^(2))`

Explanation

To solve the problem, we need to find the lowest frequency of transverse vibration in a wire that is stretched between two clamps. The given parameters are:

- Density of the wire, \( \rho = 9 \times 10^3 \, \text{kg/m}^3 \)
- Extension of the wire, \( \Delta L = 4.9 \times 10^{-4} \, \text{m} \)
- Young's modulus, \( Y = 9 \times 10^{10} \, \text{N/m}^2 \)
- Length of the wire, \( L = 1 \, \text{m} \)

### Step 1: Calculate the tension in the wire
The tension \( T \) in the wire can be calculated using the relationship between stress, strain, and Young's modulus:

\[ \text{Stress} = \frac{T}{A} = Y \cdot \text{Strain} \]

Where:
- Strain \( = \frac{\Delta L}{L} \)

Substituting the values:

\[ \text{Strain} = \frac{4.9 \times 10^{-4}}{1} = 4.9 \times 10^{-4} \]

Now, substituting into the stress equation:

\[ \frac{T}{A} = Y \cdot \text{Strain} \implies T = A \cdot Y \cdot \text{Strain} \]

### Step 2: Express mass per unit length \( \mu \)
The mass per unit length \( \mu \) can be expressed in terms of density and cross-sectional area \( A \):

\[ \mu = \rho \cdot A \]

### Step 3: Substitute into the frequency formula
The lowest frequency \( f \) of transverse vibration for a wire fixed at both ends is given by:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

Substituting \( \mu \) into the equation gives:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\rho A}} \]

### Step 4: Substitute \( T \) into the frequency equation
From Step 1, we have \( T = A \cdot Y \cdot \text{Strain} \):

\[ f = \frac{1}{2L} \sqrt{\frac{A \cdot Y \cdot \text{Strain}}{\rho A}} = \frac{1}{2L} \sqrt{\frac{Y \cdot \text{Strain}}{\rho}} \]

### Step 5: Substitute known values
Now we can substitute the known values into the frequency formula:

\[ f = \frac{1}{2 \cdot 1} \sqrt{\frac{9 \times 10^{10} \cdot 4.9 \times 10^{-4}}{9 \times 10^3}} \]

### Step 6: Calculate the frequency
Calculating the expression inside the square root:

\[ f = \frac{1}{2} \sqrt{\frac{9 \times 10^{10} \cdot 4.9 \times 10^{-4}}{9 \times 10^3}} = \frac{1}{2} \sqrt{\frac{4.9 \times 10^{7}}{1}} = \frac{1}{2} \sqrt{4.9 \times 10^{7}} \]

Calculating the square root:

\[ \sqrt{4.9 \times 10^{7}} = 7 \times 10^{3.5} \approx 7 \times 10^{3.5} \approx 4900 \]

Thus,

\[ f = \frac{1}{2} \cdot 70 = 35 \, \text{Hz} \]

### Final Answer
The lowest frequency of transverse vibration in the wire is \( f = 35 \, \text{Hz} \).

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