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Step-by-Step Solution
Step 1: Understand the Problem
We have two wires of the same length and radius (hence same cross-sectional area), joined end to end in series, and subjected to a common axial force. Each wire is made of a different material with Young’s moduli $Y_1$ and $Y_2$. We need to find the effective or equivalent Young’s modulus $Y$ of the combined system.
Step 2: Recall the Concept of Series Combination
When wires (or rods) are connected in series and subjected to the same tensile force $F$, the total extension $\Delta l$ is the sum of their individual extensions. That is:
$\Delta l = \Delta l_1 + \Delta l_2$.
Since each wire has length $l$ and the overall length of the combination is $2l$, we treat it as a single entity of length $l_{\mathrm{eq}} = 2l$ and an effective Young’s modulus $Y_{\mathrm{eq}}$.
Step 3: Express the Extension in Terms of Young’s Modulus
For a wire of length $l$, cross-sectional area $A$, and Young’s modulus $Y$, the extension under a force $F$ is:
$\Delta l = \frac{Fl}{AY}.$
Thus, the extension is directly proportional to $\frac{l}{Y}$ for a constant force $F$ and area $A$.
Step 4: Write the Condition for the Equivalent Wire
The total extension of the combined wire (effective length $2l$) under the same force $F$ can be written as:
$\Delta l_{\mathrm{eq}} = \Delta l_1 + \Delta l_2.$
But also, for the equivalent wire with length $l_{\mathrm{eq}} = 2l$ and Young’s modulus $Y$, the total extension is:
$\Delta l_{\mathrm{eq}} = \frac{F \cdot (2l)}{A \cdot Y}.$
Step 5: Use the Series Relation
Since each wire has length $l$, the extension of the first wire is $\Delta l_1 = \frac{Fl}{A Y_1}$ and that of the second wire is $\Delta l_2 = \frac{Fl}{A Y_2}$. Therefore:
$\Delta l_{\mathrm{eq}} = \frac{Fl}{A Y_1} + \frac{Fl}{A Y_2}.$
Equating this to the expression for the equivalent wire, we get:
$\frac{F \cdot (2l)}{A \cdot Y} = \frac{Fl}{A Y_1} + \frac{Fl}{A Y_2}.$
Step 6: Simplify to Find the Equivalent Young’s Modulus
Canceling common factors of $\frac{Fl}{A}$ on both sides, we obtain:
$\frac{2}{Y} = \frac{1}{Y_1} + \frac{1}{Y_2}.$
Taking reciprocals carefully, we get:
$\frac{1}{\frac{2}{Y}} = \frac{1}{\frac{1}{Y_1} + \frac{1}{Y_2}} \quad \Rightarrow \quad \frac{Y}{2} = \frac{1}{\frac{1}{Y_1} + \frac{1}{Y_2}}.$
Thus, after algebraic manipulation, the effective Young’s modulus turns out to be:
$Y = \frac{2\,Y_1\,Y_2}{Y_1 + Y_2}.$
Final Answer
$Y = \frac{2\,Y_1\,Y_2}{Y_1 + Y_2}.$
