For a damped spring-mass system, mass $m = 10$ kg, stiffness $k = 10^3$ N/m, and damping coefficient $c = 20$ kg/s. The ratio of the amplitude of oscillation of the first cycle to that of the fifth cycle is _________ (round off to 1 decimal place).
Official Answer Key:
12.45
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.64 / AE Q.54)**
**1. Intuitive Physical Setup & Given Parameters**:
- Mass: $m = 10\text{ kg}$
- Stiffness: $k = 10^3\text{ N/m} = 1000\text{ N/m}$
- Damping coefficient: $c = 20\text{ kg/s} = 20\text{ N s/m}$
- Target: Amplitude ratio $\frac{x_1}{x_5}$ over $n = 4$ completed cycles.
**2. Step-by-Step Vibration Derivation**:
1. **Critical Damping ($c_c$) and Damping Ratio ($\zeta$)**:
$\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{1000}{10}} = \sqrt{100} = 10\text{ rad/s}$
$c_c = 2 m \omega_n = 2 \times 10 \times 10 = 200\text{ N s/m}$
$\zeta = \frac{c}{c_c} = \frac{20}{200} = 0.10$
2. **Logarithmic Decrement ($\delta$)**:
$\delta = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}} = \frac{2\pi (0.10)}{\sqrt{1 - 0.01}} = \frac{0.2\pi}{\sqrt{0.99}} = \frac{0.6283185}{0.994987} \approx 0.63148$
3. **Amplitude Decay over 4 Cycles**:
Between cycle 1 ($x_1$) and cycle 5 ($x_5$), exactly $n = 5 - 1 = 4$ cycles elapse:
$\ln\left( \frac{x_1}{x_5} \right) = 4 \delta = 4 \times 0.63148 = 2.5259$
$\frac{x_1}{x_5} = e^{2.5259} \approx 12.502$
Rounding to one decimal place:
$\frac{x_1}{x_5} \approx 12.5$
This lies right in the center of the official acceptable interval **[12.3, 12.6]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Evaluation**: **$e^{4\delta} \approx 12.50$** (Acceptable Range: [12.3, 12.6]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **12.45** (Acceptable Range: **12.3 to 12.6**)
- **Key Formula**: $\frac{x_1}{x_{n+1}} = e^{n \delta}$ where $\delta = \frac{2\pi \zeta}{\sqrt{1-\zeta^2}}$.
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