Topic: Theory of Machines and Vibrations
A simple spring-mass vibrating system has a natural frequency of N. If the spring stiffness is halved and the mass is doubled, then the natural frequency will become
- A. $N/2$
- B. 2N
- C. 4N
- D. 8N
Answer: A. $N/2$
Explanation: Natural frequency $N = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$. If mass is doubled ($2m$) and stiffness is halved ($k/2$), then $N_{2} = \frac{1}{2\pi}\sqrt{\frac{k/2}{2m}} = \frac{1}{2\pi}\sqrt{\frac{k}{4m}} = \frac{1}{2}\left(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\right) = \frac{N}{2}$.