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SCIENCE · INVESTIGATE AND EXPLAIN

Carry timing bounds into an inferred spring constant

A period measurement must count complete cycles using the same phase and direction at both ends. Timing from the positive endpoint to the negative endpoint gives half a cycle, not a full one. Repeating a recognizable crossing can help, but a mistaken cycle count produces a systematic error that averaging does not fix.

Learning outcome: At assigned timing error +0.1 s, compare N=1,5 and10. Save inferred periods and stiffness bounds. Check whether the true20 N/m is compatible, and explain which uncertainty sources the recipe deliberately excludes.

Starting recap

Review Newton’s second law, radians, sine/cosine graphs, energy and momentum boundaries, and bounded measurement errors.

All lessons are open directly. Your saved experiments remain when you leave and return.

Try

Five cycles observed in 7.00±0.20 s give T=1.40 s with bounds 1.36 to 1.44 s. For exact m=1 kg, inferred k≈20.14 N/m, with bounds 4π²/1.44²≈19.04 and 4π²/1.36²≈21.34 N/m. The interval describes the stipulated bound, not a 95% confidence interval.

Open the experiment

Predict

If the same ±0.20 s total-time bound applies to ten rather than one cycle, what happens to the period bound?

Experiment

At assigned timing error +0.1 s, compare N=1,5 and10. Save inferred periods and stiffness bounds. Check whether the true20 N/m is compatible, and explain which uncertainty sources the recipe deliberately excludes.

Infer stiffness from bounded multi-cycle timing

Worked teaching model · synthetic values · Synthetic true oscillator m=1 kg,k=20 N/m. Observed total time=N×true period+assigned error. Each total-time reading has a stipulated ±0.2 s bound; exact mass and cycle count. Bounds are not a statistical confidence interval.

Reset keeps your prediction, captures and practice history.

Calculated evidence ledger — synthetic teaching model
QuantityValueUnit
Observed total elapsed time7.12481s
Inferred period1.42496s
Period lower bound1.38496s
Period upper bound1.46496s
Inferred spring constant19.44252N/m
Stiffness lower bound18.39528N/m
Stiffness upper bound20.5818N/m
Accessible plot coordinates
Primary plotted points
coordinatevalue

A fixed elapsed-time bound becomes a smaller per-cycle period bound when more cycles are timed.

Divide total time and its bound by N. Because k=4π²m/T² decreases with T, the largest period gives the lowest stiffness. This does not remove systematic clock calibration error or missed-cycle errors; mass uncertainty is excluded here.

A fixed elapsed-time bound becomes a smaller per-cycle period bound when more cycles are timed.

Divide total time and its bound by N. Because k=4π²m/T² decreases with T, the largest period gives the lowest stiffness. This does not remove systematic clock calibration error or missed-cycle errors; mass uncertainty is excluded here.

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

Understand

1. Frame the question

A period measurement must count complete cycles using the same phase and direction at both ends. Timing from the positive endpoint to the negative endpoint gives half a cycle, not a full one. Repeating a recognizable crossing can help, but a mistaken cycle count produces a systematic error that averaging does not fix.

2. Build the mechanism

If N cycles take total time t, the estimated period is t/N. An assigned total-time error bound ±δ gives period bounds (t−δ)/N and (t+δ)/N. Timing more cycles reduces this particular fixed endpoint-timing contribution per cycle. It does not automatically reduce a clock’s fractional calibration error, damping drift or miscounted cycles.

With independently known mass m and the ideal model, infer k=4π²m/T². The dependence decreases with T, so the upper period endpoint gives the lower stiffness endpoint. Propagating bounds by endpoint evaluation is preferable to pretending the timing uncertainty vanishes or attaching an unsupported confidence percentage. The recipe treats mass and count as exact to isolate this one source of error.

For a second approach, measure several masses and fit T² against m. The ideal slope is 4π²/k, so stiffness is 4π² divided by slope. Inspect residuals and an intercept rather than forcing agreement. If T²=2m+0.20 in SI units, slope suggests k≈19.74 N/m but the nonzero intercept demands an explanation such as omitted moving mass or systematic timing; it does not establish that one explanation is true.

3. Work an example

Five cycles observed in 7.00±0.20 s give T=1.40 s with bounds 1.36 to 1.44 s. For exact m=1 kg, inferred k≈20.14 N/m, with bounds 4π²/1.44²≈19.04 and 4π²/1.36²≈21.34 N/m. The interval describes the stipulated bound, not a 95% confidence interval.

4. Predict and investigate

If the same ±0.20 s total-time bound applies to ten rather than one cycle, what happens to the period bound?

At assigned timing error +0.1 s, compare N=1,5 and10. Save inferred periods and stiffness bounds. Check whether the true20 N/m is compatible, and explain which uncertainty sources the recipe deliberately excludes.

5. Transfer to a new case

Try a new practice case

Twenty cycles take 30.0±0.4 s. Find the magnitude of the period error bound.

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Explain the boundary

Why can repeated timing agree very closely yet remain systematically wrong?

These are bounded virtual models and supplied numerical cases, not independent measurements or an observed physical investigation. Do not construct spring, pendulum or collision apparatus from these instructions. Practical evidence requires an appropriate supervised setting.

Sources

Original explanations, worked cases and numerical models. Framework checked 30 October 2026: AP Physics 1 Course and Exam Description, topics 7.2; 7.3. Mechanism reference: OpenStax University Physics 1, simple harmonic motion and pendulums. No official assessment items reproduced.

Apply

Ten cycles take 14.0±0.2 s with an exact cycle count. The period bound is:

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

Why can repeated timing agree very closely yet remain systematically wrong?

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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