Problem Statement
Define the quality factor $Q$ for a damped harmonic oscillator. Express it in terms of (a) the natural frequency $\omega_0$ and damping coefficient $\beta$, and (b) energy stored vs. energy lost per cycle.
Given Information
- Natural (undamped) angular frequency $\omega_0 = \sqrt{k/m}$
- Damping coefficient $\beta = b/(2m)$ where $b$ is the viscous drag constant
- Damped frequency $\omega_d = \sqrt{\omega_0^2 – \beta^2}$
- Oscillation period $T = 2\pi/\omega_d$
Physical Concepts & Formulas
The quality factor $Q$ is the fundamental dimensionless measure of how lightly damped a resonant system is. A high-$Q$ oscillator (like a tuning fork or a quartz crystal) loses very little energy per cycle and rings for a long time. A low-$Q$ system (like a car’s shock absorber) is heavily damped and returns to equilibrium without oscillating. $Q$ appears in two equivalent definitions: one geometric (bandwidth of resonance peak) and one energetic (ratio of stored to dissipated energy per radian). Both definitions are widely used in physics and engineering.
- $Q = \dfrac{\omega_0}{2\beta}$ — quality factor in terms of frequencies
- $Q = 2\pi\dfrac{E_{\text{stored}}}{\Delta E_{\text{per cycle}}}$ — energetic definition
- Amplitude decay: $A(t) = A_0 e^{-\beta t}$
- Energy decay: $E(t) = E_0 e^{-2\beta t}$
- Resonance bandwidth: $\Delta\omega = \omega_0/Q$
Step-by-Step Solution
Step 1 — Equation of motion: For a mass-spring-damper system:
$$m\ddot{x} + b\dot{x} + kx = 0 \implies \ddot{x} + 2\beta\dot{x} + \omega_0^2 x = 0$$
where $\beta = b/(2m)$ and $\omega_0 = \sqrt{k/m}$.
Step 2 — Amplitude solution (underdamped, $\beta < \omega_0$):
$$x(t) = A_0 e^{-\beta t}\cos(\omega_d t + \phi)$$
Step 3 — Energy of oscillator:
$$E(t) = \tfrac{1}{2}kA^2(t) = \tfrac{1}{2}kA_0^2 e^{-2\beta t} = E_0 e^{-2\beta t}$$
Step 4 — Energy lost per cycle: The time for one cycle is $T = 2\pi/\omega_d \approx 2\pi/\omega_0$ for light damping.
$$\Delta E = E(t) – E(t+T) = E_0 e^{-2\beta t}(1-e^{-2\beta T}) \approx E_0 e^{-2\beta t} \cdot 2\beta T$$
Step 5 — Quality factor (energetic):
$$Q = 2\pi\frac{E}{\Delta E} = 2\pi\frac{E_0 e^{-2\beta t}}{E_0 e^{-2\beta t}\cdot 2\beta T} = \frac{2\pi}{2\beta T} = \frac{\omega_0}{2\beta} \cdot \frac{\omega_0}{\omega_d}\approx\frac{\omega_0}{2\beta}$$
Worked Calculation
Substituting all values with units:
Example: $\omega_0 = 100\,\text{rad/s}$, $\beta = 2\,\text{s}^{-1}$:
$$Q = \frac{\omega_0}{2\beta} = \frac{100}{2\times2} = 25$$
Energy lost per cycle $= E/Q \times 2\pi \approx 12.6\%$ of stored energy per cycle.
Answer
$$\boxed{Q = \dfrac{\omega_0}{2\beta} = 2\pi\,\dfrac{E}{\Delta E_{\text{cycle}}}}$$
Physical Interpretation
A quality factor of 25 means the system completes roughly $Q/\pi \approx 8$ full oscillations before its amplitude falls to $1/e$ of its initial value. Compare: a guitar string has $Q \sim 1000$; a critically damped door closer has $Q = 0.5$; superconducting microwave cavities reach $Q \sim 10^{10}$. The resonance peak width in frequency space is $\Delta f = f_0/Q$, so high-$Q$ resonators are extremely frequency-selective — the basis of all radio tuners.
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