Using Exponential Decay to Explain Amplitude Decreases

Consider the equation of motion of the underdamped harmonic oscillator:

$$x(t) = Ae^{-\frac{b}{2m} t} e^{ i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t} + Be^{-\frac{b}{2m} t} e^{ -i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t}.$$

This solution describes rapid oscillation within an envelope of exponentially decaying envelope. The amplitudes of the critically damped and overdamped harmonic oscillators similarly decay exponentially.

Several parameters are used throughout physics and engineering literature to describe how the amplitude of a damped harmonic oscillator decays over time. The most mathematically straightforward parameter is the $1/e$ decay time, often denoted as $\tau$. Suppose a damped harmonic oscillator starts at amplitude $x_0$; then the amplitude of the damping envelope is $x_0 e^{-\frac{b}{2m} t}$. The $1/e$ decay time is defined as the time $\tau$ for which the amplitude has decreased to $x_0 / e \approx .368 x_0$. This is equivalent to the exponent in the decay envelope taking value $-1$, i.e.:

$$-\frac{b}{2m} \tau = -1 \implies \tau = \frac{2m}{b} .$$

A more sophisticated parameter is the quality factor $Q$:

$$Q = \frac{\text{energy stored}}{\text{energy dissipated per radian}}.$$

As a mnemonic for understanding and remembering the name, a high quality crystal will ring for a very long time when struck. Damped harmonic oscillators with large quality factors are underdamped and have a slowly decaying amplitude and vice versa. Critical damping occurs at $Q = \frac12$, marking the boundary of the two damping regimes.

What is the quality factor of a damped harmonic oscillator in terms of $k$,$m$, and $b$?

---

Solution:

The stored energy in the damped harmonic oscillator is the "spring potential energy": $$E(t) = \frac12 kA(t)^2$$ where $A(t)$ is the amplitude of the harmonic oscillator. Recalling that the damped harmonic oscillator has a $e^{-\frac{b}{2m} t}$ decay envelope, this is equal to: $$E(t) = \frac12 k A^2 e^{-\frac{b}{m} t} = E_0 e^{-\frac{b}{m} t}.$$

The energy dissipated per radian is: $$\Delta E = \left|\frac{dE}{dt}\right| \Delta t,$$ with $\Delta t$ giving the time it takes to oscillate through one radian, equal to $\frac{1}{\omega}$.

The derivative is given by $\frac{dE}{dt} = -\frac{b}{m} E(t)$, so the energy dissipated is: $$\Delta E = \frac{b}{m \omega} E,$$

and finally the quality factor is: $$Q = \frac{E}{b/(m\omega) E} = \frac{m\omega}{b},$$ where $\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }$ is the frequency of the damped harmonic oscillator. For highly underdamped systems, $\omega \approx \sqrt{\frac{k}{m}}$ and the quality factor is $Q \approx \frac{\sqrt{km}}{b}$. From this, it is apparent that critical damping occurs at $Q = \frac12$ by squaring both sides and comparing to the criterion for critical damping.

A last metric that is more common in engineering literature for describing amplitude attenuation in damped oscillators is the logarithmic decrement $\delta$, defined as the

Challenge problem: show that $\delta = \frac{\pi}{Q}$.

It is important to note that the viscous damping model is a good model only for intermolecular forces in certain fluids. It is not a good model for dry friction, the usual friction force from rubbing against solid objects governed by the equation $F_f = \mu N$, with $\mu$ the coefficient of friction and $N$ the normal force. Interestingly, simple models of dry friction are solvable and demonstrate a linear damping envelope rather than an exponential.