SCIENCE · INVESTIGATE AND EXPLAIN

Recover harmonic timing from an energy integral

Use energy to obtain speed branches and an integrable turning-point time.

Learning outcome: Compare two amplitudes and the positions x/A=0,.5,1. Retain E,U,K and both velocity signs, then choose a phase branch explicitly. Re-derive the quarter-period integral and compare energy frequency with displacement frequency. Explain why equal time-average energies do not mean equal instantaneous energies.

Starting recap

Derive the restoring equation, retain initial conditions and distinguish exact harmonic results from approximation and calibration assumptions.

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Try

At A=.3 m and x=A/2=.15 m, E=.36 J,U=.09 J,K=.27 J. The two possible velocities are ±sqrt(.27)≈±.519615 m/s. With omega=2, the endpoint-to-center time is pi/4 s for every supported amplitude.

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Predict

At a given non-turning displacement, what can energy alone determine?

Your prediction and reason

Experiment

Compare two amplitudes and the positions x/A=0,.5,1. Retain E,U,K and both velocity signs, then choose a phase branch explicitly. Re-derive the quarter-period integral and compare energy frequency with displacement frequency. Explain why equal time-average energies do not mean equal instantaneous energies.

Integrate a harmonic energy and time account

Worked teaching model · synthetic values · Ideal m=2 kg,k=8 N/m oscillator. Energy gives both velocity signs at a position; a time/phase condition selects one. Time plot begins at positive amplitude with zero velocity. Quarter-period integral is exact for this quadratic potential.

Amplitude: 0.3 m Selected displacement / amplitude: 0.5 ratio

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Potential versus displacement

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Potential versus displacement: numerical coordinates

displacement (m)	potential energy (J)
-0.3	0.36
-0.29	0.3364
-0.28	0.3136
-0.27	0.2916
-0.26	0.2704
-0.25	0.25
-0.24	0.2304
-0.23	0.2116
-0.22	0.1936
-0.21	0.1764
-0.2	0.16
-0.19	0.1444
-0.18	0.1296
-0.17	0.1156
-0.16	0.1024
-0.15	0.09
-0.14	0.0784
-0.13	0.0676
-0.12	0.0576
-0.11	0.0484
-0.1	0.04
-0.09	0.0324
-0.08	0.0256
-0.07	0.0196
-0.06	0.0144
-0.05	0.01
-0.04	0.0064
-0.03	0.0036
-0.02	0.0016
-0.01	0.0004
0	0
0.01	0.0004
0.02	0.0016
0.03	0.0036
0.04	0.0064
0.05	0.01
0.06	0.0144
0.07	0.0196
0.08	0.0256
0.09	0.0324
0.1	0.04
0.11	0.0484
0.12	0.0576
0.13	0.0676
0.14	0.0784
0.15	0.09
0.16	0.1024
0.17	0.1156
0.18	0.1296
0.19	0.1444
0.2	0.16
0.21	0.1764
0.22	0.1936
0.23	0.2116
0.24	0.2304
0.25	0.25
0.26	0.2704
0.27	0.2916
0.28	0.3136
0.29	0.3364
0.3	0.36

Kinetic energy versus time from positive endpoint

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Kinetic energy versus time from positive endpoint: numerical coordinates

time (s)	kinetic energy (J)
0	0
0.0523599	0.00393343
0.10472	0.0155618
0.15708	0.0343769
0.20944	0.0595565
0.261799	0.09
0.314159	0.124377
0.366519	0.161185
0.418879	0.198815
0.471239	0.235623
0.523599	0.27
0.575959	0.300444
0.628319	0.325623
0.680678	0.344438
0.733038	0.356067
0.785398	0.36
0.837758	0.356067
0.890118	0.344438
0.942478	0.325623
0.994838	0.300444
1.0472	0.27
1.09956	0.235623
1.15192	0.198815
1.20428	0.161185
1.25664	0.124377
1.309	0.09
1.36136	0.0595565
1.41372	0.0343769
1.46608	0.0155618
1.51844	0.00393343
1.5708	1.15554e-31
1.62316	0.00393343
1.67552	0.0155618
1.72788	0.0343769
1.78024	0.0595565
1.8326	0.09
1.88496	0.124377
1.93732	0.161185
1.98968	0.198815
2.04204	0.235623
2.0944	0.27
2.14675	0.300444
2.19911	0.325623
2.25147	0.344438
2.30383	0.356067
2.35619	0.36
2.40855	0.356067
2.46091	0.344438
2.51327	0.325623
2.56563	0.300444
2.61799	0.27
2.67035	0.235623
2.72271	0.198815
2.77507	0.161185
2.82743	0.124377
2.87979	0.09
2.93215	0.0595565
2.98451	0.0343769
3.03687	0.0155618
3.08923	0.00393343
3.14159	4.62216e-31

Calculated values under the declared model
Quantity	Value	Unit
Total mechanical energy	0.36	J
Potential energy at selected position	0.09	J
Kinetic energy at selected position	0.27	J
Positive velocity branch	0.519615	m/s
Negative velocity branch	-0.519615	m/s
Quarter period	0.785398	s
Time-average kinetic energy	0.18	J
Time-average potential energy	0.18	J

Position fixes the energy split but not the direction of motion.

v²=omega²(A²−x²), with omega=2. The endpoint-to-center time is integral from0 toA of dx/[omega sqrt(A²−x²)]=pi/(2omega). The endpoint singularity is integrable. K and U oscillate at twice the displacement frequency while their sum is constant; average each over a whole cycle to obtain E/2.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

Position fixes the energy split but not the direction of motion.

Observation to attach to a capture

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Understand

Separate the total from its changing parts

For the ideal oscillator m=2 kg,k=8 N/m and amplitude A, total E=kA²/2=4A² J. At x, U=kx²/2=4x² and K=E−U. Since K=mv²/2=v² for this mass, v=plus or minus sqrt(E−U). Position fixes speed magnitude, not direction; a phase or prior-state condition selects the branch.

Optional calculus extension: derive the quarter-period integral

For general omega=sqrt(k/m), speed magnitude is omega sqrt(A²−x²). Traveling from the positive endpoint to the center takes integral from0 toA of dx/[omega sqrt(A²−x²)]. Substitute x=A sin(u), which cancels the square-root factor and gives pi/(2omega). Although the integrand diverges at the turning endpoint, the integral is finite. Treating the initial zero speed as constant would wrongly predict no motion.

Analyze energy versus time and position

Starting from positive amplitude, x=A cos(omega t), U=E cos²(omega t) and K=E sin²(omega t). Each energy has frequency2omega because squaring removes the sign of displacement or velocity. Their sum remains E. Over a full cycle each has average E/2, but their instantaneous values are generally unequal.

Predict amplitude scaling

Doubling amplitude at the same mass and stiffness quadruples total energy and doubles maximum speed. The ideal linear period remains unchanged because the amplitude cancels from the time integral. This cancellation is a property of the quadratic potential, not a general statement about every periodic system.

Worked original case

At A=.3 m and x=A/2=.15 m, E=.36 J,U=.09 J,K=.27 J. The two possible velocities are ±sqrt(.27)≈±.519615 m/s. With omega=2, the endpoint-to-center time is pi/4 s for every supported amplitude.

Investigation

Public numerical transfer practice

Try a new practice case

An oscillator has k=8 N/m and amplitude .5 m. What is its total mechanical energy in joules?

Numerical result in the requested units

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Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax harmonic motion and pendulums. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.

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Which feature of the potential makes the quarter-period independent of amplitude?

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