The differential equation describing the motion of a damped driven simple harmonic oscillator is \begin{equation}\label{eq:DDSHO} m\ddot{x} + b\dot{x} + m\omega_0^2 x = F_0(t) \end{equation} where \(b\) is the linear damping constant and \(m\omega_0^2 = k\) is the spring constant. Dividing through by the mass and letting \(\beta \equiv b/2m\), we get \begin{equation}\label{eq:DDSHO2} \ddot{x} + 2\beta \dot{x} + \omega_0^2 x = \frac{F_0(t)}{m} \end{equation}
In Engineering 79, you may have solved this equation using Laplace transforms. Here, we will use the Fourier transform. To take the Fourier transform, let us first take the transform of \(\dot{x}\): \[ \int_{-\infty}^{\infty} \dot{x} e^{i\omega t} \dd{t} = x e^{i\omega t} \bigg|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} x (i\omega) e^{i\omega t}\dd{t} \] Assuming that \(x(t)\) goes to zero as \(t \to \pm \infty\), the integrated term vanishes and we get \[ \text{FT} (\dot{x}) = -i\omega \tilde{x}(\omega) \] You can readily verify that each time derivative brings down another factor of \((-i\omega)\), so that the Fourier transform of Eq. (\ref{eq:DDSHO2}) is \begin{equation}\label{eq:FTDDSHO} [(-i\omega)^2 - 2 \beta i \omega + \omega_0^2] \tilde{x}(\omega) = \frac1m \tilde{F}_0(\omega) \end{equation} Solving for \(\tilde{x}\), we have \begin{equation}\label{eq:xtilde} \tilde{x}(\omega) = \frac{\tilde{F_0}/m}{\omega_0^2 - \omega^2 - 2 \beta i \omega} = \tilde{F}_0(\omega) \tilde{G}(\omega) \end{equation} where \[ \tilde{F}_0(\omega) = \int_{-\infty}^{\infty} F_0(t) e^{i\omega t}\dd{t} \qquad\text{and}\qquad \tilde{G}(\omega) = \frac{1/m}{\omega_0^2 - \omega^2 - 2 \beta i \omega} \]
Equation (\ref{eq:xtilde}) gives the Fourier transform of the oscillator’s position in terms of the Fourier transform of the forcing function \(F_0(t)\). To see the behavior of the oscillator in the time domain, we now need to take the inverse Fourier transform: \begin{align} x(t) &= \frac{1}{2\pi} \int_{-\infty}^\infty e^{-i\omega t} \int_{-\infty}^{\infty} \frac{F_0(t’)/m}{\omega_0^2 - \omega^2 - 2 \beta i \omega} e^{i\omega t’} \dd{t’} \dd{\omega} \\ &= \int_{-\infty}^{\infty}\dd{t’} F_0(t’) \; \underbrace{ \frac{1}{2\pi m} \int_{-\infty}^{\infty} \dd{\omega} \frac{e^{-i\omega(t-t’)}}{\omega_0^2 - \omega^2 - 2 \beta i \omega}}_{G(t-t’)} \end{align} where I have interchanged the order of integration. To perform the integration over \(\omega\), we can use a contour that closes on a semicircle at \(R \to \infty\), either in the upper half-plane (if \(t' > t\)) or the lower half-plane (if \(t' < t\)).
Where are the poles? At the zeros of \(\omega^2 + 2 \beta i \omega - \omega_0^2\), which are \[ \omega_{\pm} = -i\beta \pm \sqrt{-\beta^2 + \omega_0^2} \] Let \(\omega_1 \equiv \sqrt{\omega_0^2 - \beta^2}\). For an under-damped system, the radical is real, so the poles lie in the LHP at \(-i\beta \pm \omega_1\).
Hence, if \(t' > t\), we can close in the UHP and get \(G(t'-t) = 0\). On the other hand, if \(t > t'\), we must close the contour in the LHP, which means that we traverse the contour in the negative direction and that we enclose the two poles. Hence, for \(t > t'\) we get \begin{equation}\label{eq:G} G(t-t’) = \frac{-2 \pi i}{2\pi m} \times \sum \text{residues} \end{equation} We may write the integrand as \[ -\frac{e^{-i\omega(t-t’)}}{(\omega - \omega_+)(\omega - \omega_-)} \] from which we infer that the residues are \[ -\frac{e^{-i\omega_+(t-t’)}}{(\omega_+ -\omega_-)} = -\frac{e^{-\beta(t-t’)} e^{-i\omega_1(t-t’) }}{2\omega_1} \qquad\text{and}\qquad -\frac{e^{i\omega_-(t-t’)}}{\omega_- - \omega_+} = \frac{e^{-\beta(t-t’)} e^{i\omega_1(t-t’)}}{2\omega_1} \] with sum \[ \frac{e^{-\beta(t-t’)}}{\omega_1} \frac{e^{i\omega_1(t-t’)} - e^{-i\omega_1(t-t’)}}{2} = i \frac{e^{-\beta(t-t’)}}{\omega_1} \sin\big[\omega_1(t-t’)\big] \] so that \begin{equation} G(t-t’) = \begin{cases} \displaystyle\frac{e^{-\beta(t-t’)}}{m\omega_1} \sin\big[\omega_1(t-t’)\big] & t > t’ \\ 0 & t < t’ \end{cases} \end{equation} and \begin{equation}\label{eq:convo} x(t) = \int_{-\infty}^{t} F_0(t’) G(t-t’) \dd{t’} \end{equation}
In other words, the actual response of the oscillator, \(x(t)\), is the convolution of \(F_0\) and \(G\). The Green’s function, \(G(t-t')\) describes the contribution to the motion of the oscillator at time \(t\) after it has been given a unit impulse at (the earlier) time \(t'\).
