Question

A wire of negligible mass having uniform area of cross section \( A \) and Young modulus \( Y \) is used to suspend a point mass \( m \). The point mass executes simple harmonic motion in a vertical plane with a period \( T \), then the length of the wire is:

• A. \( L=\frac{TY^2A}{4\pi^2m} \)
• B. \( L=\frac{T^2YA}{4\pi m^2} \)
• C. \( L=\frac{TY^2A}{4\pi m^2} \)
• D. \( L=\frac{T^2YA}{4\pi^2m} \)

Hint:

For oscillations of a mass suspended from an elastic wire, the effective spring constant is \( k=\frac{YA}{L} \).

Answer 1

The Correct Option is D

Step 1: Relate wire extension with restoring force.
For a wire of length \(L\), area \(A\), and Young's modulus \(Y\):
\[ Y=\frac{\text{stress}}{\text{strain}} \]
\[ Y=\frac{F/A}{x/L} \]

Step 2: Rearrange the formula.
\[ Y=\frac{FL}{Ax} \]
So:
\[ F=\frac{YA}{L}x \]

Step 3: Compare with Hooke's law.
For SHM:
\[ F=kx \]
Thus, effective force constant is:
\[ k=\frac{YA}{L} \]

Step 4: Use time period formula of SHM.
\[ T=2\pi\sqrt{\frac{m}{k}} \]

Step 5: Substitute value of \(k\).
\[ T=2\pi\sqrt{\frac{m}{YA/L}} \]
\[ T=2\pi\sqrt{\frac{mL}{YA}} \]

Step 6: Square both sides.
\[ T^2=4\pi^2\frac{mL}{YA} \]

Step 7: Solve for \(L\).
\[ L=\frac{T^2YA}{4\pi^2m} \]
\[ \boxed{\frac{T^2YA}{4\pi^2m}} \]