Question

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. 0.10
• B. 0.31
• C. 0.63
• D. 3.14

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.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are asked to calculate the logarithmic decrement (\(\delta\)) given the actual damping coefficient (\(C\)) and the critical damping coefficient (\(C_c\)). Step 2: Key Formulas:
1. Damping Ratio (\(\zeta\)): \[ \zeta = \frac{C}{C_c} \] 2. Logarithmic Decrement (\(\delta\)): \[ \delta = \frac{2\pi \zeta}{\sqrt{1 - \zeta^2}} \] Step 3: Calculation:
Calculate the damping ratio: \[ \zeta = \frac{35}{350} = 0.1 \] Calculate the logarithmic decrement: \[ \delta = \frac{2\pi (0.1)}{\sqrt{1 - (0.1)^2}} \] \[ \delta = \frac{0.6283}{\sqrt{1 - 0.01}} \] \[ \delta = \frac{0.6283}{\sqrt{0.99}} \] \[ \delta = \frac{0.6283}{0.99498} \] \[ \delta \approx 0.631 \] Alternatively, for small damping (\(\zeta<0.1\)), we can approximate \(\delta \approx 2\pi\zeta\): \[ \delta \approx 2 \times 3.1416 \times 0.1 = 0.628 \] Both methods yield approximately 0.63. Step 4: Final Answer:
The logarithmic decrement is 0.63.