Let’s say that at \(t = 0\) an oscillatory forcing function begins and operates for \(N\) cycles: \[ F(t) = \begin{cases} F_0 \sin(\Omega t) & 0 < t < \frac{2 \pi N}{\Omega} \\ 0 & \text{otherwise} \end{cases} \] Its Fourier transform is \begin{align} \tilde{F}(\omega) &= \int_{0}^{2 \pi N/\Omega} F_0 \frac{e^{i\Omega t} - e^{-i\Omega t}}{2i} e^{i\omega t}\dd{t} \\ &= \frac{F_0}{2i} \bigg[ \frac{e^{i(\omega+\Omega)t}}{i(\omega+\Omega)} - \frac{e^{i(\omega-\Omega)t}}{i(\omega-\Omega)} \bigg]_0^{2\pi N/\Omega} \\ &= \frac{F_0}{2} \bigg[ \frac{e^{i 2\pi N \omega/\Omega}-1}{\omega-\Omega} - \frac{e^{i 2\pi N \omega/\Omega}-1}{\omega + \Omega}\bigg] \\ &= F_0 \Omega \frac{e^{i 2 \pi N \omega/\Omega}-1}{\omega^2 - \Omega^2} \end{align}
By the convolution theorem, then, \begin{align} x(t) &= \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{F_0 \Omega}{m} \frac{e^{-i\omega t}}{\omega_0^2-\omega^2-2\beta i \omega} \frac{e^{i2\pi N\omega/\Omega}-1}{\omega^2 - \Omega^2} \dd{\omega} \notag \end{align}
Evaluating this integral is somewhat tricky. Let’s first clean up the integrand slightly: \begin{equation} x(t) = \frac{F_0 \Omega}{2 \pi m} \int_{-\infty}^{\infty} \frac{e^{-i\omega t} - e^{i(2\pi N/\omega -t)\omega}} {(\omega-\omega_{+})(\omega-\omega_{-})(\omega-\Omega)(\omega+\Omega)} \dd{\omega} \end{equation} where \(\omega_{\pm} = \pm\omega_1 - i\beta\). Writing the integrand this way shows that it has four first-order poles, two on the real axis and in the path of integration, and two in the lower half-plane. It is straightforward to show that the residue at the poles on the real axis vanishes, because at those points, the numerator vanishes. Hence, we are left with worrying about the poles at \(\omega_\pm\). For \(t < 0\), we may close the contour along a semicircular path at \(R\to\infty\) in the UHP. Since that path contains no poles, we get zero, as we must expect: the oscillator is quiet before we start the forcing function at \(t = 0\).
For \(t > 2 \pi N/\omega\), we must close in the LHP for the exponentials in the numerators to go to zero. But if \(0 < t < 2 \pi N/\omega\), we have to separate numerator into two distinct integrals, because we have to close in opposite half planes for the two terms. In either case, the denominator goes to the same expression (apart from the term going to zero), so we can evaluate it first. When \(\omega \to \omega_+\), the denominator becomes \[ \Delta_+ = (\omega_+ - \omega_-)(\omega_+^2-\Omega^2) = 2\omega_1 (\omega_1^2 - \Omega^2 - \beta^2 -2 \beta \omega_1 i) = 2 \omega_1 (\Gamma^2 - 2 \beta \omega_1 i) \] where I have defined \(\Gamma^2 \equiv \omega_1^2 - \Omega^2 - \beta^2\), and when \(\omega \to \omega_-\), it becomes \[ \Delta_- = -(\omega_+ - \omega_-)(\omega_-^2 - \Omega^2) = -2\omega_1 (\omega_1^2 - \Omega^2 - \beta^2 + 2 \beta \omega_1 i) = - 2 \omega_1(\Gamma^2 + 2 \beta \omega_1 i) \]
When \(0 < t < 2 \pi / N\), we get a contribution from the two poles in the LHP from the term proportional to \(e^{-i\omega t}\) and nothing from the other exponential, whose contour we close in the UHP. Hence, the residue theorem yields \begin{align} x(t) &= \frac{F_0 \Omega}{2\pi m} (-2 \pi i) \bigg( \frac{e^{-i\omega_+ t}}{2\omega_1 (\Gamma^2-2\beta\omega_1 i)} - \frac{e^{-i\omega_- t}}{2 \omega_1 (\Gamma^2 + 2 \beta \omega_1 i)} \bigg)\notag \\ &= - \frac{F_0 \Omega i}{2 \omega_1 m} e^{-\beta t} \left( \frac{e^{-i\omega_1 t}}{\Gamma^2-2\beta\omega_1 i } - \frac{e^{i\omega_1 t}}{\Gamma^2+2\beta\omega_1 i} \right) \notag \\ &= - \frac{F_0 \Omega i}{2 \omega_1 m} \frac{ e^{-\beta t} }{\Gamma^4 + 4 \beta^2 \omega_1^2} \bigg( e^{-i \omega_1 t}(\Gamma^2 + 2 \beta \omega_1 i) - e^{i\omega_1 t}(\Gamma^2 - 2 \beta \omega_1 i) \bigg) \notag \\ &= - \frac{F_0 \Omega i}{2 \omega_1 m} \frac{ e^{-\beta t} }{\Gamma^4 + 4 \beta^2 \omega_1^2} \bigg( 4\beta\omega_1 i \cos(\omega_1 t) - 2 \Gamma^2 i \sin (\omega_1 t) \bigg) \notag \\ &= \frac{F_0 \Omega}{\omega_1 m} \frac{e^{-\beta t}}{\Gamma^4 + 4 \beta^2 \omega_1^2} \left[ \Gamma^2 \sin(\omega_1 t) - 2\beta\omega_1 \cos(\omega_1 t)\right] \end{align}