Problem Statement
Find the half-power bandwidth $\Delta\omega$ of a forced oscillator and express $Q$ in terms of $\omega_0$ and $\Delta\omega$.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: The power absorbed by the oscillator (proportional to $A^2$) drops to half its maximum value at two frequencies $\omega_\pm$. Setting $A^2 = A_{\max}^2/2$:
Step 2 — Apply the relevant physical law or equation: $$(\omega_0^2 – \omega^2)^2 + 4\beta^2\omega^2 = 8\beta^2\omega_0^2$$
Step 3 — Solve algebraically for the unknown: For light damping ($\beta \ll \omega_0$), $\omega_\pm \approx \omega_0 \pm \beta$, giving half-power bandwidth:
Step 4 — Substitute numerical values with units: $$\Delta\omega = \omega_+ – \omega_- = 2\beta$$
$$\boxed{Q = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0}{2\beta}}$$
Step 5 — Compute and check the result: This is the standard operational definition of $Q$: resonance frequency divided by bandwidth. Sharper resonance = higher $Q$.
Worked Calculation
$$(\omega_0^2 – \omega^2)^2 + 4\beta^2\omega^2 = 8\beta^2\omega_0^2$$
$$\Delta\omega = \omega_+ – \omega_- = 2\beta$$
$$\boxed{Q = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0}{2\beta}}$$
The power absorbed by the oscillator (proportional to $A^2$) drops to half its maximum value at two frequencies $\omega_\pm$. Setting $A^2 = A_{\max}^2/2$:
$$(\omega_0^2 – \omega^2)^2 + 4\beta^2\omega^2 = 8\beta^2\omega_0^2$$
For light damping ($\beta \ll \omega_0$), $\omega_\pm \approx \omega_0 \pm \beta$, giving half-power bandwidth:
$$\Delta\omega = \omega_+ – \omega_- = 2\beta$$
$$\boxed{Q = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0}{2\beta}}$$
This is the standard operational definition of $Q$: resonance frequency divided by bandwidth. Sharper resonance = higher $Q$.
Answer
$$\boxed{Q = \frac{\omega_0}{\Delta\omega} = \frac{\omega_0}{2\beta}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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