Resonance Frequency is less than Natural Frequency

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 the resonance frequency of $S$ is smaller than the natural frequency of the associated second order ODE:
 * $(2): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = 0$

Proof
From Natural Frequency of Underdamped System, the natural frequency of $(2)$ is:
 * $\nu = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$

From Condition for Resonance in Forced Vibration of Underdamped System, the resonance frequency of $S$ is:
 * $\omega = \sqrt {a^2 - 2 b^2}$


 * $\nu_R = \dfrac {\sqrt {a^2 - 2 b^2} } {2 \pi}$

We have that:

Thus $\nu_R < \nu$ $a^2 - 2 b^2 > 0$.

But from Condition for Resonance in Forced Vibration of Underdamped System, in order for there to be resonance, $a^2 - 2 b^2 > 0$.

The two conditions are compatible, and so hence the result.