Access to Accelerometers

a complex subject in a simple way

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Natural Frequency of a Spring-Mass Oscillator


Spring-Mass Oscillator without Damping

The base of the spring-mass oscillator model is considered fixed in space. In the equilibrium position let x = 0. The mass is m [kg] and the spring constant is k [N/m].

To obtain an equation of motion one identifies all static and dynamic forces that occur.

The force acting on the mass due to the acceleration is .

The second force is caused by the spring and is k·x.

Since no other forces are involved we can write:

mx ̈+kx=0

or

x ̈+k⁄m x=0

This is a homogeneous differential equation.

For the solution one makes the ansatz

x(t)=Ce^αt

The characteristic equation is

α^2+k/m=0

with the solutions

α_1,2=± i √(k⁄m)
Sketch showing the undamped spring-mass oscillator

Undamped spring-mass oscillator

Picture9A.png

With                                         we find for the solution of the differential equation

x=A cos⁡(√(k⁄m)  ∙t)+B sin⁡(√(k⁄m)  ∙t)

A and B are given by the initial conditions. If for instance the deflection x at time 0 is equal to X 0, then follows

x=X_0   cos⁡〖ω_0 t〗

This is a continuous, harmonic oscillation with the frequency ω₀. 

ω_0=√(k⁄m)

is called the natural frequency of the system.


Spring-Mass Oscillator with Damping

We add a damping "dashpot" with the damping coefficient c.

The damping force is proportional to the velocity and is thus c · .

The dimension of [c ] is N/m/s or kg/s.  

The summation of all forces leads to

mx ̈+cx ̇+kx=0

Here it is useful to introduce the relative damping ζ which is defined as

ζ=c/(2√mk)=c/(2mω_0 )

I.e. the damping is set in relation to m and ω0 . The "2" is an arbitrary factor.

Sketch showing the spring-mass oscillator with damping

Spring-mass oscillator with damping

Using                       again, the differential equation can be written in the form

ω_0=√(k⁄m)
x ̈+2ζω_0 x ̇+ω_0^2 x=0

The differential equation is homogenous again and we make the same ansatz

x(t)=Ce^αt

The characteristic equation is then

α^2+2ζω_0 α+ω_0^2=0

with the solutions

α_1,2=[-ζ±√(ζ^2-1)  ] ω_0

For ζ = 0, the result for the undamped oscillator is obtained again.
We are more interested here in the case of a weakly damped oscillator, i.e. ζ < 1. The root expression then becomes negative and the solutions of the characteristic equation will be

α_1,2=-ζω_0±i√(〖1-ζ〗^2 )  ω_0

They are conjugate complex
The total solution is equal to the sum of the particular solutions and becomes

x(t)=C_1 e^(α_1 t)+C_2 e^(α_2 t)

If one sets the imaginary part                               so one gets

√(〖1-ζ〗^2 )  ω_0=ω_D
x= e^(-ζω_0 t) (A cos⁡〖ω_D t〗+B sin⁡〖ω_D t〗 )

A and B are again determined by the initial conditions. With x(t=0)=0 for example, the solution is

x= A∙e^(-ζω_0 t)  〖∙cos〗⁡〖ω_D t〗

This is a sinusoidal oscillation with the frequency ωD and the initial amplitude A.

The oscillation frequency ωis the damped natural frequency. Note that it is slightly different from ω0 of the undamped oscillation, depending on ζ .

Derivation of the Frequency Characteristics


Frequency Characteristics of the Seismic Transducer


Derivation of the Transfer Function

In contrast to the above we apply now a movement u=g(t) to the base. Thus, we force the oscillator to vibrate. The position of the mass m is given by u+x.

