Q: An object of mass m is suspended at the end of a massless wire of length L and area of cross-section A . Young modulus of the material of wire is Y . If the mass is pulled down slightly its frequency of oscillation along the vertical direction is

(a) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{m L}{Y A}}$

(b) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{Y A}{m L}}$

(c) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{m A}{Y L}}$

(d) $\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{Y L}{m A}}$

**Click to See Solution : **

$\displaystyle Y = \frac{Stress}{Strain}$

$\displaystyle Y = \frac{F/A}{(\Delta L/L)}$

$\displaystyle F = (\frac{Y A}{L})\Delta L$

Equivalent Stiffness constant $\displaystyle k_{eq} = \frac{Y A}{L}$

$\displaystyle f = \frac{1}{2 \pi}\sqrt{\frac{Y A}{m L}}$