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:

$(2): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = 0$

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

$\omega = \sqrt {a^2 - 2 b^2}$

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

We have that:

$\ds \dfrac \nu {\nu_R}$

$=$

$\ds \dfrac {\sqrt {a^2 - b^2} } {\sqrt {a^2 - 2 b^2} }$

$\ds \leadsto \ \ $

$\ds \dfrac {\nu^2} { {\nu_R}^2}$

$=$

$\ds \dfrac {a^2 - b^2} {a^2 - 2 b^2}$

$\ds $

$=$

$\ds \dfrac {a^2 - 2 b^2 + b^2} {a^2 - 2 b^2}$

$\ds $

$=$

$\ds 1 + \dfrac {b^2} {a^2 - 2 b^2}$

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

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

$\blacksquare$