14. The memory of two bells

Two bells have the same natural frequency $f_0$ and are struck so that each initially stores the same mechanical energy $E_0$. One has quality factor $Q=100$; the other, $Q=5$. After the blows, no further energy is supplied. How much energy remains in each bell after ten natural periods? When the low-$Q$ bell has fallen to one percent of $E_0$, how much remains in the high-$Q$ bell — and what fraction of its original amplitude does each bell still have?

Solution

After ten cycles the $Q=100$ bell still holds about $53\%$ of its energy; the $Q=5$ bell holds only $3.5\times10^{-6}$ of its energy and is effectively still. When the low-$Q$ bell reaches $1\%$, the high-$Q$ bell retains about $79\%$. Their amplitudes are then $10\%$ and $89\%$ of their initial values.

Figure 1. The bells begin alike; only their dissipation differs. The high-$Q$ bell remembers a blow for many cycles, while the felt damper gives the low-$Q$ bell a short memory.

For a lightly damped oscillator, the definition of quality factor gives the cycle-averaged power loss as

$$\left\langle P_{\mathrm{loss}}\right\rangle=\frac{\omega_0}{Q}E.$$

Once the blow is over, that loss is the whole energy balance:

$$\frac{dE}{dt}=-\frac{\omega_0}{Q}E, \qquad E(t)=E_0e^{-\omega_0t/Q}.$$

Count time in natural periods, $N=t/T_0=\omega_0t/2\pi$. The energy envelope then becomes

$$\boxed{\frac{E(N)}{E_0}=e^{-2\pi N/Q}}.$$

Ten cycles later

Substitution at $N=10$ gives

$$\begin{aligned} \frac{E_{100}}{E_0}&=e^{-2\pi(10)/100}\approx0.533,\\ \frac{E_{5}}{E_0}&=e^{-2\pi(10)/5}\approx3.49\times10^{-6}. \end{aligned}$$

The high-$Q$ bell has lost not quite half the blow. The low-$Q$ bell has less than four millionths of it left: at the same cycle count, the former carries roughly $1.5\times10^5$ times as much energy as the latter.

Catch them at the same instant

Let the low-$Q$ bell fall to $E_5/E_0=0.01$. Its cycle count is

$$N=\frac{5}{2\pi}\ln 100\approx3.66.$$

At that very instant the high-$Q$ bell retains

$$\begin{aligned} \frac{E_{100}}{E_0} &=e^{-2\pi(3.66)/100}\\ &=(0.01)^{5/100}\\ &\approx0.794. \end{aligned}$$

Figure 2. Equal pitch makes cycle count a common clock. The two energy envelopes separate almost immediately.

Energy is quadratic in amplitude, $E\propto A^2$. Thus at this snapshot

$$\begin{aligned} \frac{A_5}{A_0}&=\sqrt{0.01}=0.10,\\ \frac{A_{100}}{A_0}&=\sqrt{0.794}\approx0.89. \end{aligned}$$

One bell has only a tenth of its original swing while the other still swings at nearly nine-tenths. Energy decays twice as fast in the exponent because it is the square of the amplitude.

A caution about clocks. Quality factor measures memory in cycles, not directly in seconds. The energy e-folding time is

$$\tau_E=\frac{Q}{\omega_0}=\frac{Q}{2\pi f_0}.$$

If the frequencies differ, a high-$Q$ resonator can therefore die sooner by the clock. A $Q=100$ resonator at $1\,\mathrm{kHz}$ has $\tau_E\approx16\,\mathrm{ms}$, whereas a $Q=5$ oscillator at $1\,\mathrm{Hz}$ has $\tau_E\approx0.80\,\mathrm{s}$. High $Q$ means long memory per cycle; the frequency decides how quickly the cycles pass.

Deeper in the notebook: the Classical Mechanics shelf — still being bound; damped motion and resonance will live there.