Quality Factor

The energy loss rate of a weakly damped (i.e., $\nu\ll 2\,\omega_0$) harmonic oscillator is conveniently characterized in terms of a parameter, $Q_f$, which is known as the quality factor. (Note that the standard symbol for the quality factor is $Q$. We are only using $Q_f$ here to avoid confusion with electrical charge.) This quantity is defined to be $2\pi$ times the energy stored in the oscillator, divided by the energy lost in a single oscillation period. If the oscillator is weakly damped then the energy lost per period is relatively small, and $Q_f$ is therefore much larger than unity. Roughly speaking, $Q_f$ is the number of oscillations that the oscillator typically completes, after being set in motion, before its amplitude decays to a negligible value. (To be more exact, $Q_f$ is the number of oscillations that the oscillator completes before its amplitude decays to about $4\%$ of its original value. See Exercise 6.) For instance, the quality factor for the damped oscillation shown in Figure 2.1 is $12.6$. Let us find an expression for $Q_f$.

As we have seen, the motion of a weakly damped harmonic oscillator is specified by [see Equation (2.12)]

$$x = a\,{\rm e}^{-\nu\,t/2}\,\cos(\omega_1\,t-\phi).$$ (2.23)

It follows that

$$\dot{x} =- \frac{a\,\nu}{2}\,{\rm e}^{-\nu\,t/2}\,\cos(\omega_1\,t-\phi)- a\,\omega_1\,{\rm e}^{-\nu\,t/2}\,\sin(\omega_1\,t-\phi).$$ (2.24)

Thus, making use of Equation (2.20), the energy lost during a single oscillation period is

$${\mit\Delta} E = -\int_{\phi/\omega_1}^{(2\pi+\phi)/\omega_1} \frac{dE}{dt}\,dt$$

$$= m\,\nu\,a^{\,2}\int_{\phi/\omega_1}^{(2\pi+\phi)/\omega_1}{\rm ... ...{2}\,\cos(\omega_1\,t-\phi) + \omega_1\,\sin(\omega_1\,t-\phi)\right]^{\,2} dt.$$ (2.25)

In the weakly damped limit, $\nu\ll 2\,\omega_0$, the exponential factor is approximately unity in the interval $t=\phi/\omega_1$ to $(2\pi+\phi)/\omega_1$, so that

$${\mit\Delta} E \simeq \frac{m\,\nu\,a^{\,2}}{\omega_1}\int_0^{2\p... ...,\omega_1\,\cos\theta\,\sin\theta + \omega_1^{\,2}\,\sin^2\theta\right)d\theta,$$ (2.26)

where $\theta=\omega_1\,t-\phi$. Thus,

$${\mit\Delta} E \simeq \frac{\pi\,m\,\nu\,a^{\,2}}{\omega_1}\left(... ...\,2}\right) = \pi\,m\,\omega_0^{\,2}\,a^{\,2}\left(\frac{\nu}{\omega_1}\right),$$ (2.27)

because, as is readily demonstrated,

$$\int_0^{2\pi}\cos^2\theta\,d\theta = \int_0^{2\pi}\sin^2\theta\,d\theta =\pi,$$ (2.28)

$$\int_0^{2\pi}\cos\theta\,\sin\theta\,d\theta =0.$$ (2.29)

The energy stored in the oscillator (at $t=0$) is [cf., Equation (1.16)]

$$E = \frac{1}{2}\,m\,\omega_0^{\,2}\,a^{\,2}.$$ (2.30)

Hence, we obtain

$$Q_f = 2\pi\,\frac{E}{{\mit\Delta} E} = \frac{\omega_1}{\nu}\simeq \frac{\omega_0}{\nu}.$$ (2.31)