SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish free decay from a driven resonance record

A freely oscillating damped system loses mechanical energy to other forms. A periodically driven system continually receives energy from an external source. These are different experimental boundaries: a large steady driven amplitude is not evidence that damping vanished or that the oscillator created energy.

Learning outcome: Compare two damping values at the same drive, then sweep drive frequency with damping fixed. Keep free decay and steady response records separate and identify where energy enters and leaves each system.

Starting recap

Declare temperature and time references, system boundaries and model regimes. Use absolute temperatures for ratios; distinguish state, path, oscillator phase and wave propagation.

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Try

At c=0.8 kg/s and Ω=2, steady amplitude is 0.125 m, half the value for c=0.4 at the same drive. The natural frequency remains 2 rad/s because m and k did not change.

Open the experiment

Predict

A sustained steady driven response despite damping means:

Your prediction and reason

Experiment

Compare two damping values at the same drive, then sweep drive frequency with damping fixed. Keep free decay and steady response records separate and identify where energy enters and leaves each system.

Distinguish free decay from a driven response

Worked teaching model · synthetic values · Separate disclosed linear-oscillator models m=1 kg,k=4 N/m: free-decay displacement envelope A0 exp(−ct/2m) with A0=.1 m, and steady sinusoidal driving-force amplitude .2 N. Supplied algebraic steady response is enrichment, not a derived calculus requirement. All selected states underdamped.

Viscous damping coefficient: 0.4 kg/s Driving angular frequency: 2 rad/s

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Supplied steady displacement response

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

Equivalent numerical coordinates for Supplied steady displacement response

linear viscous response

driving angular frequency (rad/s)	steady amplitude (m)
0	0.05
0.05	0.0500306
0.1	0.0501228
0.15	0.0502771
0.2	0.0504947
0.25	0.0507773
0.3	0.0511268
0.35	0.051546
0.4	0.0520382
0.45	0.0526072
0.5	0.0532576
0.55	0.0539951
0.6	0.054826
0.65	0.0557579
0.7	0.0567996
0.75	0.0579615
0.8	0.0592557
0.85	0.0606964
0.9	0.0623005
0.95	0.0640877
1	0.0660819
1.05	0.0683111
1.1	0.0708094
1.15	0.073618
1.2	0.0767869
1.25	0.0803777
1.3	0.0844664
1.35	0.0891481
1.4	0.0945418
1.45	0.100798
1.5	0.108108
1.55	0.116714
1.6	0.126918
1.65	0.13909
1.7	0.153642
1.75	0.17094
1.8	0.19104
1.85	0.213067
1.9	0.23413
1.95	0.248566
2	0.25
2.05	0.236789
2.1	0.213968
2.15	0.188386
2.2	0.164399
2.25	0.143632
2.3	0.126226
2.35	0.111775
2.4	0.0997609
2.45	0.0897086
2.5	0.0812277
2.55	0.0740086
2.6	0.0678095
2.65	0.0624418
2.7	0.0577579
2.75	0.0536415
2.8	0.05
2.85	0.0467593
2.9	0.0438595
2.95	0.0412517
3	0.0388955
3.05	0.0367577
3.1	0.0348104
3.15	0.0330303
3.2	0.0313975
3.25	0.0298953
3.3	0.0285091
3.35	0.0272266
3.4	0.0260371
3.45	0.0249312
3.5	0.0239007
3.55	0.0229386
3.6	0.0220386
3.65	0.0211951
3.7	0.0204032
3.75	0.0196586
3.8	0.0189572
3.85	0.0182957
3.9	0.0176709
3.95	0.01708
4	0.0165205

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
Undamped natural angular frequency	2	rad/s
Free damped angular frequency	1.98997	rad/s
Steady driven displacement amplitude	0.25	m
Displacement resonance angular frequency	1.9799	rad/s
Free displacement envelope after 4 s	0.0449329	m
Squared envelope ratio after 4 s	0.201897	ratio

Free decay, natural frequency and driven displacement resonance are distinct quantities.

The supplied steady amplitude is F0/sqrt((k−m Omega²)²+(c Omega)²). This optional algebraic model predicts a displacement peak below the undamped natural frequency for nonzero damping. The free envelope is A0 exp(−ct/(2m)); its square tracks peak-energy scaling, not exact instantaneous total energy at every phase. Driving continually supplies energy that damping removes; it is not free decay with a larger initial amplitude.

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

Free decay, natural frequency and driven displacement resonance are distinct quantities.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Declare the state, system and observation

Build the algebraic model

The packet stipulates m=1 kg,k=4 N/m and positive viscous damping c. Its undamped natural angular frequency is 2 rad/s. For the underdamped free model, the supplied displacement envelope is A0exp(−ct/(2m)) and damped angular frequency is √(k/m−c²/(4m²)). At c=0.4 kg/s, the envelope at 4 s is 0.1exp(−0.8)≈0.0449329 m. The envelope is a bound on displacement, not the signed instantaneous position.

Check the original evidence packet

For a separate steady force of amplitude 0.2 N at angular frequency Ω, the supplied optional algebraic response is A=0.2/√((4−Ω²)²+(cΩ)²). At c=0.4 and Ω=2, amplitude is 0.25 m. This particular displacement response peaks at √(4−c²/2)≈1.97990 rad/s, slightly below the undamped natural frequency. Saying every damped response peaks exactly at the natural frequency would hide the response quantity and damping dependence.

Challenge the domain and transfer

Increasing damping lowers and broadens the response peak in this model. At critical damping c=2√(mk)=4 kg/s, the free solution stops oscillating; overdamped motion returns without free oscillations. Those regimes are explained but lie outside the selected underdamped control range. Squaring the free envelope gives a peak-energy scaling, not necessarily the exact instantaneous sum of kinetic and potential energy at every phase. No physical vibration, resonance loading or safety limit is inferred.

Worked original case

At c=0.8 kg/s and Ω=2, steady amplitude is 0.125 m, half the value for c=0.4 at the same drive. The natural frequency remains 2 rad/s because m and k did not change.

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

Using a supplied free-amplitude envelope A=A0exp(−ct/(2m)), take c=0.5 kg/s,m=1 kg,t=4 s. Find A/A0.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and limits

Original content checked 30 October 2026 against OpenStax College Physics 2e. Packets are supplied or generated teaching evidence, not measured apparatus. No heating, pressure vessels, physical oscillators, sound exposure or construction is requested. Public checks and saved reports do not establish supervised practical skill, equipment safety or independent mastery.

Source for the supplied response equation

The optional steady viscous-response amplitude uses OpenStax University Physics 1, section 15.6, equation 15.29. Minimizing its squared denominator gives the displacement maximum at Ω²=k/m−c²/(2m²), when positive. This precise result comes from the supplied amplitude equation; schematic descriptions placing the peak at the undamped natural frequency are not used as an exact finite-damping equality. No differential-equation derivation is required in this algebra-based course.

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Public practice checks a specific claim. It does not establish independent mastery.

Why must the measured response quantity be named before discussing resonance frequency?

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