05 / CHAPTER 01
The simple harmonic oscillator
Recognize the restoring-force model and its sinusoidal solution.
ANCHOR EQUATION
Simple harmonic motion
In symbols: mẍ + kx = 0
The restoring force is proportional to displacement and points back toward equilibrium.
SCRATCH-PAPER LAB
Change one parameter; watch the prediction move.
Change mass or stiffness and the natural frequency changes as ω₀ = √(k/m). The trace shows displacement, not an animation pretending the spring law is universal.
A harmonic oscillator is a local restoring law
Near a stable equilibrium, many systems exert a restoring force approximately proportional to displacement. The negative sign means the force points back toward equilibrium, not that displacement is always negative.
Derive the frequency scale
For a mass m attached to a spring, Newton’s law gives m d²x/dt² = −kx, or x¨ + (k/m)x = 0. Try x = A cos(ωt + φ). Two derivatives give x¨ = −ω²x; substitution requires ω² = k/m, so ω₀ = √(k/m). The sinusoid is selected because its curvature is proportional to its negative value.
With m = 0.50 kg and k = 200 N/m, ω₀ = √400 = 20 rad/s and period T = 2π/ω₀ = 0.314 s. If amplitude is 0.04 m, maximum speed is Aω₀ = 0.80 m/s.
The spring law is local
A real spring becomes nonlinear at large extension, and friction adds damping. The harmonic result is a small-displacement approximation around a stable equilibrium.
Work it: A 2 kg mass on k = 50 N/m is pulled 0.10 m and released. Find ω₀, period, maximum acceleration, and total energy.