Topic: Theory of Machines / Vibrations
A mass of 2 kg is attached to two identical springs each with stiffness k = 40 kN/m as shown in the figure. Under frictionless condition, the natural frequency of the system in Hz is close to
- A. 32
- B. 16
- C. 24
- D. 200
Answer: A. 32
Explanation: The two springs are in parallel, so equivalent stiffness $k_{eq} = k + k = 2k = 2 \times 40\times 10^{3} = 80\times 10^{3}~N/m$. The natural frequency in Hz is $f_{n} = \frac{1}{2\pi}\sqrt{\frac{k_{eq}}{m}} = \frac{1}{2\pi}\sqrt{\frac{80000}{2}} = \frac{1}{2\pi}\sqrt{40000} = \frac{200}{2\pi} = \frac{100}{\pi} \approx 31.83~Hz$, which is close to 32 Hz.