HARD · DAMPED AND DRIVEN OSCILLATIONS
RESONANT AMPLITUDE WITH Q
A driven oscillator has mass m = 0.5 kg, natural frequency ω₀ = 10 rad/s, and quality factor Q = 8. It is driven by an external force F(t) = F₀ cos(ωd t) with amplitude F₀ = 4 N. (a) Find the steady-state amplitude at resonance (ωd = ω₀). (b) Find the steady-state amplitude when driven off-resonance at ωd = 6 rad/s. (c) Compute the ratio of the two amplitudes.
Linked equations:A(\omega_d) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega_d^2)^2+(\gamma\omega_d)^2}}\omega_d \approx \omega_0Q = \frac{\omega_0}{\gamma}
§ 01
Step-by-step solution
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Step 1
Recover the damping coefficient γ from Q and ω₀.
Hint
Q = ω₀/γ, so γ = ω₀/Q.
Step 2
Compute the specific force F₀/m (force per unit mass).
Step 3
Evaluate the steady-state amplitude at exact resonance ωd = ω₀.
Step 4
Compute the steady-state amplitude at ωd = 6 rad/s using the full driven-amplitude formula.
Solution walkthrough
Starting from Q = 8 and ω₀ = 10 rad/s, we recover γ = ω₀/Q = 10/8 = 1.25 rad/s. The specific force is F₀/m = 4/0.5 = 8 m/s². At resonance the denominator of the amplitude formula is √((0)² + (γ ω₀)²) = γ ω₀ = 1.25 × 10 = 12.5 (rad/s)², giving A_res = 8/12.5 = 0.64 m. At ωd = 6 rad/s, ω₀² − ωd² = 100 − 36 = 64 (rad/s)², and γ ωd = 1.25 × 6 = 7.5 (rad/s)². The denominator is √(64² + 7.5²) = √(4096 + 56.25) = √4152.25 ≈ 64.44 (rad/s)², giving A_off = 8/64.44 ≈ 0.124 m. The ratio A_res/A_off ≈ 5.15 — the oscillator responds five times more strongly at resonance. Note that for very low damping the resonant amplitude approaches Q × F₀/(m ω₀²), and indeed 8 × 8 / (0.5 × 100) = 1.28 m is approximately Q times the static displacement F₀/(k) = F₀/(m ω₀²) = 8/100 = 0.08 m, so the ratio is close to Q = 8 for very off-resonance comparisons.
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