Aperiodic borderline case

Normalized deflection in the aperiodic limit case as a function of normalized time . Three courses with different initial speeds are shown . ${\displaystyle x/x_{0}}$ ${\displaystyle t\cdot \delta }$ ${\displaystyle v_{0}}$

The aperiodic borderline case describes a damping state of a harmonic oscillator . It is the smallest damping at which the deflection without overshoot, i.e. H. a change of direction that tends towards equilibrium if it is released from a deflected state without an initial speed. The approach to equilibrium takes place in a very short time. If the oscillator has an initial speed, a zero crossing can occur in the aperiodic borderline case. If the damping is even greater, one speaks of an overperiodic fall or creep fall .

The aperiodic limit case corresponds to a Lehr's damping of D = 1 or a quality factor of Q = 0.5.

Linear damped harmonic oscillator

The equation of motion of a damped oscillating mass is:

${\displaystyle m{\ddot {x}}+d\;{\dot {x}}+kx=0}$

with the deflection x , the damping constant d , the mass m and the spring constant k .

Usually one identifies as the undamped natural angular frequency of the harmonic oscillator and as the decay constant , so that the following form results for the equation of motion of a damped harmonic oscillator: ${\displaystyle \omega _{0}={\sqrt {\frac {k}{m}}}}$ ${\displaystyle \delta ={\frac {d}{2m}}}$

${\displaystyle {\ddot {x}}+2\;\delta \;{\dot {x}}+\omega _{0}^{2}\;x=0}$

This equation can be solved with the exponential approach . The result is the characteristic equation : ${\displaystyle x(t)\sim e^{\lambda \cdot t}}$

${\displaystyle \lambda ^{2}+2\;\delta \;\lambda \;+\omega _{0}^{2}=0}$

With the solution:

${\displaystyle \lambda _{1,2}=-\delta \pm {\sqrt {\delta ^{2}-\omega _{0}^{2}}}}$

The aperiodic borderline case results for, since the discriminant of this equation then becomes 0. Therefore, the oscillator does not oscillate periodically, but rather returns to the rest position in a minimal time. ${\displaystyle \delta =\omega _{0}}$

It then holds and the general solution for the case of a double zero has the following form: ${\displaystyle \lambda =-\delta }$

${\displaystyle x(t)=c_{1}\cdot e^{-\delta \;t}+c_{2}\cdot t\cdot e^{-\delta \;t}}$

If the oscillator is released at the point in time zero at the point with the speed zero, then and applies , so that the following special solution results: ${\displaystyle x_{0}}$ ${\displaystyle x(0)=x_{0}}$ ${\displaystyle {\dot {x}}(0)=0}$

${\displaystyle x(t)=x_{0}\cdot (1+\delta \;t)\cdot e^{-\delta \;t}}$

If, on the other hand, the transducer is "kicked" at the point with the velocity at the point in time zero , the solution is: ${\displaystyle x(0)=0}$ ${\displaystyle v_{0}={\dot {x}}(0)}$

${\displaystyle x(t)=v_{0}\cdot t\cdot e^{-\delta \;t}}$

If both initial conditions are specified, these solutions can also be superimposed linearly, so that, overall , the solution for the initial data ${\displaystyle x(0)=x_{0}}$ ${\displaystyle {\dot {x}}(0)=v_{0}}$

${\displaystyle x(t)={\bigl (}x_{0}+(v_{0}+x_{0}\cdot \delta )\cdot t{\bigr )}\cdot e^{-\delta \;t}}$

reads.

application

• Shock absorbers
• Moving coil measuring instrument