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Mechanical

Natural frequency of a mass on a spring

Frequency at which a spring-mass system oscillates freely.

Ref
MEC-05
Rev
v1
Field
Mechanical
Result
fn [Hz]

fn in hertz, k stiffness in N/m, m mass in kg. The angular natural frequency is ω_n = √(k/m) in rad/s. Undamped system.

A mass on a spring bounces at its own natural rate. Stiffer springs make it bounce faster; heavier masses make it bounce slower.

Calculator

Solve for
Hz
N/m
1001,000,000
kg
0.11,000
  • Angular frequency ωn20 rad/s
  • Period0.3142 s

What drives the result

Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.

Natural frequency (Hz) current value 3.183 Hz
kStiffnessKey driver
0.50350.3
mMass
0.31831.8
  • Because of the square root, frequency changes slowly: to halve the frequency you need four times the mass or a quarter of the stiffness.
  • Excitation near fn causes resonance and large vibration; isolation mounts need a natural frequency well below the disturbance.
  • The static deflection under the weight, mg/k, gives the same frequency as 15.8/√(δ in mm) hertz, a useful field check.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (Hz)Local stepChange in result
k Stiffness100 to 1,000,000 N/m0.503350.3349.83+10%+4.88 %
m Mass0.1 to 1,000 kg0.318331.8331.51+10%−4.65 %

Sweep of Stiffness

Stiffness (N/m)Natural frequency (Hz)
1000.5033
316.20.895
1,0001.592
3,1622.83
10,0005.033
31,6208.95
100,00015.92
316,20028.3
1,000,00050.33

Sweep one variable

Natural frequency (Hz)Hover, touch, or use the arrow keys on the chart
01020304050 100100010k100k1M Stiffness (N/m), log scale now 3.183

Variables and how each one contributes

fn

Natural frequency

Result · Hz

What the equation gives you. Everything else feeds it.

k

Stiffness

N/m · typical 100 to 1,000,000

Spring rate: force per unit deflection. Raises frequency with the square root: 4× the stiffness doubles the frequency.

m

Mass

kg · typical 0.1 to 1,000

Mass that vibrates. Lowers frequency with the square root: 4× the mass halves the frequency.

Worked example with the values above

  1. ωn = √(k/m) = √(4,000 / 10) = 20 rad/s.
  2. Divide by 2π to convert to hertz.
  3. fn = 3.183 Hz.

Assumptions and limits

  • Linear spring, rigid mass, one degree of freedom.
  • Damping neglected (small damping changes f only slightly).
  • Spring mass much smaller than the supported mass.

Common mistakes

  • Reporting ω in rad/s when hertz is asked for.
  • Using mass in grams.
  • Using the stiffness of one spring when two act in parallel.

Statistics and uncertainty

With a square root, the relative error is halved: δf/f ≈ ½ √[(δk/k)² + (δm/m)²]. A ±10% stiffness uncertainty, common for rubber mounts, gives about ±5% on frequency.

Where it is used

  • Equipment mounting and shock isolation.
  • Avoiding resonance in structures and shafts.
  • Vehicle suspension design.

Maths you need

  • Hooke’s law: spring force F = kx
  • Frequency: cycles per second in hertz

Related formulas

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