[Physics] RLC circuit – calculating resonant frequency

dissipationelectric-circuitselectrical-resistanceoscillatorsresonance

If I take a series RLC circuit connected to a battery, the impedance is minimized when $\omega = \frac{1}{\sqrt{LC}}$.

I also know that the series RLC circuit is analogous to a damped driven harmonic oscillator. However, the resonant frequency of a damped driven harmonic oscillator is reduced due to the damping. It is given by $\omega = \sqrt{\omega_0^2 – \gamma^2}$ where $\gamma$ is a damping parameter.

I am unable to see why the analogy fails here: How come the RLC circuit's resonant frequency has no dependence on $R$ but the harmonic oscillator does?

For definition – resonant frequency is the frequency of the driving force (or voltage) that maximized the amplitude (or current).

Best Answer

The anomaly is explained when it is realised that there are different types of resonance.
Once steady state has been reached a driver of constant amplitude makes a driven system oscillate at the frequency of the driver.
The frequency at which the response of the driven system is a maximum is called the resonant frequency.
For a mechanical system the maximum response of the driven system can be found by measuring by displacement amplitude, velocity amplitude, energy amplitude.
For an electrical system the responses can be found by measuring change, current, energy.

Often the easiest response to measure for a mechanical system is the displacement amplitude and for this type of resonance - displacement resonance - the peaks occurs at frequencies given by $\omega = \sqrt{\omega_0^2 - \gamma^2}$.
This is also true for the charge resonance of an LCR circuit.

For an electrical circuit it is often easier to measure the current and the current resonance occurs when $\omega = \omega_o$. For the mechanical case this would be called velocity resonance.