Эротические рассказы

Vibroacoustic Simulation. Alexander PeifferЧитать онлайн книгу.

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to the overdamped oscillator the equilibrium is reached as can be seen in Figure 1.4.

Figure 1.4 Motion of the critically damped oscillator. Source: Alexander Peiffer.

Let us summarize some facts and observations about free damped oscillators:

1 Oscillation occurs only if the system is underdamped.

2 ωd is always less than ω0.

3 The motion will decay.

4 The frequency ωd and the decay rate are properties of the system and independent from the initial conditions.

5 The amplitude of the damped oscillator is u^(t)=u^0e−βt with β=ζω0. β is called the decay rate of the damped oscillator.

The decay rate is related to the decay time τ. This is the time interval where the amplitude decreases to e−1 of the initial amplitude. Thus, the decay time is:

tau equals StartFraction 1 Over beta EndFraction equals StartFraction 1 Over zeta omega 0 EndFraction (1.22)

1.2 Forced Harmonic Oscillator

When an external force F^xcos⁡(ωt) is exciting the damped oscillator as shown in Figure 1.1 b), applying Newton’s second law we get for the equation of motion:

m ModifyingAbove u With two-dots plus c Subscript v Baseline ModifyingAbove u With dot plus k Subscript s Baseline u equals ModifyingAbove upper F With caret Subscript x Baseline cosine left-parenthesis omega t right-parenthesis (1.23)

This is an inhomogeneous, linear, second-order equation for u. The solution of this equation is given by a particular solution uP(t) and the solutions of the homogeneous Equation (1.1) uH(t).

u left-parenthesis t right-parenthesis equals u Subscript upper H Baseline left-parenthesis t right-parenthesis plus u Subscript upper P Baseline left-parenthesis t right-parenthesis (1.24)

Any linear combination of the homogeneous solution can be added to the particular solution because it equals zero.

1.2.1 Frequency Response

There are several methods to determine the particular solutions, like Laplace and Fourier transforms. Here, complex algebra will be used¹. Amplitude and phase are given by a complex pointer denoted by bold italic type as depicted in Figure 1.5.

ModifyingAbove upper F With caret Subscript x Baseline cosine left-parenthesis omega t plus phi right-parenthesis equals upper R e left-parenthesis bold-italic upper F Subscript x Baseline e Superscript j omega t Baseline right-parenthesis (1.25)

Figure 1.5 Complex pointer, amplitude and phase relationship. Source: Alexander Peiffer.

F_x is the complex amplitude of the force, and the Re(⋅) expression is usually omitted. The displacement and velocity response is then given by

StartLayout 1st Row 1st Column u left-parenthesis t right-parenthesis 2nd Column equals bold-italic u e Superscript j omega t Baseline 3rd Column v Subscript x Baseline left-parenthesis t right-parenthesis 4th Column equals j omega bold-italic u e Superscript j omega t Baseline equals bold-italic v Subscript x Baseline e Superscript j omega t EndLayout (1.26)

with u and v_x as complex amplitudes of the displacement and velocity, respectively. Introducing this into Equation (1.23).

minus m omega squared bold-italic u e Superscript j omega t Baseline plus j c Subscript v Baseline omega bold-italic u e Superscript j omega t Baseline plus k Subscript s Baseline bold-italic u e Superscript j omega t Baseline equals bold-italic upper F e Superscript j omega t (1.27)

and solving this for u gives:

bold-italic u equals StartFraction bold-italic upper F Subscript x Baseline Over k Subscript s Baseline minus m omega squared plus j c Subscript v Baseline omega EndFraction (1.28)

The magnitude u^ and the phase ϕ of u are:

StartLayout 1st Row ModifyingAbove u With caret equals StartFraction ModifyingAbove upper F With caret Subscript x Baseline Over StartRoot left-parenthesis k Subscript s Baseline minus m omega squared right-parenthesis squared plus left-parenthesis c Subscript v Baseline omega right-parenthesis Subscript 2 Baseline EndRoot EndFraction EndLayout (1.29)

StartLayout 1st Row phi 0 equals arc tangent left-parenthesis StartFraction c Subscript v Baseline omega Over k Subscript s Baseline minus m omega squared EndFraction right-parenthesis EndLayout (1.30)

At ω = 0 the static displacement amplitude is u^0=F^x/ks. Using the definitions from (1.6) and dividing u^ by u^0 gives the normalized amplitude

StartFraction ModifyingAbove u With caret Over ModifyingAbove u With caret Subscript 0 Baseline EndFraction equals StartFraction 1 Over StartRoot left-bracket 1 minus left-parenthesis omega slash omega 0 right-parenthesis squared right-bracket squared plus left-parenthesis 2 zeta omega slash omega 0 right-parenthesis squared EndRoot EndFraction (1.31)

and phase

phi 0 equals arc tangent minus StartFraction 2 zeta omega slash omega 0 Over 1 minus left-parenthesis omega slash omega 0 right-parenthesis squared EndFraction (1.32)

It can be shown that the maximum of u^ is at

StartFraction omega Subscript r Baseline Over omega 0 EndFraction equals StartRoot 1 minus 2 zeta squared EndRoot (1.33)

and the maximum value is

ModifyingAbove u With caret Subscript r Baseline equals StartFraction ModifyingAbove u With caret Subscript 0 Baseline Over 2 zeta StartRoot 1 minus zeta squared EndRoot EndFraction (1.34)

with the corresponding phase

phi Subscript r Baseline equals arc tangent left-parenthesis minus StartFraction StartRoot 1 minus 2 zeta squared EndRoot Over zeta EndFraction right-parenthesis (1.35)

The

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