A wire of radius `10^(-3)m` and length 2m is stretched by a force of 50 N. Calculate the fundamental frequency of the note emitted by it. Density of wire is `1.6. xx 10^(3)" kg m"^(-3)`.

Answer

To calculate the fundamental frequency of the note emitted by the wire, we can follow these steps:

### Step 1: Identify the given values
- Radius of the wire, \( r = 10^{-3} \, \text{m} \)
- Length of the wire, \( L = 2 \, \text{m} \)
- Force applied, \( F = 50 \, \text{N} \)
- Density of the wire, \( \rho = 1.6 \times 10^{3} \, \text{kg/m}^3 \)

### Step 2: Calculate the cross-sectional area of the wire
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the mass per unit length (\( \mu \))
The mass per unit length \( \mu \) can be calculated using the formula:
\[ \mu = \frac{m}{L} \]
Where \( m \) is the mass of the wire. The mass can be calculated from the density and volume:
\[ m = \rho \times V \]
The volume \( V \) of the wire is given by:
\[ V = A \times L \]
Thus,
\[ m = \rho \times A \times L \]
Substituting the values:
\[ m = (1.6 \times 10^{3}) \times (\pi \times 10^{-6}) \times 2 \]
Now, substituting this back into the equation for \( \mu \):
\[ \mu = \frac{(1.6 \times 10^{3}) \times (\pi \times 10^{-6}) \times 2}{2} \]
The \( L \) cancels out:
\[ \mu = (1.6 \times 10^{3}) \times (\pi \times 10^{-6}) \]

### Step 4: Use the formula for fundamental frequency
The formula for the fundamental frequency \( f \) of a stretched wire is given by:
\[ f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \]
Substituting the values we have:
\[ f = \frac{1}{2 \times 2} \sqrt{\frac{50}{(1.6 \times 10^{3}) \times (\pi \times 10^{-6})}} \]
This simplifies to:
\[ f = \frac{1}{4} \sqrt{\frac{50}{(1.6 \times 10^{3}) \times (\pi \times 10^{-6})}} \]

### Step 5: Calculate the numerical value
First, calculate \( \mu \):
\[ \mu = (1.6 \times 10^{3}) \times (3.14 \times 10^{-6}) \approx 5.03 \times 10^{-3} \, \text{kg/m} \]
Now substitute \( \mu \) back into the frequency formula:
\[ f = \frac{1}{4} \sqrt{\frac{50}{5.03 \times 10^{-3}}} \]
Calculating the inside of the square root:
\[ \frac{50}{5.03 \times 10^{-3}} \approx 9.94 \times 10^{3} \]
Taking the square root:
\[ \sqrt{9.94 \times 10^{3}} \approx 99.7 \]
Now calculate \( f \):
\[ f = \frac{1}{4} \times 99.7 \approx 24.93 \, \text{Hz} \]

### Final Result
The fundamental frequency of the note emitted by the wire is approximately \( 25 \, \text{Hz} \).

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