Second-order differential equations describe an enormous variety of physical phenomena. The first key example is the ideal mass-spring.

Example 1 — Mass-spring without friction

A mass mm attached to a spring of constant kk obeys Newton’s second law mx¨=kxm\ddot x = -kx, that is: x¨+ω02x=0,ω0=k/m.\ddot x + \omega_0^2 x = 0, \qquad \omega_0 = \sqrt{k/m}. Characteristic equation λ2+ω02=0\lambda^2 + \omega_0^2 = 0, roots λ=±iω0\lambda = \pm i\omega_0. General solution: x(t)=Acos(ω0t)+Bsin(ω0t)=Ccos(ω0t+φ).x(t) = A\cos(\omega_0 t) + B\sin(\omega_0 t) = C\cos(\omega_0 t + \varphi). With initial conditions x(0)=x0x(0)=x_0, x˙(0)=0\dot x(0)=0: A=x0A = x_0, B=0B=0, hence x(t)=x0cos(ω0t)x(t) = x_0\cos(\omega_0 t).

The period is T=2π/ω0=2πm/kT = 2\pi/\omega_0 = 2\pi\sqrt{m/k}: larger mass \Rightarrow longer period; stiffer spring \Rightarrow shorter period. Numerically, with m=0,5m=0{,}5 kg and k=200k=200 N/m we have ω0=20\omega_0 = 20 rad/s and T0,31T \approx 0{,}31 s.

Topics: Differential equations
Concepts: Initial condition · Second-order linear ODE · Characteristic equation · Harmonic oscillator
Functions: Cosine
Skills: Modelling · Solving equations