Transfer function pole on the Imaginary axis indicates that the system is marginally stable which in time domain can be represented as a sinusoidal motion with constant amplitude and frequency of the Imaginary axis pole. In some applications, oscillations with small amplitude might be acceptable.
Is amplitude of oscillation a (only) function of initial condition of the system?
No. For a system with a transfer function of the form $H(s) = \frac{\cdots}{s^2 + \omega_0^2}$, the amplitude of oscillation will increase any time it is excited by a signal that has a component at $x(t) = \cos \omega_0 t + \phi$. Even if you're not exciting it intentionally, random noise will excite that pole. Basically, the output will be of the form $a(t) \cos \omega_0 t + b(t) \sin \omega_0 t$, where $a(t)$ and $b(t)$ will be Wiener (random-walk) processes. Such processes have variances that tend to infinity as $t \to \infty$.
How to intuitively quantify amplitude of the oscillations for marginally stable system?
"Big and growing, boss, and I don't think they'll get smaller!"
Possibly by how fast they grow. But grow they will.
In some applications, oscillations with small amplitude might be acceptable.
If you have a system that's exhibiting persistently small oscillations, then you're seeing the signs of a nonlinear phenomenon called a "limit cycle". Basically, there's some oscillator in there that's either inherently small-valued (like the oscillation you may see around the least significant bit of a DAC or ADC), or there's a big oscillation that's mostly not getting to the output. Either way, if it persists at some amplitude, there's some nonlinear process that's keeping it that way.
I have to disagree with Tim here. And it's because I once took that position and was shown wrong by James McCartney, the author of SuperCollider.
There's a difference between "marginally stable" and either "stable" or "unstable".
For the marginally-stable oscillator:
$$\begin{align} y[n] &= 2 \cos(\omega_0) y[n-1] - y[n-2] \\ \\ y[-1] &= A \cos(-\omega_0 + \phi) \\ y[-2] &= A \cos(-2\omega_0 + \phi) \\ \end{align}$$
will result in this output:
$$ y[n] = A \cos(\omega_0 n + \phi) \qquad \forall n \in \mathbb{Z} \ge 0 $$
and with double-precision floating point is stable to the extent that I have run it overnight and no change in amplitude.
The oscillation of marginally stable system is causesing by feed-back system with totall phase shift of 2pi and total loop gain of -1 at ultimate frequency. The oscillition will not occur with zero constant input so we need to feed step input or impulse input to the system and the system will oscilate at sustain or constant amplitude.
For Q1. if there are no change in input afterward so the amplitude of the input will only depended on initial condition. But if there is some input it another story. The input of marginal stable system can either increse or decrese the amplitude of the oscillation. You can think about pushing the swing, if you push in phase in will swing further otherwise it will slow down.
For Q2 it make no sense. The control system that we use don't rely on marginal stabillity. we design it to be robust so the osilation will be dampen down. However we can find the overshoot and decay rate the the oscilation from step response to determine performance of the system.
