A 40 cm wire having a mass of 3.2 g is stretched between two fixed supports 40.05 cm apart. In its fundamental mode, the wire vibrates at 220 Hz. If the area of cross section of the wire is `1.0 mm^2`, find its Young modulus.

Explanation

To find the Young's modulus of the wire, we will follow these steps:

### Step 1: Calculate the Change in Length (ΔL)
The original length of the wire is given as 40 cm, and the stretched length is 40.05 cm.

\[ \Delta L = \text{Stretched Length} - \text{Original Length} = 40.05 \, \text{cm} - 40 \, \text{cm} = 0.05 \, \text{cm} = 0.0005 \, \text{m} \]

### Step 2: Calculate the Strain (ε)
Strain is defined as the change in length divided by the original length.

\[ \text{Strain} \, (\epsilon) = \frac{\Delta L}{L} = \frac{0.0005 \, \text{m}}{0.4 \, \text{m}} = 0.00125 \]

### Step 3: Calculate the Mass per Unit Length (μ)
The mass of the wire is given as 3.2 g, which needs to be converted to kg.

\[ \text{Mass} = 3.2 \, \text{g} = 3.2 \times 10^{-3} \, \text{kg} \]

The length of the wire is 40 cm or 0.4 m. Thus, the mass per unit length (μ) is:

\[ \mu = \frac{\text{Mass}}{L} = \frac{3.2 \times 10^{-3} \, \text{kg}}{0.4 \, \text{m}} = 8 \times 10^{-3} \, \text{kg/m} \]

### Step 4: Calculate the Velocity (v)
The frequency of vibration is given as 220 Hz. The wavelength (λ) for the fundamental mode is given by:

\[ \lambda = 2L = 2 \times 0.4 \, \text{m} = 0.8 \, \text{m} \]

Now, we can calculate the velocity (v) using the formula:

\[ v = f \cdot \lambda = 220 \, \text{Hz} \times 0.8 \, \text{m} = 176 \, \text{m/s} \]

### Step 5: Calculate the Tension (T)
Using the relationship between velocity, tension, and mass per unit length:

\[ v = \sqrt{\frac{T}{\mu}} \implies T = \mu v^2 \]

Substituting the values:

\[ T = (8 \times 10^{-3} \, \text{kg/m}) \cdot (176 \, \text{m/s})^2 \]

Calculating \(v^2\):

\[ v^2 = 176^2 = 30976 \, \text{m}^2/\text{s}^2 \]

Now substituting back:

\[ T = 8 \times 10^{-3} \cdot 30976 = 248.81 \, \text{N} \]

### Step 6: Calculate the Young's Modulus (Y)
Young's modulus is defined as:

\[ Y = \frac{T/A}{\epsilon} \]

Where A is the area of cross-section. The area is given as \(1.0 \, \text{mm}^2\), which needs to be converted to \(m^2\):

\[ A = 1.0 \, \text{mm}^2 = 1.0 \times 10^{-6} \, \text{m}^2 \]

Now substituting the values into the Young's modulus formula:

\[ Y = \frac{T/A}{\epsilon} = \frac{248.81 \, \text{N}/(1.0 \times 10^{-6} \, \text{m}^2)}{0.00125} \]

Calculating:

\[ Y = \frac{248.81 \times 10^{6}}{0.00125} = 198928800 \, \text{N/m}^2 \approx 1.99 \times 10^{11} \, \text{N/m}^2 \]

### Final Answer
The Young's modulus of the wire is approximately:

\[ Y \approx 1.99 \times 10^{11} \, \text{N/m}^2 \]

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