Resonance gets sharper just by increasing the resonance freq, why?

[Aaron121]

The $Q$ factor of an oscillating system is defined as $\omega_{r}/\Delta \omega$, where $\omega_{r}$ is the resonant frequency, and $\Delta \omega$ the resonance width. As I understand, $Q$ measures how sharp the resonance curve is.

Why is it that the resonance curve gets sharper (higher $Q$) just by virtue of having a higher resonant frequency, while keeping the same width $\Delta \omega$ ?

 

[BvU]

Who says $\Delta \omega$ does not change ?

 

[Aaron121]

@BvU For example, for a damped, driven harmonic oscillator with a natural frequency $\omega_{0}$, the cutoff frequencies are at $\sqrt{\omega_{0}^2+\frac{\gamma ^2}{4}}\pm \frac{\gamma}{2}$, where $\gamma$ is the damping coefficient. The equation of motion of the oscillator is given by $\ddot{x}(t)+\gamma \dot{x}(t)+\omega_{0}^2 x(t)=F(t)/m$, (see page $10$ of this document). $\Delta \omega$, in this case, is simply $\gamma$, which doesn't depend on the resonant frequency $\omega _{r}=\omega _{0}$.

 

[sophiecentaur]

You need to look closely at the damping coefficient and see how it is actually defined. I think you will find it refers to number of oscillations to reach 1/e or whatever. The scaling is implicit.

 

[Aaron121]

$\gamma$ is the proportionality factor between the friction force $\textbf{f}_{r}$ and the velocity $\dot{\textbf{x}}(t)$, i.e., $\textbf{f}_{r}=-\gamma \dot{\textbf{x}}(t)$. The friction force is a viscous one.

 

[sophiecentaur]

$\gamma$ is the proportionality factor between the friction force $\textbf{f}_{r}$ and the velocity $\dot{\textbf{x}}(t)$, i.e., $\textbf{f}_{r}=-\gamma \dot{\textbf{x}}(t)$. The friction force is a viscous one.

That has the rate of change of x in it - aka proportional to FREQUENCY boom boom.