The length of a sonometer wire is 0.75 m and density `9xx 10^(3) Kg//m^(3)`. It can bear a stress of `8.1 xx 10^(8) N//m^(2)` without exceeding the elastic limit . The fundamental that can be produced in the wire , is

Options

A. 200 Hz
B. 150 Hz
C. 600 Hz
D. 450Hz

Answer

To find the fundamental frequency that can be produced in the sonometer wire, we will follow these steps:

### Step 1: Identify the given values
- Length of the sonometer wire, \( L = 0.75 \, \text{m} \)
- Density of the wire, \( \rho = 9 \times 10^3 \, \text{kg/m}^3 \)
- Maximum stress, \( \sigma = 8.1 \times 10^8 \, \text{N/m}^2 \)

### Step 2: Calculate the tension in the wire
The maximum stress is given by the formula:
\[ \sigma = \frac{F}{A} \]
Where \( F \) is the force (tension) and \( A \) is the cross-sectional area. Rearranging gives:
\[ F = \sigma \cdot A \]
However, we will not need the area directly, as we will find the wave velocity using the tension and mass per unit length.

### Step 3: Calculate the mass per unit length (\( \mu \))
The mass per unit length is given by:
\[ \mu = \frac{m}{L} \]
Where \( m \) is the mass of the wire. The mass can be calculated using the density and volume:
\[ m = \rho \cdot V = \rho \cdot (A \cdot L) \]
Thus,
\[ \mu = \frac{\rho \cdot (A \cdot L)}{L} = \rho \cdot A \]

### Step 4: Calculate the velocity of the wave in the wire
The velocity \( v \) of a wave in a stretched string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting \( \mu = \rho \cdot A \):
\[ v = \sqrt{\frac{F}{\rho \cdot A}} \]
Using the maximum tension \( F = \sigma \cdot A \):
\[ v = \sqrt{\frac{\sigma \cdot A}{\rho \cdot A}} = \sqrt{\frac{\sigma}{\rho}} \]

### Step 5: Substitute the values to find the velocity
Now substituting the values of stress and density:
\[ v = \sqrt{\frac{8.1 \times 10^8 \, \text{N/m}^2}{9 \times 10^3 \, \text{kg/m}^3}} \]
Calculating:
\[ v = \sqrt{\frac{8.1 \times 10^8}{9 \times 10^3}} = \sqrt{9 \times 10^4} = 300 \, \text{m/s} \]

### Step 6: Calculate the fundamental frequency
The fundamental frequency \( f \) is given by the formula:
\[ f = \frac{v}{2L} \]
Substituting the values:
\[ f = \frac{300 \, \text{m/s}}{2 \times 0.75 \, \text{m}} = \frac{300}{1.5} = 200 \, \text{Hz} \]

### Final Answer
The fundamental frequency that can be produced in the wire is \( 200 \, \text{Hz} \).

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