The acceleration force therefore becomes

m∙(d^2 (u+x))/(dt^2 )

For the equilibrium of the forces we find the equation

m (d^2 (u+x))/(dt^2 )+c dx/dt+kx=0
m (d^2 x)/(dt^2 )+c dx/dt+kx=-m (d^2 u)/(dt^2 )

                                                   or

〖ω_0〗^2=k/m   and    ζ=c/(2mω_0 )

dividing by m , introducing                      and

we obtain the same differential equation as for the free oscillation, but now it is inhomogeneous due to the term to the right

(d^2 x)/(dt^2 )+2ζω_0  dx/dt+ω_0^2 x=-(d^2 u)/(dt^2 )
Sketch showing spring-mass oscillator under forced vibration

Spring-mass oscillator under forced vibration

The solution consists of the superposition of the solution of the homogeneous differential equation and a particular solution of the inhomogeneous differential equation.

Hence, we obtain a decaying first part, as in the case of free oscillation, as well as a second part, which follows the excitation u=g(t).

As excitation function we choose a harmonic oscillation (with constant displacement amplitude)

u= U_0  cos⁡ωt
du/dt= -ωU_0  sin⁡ωt
(d^2 u)/(dt^2 )= 〖-ω〗^2 U_0  cos⁡ωt

With the derivations                                    and

the differential equation therefore becomes

(d^2 x)/(dt^2 )+2ζω_0  dx/dt+ω_0^2 x=ω^2 U_0  cos⁡ωt

For the solution we take the ansatz

x(t)= A cos⁡ωt+B sin⁡ωt

By deriving and substituting one obtains

[(ω_0^2-ω^2 )A+2ζω_0 ωB]  cos⁡ωt+[-2ζω_0 ωA+(ω_0^2-ω^2 )B]  sin⁡ωt=ω^2 U_0  cos⁡ωt

A comparison of coefficients leads to the following system of equations for A and B

■((ω_0^2-ω^2 )A+      2ζω_0 ω B=ω^2 U_0@  -2ζω_0 ωA+(ω_0^2-ω^2 )B=0      )

The determinant of the coefficients is calculated to be

∆ =(ω_0^2-ω^2 )^2+(2ζω_0 ω)^2

The two solutions of the system of equations become

A=(ω_0^2-ω^2)/∆∙ω^2 U_0         B=(2ζω_0 ω)/∆∙ω^2 U_0

and therfore

x(t)= A cos⁡ωt+B sin⁡ωt=(U_0 ω^2)/∆ [(ω_0^2-ω^2 )  cos⁡ωt+(2ζω_0 ω)  sin⁡ωt ]

We transform                                                 into

x(t)= A cos⁡ωt+B sin⁡ωt           x(t)=X_0  cos⁡(ωt-φ)

For the amplitude and phase angle we get

▁(X_0 )=√(A^2+B^2 )=▁(ω^2/√∆∙U_0 )
〖▁(tan⁡φ )=B/A〗⁡=  (2ζω_0 ω)/(ω_0^2-ω^2 )=▁((2ζ ω/ω_0 )/(1-(ω/ω_0 )^2 ))

and thereby the final result of the

general transfer function:

x(t)=ω^2/√((ω_0^2-ω^2 )^2+(2ζω_0 ω)^2 )∙U_0  cos⁡(ωt-φ)


Frequency Response of Displacement, Velocity and Acceleration

To make it easier and for transferable results it is better to use the frequency response representation. For the moment we only consider the amplitude response i.e. the ratio of the output amplitude to the input amplitude as a function of the frequency.


Displacement Frequency Response

X_0/U_0 =ω^2/√((ω_0^2-ω^2 )^2+(2ζω_0 ω)^2 )

We introduce also the dimensionless relative frequency ωR  i.e. the excitation frequency in relation to the natural frequency

ω_R=ω/ω_0

For this we expand the expression of the frequency response with 1/ω0² to obtain the (dimensionless) displacement response

Φ_d=X_0/U_0 =〖ω_R〗^2/√((1-〖ω_R〗^2 )^2+(2ζω_R )^2 )

Both the input U and output in our model above are displacements and the frequency response is therefore denoted by displacement amplitude response Φd.

The double logarithmic diagram shows the function for different values of the relative damping ζ .

