Critical vs Underdamped Ringdown

Work through one named subgoal at a time. Each step is checked deterministically against the canonical solver — no AI required to verify correctness. Get an AI explanation when you're stuck.

Step 1

Compute the natural angular frequency ω₀ from k and m.

Hint

ω₀ = √(k/m) for a spring–mass system.

Step 2

Find the critical damping coefficient γ_crit.

Step 3

Estimate the fraction of initial mechanical energy remaining at t = 2 s for the underdamped system.

Step 4

For the critically-damped system (γ = 8 rad/s), evaluate x(t = 2 s).

Solution walkthrough

With k = 16 N/m and m = 1 kg, ω₀ = 4 rad/s. Critical damping requires γ = 2ω₀ = 8 rad/s. For the underdamped system (γ = 1 rad/s), ωd = √(16 − 0.25) = √15.75 ≈ 3.969 rad/s. At t = 2 s: x ≈ 0.25 e^(−1) [cos(3.969 × 2) + (1/(2 × 3.969)) sin(3.969 × 2)] ≈ 0.25 × 0.368 × (cos 7.938 + 0.126 sin 7.938). The energy fraction is approximately (e^(−γt/2))² = e^(−γt) = e^(−2) ≈ 0.135, so about 13.5% of the initial energy remains. For the critically-damped system, x(2) = 0.25 × e^(−8) × (1 + 8) = 0.25 × 9 × e^(−8) ≈ 0.25 × 9 × 3.35 × 10^(−4) ≈ 7.5 × 10^(−4) m — already extremely close to zero. The bisection for t* where x = 0.01 × 0.25 = 0.0025 m converges to a time well under 1 s, confirming that critical damping achieves the fastest possible return without overshoot.

Step-by-step solution

Compute the natural angular frequency ω₀ from k and m.

Find the critical damping coefficient γ_crit.

Estimate the fraction of initial mechanical energy remaining at t = 2 s for the underdamped system.

For the critically-damped system (γ = 8 rad/s), evaluate x(t = 2 s).

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