In driven oscillator it can be explained by the following differential equation $$\ddot{x} + 2\beta \dot {x} + \omega_0 ^2x = A \cos(\omega t)$$ where the $2\beta$ is coefficient of friction, the $\omega_0$ is the frequency of simple harmonic oscillator, and $A \cos(\omega t)$ is the driven force divided by the mass of oscillating object.

The particular solution $x_p$ of the equation is

\begin{align} x_p &= \frac{A}{\sqrt{(\omega_0 ^2 - \omega ^2)^2 + 4 \omega ^2 \beta ^2}}\cos(\omega t - \delta) \\ \tan \delta &= \frac{2\omega \beta}{\omega_0 ^2 - \omega ^2} \end{align}

Now, in classical mechanics of particles and systems(Stephen T. Thornton, Jerry B. Marrion) it finds the amplitude's maximum by \begin{align} \left . \frac{\mathrm{d}}{\mathrm{d}\omega}\frac{A}{\sqrt{(\omega_0 ^2 - \omega ^2)^2 + 4 \omega ^2 \beta ^2}} \right | _{\omega = \omega_R} = 0 \\ \therefore \omega_R = \sqrt{\omega_0^2 - 2\beta ^2} \qquad (\omega_0 ^2 -2\beta^2 >0) \end{align}

and defines Q factor in driven oscillator by $$Q \equiv \frac{\omega_R}{2\beta}$$

Here I have some questions about calculating Q factor in lightly damped driven oscillator. $$Q = \frac{\omega_R}{2\beta} \simeq \frac{\omega_0}{\Delta \omega}$$ $\Delta \omega$ is the width of $\frac{1}{\sqrt{2}}$(amplitude maximum).

  1. I searched Q factor in google, but there are so much confusion on understanding the condition "lightly damped". One says it means $\omega_0 >> 2\beta$, and the other says $\omega_0 >> \beta$. Which is right?

  2. In google, they calculate this very absurdly. They assume that $\omega \simeq \omega_0$ and change the part of amplitude denominator by $$(\omega_0 ^2 - \omega ^2) = (\omega_0 + \omega)(\omega_0 - \omega) \simeq 2\omega_0(\omega_0 - \omega)$$ I don't understand this absurd approximation. Why $(\omega_0 + \omega) \simeq 2\omega_0$ is possible and $(\omega_0 - \omega) \simeq 0$ is not? Also, how can we assume $\omega \simeq \omega_0$?

  3. I want to know how to derive the $Q \simeq \frac{\omega_0}{\Delta \omega}$


2 Answers 2


Let me answer your questions one by one.

  1. $\omega_0\gg 2\beta$ means the same thing with $\omega_0\gg\beta$. For example, we know that $10^4\gg 2$, so $10^4\gg2\times1$ and $10^4\gg1$. For $10^4$, $2$ has no differnece with $1$. Hence for $\omega_0$, $2\beta$ have no difference with $\beta$.
  2. Because of $\omega=\sqrt{\omega_0^2-2\beta^2}$ and $\omega_0\gg 2\beta$, we can get $\omega=\omega_0+\Delta\omega$, where $\Delta\omega$ is a small quantity. And then \begin{align} \omega_0^2-\omega^2 & =(\omega_0+\omega)(\omega_0-\omega) \\ & =(2\omega_0+\Delta\omega)\Delta\omega \end{align}

On the right side of the equation, because of $\Delta\omega\ll 2\omega_0$, we can give up $\Delta\omega$. But we shouldn't give up the second term $\Delta \omega$. As a result, we have $\omega_0^2-\omega^2=2\omega_0\Delta\omega$.

  1. We start from $Q=\frac{\omega}{2\beta}$, \begin{align} \omega & =\sqrt{\omega_0^2-2\beta^2}\\ \omega_0^2-\omega^2 &=2\beta^2=2\omega_0\Delta\omega \end{align}

From the knowledge of calculus, we know $2xdx=d(x^2)$,so $2\omega_0\Delta\omega=\Delta\omega^2$, we get, $$ 2\beta=\sqrt{2}\Delta\omega\approx\Delta\omega $$ since $\Delta\omega$ is so small. And $\omega\approx\omega_0$, as a result, $$ Q=\frac{\omega_0}{\Delta\omega} $$

  • $\begingroup$ Completely agree with part 1. As for part 2, I suggest you are more explicit about ignoring small difference when adding, but not ignoring them when subtracting (since the difference is "all you have" it matters enormously whether it's zero or just "small"). You are kind-of saying that... but could be more obvious about it. Upvoting all the same. $\endgroup$
    – Floris
    Jan 18, 2016 at 5:32
  • $\begingroup$ In part2, does $\omega$ mean that the frequency of the $\frac{1}{\sqrt{2}}$ of maximum amplitude? Then how one verify the $\Delta \omega$ is very small quantity? And also I don't understand $\omega = \sqrt{\omega_0 ^2 -2\beta ^2}$. I think it is $\omega_R$, not $\omega$. $\endgroup$ Jan 18, 2016 at 5:59
  • $\begingroup$ @cokecokecoke Yeah , it is $\omega_R$, I just ignored the subscript, I hope it won't make you confused. $\endgroup$
    – Wang Yun
    Jan 18, 2016 at 6:12
  • $\begingroup$ @cokecokecoke Because every thing here occured in around the resonance i.e. $\omega=\omega_0$, we can asumme $\omega=\omega_0+\Delta\omega$, where $\Delta\omega$ is a very small quantity. $\endgroup$
    – Wang Yun
    Jan 18, 2016 at 6:15

So we can assume that $\omega$ $\approx$ $\omega_0$, since in the derivation we are near resonance (i.e. during maximum amplitude). In other words, Q-factor is concerned only with the behavior near resonance, which implies $\omega$ $\approx$ $\omega_0$.

As for the second part of your question as to why $(\omega_0 - \omega) \approx 0$ is not possible, well this is where the bit of hand waving comes in, maybe. Most physicists do approximations to lowest order of reasonability. That is, if you get something absurd (like 0 in this case) in making your approximation, you need to expand another term or be a bit more careful. I wasn't satisfied with this either when my professor first told me, but it's how most physicists do their work. However, there is an explanation that is more precise that involves considering the percent difference between considering $(\omega_0 - \omega)$ and $(\omega_0 + \omega)$ and their respective approximations, nothing that the percent difference in the former (saying it's zero) is larger than saying that the latter is $2\omega_0$.


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