SCIENCE · INVESTIGATE AND EXPLAIN

Use the driven phase to close the average-power account

A continuing sinusoidal force supplies energy to a damped oscillator. Consider xddot + c xdot + 4x = 0.2 cos(Ωt) after transients have decayed. A steady response can be written A cos(Ωt − φ), with amplitude and phase determined together by substitution or a complex-amplitude calculation.

Learning outcome: Derive both quadratures, verify the average input/loss identity and distinguish a displacement maximum from other response measures.

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 c = 2 and Ω = 2, amplitude is 0.05 m and average power is 0.01 W. Increasing damping at this fixed drive frequency reduces the amplitude and the absorbed power for this particular fixed-force packet.

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Predict

At c = 1 and Ω = 2, the displacement phase lag is:

Your prediction and reason

Experiment

Derive both quadratures, verify the average input/loss identity and distinguish a displacement maximum from other response measures.

Reconcile driven phase with average power

Worked teaching model · synthetic values · Steady-state linear oscillator m=1 kg,k=4 N/m driven by .2 cos(Omega t) N. Transients have decayed; phase lag is measured relative to that force. No physical resonance loading.

Viscous damping: 1 kg/s Drive angular frequency: 2 rad/s

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Steady displacement-amplitude response

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

Equivalent numerical coordinates for Steady displacement-amplitude response

force over impedance magnitude

drive angular frequency (rad/s)	amplitude (m)
0	0.05
0.05	0.05002736
0.1	0.05010958
0.15	0.05024712
0.2	0.05044076
0.25	0.05069158
0.3	0.051001
0.35	0.05137077
0.4	0.05180304
0.45	0.05230031
0.5	0.05286549
0.55	0.05350194
0.6	0.05421348
0.65	0.05500444
0.7	0.05587966
0.75	0.05684455
0.8	0.05790513
0.85	0.05906801
0.9	0.06034041
0.95	0.06173014
1	0.06324555
1.05	0.06489538
1.1	0.06668853
1.15	0.06863373
1.2	0.07073898
1.25	0.07301066
1.3	0.07545236
1.35	0.07806305
1.4	0.08083464
1.45	0.0837487
1.5	0.08677218
1.55	0.08985243
1.6	0.09291177
1.65	0.09584303
1.7	0.09850783
1.75	0.1007406
1.8	0.1023611
1.85	0.1031969
1.9	0.1031133
1.95	0.1020421
2	0.1
2.05	0.09708845
2.1	0.09347325
2.15	0.08935335
2.2	0.08492894
2.25	0.08037766
2.3	0.07584196
2.35	0.07142627
2.4	0.06720042
2.45	0.06320604
2.5	0.05946353
2.55	0.05597827
2.6	0.05274567
2.65	0.04975493
2.7	0.04699175
2.75	0.04444016
2.8	0.04208373
2.85	0.03990633
2.9	0.03789259
2.95	0.03602814
3	0.03429972
3.05	0.03269519
3.1	0.03120353
3.15	0.02981473
3.2	0.02851979
3.25	0.02731055
3.3	0.02617969
3.35	0.02512059
3.4	0.0241273
3.45	0.02319443
3.5	0.02231713
3.55	0.02149103
3.6	0.02071215
3.65	0.01997689
3.7	0.01928199
3.75	0.01862449
3.8	0.01800169
3.85	0.01741113
3.9	0.01685056
3.95	0.01631794
4	0.01581139

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
Steady displacement amplitude	0.1	m
Displacement phase lag relative to drive	1.570796	rad
Average dissipated power	0.02	W
Average drive power	0.02	W
Displacement peak angular frequency	1.870829	rad/s

The phase lag is needed to reconcile drive work with viscous dissipation.

Substitute x=A cos(Omega t−phi) into xddot+c xdot+4x=.2cos(Omega t), or solve the complex amplitude. The lag is atan2(cOmega,4−Omega²), preserving the quadrant. Average dissipation c<v²>=cOmega²A²/2 equals average input F0 Omega A sinphi/2. Displacement peak follows minimizing the squared denominator; it is not exactly the undamped frequency at finite 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 phase lag is needed to reconcile drive work with viscous dissipation.

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Understand

Use the driven phase to close the average-power account

Matching cosine and sine components gives A = 0.2/sqrt((4 − Ω²)² + (cΩ)²) and lag φ = atan2(cΩ, 4 − Ω²). The two-argument angle is important: above the natural frequency, the stiffness-minus-inertia term is negative and the lag lies beyond π/2.

Average viscous power is c times the mean squared velocity, cΩ²A²/2. Average drive power is 0.2 ΩA sinφ/2. They agree in steady state. At c = 1 and Ω = 2, A = 0.1 m, φ = π/2 and each average power is 0.02 W; the drive does not create energy without a matching transfer.

Minimizing the squared amplitude denominator gives the displacement-response peak Ω² = 4 − c²/2 when positive. At c = 1 this is sqrt(3.5) ≈ 1.87083 rad/s, below the undamped frequency 2. Peak displacement, peak power and free damped frequency must not be called the same quantity without checking their definitions.

Changed-case check

Predict before reading the comparison. At c = 2 and Ω = 2, amplitude is 0.05 m and average power is 0.01 W. Increasing damping at this fixed drive frequency reduces the amplitude and the absorbed power for this particular fixed-force packet.

Retained investigation

Derive both quadratures, verify the average input/loss identity and distinguish a displacement maximum from other response measures. Why does an amplitude-only account omit information needed for average work?

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 an amplitude-only account omit information needed for average work?

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Previous: Keep the same initial conditions across three damping regimes Next: Derive independent modes before interpreting energy exchange

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