SCIENCE · INVESTIGATE AND EXPLAIN

Challenge a period-based spring inference with an independent clock

A simple harmonic oscillator experiences a restoring force proportional to displacement from equilibrium: F=−kx. Its acceleration is −(k/m)x, always toward equilibrium. This ideal linear relation gives angular frequency ω=√(k/m) and period T=2π√(m/k). The amplitude sets excursion and energy, not the period within this model.

Learning outcome: Compare two clock gains at one amplitude, then two amplitudes at one gain. Retain period-derived and independent static stiffness, quarter-cycle signs and energy partition. Explain why a shared instrument fault can survive repeated trials.

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

Doubling amplitude from 0.1 to 0.2 m doubles maximum speed and quadruples energy from 0.04 to 0.16 J, while the ideal period remains π s.

Open the experiment

Predict

At equilibrium during nonzero-amplitude ideal harmonic motion:

Your prediction and reason

Experiment

Compare two clock gains at one amplitude, then two amplitudes at one gain. Retain period-derived and independent static stiffness, quarter-cycle signs and energy partition. Explain why a shared instrument fault can survive repeated trials.

Separate stiffness evidence from clock calibration

Worked teaching model · synthetic values · Ideal horizontal oscillator m=2 kg,k=8 N/m, released at positive amplitude with zero velocity. Indicated time is a×true time. An independent static force/extension record uses exact units and gives the same actual k. No apparatus or inferred real calibration.

Time-reading scale factor: 1.05 ratio Initial amplitude: 0.1 m

Reset keeps your prediction, captures and practice history.

Supplied harmonic displacement versus indicated time

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: x=A cos(2 indicatedTime/clockGain) (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Supplied harmonic displacement versus indicated time

x=A cos(2 indicatedTime/clockGain)

indicated time (s)	displacement (m)
0	0.1
0.0824668	0.0987688
0.164934	0.0951057
0.2474	0.0891007
0.329867	0.0809017
0.412334	0.0707107
0.494801	0.0587785
0.577268	0.045399
0.659734	0.0309017
0.742201	0.0156434
0.824668	6.12323e-18
0.907135	-0.0156434
0.989602	-0.0309017
1.07207	-0.045399
1.15454	-0.0587785
1.237	-0.0707107
1.31947	-0.0809017
1.40194	-0.0891007
1.4844	-0.0951057
1.56687	-0.0987688
1.64934	-0.1
1.7318	-0.0987688
1.81427	-0.0951057
1.89674	-0.0891007
1.9792	-0.0809017
2.06167	-0.0707107
2.14414	-0.0587785
2.2266	-0.045399
2.30907	-0.0309017
2.39154	-0.0156434
2.474	-1.83697e-17
2.55647	0.0156434
2.63894	0.0309017
2.7214	0.045399
2.80387	0.0587785
2.88634	0.0707107
2.96881	0.0809017
3.05127	0.0891007
3.13374	0.0951057
3.21621	0.0987688
3.29867	0.1
3.38114	0.0987688
3.46361	0.0951057
3.54607	0.0891007
3.62854	0.0809017
3.71101	0.0707107
3.79347	0.0587785
3.87594	0.045399
3.95841	0.0309017
4.04087	0.0156434
4.12334	3.06162e-17
4.20581	-0.0156434
4.28827	-0.0309017
4.37074	-0.045399
4.45321	-0.0587785
4.53567	-0.0707107
4.61814	-0.0809017
4.70061	-0.0891007
4.78307	-0.0951057
4.86554	-0.0987688
4.94801	-0.1
5.03048	-0.0987688
5.11294	-0.0951057
5.19541	-0.0891007
5.27788	-0.0809017
5.36034	-0.0707107
5.44281	-0.0587785
5.52528	-0.045399
5.60774	-0.0309017
5.69021	-0.0156434
5.77268	-4.28626e-17
5.85514	0.0156434
5.93761	0.0309017
6.02008	0.045399
6.10254	0.0587785
6.18501	0.0707107
6.26748	0.0809017
6.34994	0.0891007
6.43241	0.0951057
6.51488	0.0987688
6.59734	0.1

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
True period	3.14159	s
Indicated period	3.29867	s
Stiffness inferred from uncorrected period	7.25624	N/m
Independent static stiffness	8	N/m
Maximum speed	0.2	m/s
Total mechanical energy	0.04	J
Displacement after true quarter period	0	m
Velocity after true quarter period	-0.2	m/s

An amplitude change changes energy, while a common clock gain biases a period-based stiffness estimate.

For an ideal spring oscillator omega=sqrt(k/m), T=2pi sqrt(m/k), x=A cos(omega t), v=−omega A sin(omega t), and acceleration=−omega² x. The supplied trigonometric laws require no derivative calculation here. A vertical spring shifts equilibrium by mg/k without changing this small-oscillation frequency when k remains constant. Independent static calibration, rather than another reading using the same clock, tests the inferred stiffness.

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

An amplitude change changes energy, while a common clock gain biases a period-based stiffness estimate.

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 disclosed oscillator has mass 2 kg and stiffness 8 N/m, giving ω=2 rad/s and T=π s. It starts at positive amplitude A with zero velocity. The supplied phase laws are x=Acos(ωt), v=−ωAsin(ωt) and a=−ω²x; no derivative calculation is needed to use them here. At a true quarter period the object passes equilibrium with velocity −2A, not zero velocity.

Check the original evidence packet

Total energy is kA²/2. At A=0.1 m it is 0.04 J and maximum speed is 0.2 m/s. At displacement A/2, potential energy is one quarter of the total and kinetic energy is three quarters. A vertical spring instead has equilibrium extension mg/k, but displacement measured from that equilibrium still follows the same ideal small-oscillation law if stiffness stays constant.

Challenge the domain and transfer

The time readout is a times true elapsed time. At clock gain 1.05, an uncorrected period gives inferred stiffness 8/(1.05²)≈7.25624 N/m. Changing amplitude does not repair that calibration fault. An independently calibrated static force-extension record yields 8 N/m and can challenge the dynamic inference. Another oscillation record using the same faulty clock is not independent timing evidence. Real nonlinear springs, damping and finite sensor resolution need additional models.

Worked original case

Doubling amplitude from 0.1 to 0.2 m doubles maximum speed and quadruples energy from 0.04 to 0.16 J, while the ideal period remains π s.

Bounded investigation

Changed-case numerical practice

Try the changed case before feedback

An ideal oscillator has m=2 kg,k=18 N/m and amplitude 0.2 m. Find maximum speed in m/s.

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.

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

Why does the same period not imply the same oscillator energy?

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Previous: Reject an energy-balanced engine that violates the second law Next: Separate a pendulum length offset from a clock-scale error

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