2.1 Simple harmonic motion

A mass mm on a spring with stiffness kk, free to move along one axis, displaced from equilibrium by a small amount x(t)x(t). The spring exerts a restoring force kx-k x. Newton’s second law gives

mx¨  =  kx.m \ddot x \;=\; -k x.

Divide by mm, define ω02k/m\omega_0^2 \equiv k/m, and write the equation in canonical form:

x¨+ω02x  =  0.\ddot x + \omega_0^2 x \;=\; 0.

This is the equation of simple harmonic motion — SHM. Its general solution is

x(t)  =  Acos(ω0t)+Bsin(ω0t)  =  Xcos(ω0t+φ),x(t) \;=\; A \cos(\omega_0 t) + B \sin(\omega_0 t) \;=\; X \cos(\omega_0 t + \varphi),

with AA, BB, XX, φ\varphi determined by initial conditions. The motion is exactly sinusoidal, with angular frequency ω0=k/m\omega_0 = \sqrt{k/m} (radians per second), frequency f0=ω0/2πf_0 = \omega_0 / 2\pi (cycles per second, hertz), and period T0=1/f0=2π/ω0T_0 = 1/f_0 = 2\pi/\omega_0.

Why try $x = X \cos(\omega t + \varphi)$ and where $\omega_0 = \sqrt{k/m}$ comes from

The ODE is second-order, linear, with constant coefficients. The standard trick is to guess x(t)=eλtx(t) = e^{\lambda t}, substitute, and solve for λ\lambda:

λ2eλt+ω02eλt=0        λ2=ω02        λ=±iω0.\lambda^2 e^{\lambda t} + \omega_0^2 e^{\lambda t} = 0 \;\;\Longrightarrow\;\; \lambda^2 = -\omega_0^2 \;\;\Longrightarrow\;\; \lambda = \pm i\omega_0.

Two complex roots ±iω0\pm i\omega_0. The general solution is

x(t)  =  C+eiω0t+Ceiω0t.x(t) \;=\; C_+ e^{i\omega_0 t} + C_- e^{-i\omega_0 t}.

For real xx we need C=C+C_- = C_+^*, which gives x(t)=2Re[C+eiω0t]=Xcos(ω0t+φ)x(t) = 2 \operatorname{Re}[C_+ e^{i\omega_0 t}] = X \cos(\omega_0 t + \varphi) with C+=12XeiφC_+ = \tfrac12 X e^{i\varphi}. Identifying ω02=k/m\omega_0^2 = k/m from the original equation closes the derivation.

Energy

The mechanical energy is

E  =  12mx˙2+12kx2,E \;=\; \tfrac12 m \dot x^2 + \tfrac12 k x^2,

kinetic plus potential. Substituting x=Xcos(ω0t+φ)x = X \cos(\omega_0 t + \varphi), x˙=Xω0sin(ω0t+φ)\dot x = -X \omega_0 \sin(\omega_0 t + \varphi):

E  =  12mX2ω02sin2()+12kX2cos2()  =  12kX2,E \;=\; \tfrac12 m X^2 \omega_0^2 \sin^2(\cdots) + \tfrac12 k X^2 \cos^2(\cdots) \;=\; \tfrac12 k X^2,

using mω02=km \omega_0^2 = k and sin2+cos2=1\sin^2 + \cos^2 = 1. The total energy is constant — as it must be for a frictionless system — and proportional to the amplitude squared. Energy sloshes between kinetic and potential at twice the oscillation frequency, but the sum stays put.

Why this matters for sound

Every acoustic mode is a simple harmonic oscillator. The wave equation in a bounded domain (a tube, a room, a cavity) separates into a sum of mode amplitudes, each obeying a¨n+ωn2an=0\ddot a_n + \omega_n^2 a_n = 0. The frequencies ωn\omega_n are determined by the geometry and boundary conditions; the dynamics of each mode is exactly what we have here. SHM is the atomic unit of linear acoustics.

It is also a strong simplification. Real systems lose energy to friction and gain energy from forcing. We add those next.