Difficulty: Medium
Correct Answer: 52.92 N-s/m
Explanation:
Introduction / Context: The critical damping coefficient defines the threshold between oscillatory (underdamped) and non-oscillatory (overdamped) responses in a mass–spring–damper system. It is a key design target when fast return without overshoot is desired (e.g., measuring instruments).
Given Data / Assumptions:
Concept / Approach: For a single-DOF system, cc=2√(km). Substituting k and m gives the required value. Units must be consistent (N·s/m).
Step-by-Step Solution:
1) Convert stiffness: 0.7 N/mm = 0.7×1000 = 700 N/m.2) Compute √(k m) = √(700×1) = √700 ≈ 26.4575.3) Compute cc = 2×26.4575 ≈ 52.915 ≈ 52.92 N·s/m.Verification / Alternative check: Natural frequency ωn=√(k/m)=√700≈26.46 rad/s. Since cc=2mωn=2×1×26.46≈52.92 N·s/m, which matches.
Why Other Options Are Wrong:
1.4 N·s/m — far below critical; implies very light damping.
18.52 N·s/m — still underdamped relative to cc.
529.2 N·s/m — ten times the correct value; overdamped by a wide margin.
Common Pitfalls: Forgetting to convert N/mm to N/m; using c=2k instead of c=2√(km).
Final Answer: 52.92 N-s/m.
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