Critical damping coefficient — a 1 kg mass is attached to a spring of stiffness 0.7 N/mm. What is the critical damping coefficient c<sub>c</sub> for this system?

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:

  • Mass m = 1 kg.
  • Stiffness k = 0.7 N/mm = 700 N/m.
  • Viscous damping model; find cc.

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.

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion