2.1.2. Responses

The free response, meaning that the response with *e*(*t*) = 0, is driven by initial conditions only. It is given by

$$q(t) = e^{-\zeta\omega\_n t} \left( q\_0 \cos\omega\_d t + \frac{v\_0 + \zeta\omega\_n q\_0}{\omega\_d} \sin\omega\_d t \right), t \ge 0. \tag{9}$$

If *e*(*t*) = *δ*(*t*), where *δ*(*t*) is the Dirac-delta function, the response with zero initial conditions is called the impulse response. In the theory of linear systems (Gabel and Roberts [65], Zheng et al. [66]), the symbol *h*(*t*) is used for the impulse response. Thus, consider the equation

$$\frac{d^2h(t)}{dt^2} + 2\xi\omega\_n \frac{dh(t)}{dt} + \omega\_n^2 h(t) = \frac{\delta(t)}{m}.\tag{10}$$

One has

$$h(t) = \frac{e^{-\xi\omega\_n t}}{m\omega\_d} \sin \omega\_d t, t \ge 0. \tag{11}$$

Let *u*(*t*) be the Heaviside unit step (unit step for short) function. Then, the response to (3) with zero initial conditions is called the unit step response. As usual, it is denoted by *g*(*t*) in practice. Thus, consider

$$\frac{d^2\mathcal{S}(t)}{dt^2} + 2\xi\omega\_n\frac{d\mathcal{S}(t)}{dt} + \omega\_n^2\mathcal{g}(t) = \frac{u(t)}{m}.\tag{12}$$

One has

$$\log(t) = \int\_0^t h(\tau)d\tau = \frac{1}{k} \left[ 1 - \frac{e^{-\zeta\omega\_n t}}{\sqrt{1-\zeta^2}} \cos(\omega\_d t - \phi) \right],\tag{13}$$

where

$$\phi = \tan^{-1} \frac{\xi}{\sqrt{1 - \xi^2}}.\tag{14}$$

Denote by *H*(*ω*) the Fourier transform of *h*(*t*). Then, *H*(*ω*) is usually called the frequency response to the oscillator described by (3). It is in the form

$$H(\omega) = \frac{1}{m(\omega\_n^2 - \omega^2 + i2\xi\omega\_n\omega)} = \frac{1}{m\omega\_n^2 \left(1 - \frac{\omega^2}{\omega\_n^2} + i2\xi\frac{\omega}{\omega\_n}\right)}.\tag{15}$$

With the parameter *γ* defined by

$$
\gamma = \frac{\omega}{\omega\_n},\tag{16}
$$

which is called frequency ratio, *H*(*ω*) may be rewritten by

$$H(\omega) = \frac{1}{m\omega\_n^2(1-\gamma^2+i2\zeta\gamma)}.\tag{17}$$

The amplitude of *H*(*ω*) is called the amplitude frequency response. It is in the form

$$|H(\omega)| = \frac{1}{m\omega\_n^2\sqrt{\left(1-\gamma^2\right)^2 + \left(2\zeta\gamma\right)^2}}.\tag{18}$$

Its phase is termed the phase frequency response given by

$$\varphi(\omega) = \tan^{-1} \frac{2\zeta\gamma}{1 - \gamma^2}.\tag{19}$$

When the oscillator is excited by a sinusoidal function, the solution to (3) is termed the sinusoidal or simple harmonic response. Suppose the sinusoidal excitation function is *A*cos*ωt*, where *A* is a constant. Then, the solution to

$$\begin{cases} \frac{d^2q(t)}{dt^2} + 2\xi\omega\nu\_n\frac{dq(t)}{dt} + \omega\_n^2q(t) = \frac{A\cos\omega t}{m} \\ \quad q(0) = q\_0, q'(0) = v\_0 \end{cases} \tag{20}$$

is the sinusoidal response in the form

$$q(t) = \frac{\frac{A}{\hbar \omega\_d}}{\left(\omega\_n^2 - \omega^2\right)^2 + \left(2\zeta\omega\_n\omega\right)^2} \left\{ \begin{array}{l} \left(\omega\_n^2 - \omega^2\right)\cos\omega t + 2\zeta\omega\_n\omega\sin\omega t\\ + e^{-\zeta\omega\_n t} \left[\left(\omega\_n^2 - \omega^2\right)\cos\omega\_d t - \frac{\zeta}{\sqrt{1-\zeta^2}}\left(\omega\_n^2 + \omega^2\right)\sin\omega\_d t \right] \end{array} \right\}. \tag{21}$$

The responses mentioned above are essential to linear oscillators. We shall give our results for three classes of fractional oscillators with respect to those responses in this research.
