GATE AE 2019 • Q65 NAT • 2 Marks Aerospace Structures

Aerospace Structures

Structural Dynamics & Aeroelasticity

Subtopic: Free & Forced Vibrations of Continuous Beams

Question Statement:

For the system of springs and masses shown below, $k = 1250$ N/m and $m = 10$ kg. The highest natural frequency, $\omega$, of the system is _________ radians/s (round off to the nearest integer).

Official Answer Key:

25.0

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2019 Q.65 / AE Q.55)** **1. Intuitive Physical Setup & Given Parameters**: - 2-DOF spring-mass system: - Spring stiffness: $k = 1250\text{ N/m}$ - Mass: $m = 10\text{ kg}$ - Base ratio: $\frac{k}{m} = \frac{1250}{10} = 125\text{ s}^{-2}$. **2. Step-by-Step Vibration Derivation**: 1. **Equations of Motion & Characteristic Determinant**: For the standard 2-DOF symmetric/coupled spring-mass configuration with springs $k, 2k, k$: [M] = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix}, \quad [K] = \begin{bmatrix} 3k & -2k \\ -2k & 3k \end{bmatrix} The characteristic equation is: \det([K] - \omega^2 [M]) = 0 \implies \begin{vmatrix} 3k - m\omega^2 & -2k \\ -2k & 3k - m\omega^2 \end{vmatrix} = 0 2. **Solving the Eigenvalue Problem**: (3k - m\omega^2)^2 - 4k^2 = 0 (3k - m\omega^2 - 2k)(3k - m\omega^2 + 2k) = 0 (k - m\omega^2)(5k - m\omega^2) = 0 3. **Two Natural Frequencies**: - Lower natural frequency (in-phase mode): \omega_1^2 = \frac{k}{m} = 125 \implies \omega_1 = \sqrt{125} \approx 11.18\text{ rad/s} - Highest natural frequency (out-of-phase mode): \omega_2^2 = \frac{5k}{m} = 5 \times 125 = 625\text{ rad/s}^2 \omega_2 = \sqrt{625} = 25\text{ rad/s} The highest natural frequency is exactly **25 rad/s**. **3. Comprehensive Options & Distractor Analysis**: - **Question Type**: Numerical Answer Type (NAT). - **Exact Analytical Frequency**: **25 rad/s** (Acceptable Range: [25, 25]). **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **25.0** - **Exam Rule**: In symmetric 2-DOF systems, the anti-symmetric (out-of-phase) mode always gives the highest natural frequency: $\omega_{\max} = \sqrt{\frac{k + 2k_c}{m}}$.
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