Solve and verify a harmonic initial-value equation

Identify the restoring law

Simple harmonic motion requires acceleration proportional to displacement from equilibrium and oppositely directed: x double-dot=−omega²x with positive constant omega². For an ideal mass-spring system, Newton law gives m x double-dot=−kx, so omega=sqrt(k/m). A repeating motion is periodic, but need not be harmonic: a nonlinear restoring force generally fails this proportionality test.

Construct the solution with both constants

Recognizing the harmonic equation and using its sinusoidal solution are core requirements. The explicit second-derivative verification here is optional calculus enrichment, not an additional AP expectation to prove the general differential-equation solution.

The functions cos(omega t) and sin(omega t) each have second derivative −omega² times themselves. Thus x=C cos(omega t)+D sin(omega t) solves the linear equation. At t=0, x0=C; differentiating gives v0=omega D. Therefore x=x0 cos(omega t)+(v0/omega)sin(omega t). Initial displacement alone cannot determine the subsequent motion.

Translate to amplitude and phase

Writing x=A cos(omega t+phi) gives A=sqrt(x0²+(v0/omega)²) and phi=atan2(−v0/omega,x0). The two-argument angle retains quadrant information. The period is T=2pi/omega and frequency f=1/T, whereas angular frequency is omega=2pi f. Radians in phase are not cycles; the numerical factors matter.

Check derivatives rather than curve appearance

For the selected m=2 kg and k=8 N/m, omega=2 rad/s. Differentiate the constructed x twice, then compare a/x wherever x is nonzero. It must equal −4 s⁻². At equilibrium x=0, acceleration is zero but speed can be maximal. At a turning point v=0 while the restoring acceleration is nonzero. Smoothness alone does not establish either relationship.

The same derivatives give maximum speed A omega and maximum acceleration magnitude A omega². For a changed amplitude .5 m at omega=2, predict these maxima before checking 1 m/s and 2 m/s². Doubling amplitude doubles both maxima but leaves the period unchanged.

Worked original case

For x0=.3 m and v0=.4 m/s, x=.3 cos(2t)+.2 sin(2t), amplitude sqrt(.13)≈.360555 m and phase≈−.588003 rad. At t=pi/4 s, x=.2 m, v=−.6 m/s and a=−.8 m/s². This written packet uses a different initial velocity from the default control.

Investigation

Choose two supported initial velocities at the same initial displacement. Derive both functions and verify their initial conditions, amplitudes, phases and second derivatives. Compare their identical periods despite different amplitudes. Explain why a periodic triangle-wave position would not satisfy the same harmonic differential equation.

Public numerical transfer practice

Try a new practice case

For m=2 kg,k=8 N/m,x0=.3 m and v0=.4 m/s, what is x at t=pi/4 s in metres?

Numerical result in the requested units Check this practice result

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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.