Amplitude-Frequency curve Given a resonance curve just like this:

Could someone explain to me what the physical meaning of the intersection with the ordinate is? 
At first glance I would say it has to be $(0 | 0) $ since if I have given no external excitation, I have no amplitude. But many diagrams show it differently, are these just poorly done or is there any background?
 A: It's the response of the system to a stimulation at zero frequency.
In other words, it tells you the displacement of the system in equilibrium under a time independent force.
Let me give an example.
Consider a mass on a spring with friction and an external force $F_{\text{ext}}(t)$.
The friction force is
$$F_{\text{friction}} = -\mu \dot{x}$$
so the equation of motion is
\begin{align}
F(t) &= m \ddot{x}(t) \\
F_{\text{ext}}(t) + F_{\text{spring}}(t) + F_{\text{friction}}(t) &= m \ddot{x}(t) \\
F_{\text{ext}}(t) -k x(t) - \mu \dot{x}(t) &= m\ddot{x}(t) \\
\ddot{x}(t) + (\mu/m) \dot{x}(t) + (k/m) x(t) = F_{\text{ext}}(t)/m
\end{align}
where $k$ is the spring constant and $m$ is the mass.
Define $\omega_0^2 = k/m$ and $\beta \equiv \mu/m$ to get
$$\ddot{x}(t) + \beta \dot{x}(t) + \omega_0^2 x(t) = F_{\text{ext}}(t)/m \, .$$
This is the standard equation for a forced, damped oscillator.
If you Fourier transform this equation you get
\begin{align}
(-\omega^2 - i\omega \beta + \omega_0^2)\tilde{x}(\omega) &= \tilde{F}_{\text{ext}}(\omega)/m \\
\tilde{x}(\omega) &= \frac{\tilde{F}_{\text{ext}}(\omega)/m}{\omega_0^2 - \omega^2 - i \omega \beta} \, .
\end{align}
If you plot the modulus of this you get a peak as a function of $\omega$, like the one in the question.
The peak happens near $\omega = \omega_0$ (as long as $\beta$ isn't too big).
Now let's put in $\omega=0$:
$$\tilde{x}(0) = \frac{\tilde{F}_{\text{ext}}(0)/m}{\omega_0^2} = \tilde{F}_{\text{ext}}(0)/k \, .$$
This just says that the average position is the average external force divided by the spring constant.
We could have guessed this by considering a time independent force $F_{\text{ext}}$ and solving Newton's equation (no friction if there's no motion in equilibrium)
\begin{align}
F_{\text{total}} &= 0 \\
F_{\text{ext}} - k x_{\text{equilibrium}} &= 0 \\
x_{\text{equilibrium}} = F_{\text{ext}} / k \, .
\end{align}
