EXAM · DAMPED AND DRIVEN OSCILLATIONS

DRIVEN AMPLITUDE VS FREQUENCY

A mechanical oscillator has mass m = 2 kg, natural frequency ω₀ = 8 rad/s, and damping coefficient γ = 2 rad/s. It is driven by F(t) = 20 cos(ωd t) N. (a) Find Q and classify the damping regime. (b) Compute the steady-state amplitude at ωd = 2 rad/s, ωd = ω₀ (resonance), and ωd = 14 rad/s. (c) Compute the phase lag φ at each of the three driving frequencies. (d) Find the static amplitude A_static (ωd → 0 limit) and confirm that A_res / A_static ≈ Q.

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}\omega_d = \sqrt{\omega_0^2 - \gamma^2/4}
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Step-by-step solution

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Step 1

Compute the quality factor Q and state the damping regime.

Hint

Q = ω₀/γ. The regime is determined by comparing γ to 2ω₀: underdamped if γ < 2ω₀, critically damped if γ = 2ω₀, overdamped if γ > 2ω₀.

Step 2

Compute the specific driving force F₀/m.

Step 3

Evaluate the steady-state amplitude at resonance ωd = ω₀.

Step 4

Find the static amplitude A_static, i.e. the response in the limit ωd → 0.

Step 5

Compute A_res / A_static and compare it to Q.

Solution walkthrough
Q = 8/2 = 4, so the system is underdamped (γ = 2 < 2ω₀ = 16). The specific force is F₀/m = 20/2 = 10 m/s². At ωd = 2 rad/s: denominator = √((64−4)²+(2×2)²) = √(3600+16) = √3616 ≈ 60.13, so A_low ≈ 10/60.13 ≈ 0.166 m. At resonance: denominator = γ ω₀ = 2×8 = 16, so A_res = 10/16 = 0.625 m. At ωd = 14 rad/s: ω₀²−ωd² = 64−196 = −132, denominator = √(132²+(2×14)²) = √(17424+784) = √18208 ≈ 134.9, so A_high ≈ 10/134.9 ≈ 0.074 m. Phases: φ_low = atan2(2×2, 64−4) = atan2(4,60) ≈ 0.067 rad (≈ 3.8°); φ_res = π/2 (exactly); φ_high = atan2(2×14, 64−196) = atan2(28, −132) ≈ π − atan(28/132) ≈ π − 0.210 ≈ 2.93 rad (≈ 168°). Static amplitude: A_static = 10/64 ≈ 0.156 m. Ratio: 0.625/0.156 = 4.00 = Q exactly. This elegant result — that the resonance gain equals Q — is one of the most important facts in oscillator physics.
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