Learn: Critical damping coefficient
Critical viscous damping is the dividing level between an oscillatory and non-oscillatory free response in a linear single-degree system. It is a damping coefficient, not a percentage.
What the formula is saying
Multiply stiffness k by mass m, take the square root, then multiply by 2. The equivalent expression is cc = 2mωn.
Read the symbols in plain language
- k
- Stiffness
Restoring force per unit displacement for the chosen degree of freedom; it is not an elastic modulus.
N/mUse N/m as the base unit shown here. With k in N/m and m in kg, cc is N·s/m. A viscous damper force is c times velocity, so multiplying N·s/m by m/s gives N.
- m
- Mass
Inertial mass of the selected dynamic system, not its weight force. A spectral acceleration times mass produces force.
kgUse kg as the base unit shown here. With k in N/m and m in kg, cc is N·s/m. A viscous damper force is c times velocity, so multiplying N·s/m by m/s gives N.
- cc
- Result to find
Critical damping coefficient. The linear single-degree characteristic equation gives twice the square-root product.
N·s/m
Sort out the units first
With k in N/m and m in kg, cc is N·s/m. A viscous damper force is c times velocity, so multiplying N·s/m by m/s gives N.
Assumptions before calculating
Assume a linear, single-degree-of-freedom mass–spring model with positive effective mass and stiffness, small motion and consistent units. The selected mass and stiffness must refer to the same generalized displacement.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find cc and explain the result in the stated output unit.
- k · Stiffness
- 1000000 N/m
- m · Mass
- 10000 kg
Combine stiffness and inertia
The square root of their product has the units of a viscous damping coefficient.
√((1000000) × (10000)) = 100000 N·s/mApply the critical-damping factor
The linear single-degree characteristic equation gives twice the square-root product.
2 × (100000) = 200000 N·s/m
Does this worked answer make sense?
Doubling both mass and stiffness doubles cc. An actual c equal to this result corresponds to a damping ratio of one, or 100%.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- k · Stiffness
- 2000000 N/m
- m · Mass
- 8000 kg
Combine stiffness and inertia
The square root of their product has the units of a viscous damping coefficient.
√((2000000) × (8000)) ≈ 126491.1064 N·s/mApply the critical-damping factor
The linear single-degree characteristic equation gives twice the square-root product.
2 × (126491.1064) ≈ 252982.2128 N·s/m
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- k · Stiffness
- 1500000 N/m
- m · Mass
- 12000 kg
Find: Learn: Critical damping coefficient
A hint, not the answer
Multiply stiffness k by mass m, take the square root, then multiply by 2. The equivalent expression is cc = 2mωn.
With k in N/m and m in kg, cc is N·s/m. A viscous damper force is c times velocity, so multiplying N·s/m by m/s gives N.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Combine stiffness and inertia
The square root of their product has the units of a viscous damping coefficient.
√((1500000) × (12000)) ≈ 134164.0786 N·s/mApply the critical-damping factor
The linear single-degree characteristic equation gives twice the square-root product.
2 × (134164.0786) = 268328.1573 N·s/m
Avoid the common trap
Do not confuse cc with actual c or with damping ratio ζ = c/cc. Do not omit the square root or use kg instead of N·s/m as the output unit.
When this method applies — and when it does not
The critical value applies to the ideal viscous damping model cẋ. Frictional, hysteretic or frequency-dependent damping cannot always be replaced by a single coefficient without a stated equivalence. This does not select a physical damper device.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Critical damping coefficient. Single-degree-of-freedom free vibration, natural frequency, period and viscous damping. Angular frequency and cycles per second are different quantities.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.