Topic: Theory of Machines / Vibrations
A mass of 100 kg is held between two springs as shown in figure. The natural frequency of vibration of the system of cycle/second is-
- A. $\frac{10}{\pi}$
- B. $\frac{5}{\pi}$
- C. $\frac{1}{2\pi}$
- D. $\frac{20}{\pi}$
Answer: A. $\frac{10}{\pi}$
Explanation: Both the springs are in parallel connection. Equivalent stiffness $k_{eff} = k_{1} + k_{2} = 20 + 20 = 40 \text{ kN/m} = 40 \times 10^{3} \text{ N/m}$. Natural frequency of vibration of the system in cycles/second ($f_{n}$) is given by $f_{n} = \frac{1}{2\pi}\sqrt{\frac{k_{eff}}{m}} = \frac{1}{2\pi}\sqrt{\frac{40 \times 10^{3}}{100}} = \frac{1}{2\pi}\sqrt{400} = \frac{20}{2\pi} = \frac{10}{\pi}$.