Editing Harmonic oscillator (section)

=== Steady-state solution ===
Apply the "[[complex analysis|complex variables]] method" by solving the [[auxiliary equation]] below and then finding the real part of its solution:
<math display="block">\frac{\mathrm{d}^2 q}{\mathrm{d}\tau^2} + 2 \zeta \frac{\mathrm{d}q}{\mathrm{d}\tau} + q = \cos(\omega \tau) + i\sin(\omega \tau) = e^{ i \omega \tau}.</math>

Supposing the solution is of the form
<math display="block">q_s(\tau) = A e^{i (\omega \tau + \varphi) }. </math>

Its derivatives from zeroth to second order are
<math display="block">q_s = A e^{i (\omega \tau + \varphi) }, \quad
\frac{\mathrm{d}q_s}{\mathrm{d} \tau} = i \omega A e^{i (\omega \tau + \varphi) }, \quad
\frac{\mathrm{d}^2 q_s}{\mathrm{d} \tau^2} = -\omega^2 A e^{i (\omega \tau + \varphi) } .</math>

Substituting these quantities into the differential equation gives
<math display="block">-\omega^2 A e^{i (\omega \tau + \varphi)} + 2 \zeta i \omega A e^{i(\omega \tau + \varphi)} + A e^{i(\omega \tau + \varphi)} = (-\omega^2 A + 2 \zeta i \omega A + A) e^{i (\omega \tau + \varphi)} = e^{i \omega \tau}.</math>

Dividing by the exponential term on the left results in
<math display="block">-\omega^2 A + 2 \zeta i \omega A + A = e^{-i \varphi} = \cos\varphi - i \sin\varphi.</math>

Equating the real and imaginary parts results in two independent equations
<math display="block">A (1 - \omega^2) = \cos\varphi, \quad 2 \zeta \omega A = -\sin\varphi.</math>

==== Amplitude part ====
[[Image:Harmonic oscillator gain.svg|thumb|[[Bode plot]] of the frequency response of an ideal harmonic oscillator]]
Squaring both equations and adding them together gives
<math display="block">\left. \begin{aligned}
 A^2 (1-\omega^2)^2 &= \cos^2\varphi \\
 (2 \zeta \omega A)^2 &= \sin^2\varphi
\end{aligned} \right\}
\Rightarrow A^2[(1 - \omega^2)^2 + (2 \zeta \omega)^2] = 1.</math>

Therefore,
<math display="block">A = A(\zeta, \omega) = \sgn \left( \frac{-\sin\varphi}{2 \zeta \omega} \right) \frac{1}{\sqrt{(1 - \omega^2)^2 + (2 \zeta \omega)^2}}.</math>

Compare this result with the theory section on [[resonance]], as well as the "magnitude part" of the [[RLC circuit]]. This amplitude function is particularly important in the analysis and understanding of the [[frequency response]] of second-order systems.

==== Phase part ====
To solve for {{math|''φ''}}, divide both equations to get
<math display="block">\tan\varphi = -\frac{2 \zeta \omega}{1 - \omega^2} = \frac{2 \zeta \omega}{\omega^2 - 1}~~ \implies ~~ \varphi \equiv \varphi(\zeta, \omega) = \arctan \left( \frac{2 \zeta \omega}{\omega^2 - 1} \right ) + n\pi.</math>

This phase function is particularly important in the analysis and understanding of the [[frequency response]] of second-order systems.