Topic: Theory of Machines
For what value of m (mass) will resonance occur for the spring mass system given below with the given spring constants?
Excitation: $200\sin(50t)$
Springs: $2\times 10^{5}\text{ N/m}$ and $1\times 10^{5}\text{ N/m}$
- A. 85kg
- B. 150kg
- C. 100kg
- D. 120kg
Answer: D. 120kg
Explanation: The two springs are connected in parallel, so their equivalent stiffness is $k_{eq} = k_{1} + k_{2} = 2\times 10^{5} + 1\times 10^{5} = 3\times 10^{5}\text{ N/m}$. From the excitation force $200\sin(50t)$, the excitation frequency is $\omega = 50\text{ rad/sec}$. Resonance occurs when the natural frequency $\omega_{n}$ equals the excitation frequency $\omega$, i.e., $\omega = \sqrt{\frac{k_{eq}}{m}}$. Rearranging for mass: $m = \frac{k_{eq}}{\omega²} = \frac{3\times 10^{5}}{50^{2}} = \frac{300000}{2500} = 120\text{ kg}$.