About resonance in the (undamped) harmonic oscillator

Question

Consider the undamped harmonic oscillator $x''(t) + \omega_0^2 x(t) = A cos(\omega t)$. The general (real) solution of the homogenous equation is given by $x_h(t) = a cos(\omega_0 t) + b sin(\omega_0 t)$. For the inhomogeneous we get a particular solution $x_p(t) = \frac{A}{2 \omega} t sin(\omega t)$ (via the Ansatz $x_p(t) = c t sin (\omega t)$). Why do we say that the forcing term is 'resonant' if $\omega = \omega_0$? Is it meant that the solution simply tends in this case to infinity for $t \to \infty$?

What is the difference if we consider the damped oscillator $x''(t) + r x' + \omega_0^2 x(t) = A cos(\omega t)$? I suppose that in this case $x(t)$ gets maximal if $\omega = \omega_0$. Is this correct and what more could one say about resonance here?

Answer 1

Taking the Fourier transform you get $$(\omega_{0}^{2}-\Omega^{2})\hat{x}(\Omega)=\pi{A}(\delta(\Omega-\omega)+\delta(\Omega+\omega))$$ $$\hat{x}(\Omega)=\frac{A\pi(\delta(\Omega-\omega)+\delta(\Omega+\omega))}{(\omega_{0}^{2}-\Omega^{2})}$$ $$x(t)=\frac{1}{2\pi}\int_{\mathbb{R}}\hat{x}(\Omega)e^{i\Omega{t}}d\Omega=\frac{A\cos(\omega{t})}{\omega_{0}^{2}-\omega^{2}}$$ When $\omega_{0}\rightarrow\omega$ solution diverges, in physics you say it's a resonance.

Answer 2

In the whole below, it is supposed that $\omega_0\neq 0$ . $$\text{FIRST PART}$$ $$x''(t) + \omega_0^2 x(t) = A \cos(\omega t)$$

Case $\omega_0\neq \omega$ :

$x_p(t)=c \cos(\omega t) \quad\to\quad c=\frac{A}{\omega_0^2-\omega^2}$ $$x(t)=\frac{A}{\omega_0^2-\omega^2}\cos(\omega t)+a\sin(\omega t)+b\cos(\omega t)$$

Case $\omega_0= \omega$ :

$x_p(t)=c t\sin(\omega_0 t) \quad\to\quad c=\frac{A}{2\omega_0^2}$ $$x(t)=\frac{A}{2\omega_0^2}\sin(\omega_0 t)+a\sin(\omega_0 t)+b\cos(\omega_0 t)$$

$$\text{SECOND PART}$$ $$x''(t) +r x'+ \omega_0^2 x(t) = A \cos(\omega t)$$ They are six cases to consider :

(1) : $\omega_0\neq \omega \quad$ and $\quad r^2-4\omega_0^2>0$

(2) : $\omega_0\neq \omega\quad$ and $\quad r^2-4\omega_0^2=0$

(3) : $\omega_0\neq \omega\quad$ and $\quad r^2-4\omega_0^2<0$

(4) : $\omega_0= \omega\quad$ and $\quad r^2-4\omega_0^2>0$

(5) : $\omega_0= \omega\quad$ and $\quad r^2-4\omega_0^2=0$

(6) : $\omega_0= \omega\quad$ and $\quad r^2-4\omega_0^2<0$

The three first cases are more complicated to solve (but possible, of course).

For the three last cases (with $\omega_0=\omega$) the solution is :

Case (4) : $$x(t)=\frac{A}{r\omega_0}\sin(\omega_0 t) +a\,\exp\left(\left(\sqrt{r^2-4\omega_0^2}-r\right) \frac{t}{2}\right) +b\,\exp\left(\left(-\sqrt{r^2-4\omega_0^2}-r\right) \frac{t}{2}\right)$$

Case (5) : $$x(t)=\frac{A}{r\omega_0}\sin(\omega_0 t) +a\,e^{-\omega_0 t} +b\,t\,e^{-\omega_0 t}$$

Case (6) : $$x(t)=\frac{A}{r\omega_0}\sin(\omega_0 t) +a\,\sin\left(\left(\sqrt{-r^2+4\omega_0^2}-r\right) \frac{t}{2}\right) +b\,\cos\left(\left(-\sqrt{-r^2+4\omega_0^2}-r\right) \frac{t}{2}\right)$$

Note: the same symbols $a$ and $b$ are used for all cases. This doesn't mean that those constants are equal from one case to another.