A steel wire of length `1m`, `mass 0.1kg` and uniform cross-sectional area `10^(-6)m^(2)` is rigidly fixed at both ends. The temperature of the wire is lowered by `20^(@)C`. If transverse calculate the frequency of the fundamental mode of vibration.
Given for steel `Y = 2 xx 10^(11)N//m^(2)`
`alpha = 1.21 xx 10^(-5) per ^(@)C`

Explanation

To calculate the frequency of the fundamental mode of vibration of a steel wire that is fixed at both ends and has undergone a temperature change, we can follow these steps:

### Step 1: Understand the Problem
We have a steel wire with the following parameters:
- Length (L) = 1 m
- Mass (m) = 0.1 kg
- Cross-sectional area (A) = \(10^{-6} \, m^2\)
- Young's modulus (Y) = \(2 \times 10^{11} \, N/m^2\)
- Coefficient of linear expansion (α) = \(1.21 \times 10^{-5} \, /^{\circ}C\)
- Temperature change (Δθ) = -20°C

### Step 2: Calculate the Change in Length (ΔL)
The change in length (ΔL) due to the temperature change can be calculated using the formula:
\[ \frac{\Delta L}{L} = \alpha \Delta \theta \]
Substituting the values:
\[ \frac{\Delta L}{1} = 1.21 \times 10^{-5} \times (-20) \]
\[ \Delta L = 1.21 \times 10^{-5} \times (-20) \times 1 = -2.42 \times 10^{-4} \, m \]

### Step 3: Calculate the Tension (T) in the Wire
The tension in the wire can be calculated using Young's modulus:
\[ T = Y \cdot A \cdot \frac{\Delta L}{L} \]
Substituting the values:
\[ T = 2 \times 10^{11} \cdot 10^{-6} \cdot \left(-2.42 \times 10^{-4}\right) \]
\[ T = 2 \times 10^{11} \cdot 10^{-6} \cdot (-2.42 \times 10^{-4}) = -4.84 \times 10^{1} \, N \]
(Note: The negative sign indicates a decrease in length, but tension is a magnitude, so we take the absolute value.)

### Step 4: Calculate the Mass per Unit Length (μ)
The mass per unit length (μ) is given by:
\[ \mu = \frac{m}{L} = \frac{0.1 \, kg}{1 \, m} = 0.1 \, kg/m \]

### Step 5: Calculate the Frequency of the Fundamental Mode (f)
The frequency of the fundamental mode of vibration is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ f = \frac{1}{2 \cdot 1} \sqrt{\frac{4.84 \times 10^{1}}{0.1}} \]
\[ f = \frac{1}{2} \sqrt{4.84 \times 10^{2}} = \frac{1}{2} \cdot 22 = 11 \, Hz \]

### Final Answer
The frequency of the fundamental mode of vibration of the steel wire is approximately **11 Hz**.

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