A wire of density `9 xx 10^(3) kg//m^(3)` is stretched between two clamps ` 1m` apart and is stretched to an extension of `4.9 xx 10^(-4) m`. Young's modulus of material is `9 xx 10^(10) N//m^(2)`.

Options

A. The lowest frequency of standing wave is `35 Hz`
B. The frequency of `1 st` overtone is `70 Hz`
C. The frequency of `1 st` overtone is `105 Hz`
D. The stress in the wire is `4.41 xx 10^(7) N//m^(2)`

Explanation

To solve the problem step by step, we will calculate the fundamental frequency of the wire and the first overtone frequency, as well as the stress in the wire.

### Step 1: Calculate the Mass per Unit Length (μ)
The mass per unit length (μ) of the wire can be calculated using the formula:
\[ \mu = \rho \cdot A \]
where:
- \(\rho\) = density of the wire = \(9 \times 10^3 \, \text{kg/m}^3\)
- \(A\) = cross-sectional area of the wire (not provided, but we will express it in terms of Young's modulus and other variables).

### Step 2: Calculate the Tension (T)
The tension (T) in the wire can be related to Young's modulus (Y), the extension (\(\Delta L\)), and the original length (L) using the formula:
\[ T = Y \cdot \frac{\Delta L}{L} \cdot A \]
Given:
- \(Y = 9 \times 10^{10} \, \text{N/m}^2\)
- \(\Delta L = 4.9 \times 10^{-4} \, \text{m}\)
- \(L = 1 \, \text{m}\)

### Step 3: Calculate the Velocity (v)
The velocity of the wave on the wire can be calculated using:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the expression for T from Step 2:
\[ v = \sqrt{\frac{Y \cdot \frac{\Delta L}{L} \cdot A}{\rho \cdot A}} = \sqrt{\frac{Y \cdot \Delta L}{\rho \cdot L}} \]

### Step 4: Calculate the Fundamental Frequency (f₀)
The fundamental frequency (f₀) for a wire fixed at both ends is given by:
\[ f_0 = \frac{v}{2L} \]
Substituting the expression for v from Step 3:
\[ f_0 = \frac{1}{2L} \sqrt{\frac{Y \cdot \Delta L}{\rho \cdot L}} \]
Now substituting the known values:
- \(Y = 9 \times 10^{10} \, \text{N/m}^2\)
- \(\Delta L = 4.9 \times 10^{-4} \, \text{m}\)
- \(\rho = 9 \times 10^3 \, \text{kg/m}^3\)
- \(L = 1 \, \text{m}\)

### Step 5: Simplify and Calculate
Substituting the values into the frequency formula:
\[ f_0 = \frac{1}{2 \cdot 1} \sqrt{\frac{9 \times 10^{10} \cdot 4.9 \times 10^{-4}}{9 \times 10^3 \cdot 1}} \]
\[ = \frac{1}{2} \sqrt{\frac{9 \times 4.9 \times 10^{10 - 3}}{9}} = \frac{1}{2} \sqrt{4.9 \times 10^{7}} = \frac{1}{2} \cdot \sqrt{4.9} \cdot 10^{3.5} \]
Calculating \(\sqrt{4.9} \approx 2.213\):
\[ f_0 \approx \frac{1}{2} \cdot 2.213 \cdot 10^{3.5} \approx 35 \, \text{Hz} \]

### Step 6: Calculate the First Overtone Frequency (f₁)
The first overtone frequency (f₁) is given by:
\[ f_1 = 2f_0 \]
Thus,
\[ f_1 = 2 \cdot 35 \, \text{Hz} = 70 \, \text{Hz} \]

### Step 7: Calculate the Stress in the Wire
Stress (\(\sigma\)) in the wire can be calculated using:
\[ \sigma = \frac{T}{A} \]
Using the relation from Young's modulus:
\[ \sigma = Y \cdot \frac{\Delta L}{L} \]
Substituting the values:
\[ \sigma = 9 \times 10^{10} \cdot \frac{4.9 \times 10^{-4}}{1} = 4.41 \times 10^{7} \, \text{N/m}^2 \]

### Summary of Results
- Fundamental Frequency \(f_0 \approx 35 \, \text{Hz}\)
- First Overtone Frequency \(f_1 \approx 70 \, \text{Hz}\)
- Stress in the Wire \(\sigma \approx 4.41 \times 10^{7} \, \text{N/m}^2\)