© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify Known Quantities
• Mass of wire, $m = 6.0\text{ g} = 0.006\text{ kg}$
• Length of wire, $l = 60\text{ cm} = 0.60\text{ m}$
• Area of cross-section, $A = 1.0\text{ mm}^2 = 1.0 \times 10^{-6}\text{ m}^2$
• Speed of transverse wave, $v = 90\text{ m s}^{-1}$
• Young’s modulus of wire, $Y = 16 \times 10^{11}\text{ N m}^{-2} = 1.6 \times 10^{12}\text{ N m}^{-2}$
We are asked to find the extension $\Delta l$ of the wire over its natural length.
Step 2: Calculate Linear Mass Density
The linear mass density $\mu$ (mass per unit length) is given by:
$\displaystyle \mu = \frac{m}{l}.$
Substituting the values:
$\displaystyle \mu = \frac{0.006}{0.60} = 0.01\text{ kg m}^{-1}.$
Step 3: Relate Wave Speed and Tension
The speed $v$ of a transverse wave on a stretched string (or wire) is:
$\displaystyle v = \sqrt{\frac{T}{\mu}},$
where $T$ is the tension in the wire and $\mu$ is the linear mass density.
Rearranging for $T$:
$\displaystyle T = v^{2} \,\mu.$
Substitute $v = 90\text{ m s}^{-1}$ and $\mu = 0.01\text{ kg m}^{-1}$:
$\displaystyle T = (90)^{2} \times 0.01 = 8100 \times 0.01 = 81\text{ N}.$
Step 4: Use Young’s Modulus to Find Extension
Young’s modulus $Y$ for a wire is given by:
$\displaystyle Y = \frac{\frac{F}{A}}{\frac{\Delta l}{l}} \quad \Longrightarrow \quad Y \cdot \frac{\Delta l}{l} = \frac{F}{A},$
where $F$ is the stretching force (in this case, the tension $T$), $A$ is the cross-sectional area, and $\Delta l$ is the extension in length from its natural length $l$.
Thus,
$\displaystyle \Delta l = \frac{T \, l}{A \, Y}.$
Step 5: Substitute Numerical Values
Now substitute $T = 81\text{ N}$, $l = 0.60\text{ m}$, $A = 1.0 \times 10^{-6}\text{ m}^2$, and $Y = 1.6 \times 10^{12}\text{ N m}^{-2}$:
$\displaystyle \Delta l = \frac{(81)\,(0.60)}{\bigl(1.0 \times 10^{-6}\bigr)\,\bigl(1.6 \times 10^{12}\bigr)}.$
First compute the numerator:
$\displaystyle \text{Numerator} = 81 \times 0.60 = 48.6.$
Then the denominator:
$\displaystyle \text{Denominator} = \bigl(1.0 \times 10^{-6}\bigr)\,\bigl(1.6 \times 10^{12}\bigr) = 1.6 \times 10^{6}.$
So,
$\displaystyle \Delta l = \frac{48.6}{1.6 \times 10^{6}} = 3.0375 \times 10^{-5}\text{ m}.$
Converting to millimeters:
$\displaystyle 3.0375 \times 10^{-5}\text{ m} = 3.0375 \times 10^{-2}\text{ mm} \approx 0.03\text{ mm}.$
Step 6: Final Answer
The extension of the wire over its natural length is approximately $0.03\text{ mm}.$
