An aircraft wing model is idealized as a cantilever beam with tip mass $M = 50\text{ kg}$. The equivalent bending stiffness at the tip is $k = 20000\text{ N/m}$. What is the undamped natural frequency in Hertz (Hz)? (Take $\pi \approx 3.1416$, round to two decimal places).

Answer Key:

3.18

### Solution: Circular natural frequency $\omega_n = \sqrt{\frac{k}{M}} = \sqrt{\frac{20000}{50}} = \sqrt{400} = 20\text{ rad/s}$. Natural frequency in Hz: $f_n = \frac{\omega_n}{2\pi} = \frac{20}{2 \times 3.1416} = \frac{20}{6.2832} \approx 3.183 \approx 3.18\text{ Hz}$. Correct: 3.18 (range 3.15 to 3.22).

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