To solve the problem of finding the extension of the wire over its natural length, we will follow these steps:
### Step 1: Understand the relationship between wave speed, tension, and mass per unit length
The speed of a transverse wave on a wire is given by the formula:
\[
v = \sqrt{\frac{T}{\mu}}
\]
where:
- \( v \) is the speed of the wave,
- \( T \) is the tension in the wire,
- \( \mu \) is the mass per unit length of the wire.
### Step 2: Calculate the mass per unit length (\( \mu \))
The mass of the wire is given as 6 g, which we convert to kilograms:
\[
\text{Mass} = 6 \, \text{g} = 6 \times 10^{-3} \, \text{kg}
\]
The length of the wire is given as 120 cm, which we convert to meters:
\[
\text{Length} = 120 \, \text{cm} = 1.2 \, \text{m}
\]
Now, we can calculate \( \mu \):
\[
\mu = \frac{\text{Mass}}{\text{Length}} = \frac{6 \times 10^{-3} \, \text{kg}}{1.2 \, \text{m}} = 5 \times 10^{-3} \, \text{kg/m}
\]
### Step 3: Rearrange the wave speed formula to find tension (\( T \))
We know the speed \( v = 100 \, \text{m/s} \). Squaring both sides of the wave speed equation gives:
\[
v^2 = \frac{T}{\mu} \implies T = \mu v^2
\]
Substituting the values we have:
\[
T = (5 \times 10^{-3} \, \text{kg/m}) \times (100 \, \text{m/s})^2 = (5 \times 10^{-3}) \times (10^4) = 50 \, \text{N}
\]
### Step 4: Use Young's modulus to find the extension (\( \Delta L \))
Young's modulus (\( Y \)) is given as \( 10^{12} \, \text{N/m}^2 \). The formula for extension in terms of Young's modulus is:
\[
Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta L/L_0}
\]
Rearranging gives:
\[
\Delta L = \frac{T \cdot L_0}{A \cdot Y}
\]
Where:
- \( A = 1.2 \, \text{mm}^2 = 1.2 \times 10^{-6} \, \text{m}^2 \)
- \( L_0 = 1.2 \, \text{m} \)
Substituting the values:
\[
\Delta L = \frac{50 \, \text{N} \cdot 1.2 \, \text{m}}{1.2 \times 10^{-6} \, \text{m}^2 \cdot 10^{12} \, \text{N/m}^2}
\]
Calculating the denominator:
\[
1.2 \times 10^{-6} \cdot 10^{12} = 1.2 \times 10^6
\]
Now substituting:
\[
\Delta L = \frac{60}{1.2 \times 10^6} = 5 \times 10^{-5} \, \text{m}
\]
### Step 5: Convert the extension to mm
To convert meters to millimeters:
\[
\Delta L = 5 \times 10^{-5} \, \text{m} \times 1000 = 0.05 \, \text{mm}
\]
### Final Answer
The extension of the wire over its natural length is:
\[
\Delta L = 0.05 \, \text{mm}
\]
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