# For DE, Is the natural frequency always the resonant frequency?

Let's say I have a DE of the form:

$$a y''(t) + b y'(t) + c y(t) = A \cos(\omega t)$$

The homogeneous DE has complex conjugate roots, so the system oscillates without the presences of a forcing function: $$A \cos(\omega t)$$

Let's say I remove the forcing function $$A \cos(\omega t)$$ from the DE by setting RHS to zero, and i'm left with a homogenous DE equation:

$$a y''(t) + b y'(t) + c y(t) = 0$$

Then, the solution would of this system is the zero input response of the system, and the frequency of oscillation system of this undriven system is called the natural response $$\omega_0$$ of the system.

QUESTION: Is it always the case that the resonate frequency of the inhomogenous DE with a forcing function:

$$a y''(t) + b y'(t) + c y(t) = A \cos(\omega t)$$

is equal to the natural response frequency $$\omega_0$$ of the undriven system?

By resonant frequency, I mean the value of $$\omega$$ that I can select a value such that the forcing function $$A \cos(\omega t)$$ will produce a y(t) output the has the largest average peak amplitude for the inhomogenous system equation.

• intuitively, for the total solution, $y(t) = y_h(t) + y_p(t)$, it seems like if $\omega$ is set to the natural frequency $\omega_0$ then the homogeneous solution and particular solution add together to generate the largest peak amplitude... except, i can't really assume that the particular solution is in phase with the homogeneous solution..so maybe they don't add together constructively... – pico Jan 16 at 16:05
• the case of the LC circuit: $LCy'' + y = A \cos(\omega t)$ ... has a total solution of: $y = K \cos(\frac{1}{\sqrt{LC}}t + \phi) + \frac{A}{1 - LC\omega^2} \cos(\omega t)$... where the natural frequency is: $\omega_0 = \frac{1}{\sqrt{LC}}$... rearranging, $y = \cos(\omega_0 t + \phi) + LC \bigg(\frac{A}{\omega_0^2 - \omega^2}\bigg) \cos(\omega t)$... in this case, natural frequency = resonate frequency....becuse $y = \infty$ when $\omega = \omega_0$ ... I don't know what makes this case special though... you say damping coefficient in characteristic equation... not sure... – pico Jan 16 at 16:19

Using complex numbers, the admittance (inverse of the transmittance) of the system is

$$-a\omega^2+ib\omega+c$$ and its squared modulus is

$$(c-a\omega^2)^2+(b\omega)^2.$$

It has a minimum at

$$\omega_r=\frac{\sqrt{4ac-\color{red}2b^2}}{2a}$$ which is the resonant pulsation.

By contrast, the natural pulsation is the imaginary part of the roots of $$az^2+bz+c$$, or

$$\omega_0=\frac{\sqrt{4ac-b^2}}{2a}.$$

As you see, these two puslations are not identical, and this is due to the damping term.

• which one is the damping term? – pico Jan 16 at 15:55
• what damping value would give you natural freq = resonate freq? – pico Jan 16 at 15:56
• @pico: no damping. – Yves Daoust Jan 16 at 15:57
• @pico The damping term is $b$. – Milo Brandt Jan 16 at 16:04
• if b = 0, then natural frequency = resonate frequency? – pico Jan 16 at 16:21