A metal wire of length `L_1` and area of cross section A is ttached to a rigid support. Another metal wire of length `L_2` and of the same cross sectional area is attached to the free end of the first wire. A body of mass M is then suspended from the free end of the second wire, if `Y_1` and `Y_2` are the Young's moduli of the wires respectively the effective force constant of the system of two wires is

Options

A. `(Y_1Y_2A)/(2(Y_1L_2+Y_2L_1))`
B. `(Y_1Y_2A)/(L_1L_2)^(1//2)`
C. `(Y_1Y_2A)/((Y_1L_2+Y_2L_1))`
D. `((Y_1Y_2)^(1//2)A)/((L_1L_2)^(1//2))`

Answer

To find the effective force constant of the system of two wires, we can follow these steps:

### Step 1: Understand the System
We have two metal wires of lengths \(L_1\) and \(L_2\) with the same cross-sectional area \(A\). The Young's moduli for these wires are \(Y_1\) and \(Y_2\) respectively. A mass \(M\) is suspended from the free end of the second wire.

### Step 2: Define the Force Constants
The force constant \(k\) for a wire can be derived from the Young's modulus formula:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Rearranging gives:
\[ Y = \frac{F L}{A \Delta L} \]
From this, we can express the force constant \(k\) as:
\[ k = \frac{Y A}{L} \]

### Step 3: Calculate Force Constants for Each Wire
For the first wire (length \(L_1\) and Young's modulus \(Y_1\)):
\[ k_1 = \frac{Y_1 A}{L_1} \]
For the second wire (length \(L_2\) and Young's modulus \(Y_2\)):
\[ k_2 = \frac{Y_2 A}{L_2} \]

### Step 4: Combine the Force Constants
Since the wires are in series, the effective force constant \(k_{\text{equiv}}\) can be calculated using the formula for series springs:
\[ \frac{1}{k_{\text{equiv}}} = \frac{1}{k_1} + \frac{1}{k_2} \]
Substituting \(k_1\) and \(k_2\):
\[ \frac{1}{k_{\text{equiv}}} = \frac{L_1}{Y_1 A} + \frac{L_2}{Y_2 A} \]

### Step 5: Simplify the Equation
Combining the fractions:
\[ \frac{1}{k_{\text{equiv}}} = \frac{L_1 Y_2 + L_2 Y_1}{Y_1 Y_2 A} \]
Taking the reciprocal gives:
\[ k_{\text{equiv}} = \frac{Y_1 Y_2 A}{L_1 Y_2 + L_2 Y_1} \]

### Final Result
Thus, the effective force constant of the system of two wires is:
\[ k_{\text{equiv}} = \frac{Y_1 Y_2 A}{L_1 Y_2 + L_2 Y_1} \]

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