Topic: Theory of Machines
A cantilever beam of negligible mass has a mass m at its free end. If the length of the cantilever is halved, what is the factor by which its natural frequency is increased?
- A. $\frac{1}{2\sqrt{2}}$
- B. $\frac{1}{2}$
- C. 2
- D. $2\sqrt{2}$
Answer: D. $2\sqrt{2}$
Explanation: Equivalent stiffness of a cantilever beam with a mass at the free end is $k_{eq} = \frac{3EI}{L^3}$.
Natural frequency $\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{3EI}{mL^3}}$, so $\omega_n \propto L^{-1.5}$.
When length is halved ($L' = L/2$), the new natural frequency is $\omega_n' \propto (L/2)^{-1.5} = L^{-1.5} \times 2^{1.5} = \omega_n \times \sqrt{8} = 2\sqrt{2} \omega_n$.