Trace lost oscillator energy through a finite thermal receiver

Declare the state, system and observation

A mechanical oscillator that settles loses its organized mechanical energy, but that energy is not destroyed. To infer a temperature rise, first name the receiver and the rest of the environment. The chosen receiver may capture only part of the energy, and its heat capacity is independent information, not automatically equal to the oscillator’s mass.

Build the algebraic model

The ideal initial spring-mass system has m=2 kg,k=200 N/m and starts at rest at displacement A. Initial mechanical energy is kA²/2=100A² J. A receiver of heat capacity 20 J/K starts at 300 K. The selected fraction f of lost energy reaches it, while (1−f) reaches the external environment. The model describes endpoints and does not prescribe how rapidly damping occurs.

Check the original evidence packet

At A=0.2 m and f=0.8, initial energy is 4 J, receiver gain is 3.2 J and external gain is 0.8 J. Receiver rise is 3.2/20=0.16 K, giving final temperature 300.16 K. The undamped angular frequency is √(200/2)=10 rad/s and period 2π/10≈0.628319 s. Doubling amplitude quadruples initial energy and receiver warming at the same fraction, but leaves this ideal period unchanged.

Challenge the domain and transfer

An inferred capture fraction CΔT/E is meaningful only under the stated initial-state, heat-capacity and no-other-input assumptions. A biased thermometer, wrong receiver capacity or external heating can mimic a different fraction. For a changed packet with observed rise 0.10 K at A=0.2 m, the conditional fraction is 20×0.10/4=0.5, not proof of a specific loss mechanism. Retain a separately justified temperature calibration or receiver-capacity check before choosing between those explanations. No physical heating or vibration experiment is requested.

Worked original case

At A=0.1 m and f=0.6, initial energy is 1 J and receiver gain 0.6 J, so temperature rise is 0.03 K. The environment gains the remaining 0.4 J.

Bounded investigation

Compare amplitude changes with capture-fraction changes. Retain the whole-system energy closure and receiver-only thermal account, then design an independent observation distinguishing a capture change from a temperature or heat-capacity error.

Changed-case numerical practice

Try the changed case before feedback

A spring with k=100 N/m starts at rest at amplitude 0.2 m. All its initial mechanical energy reaches a receiver of heat capacity 10 J/K. Find temperature rise in K.

Numerical result in the requested unitsCompare this result

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.