Subtopic: Free & Forced Vibrations of Continuous Beams
Question Statement:
The natural frequency of the system suspended by two identical vertical springs each of stiffness $k = 1000$ N/m carrying a mass $m = 2$ kg in parallel is _________ rad/s (round off to 1 decimal place).
Official Answer Key:
31.6
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2017 Q.55)**
**1. Intuitive Physical Setup & Given Parameters**:
- Two identical springs in parallel: $k_1 = k_2 = 1000\text{ N/m}$.
- Suspended mass: $m = 2\text{ kg}$.
**2. Step-by-Step Vibration Derivation**:
1. **Equivalent Spring Stiffness ($k_{eq}$)**:
For two springs connected in parallel:
k_{eq} = k_1 + k_2 = 1000 + 1000 = 2000\text{ N/m}
2. **Angular Natural Frequency ($\omega_n$)**:
\omega_n = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{2000}{2}} = \sqrt{1000} \approx 31.62277\text{ rad/s}
Rounding to one decimal place:
\omega_n \approx 31.6\text{ rad/s}
This falls right in the middle of the official acceptable interval **[31.0, 32.0] rad/s**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: **$\sqrt{1000} \approx 31.62$ rad/s** (Official Range: [31.0, 32.0]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **31.6**
- **Key Formula**: $\omega_n = \sqrt{\frac{k_1 + k_2}{m}}$ for parallel springs.