Vibroacoustic Simulation. Alexander Peiffer

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:

(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:

(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).

       (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.

       (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

       (1.26)

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

(1.27)

and solving this for u gives:

(1.28)

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

       (1.29)

(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

(1.31)

      and phase

(1.32)

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

       (1.33)

      and the maximum value is

(1.34)

      with the corresponding phase

       (1.35)

The