Condition for Resonance in Forced Vibration of Underdamped System

Theorem
Consider a physical system $S$ whose behaviour is defined by the second order ODE:
 * $(1): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$

where:
 * $K \in \R: k > 0$
 * $a, b \in \R_{>0}: b < a$

Then $S$ is in resonance when:
 * $\omega = \sqrt {a^2 - 2 b^2}$

and thus the resonance frequency is:
 * $\nu_R = \dfrac {\sqrt {a^2 - 2 b^2} } {2 \pi}$

This resonance frequency exists only when $a^2 - 2 b^2 > 0$.

Proof
From Linear Second Order ODE: $y'' + 2 b y' + a^2 y = K \cos \omega x$ where $b < a$ the general solution of $(1)$ is:
 * $(2): \quad y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$

where:
 * $\alpha = \sqrt {a^2 - b^2}$
 * $\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$

Consider the steady-state component of $(2)$:


 * $y_s = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$

By definition, the amplitude of $y_s$ is:
 * $A = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} }$

We have that both $a$ and $b$ are greater than $0$.

It is also assumed that $\omega$ is also greater than $0$.

Thus:
 * $4 b^2 \omega^2 > 0$
 * $a^2 > 0$
 * $\omega^2 > 0$

Then $y_s$ is at a maximum when:
 * $\map f \omega = 4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2$

is at a minimum.

Differentiating $\map f \omega$ $\omega$:

Setting $\map {f'} \omega = 0$:

As $\omega > 0$ the first of these cannot apply.

If $a^2 < 2 b^2$ then $a^2 - 2 b^2 < 0$ and so $\omega^2$ has no real roots.

Hence the resonance frequency is given by:
 * $\omega = \sqrt {a^2 - 2 b^2}$

$a^2 > 2 b^2$.