We can see that for frequencies well above the natural frequency the transfer ratio becomes 1. This means in this region we have a 1 to 1

Graph showing displacement amplitude response

Displacement amplitude response

displacement behaviour. Near to the natural frequency we find a more ore less pronounced amplification depending on ζ with a maximum atthe resonance frequency. Below the resonance the curve leads into slope of 40dB / decade.


Acceleration Frequency Response

For the frequency response with an acceleration signal as input we have to derive the input function twice.

(d^2 u)/(dt^2 )= 〖-ω〗^2 U_0  cos⁡ωt

The negative sign actually concerns the phase angle and we could also write

(d^2 u)/(dt^2 )= ω^2 U_0  cos⁡(ωt-π)

Because we are not interested in the phase we can set it to zero and obtain for the

acceleration amplitude response Φ :

Φ_a=X_0/U_A =1/〖ω_0〗^2 ∙1/√((1-〖ω_R〗^2 )^2+(2ζω_R )^2 )

The dimension of the function is [s²] because it is [displacement X/ acceleration Ü0 ].

Graph showing acceleration amplitude response

Acceleration amplitude response

Due to the double derivative the function now tilts to the other side. I.e. in acceleration terms we find a direct transfer in the frequency domain below the natural frequency, while we find a roll-off rate of -40dB per decade for high frequencies. The resonance region looks similar to the displacement response.


Velocity Frequency Response

Finally we can calculate and plot the amplification function for an input in velocity terms and we get the

velocity amplitude response Φv .

Φ_v=1/ω_0 ∙ω_R/√((1-〖ω_R〗^2 )^2+(2ζω_R )^2 )

The function is now symmetric around the natural frequency. The dimension of the function is [displacement / velocity] = [s]

The phase response is the same in all three cases, displacement velocity and acceleration.

φ⁡=  arctan⁡〖(2ζω_R)/(1-〖ω_R〗^2 )〗

The change in phase through the resonance must not be confused with the fact that displacement, velocity and acceleration are each offset by π/2.

Graph showing velocity amplitude response

Velocity amplitude response

Graph showing phase angle frequency response

Phase angle frequency response


Resonance Frequency

In sensor technology, the resonance is defined as the maximum of the output amplitude for a constant imput amplitude.

Therefore, in view of the three different amplitude response functions, we also obtain different resonant frequencies for displacement, velocity and acceleration. We can obtain the values by deriving the function Φ  and setting it to 0.

For the acceleration resonance as an example we obtain

(dΦ_a)/(dω_R )=1/〖ω_0〗^2 ∙(4ζ^2 ω_R-2ω_R (1-〖ω_R〗^2))/[(1-〖ω_R〗^2 )^2+4ζ^2 〖ω_R〗^2 ]^(3⁄2)

and we find the zero at

ω_R=√(1-〖2ζ〗^2 )

Thus one obtains for the different resonance frequencies in an analogous way:

Acceleration resonance frequency:

[ω_res ]_a=ω_0∙√(1-〖2ζ〗^2 )

Velocity resonance frequency:

[ω_res ]_v=ω_0

Displacement resonance frequency:

[ω_res ]_d=ω_0∙1/√(1-〖2ζ〗^2 )


How can it be that you get three different solutions for the same physical phenomenon?

At the resonance frequency, the oscillator absorbs the maximum energy. Since the power is equal to force times velocity, this maximum is found when the phase angle of the output signal x is equal to π/2. In this situation, the amplitude of the system increases continuously until the energy absorption is in equilibrium with the energy loss due to damping.
This means that the physical phenomenon of the resonance is always at a phase angle of π/2 and this is always the case at ω= ω0 .
So in order to see the "correct" (physical) resonance we need to keep the velocity amplitude of the input signal constant. We note that then the resonance is at ω0.
With constant acceleration, the velocity amplitude decreases with increasing frequency with the factor 10 per decade. This means that the energy applied becomes greater at frequencies below ω0  and therefore the maximum-amplitude resonance shifts downwards.

The same, but in opposite, is true for constant displacement. Here, the maximum-amplitude resonance is shifted upwards.