Question:medium
A vibrating system has a critical damping coefficient \(C_c = 350 \, \text{N}\cdot\text{s/m}\) and an actual damping coefficient \(C = 35 \, \text{N}\cdot\text{s/m}\). The logarithmic decrement of the system is approximately:
A vibrating system has a critical damping coefficient \(C_c = 350 \, \text{N}\cdot\text{s/m}\) and an actual damping coefficient \(C = 35 \, \text{N}\cdot\text{s/m}\). The logarithmic decrement of the system is approximately:
Show Hint
For small damping ratios (\(\zeta<0.1\)), the logarithmic decrement can be quickly estimated using \(\delta \approx 2\pi\zeta\). The error in this approximation is negligible for calculation-heavy exams.
Updated On: Feb 14, 2026
- 0.10
- 0.31
- 0.63
- 3.14
Show Solution
The Correct Option is C
Solution and Explanation
Step 1: Calculate the Damping Ratio (\(\zeta\)). \[ \zeta = \frac{C}{C_c} = \frac{35}{350} = 0.1 \]
Step 2: Apply the formula for Logarithmic Decrement. \[ \delta = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}} \]
Step 3: Substitute and solve. \[ \delta = \frac{2 \times 3.14159 \times 0.1}{\sqrt{1 - (0.1)^2}} = \frac{0.6283}{\sqrt{0.99}} = \frac{0.6283}{0.9949} \approx 0.631 \]
Step 2: Apply the formula for Logarithmic Decrement. \[ \delta = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}} \]
Step 3: Substitute and solve. \[ \delta = \frac{2 \times 3.14159 \times 0.1}{\sqrt{1 - (0.1)^2}} = \frac{0.6283}{\sqrt{0.99}} = \frac{0.6283}{0.9949} \approx 0.631 \]
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