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

Answer 1

Answer: (D)

L = \frac{T^2YA}{4\pi^2m}

Answer 2

The wire acts as a spring. The effective spring constant 'k' of the wire can be determined from Young's Modulus (Y), cross-sectional area (A), and length (L).

Young's Modulus is given by:

$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}$.

From this, the restoring force $F = \left(\frac{YA}{L}\right)\Delta L$.

Comparing with Hooke's Law $F = kx$, the spring constant $k = \frac{YA}{L}$.

For a mass-spring system, the period of simple harmonic motion (T) is given by:

$T = 2\pi\sqrt{\frac{m}{k}}$

Substitute the value of k:

$T = 2\pi\sqrt{\frac{m}{YA/L}}$

$T = 2\pi\sqrt{\frac{mL}{YA}}$

Square both sides:

$T^2 = 4\pi^2 \frac{mL}{YA}$

Rearrange to solve for L:

$L = \frac{T^2 YA}{4\pi^2 m}$