We have a system represented by this differential equation:
$$\frac{d^2y(t)}{dt^2} + b\frac{dy(t)}{dt} + cy(t) = u(t)$$
Our objective is to determine the condition where this system just begins to oscillate when disturbed.
Here is how I have interpreted this question:
My first move is to take the Laplace Transform of the differential equation.
$$\mathscr{L}\lbrace\frac{d^2y(t)}{dt^2}\rbrace + \mathscr{L}\lbrace b\frac{dy(t)}{dt}\rbrace + \mathscr{L}\lbrace cy(t)\rbrace = \mathscr{L}\lbrace u(t)\rbrace$$
$$S^2Y(S) + bSY(S) + cY(S) = U(S)$$
$$Y(S)[S^2 + bS + c] = U(S)$$
and so, the transfer function $H(S) = \frac{1}{S^2 + bS + c}$.
Do I need to take the Laplace here?
Since we are looking to derive the condition where a disturbance leads to oscillation, we can model this situation as an Harmonic Oscillator of the form:
$$\frac{d^2y(t)}{dt^2} + 2 \zeta \omega_{n}\frac{dy(t)}{dt} + \omega_{n}^{2}y(t) = 0$$
Also, there are 4 flavours of damping:
1. Undamped ($\zeta = 0$)
2. Underdamped ($0 < \zeta < 1$)
3. Overdamped ($\zeta > 1$)
4. Critically Damped ($\zeta = 1$)
The only flavour that exhibits oscillation is the Underdamped case.
So using the condition $0 < \zeta < 1$ and the fact that $b = 2\zeta\omega_{n}$ and $c = \omega_{n}^{2}$
So, $0 < \frac{b}{2\omega_{n}} < 1 \implies 0 < b < 2\omega_{n}$.
$\omega_{n} = \sqrt{c} \implies 0 < b < 2\sqrt{c}$.
Is this the correct way to answer this question?
All feedback is welcome.