A sonometer wire of length `1.5m` is made of steel. The tension in it produces an elastic strain of `1%`. What is the fundamental frequency of steel if density and elasticity of steel are `7.7 xx 10^(3) kg//m^(3)` and `2.2 xx 10^(11) N//m^(2)` respectively ?

Options

A. 178.2 Hz
B. 200.5 Hz
C. 770 Hz
D. 188.5 Hz

Answer

To find the fundamental frequency of the sonometer wire made of steel, we will follow these steps:

### Step 1: Understand the Formula
The fundamental frequency \( f \) of a vibrating string (or wire) is given by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( L \) is the length of the wire,
- \( T \) is the tension in the wire,
- \( \mu \) is the linear mass density of the wire.

### Step 2: Calculate the Linear Mass Density \( \mu \)
The linear mass density \( \mu \) can be expressed in terms of the density \( \rho \) and the cross-sectional area \( A \) of the wire:
\[ \mu = \frac{m}{L} = \rho \cdot A \]
However, we do not have the area \( A \) directly, but we can use the relationship of tension and stress.

### Step 3: Calculate the Tension \( T \)
The tension \( T \) in the wire can be related to the stress and strain. The stress \( \sigma \) is given by:
\[ \sigma = \frac{T}{A} \]
And stress is also defined as:
\[ \sigma = E \cdot \text{strain} \]
where \( E \) is the Young's modulus of the material, and the strain is given as \( 1\% = 0.01 \).

Thus, we can write:
\[ \frac{T}{A} = E \cdot \text{strain} \]
From this, we can express \( T \):
\[ T = E \cdot \text{strain} \cdot A \]
Substituting the values:
- \( E = 2.2 \times 10^{11} \, \text{N/m}^2 \)
- Strain = \( 0.01 \)

This gives:
\[ T = 2.2 \times 10^{11} \cdot 0.01 \cdot A = 2.2 \times 10^{9} \cdot A \]

### Step 4: Substitute \( T \) and \( \mu \) into the Frequency Formula
Now substituting \( T \) and \( \mu \) into the frequency formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} = \frac{1}{2L} \sqrt{\frac{2.2 \times 10^{9} \cdot A}{\rho \cdot A}} \]
The area \( A \) cancels out:
\[ f = \frac{1}{2L} \sqrt{\frac{2.2 \times 10^{9}}{\rho}} \]

### Step 5: Substitute Known Values
Now substituting the known values:
- \( L = 1.5 \, \text{m} \)
- \( \rho = 7.7 \times 10^{3} \, \text{kg/m}^3 \)

Thus:
\[ f = \frac{1}{2 \cdot 1.5} \sqrt{\frac{2.2 \times 10^{9}}{7.7 \times 10^{3}}} \]

### Step 6: Calculate the Frequency
Calculating the values:
1. Calculate \( \frac{2.2 \times 10^{9}}{7.7 \times 10^{3}} \):
\[ \frac{2.2 \times 10^{9}}{7.7 \times 10^{3}} \approx 285714.29 \]
2. Now take the square root:
\[ \sqrt{285714.29} \approx 534.52 \]
3. Finally calculate \( f \):
\[ f = \frac{1}{3} \cdot 534.52 \approx 178.17 \, \text{Hz} \]

### Final Answer
Thus, the fundamental frequency of the steel sonometer wire is approximately:
\[ \boxed{178.2 \, \text{Hz}} \]