Simple harmonic motion basics
A restoring force proportional to stretch makes any mass oscillate sinusoidally, at a frequency set by stiffness and mass alone.
The spring-mass ideal
A spring that pulls back proportionally to stretch, F = -kx, makes a mass oscillate in simple harmonic motion: x(t) = A cos(omega t). The acceleration peaks at the turning points (maximum stretch) and vanishes at equilibrium, exactly the opposite of the speed, which peaks at the center. Any restoring force that grows linearly with displacement produces this same motion.
x(t) = A\cos(\omega t), \quad \omega = \sqrt{\frac{k}{m}}Period and frequency
The angular frequency omega = sqrt(k/m) sets the pace: T = 2 pi/omega = 2 pi sqrt(m/k) and f = 1/T. Notice what is missing: amplitude. Pull the spring twice as far and the mass travels farther each cycle but finishes each cycle in the same time — a heavier mass or a lazier spring changes the period, amplitude never does.
T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}Worked example
A 0.50 kg mass hangs from a stiff spring, k = 200 N/m. omega = sqrt(200/0.50) = sqrt(400) = 20 rad/s, so the period is 2 pi/20 = 0.31 s and the frequency is 20/2 pi = 3.2 Hz. If it oscillates with amplitude 0.10 m, its top speed is A omega = (0.10)(20) = 2.0 m/s — fastest at equilibrium, dead stop at the extremes.
\omega = \sqrt{\frac{200}{0.50}} = 20\ \text{rad/s}, \quad T = \frac{2\pi}{20} \approx 0.31\ \text{s}Flashcard
Linear restoring force defines SHM
Check your understanding
A 0.25 kg mass on a spring with k = 100 N/m is pulled aside and released. What is its period of oscillation?