A critically damped single degree of freedom spring-mass-damper system used in a door closing mechanism becomes overdamped due to softening of the spring with extended use. If the new damping ratio ($\zeta_{new}$) for the overdamped condition is 1.2, the ratio of the original spring stiffness to the new spring stiffness ($k_{orig} / k_{new}$), assuming that the other parameters remain unchanged, is _________ (round off to two decimal places).
Official Answer Key:
1.44
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2020 Q.51)**
**1. Intuitive Physical Setup & Given Parameters**:
- Original state: Critically damped $\implies \zeta_{orig} = 1.0$.
- New state: Overdamped $\implies \zeta_{new} = 1.2$.
- Mass $m$ and damper coefficient $c$ remain constant ($c_{orig} = c_{new} = c$, $m_{orig} = m_{new} = m$).
- Spring softens from $k_{orig}$ to $k_{new}$.
**2. Step-by-Step Vibration Mechanics Derivation**:
1. **Damping Ratio Formulation**:
For a single degree of freedom system:
$\zeta = \frac{c}{c_c} = \frac{c}{2\sqrt{k m}}$
2. **Relating Damping Ratio to Stiffness $k$**:
Since $c$ and $m$ are constant:
$\zeta \propto \frac{1}{\sqrt{k}} \implies \zeta^2 \propto \frac{1}{k} \implies k \propto \frac{1}{\zeta^2}$
3. **Ratio of Original to New Stiffness**:
$\frac{k_{orig}}{k_{new}} = \left( \frac{\zeta_{new}}{\zeta_{orig}} \right)^2$
4. **Substituting Given Numerical Values**:
$\frac{k_{orig}}{k_{new}} = \left( \frac{1.2}{1.0} \right)^2 = (1.2)^2 = 1.44$
The stiffness ratio $k_{orig} / k_{new}$ is exactly **1.44**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Exact Analytical Evaluation**: **1.44** (Acceptable Range: [1.43, 1.45]).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **1.44**
- **Key Formula**: $\zeta = \frac{c}{2\sqrt{km}} \implies \frac{k_{orig}}{k_{new}} = \left(\frac{\zeta_{new}}{\zeta_{orig}}\right)^2$.
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