Learn: Natural period
Natural period is the time a linear mass–spring system needs to complete one undamped vibration cycle. A more flexible system takes longer, while a stiffer one vibrates more quickly.
What the formula is saying
Divide effective mass by stiffness, take the square root and multiply by 2π. This is equivalent to Tn = 2π/ωn with ωn = √(k/m).
Read the symbols in plain language
- 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. Mass is kg and stiffness is N/m. The ratio m/k has unit s²; its square root has unit s. Multiplying by 2π produces seconds per complete cycle.
- 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. Mass is kg and stiffness is N/m. The ratio m/k has unit s²; its square root has unit s. Multiplying by 2π produces seconds per complete cycle.
- T
- Result to find
Natural period. A complete cycle spans two pi radians rather than one radian.
s
Sort out the units first
Mass is kg and stiffness is N/m. The ratio m/k has unit s²; its square root has unit s. Multiplying by 2π produces seconds per complete cycle.
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 T and explain the result in the stated output unit.
- m · Mass
- 10000 kg
- k · Stiffness
- 1000000 N/m
Calculate the basic inertial time scale
The square root of mass divided by stiffness gives the time scale for one radian of phase.
√((10000) ÷ (1000000)) = 0.1 sScale to one full vibration cycle
A complete cycle spans two pi radians rather than one radian.
2 × π × (0.1) ≈ 0.6283185307 s
Does this worked answer make sense?
Four times the mass doubles Tn, while four times the stiffness halves it. The reciprocal 1/Tn is ordinary natural frequency in Hz.
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.
- m · Mass
- 8000 kg
- k · Stiffness
- 2000000 N/m
Calculate the basic inertial time scale
The square root of mass divided by stiffness gives the time scale for one radian of phase.
√((8000) ÷ (2000000)) ≈ 0.0632455532 sScale to one full vibration cycle
A complete cycle spans two pi radians rather than one radian.
2 × π × (0.0632455532) ≈ 0.3973835306 s
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.
- m · Mass
- 12000 kg
- k · Stiffness
- 1500000 N/m
Find: Learn: Natural period
A hint, not the answer
Divide effective mass by stiffness, take the square root and multiply by 2π. This is equivalent to Tn = 2π/ωn with ωn = √(k/m).
Mass is kg and stiffness is N/m. The ratio m/k has unit s²; its square root has unit s. Multiplying by 2π produces seconds per complete cycle.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Calculate the basic inertial time scale
The square root of mass divided by stiffness gives the time scale for one radian of phase.
√((12000) ÷ (1500000)) = 0.0894427191 sScale to one full vibration cycle
A complete cycle spans two pi radians rather than one radian.
2 × π × (0.0894427191) ≈ 0.5619851785 s
Avoid the common trap
Do not invert mass and stiffness inside the square root. Do not omit the factor 2π or use weight in kN as mass in kg.
When this method applies — and when it does not
The answer is a natural period, not necessarily the observed period under damping, nonlinear response or imposed excitation. A building with many degrees of freedom has several modal periods rather than one universal value.
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: Natural period. 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.