Your Equation

Library / Vibration exposure / VIB-01

Vibration exposure

Apparent mass at a driving point

Force divided by acceleration measured at the same point where the vibration is applied, such as a seat.

Ref
VIB-01
Rev
v1
Field
Vibration exposure
Result
|Mapp| [kg]

Apparent mass is the force at a driving point divided by the acceleration at that same point, as a function of frequency. It is generally a complex frequency-response function; this entry works with the magnitudes at one frequency. It needs force and acceleration measured at the same point where the vibration is applied: a remote response acceleration gives a different transfer function. At very low frequency a seated person’s apparent mass tends to the mass carried by the seat.

Force divided by acceleration measured at the same point where the vibration is applied, such as a seat. It shows how much of a person’s mass is moving with the seat at one frequency.

Calculator

Solve for
kg
N
202,000
m/s²
0.530
  • Equivalent weight carried, Mapp g588.6 N
  • Share of a 80 kg person’s mass75 %

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.

Apparent mass (kg) current value 60 kg
|A|Acceleration amplitude at the interfaceKey driver
10600
|F|Force amplitude at the interface
4400
  • A force of 300 N for an acceleration of 5 m/s² gives 60 kg, 75 % of an 80 kg person. Near a resonance the apparent mass can rise well above the static value, and above resonance it falls.
  • If the acceleration measured is taken at a different point, for example on the head, the ratio is not an apparent mass but a transmissibility-like quantity.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (kg)Local stepChange in result
|A| Acceleration amplitude at the interface0.5 to 30 m/s²10600590+10%−9.09 %
|F| Force amplitude at the interface20 to 2,000 N4400396+10%+10 %

Sweep of Force amplitude at the interface

Force amplitude at the interface (N)Apparent mass (kg)
204
267.553.5
515103
762.5152.5
1,010202
1,258251.5
1,505301
1,753350.5
2,000400

Sweep one variable

Apparent mass (kg)Hover, touch, or use the arrow keys on the chart
0100200300400 500100015002000 Force amplitude at the interface (N) now 60

Variables and how each one contributes

|Mapp|

Apparent mass

Result · kg

What the equation gives you. Everything else feeds it.

|F|

Force amplitude at the interface

N · typical 20 to 2,000

The force amplitude at the interface (for example a seat force plate), in N. Raises the apparent mass in proportion. Raising it by 10 % (to 330) moves the result by +10 %, from 60 to 66 kg.

|A|

Acceleration amplitude at the interface

m/s² · typical 0.5 to 30

The acceleration amplitude at the same interface, in m/s², at the same frequency. Lowers the apparent mass in inverse proportion. Raising it by 10 % (to 5.5) moves the result by −9.09 %, from 60 to 54.55 kg.

Worked example with the values above

  1. Force and acceleration at the same point: |F| = 300 N and |A| = 5 m/s².
  2. Divide: 300 / 5.
  3. |Mapp| = 60 kg.

Assumptions and limits

  • Force and acceleration at the same driving point and the same frequency, with both as amplitudes.
  • Linear, time-invariant response at that frequency.

Common mistakes

  • Using the response acceleration at a remote point in the denominator.
  • Treating the magnitude as the full complex frequency response.

Statistics and uncertainty

The ratio carries the relative errors of both measurements in quadrature: ±3 % in force and ±3 % in acceleration give about ±4 %. Phase is lost in the magnitude-only form, so it cannot be used for power-flow estimates without it.

Where it is used

  • Characterising a person or a payload as a dynamic load on a seat or platform.
  • Setting up a model of the body for seat design.

Maths you need

  • Frequency response: output over input at each frequency
  • Newton’s second law: force equals mass times acceleration

Related formulas

Practise this formula