Topic: Theory of Machines
As shown in the figure a mass of 100 kg is held between two springs. The natural frequency of vibration of the system in cycle/s is (given both springs have stiffness 20 kN/m)
- A. $1/2\pi$
- B. $5/\pi$
- C. $10/\pi$
- D. $20/\pi$
Answer: C. $10/\pi$
Explanation: The mass is held between two springs which are effectively in a parallel configuration. The equivalent stiffness $k_{eq} = k_1 + k_2 = 20 + 20 = 40$ kN/m = $40000$ N/m. The natural circular frequency is $\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{40000}{100}} = 20$ rad/sec. Since $\omega_n = 2\pi f_n$, the natural frequency $f_n = \frac{20}{2\pi} = \frac{10}{\pi}$ cycles/s (Hz).