SCIENCE · INVESTIGATE AND EXPLAIN

Keep the same initial conditions across three damping regimes

The model xddot + c xdot + 4x = 0 has unit mass and stiffness 4 N/m. Trying exp(rt) gives the characteristic equation r² + cr + 4 = 0. Its discriminant c² − 16 separates oscillatory complex roots, a repeated real root and two distinct real roots.

Learning outcome: Derive roots and coefficients for one case in each regime, then check initial position, initial velocity and the energy-rate identity.

Starting recap

Derive equations with their initial or boundary conditions, integrate distributed profiles, and check model domains independently of numerical outputs.

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At critical damping c = 4 and t = 1 s, x = 3exp(−2) ≈ 0.406006 m and v = −4exp(−2) ≈ −0.541341 m/s. At c = 0, energy remains exactly 2 J and the solution reduces to cos(2t).

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Why does the underdamped release-from-rest solution need its sine term?

Your prediction and reason

Experiment

Derive roots and coefficients for one case in each regime, then check initial position, initial velocity and the energy-rate identity.

Compare underdamped, critical and overdamped roots

Worked teaching model · synthetic values · Disclosed m=1 kg,k=4 N/m free oscillator with x(0)=1 m,v(0)=0. All three damping branches use those same initial conditions. Constant linear damping; mathematical scale only, not apparatus motion.

Linear damping coefficient: 2 kg/s Evaluation time: 2 s

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Same initial conditions across damping roots

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: exact homogeneous solution (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Same initial conditions across damping roots

exact homogeneous solution

time (s)	displacement (m)
0	1
0.05	0.9951666
0.1	0.9813308
0.15	0.9594808
0.2	0.930587
0.25	0.8955945
0.3	0.8554164
0.35	0.8109282
0.4	0.762963
0.45	0.7123077
0.5	0.6597002
0.55	0.6058262
0.6	0.5513186
0.65	0.496756
0.7	0.4426624
0.75	0.3895075
0.8	0.3377073
0.85	0.2876256
0.9	0.239575
0.95	0.193819
1	0.1505744
1.05	0.1100133
1.1	0.07226609
1.15	0.03742387
1.2	0.005541484
1.25	-0.02335958
1.3	-0.04928823
1.35	-0.0722808
1.4	-0.09239806
1.45	-0.1097224
1.5	-0.1243548
1.55	-0.1364123
1.6	-0.1460254
1.65	-0.1533352
1.7	-0.1584914
1.75	-0.1616499
1.8	-0.1629709
1.85	-0.1626165
1.9	-0.1607497
1.95	-0.1575321
2	-0.1531228
2.05	-0.1476771
2.1	-0.1413455
2.15	-0.1342725
2.2	-0.1265959
2.25	-0.1184461
2.3	-0.1099456
2.35	-0.1012086
2.4	-0.09234074
2.45	-0.08343877
2.5	-0.07459057
2.55	-0.06587515
2.6	-0.05736275
2.65	-0.04911499
2.7	-0.04118511
2.75	-0.03361825
2.8	-0.02645182
2.85	-0.01971586
2.9	-0.01343344
2.95	-0.007621142
3	-0.002289494
3.05	0.00255655
3.1	0.006917124
3.15	0.01079696
3.2	0.01420493
3.25	0.01715351
3.3	0.0196584
3.35	0.02173801
3.4	0.02341301
3.45	0.024706
3.5	0.02564104
3.55	0.02624329
3.6	0.0265387
3.65	0.02655366
3.7	0.02631471
3.75	0.02584829
3.8	0.02518049
3.85	0.02433681
3.9	0.02334201
3.95	0.02221994
4	0.02099337

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Damping branch	0	0 under;1 critical;2 over
Displacement	-0.1531228	m
Velocity	0.09905976	m/s
Mechanical energy	0.05179958	J
Instantaneous energy rate	-0.01962567	W
Initial mechanical energy	2	J

The characteristic discriminant changes the form of the solution, not the initial-condition obligations.

Solve r²+c r+4=0. Underdamping uses exp(−ct/2)[cos(wd t)+(c/(2wd))sin(wd t)] for x0=1,v0=0. At c4 repeated roots give(1+2t)exp(−2t). Above4 combine both real roots with coefficients fixed by the same initial data. Energy derivative is−c v²; a zero instantaneous rate at a turning point does not mean zero damping.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The characteristic discriminant changes the form of the solution, not the initial-condition obligations.

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Understand

Keep the same initial conditions across three damping regimes

All displayed branches start at x(0) = 1 m and v(0) = 0. Below c = 4, let γ = c/2 and ωd = sqrt(4 − γ²). The solution is exp(−γt)[cos(ωdt) + (γ/ωd)sin(ωdt)]. Omitting the sine term would generally violate the specified zero initial velocity.

At c = 4, the repeated-root solution is (1 + 2t)exp(−2t), with velocity −4t exp(−2t). For c above 4, combine the two decaying exponentials with coefficients fixed by the same initial conditions. Merely switching a formula label without rematching those conditions would compare different physical problems.

Energy is v²/2 + 2x². Differentiating and using the equation of motion gives dE/dt = −cv². It is nonincreasing for c nonnegative, but its instantaneous derivative can vanish at a turning point while damping remains present. Critical and overdamped branches here do not oscillate for the supplied release conditions; arbitrary forced motion would be a different boundary.

Changed-case check

Predict before reading the comparison. At critical damping c = 4 and t = 1 s, x = 3exp(−2) ≈ 0.406006 m and v = −4exp(−2) ≈ −0.541341 m/s. At c = 0, energy remains exactly 2 J and the solution reduces to cos(2t).

Retained investigation

Derive roots and coefficients for one case in each regime, then check initial position, initial velocity and the energy-rate identity. Why does zero instantaneous dE/dt not prove that the damping coefficient is zero?

Source and boundary

Primary conceptual reference checked 30 October 2026: OpenStax University Physics Volume 1. All prose, functions and evidence packets are original. Bounded mathematical models do not establish measured apparatus, supervised practical skill, independent learner mastery or institutional credit. No physical loading, liquids, heat or sound exposure is requested.

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Why does zero instantaneous dE/dt not prove that the damping coefficient is zero?

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