[Physics] Q factor of driven oscillator

classical-mechanicsharmonic-oscillator

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 $Q \simeq \frac{\omega_0}{\Delta \omega}$

Best Answer

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, $$ Q=\frac{\omega_0}{\Delta\omega} $$

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