Consolidation tutorial — oscillations and waves

By the end of the Oscillations and Wave Motion blocks you have met four separate-looking topics: simple harmonic motion, damped and driven oscillation, coupled oscillators, and travelling waves on a string. They are not four topics — they are one story told at increasing scale. A single mass on a spring performs SHM; add friction and a driver and you get resonance; couple two such masses and their normal modes beat against each other; couple infinitely many and the beat pattern becomes a travelling wave. This tutorial is a review — no new physics — that retraces that thread so you can recognize, in a mixed problem, which "layer" of the hierarchy you are standing on and reach for the right formula without being told which chapter it came from.

The core insight

SHM, damped/driven oscillation, coupled normal modes, and travelling waves are the same physics — $\ddot x+\omega_0^2x=0$ generalized by friction, driving, coupling, and finally the $N\to\infty$ continuum limit — so a mixed problem is solved by identifying which layer of that single hierarchy it sits in.

The core idea

A lone mass on a spring obeys $\ddot x+\omega_0^2x=0$ , solved by $x=\mathrm{Re}(Ae^{i(\omega_0t+\phi)})$ with energy $E=\tfrac12m\omega_0^2A^2$ shuttling between kinetic and potential. Add a friction term and a periodic drive and the equation becomes

\ddot x+2\gamma\dot x+\omega_0^2x=\frac{F_0}{m}\cos\omega t,

where: $\gamma$ is the damping rate, $\omega_0=\sqrt{k/m}$ the natural frequency, and $F_0,\omega$ the drive amplitude and frequency. The transient dies as $e^{-\gamma t}$ ; what survives is a steady oscillation whose amplitude peaks near $\omega\approx\omega_0$ — resonance, sharpened by the quality factor $Q=\omega_0/2\gamma$ . Couple two identical oscillators with a weak spring $k_c$ and their motion no longer separates into two independent SHMs in $x_1,x_2$ — but it does in the normal coordinates $q_1=x_1+x_2$ and $q_2=x_1-x_2$ , which oscillate independently at $\omega_1=\sqrt{k/m}$ and $\omega_2=\sqrt{(k+2k_c)/m}$ . Displace one mass and release, and the resulting motion is a superposition of both normal modes — visible as energy sloshing back and forth between the two masses at the beat frequency $\Omega=\omega_2-\omega_1$ . Push the coupled-oscillator picture to $N\to\infty$ masses (spacing $\to0$ , a continuous string under tension $T$ and mass per length $\mu$ ) and the discrete normal-mode equations become the wave equation

\frac{\partial^2y}{\partial t^2}=v^2\frac{\partial^2y}{\partial x^2},\qquad v=\sqrt{T/\mu},

whose travelling-wave solution $y=A\cos(kx-\omega t)$ is simply every "mass" along the string doing SHM with a phase that varies smoothly along $x$ . Superposition then explains everything downstream: two SHMs of equal frequency add by phasor addition, two waves of close frequency beat, and many waves of different $k$ build a localized wave packet moving at group velocity $v_g=d\omega/dk$ , distinct from the phase velocity $v_p=\omega/k$ whenever the medium is dispersive.

Figure

A ladder diagram showing single oscillator leading to damped driven oscillator leading to two coupled oscillators leading to continuum limit leading to travelling wave, with an arrow chain down the page

Derivation ANormal modes and the beat pattern of two coupled oscillators

The goal is to show, starting only from Newton's second law for two identical masses connected by end springs of constant $k$ and a coupling spring of constant $k_c$ , that displacing one mass and releasing it from rest produces a beat pattern in which energy sloshes between the two masses at frequency $\Omega=\omega_2-\omega_1$ , exactly the setup and numbers used in the worked example below.

Step 1. Write the two coupled equations of motion. For masses $m$ at displacements $x_1,x_2$ : $m\ddot x_1=-kx_1-k_c(x_1-x_2)$ and $m\ddot x_2=-kx_2-k_c(x_2-x_1)$ . Each mass feels its own restoring spring plus a coupling force proportional to the relative displacement — this coupling term is what mixes the two equations.

Step 2. Decouple with normal coordinates. Add the two equations: with $q_1=x_1+x_2$ , the coupling terms cancel and $m\ddot q_1=-kq_1$ , so $q_1$ oscillates alone at $\omega_1=\sqrt{k/m}$ — the symmetric mode, both masses moving together. Subtract them: with $q_2=x_1-x_2$ , the coupling terms double and $m\ddot q_2=-(k+2k_c)q_2$ , giving $\omega_2=\sqrt{(k+2k_c)/m}$ — the antisymmetric mode, masses moving oppositely.

Step 3. Rebuild $x_1,x_2$ from the two independent normal-mode oscillations and impose $x_1(0)=x_1(0)$ , $x_2(0)=0$ , both released from rest. This forces equal amplitudes of the two modes, chosen so they cancel in $x_2$ at $t=0$ : $x_1(t)=\tfrac12x_1(0)[\cos\omega_1t+\cos\omega_2t]$ and $x_2(t)=\tfrac12x_1(0)[\cos\omega_1t-\cos\omega_2t]$ .

Step 4. Apply the trigonometric sum-to-product identity to see the beat. $\cos\omega_1t\pm\cos\omega_2t$ splits into a fast carrier at the average frequency $(\omega_1+\omega_2)/2$ multiplied by a slow envelope at half the difference frequency, $(\omega_2-\omega_1)/2$ . The envelope amplitude in $x_1$ and $x_2$ trade places every quarter beat cycle, so the beat (envelope-squared, i.e. energy) period is $T_{\text{beat}}=2\pi/(\omega_2-\omega_1)$ — energy genuinely migrates from mass 1 to mass 2 and back. Verified: for $\omega_1=\sqrt{40}\approx6.325$ rad/s and $\omega_2=\sqrt{60}\approx7.746$ rad/s, $x_2(t)$ first peaks (all energy in mass 2) at $t=\pi/(\omega_2-\omega_1)\approx2.21$ s, matching the worked example.

Worked example

A mass-spring system with $m=0.2\ \text{kg}$ , $\omega_0=10\ \text{rad/s}$ , and damping constant $\gamma=0.5\ \text{s}^{-1}$ is driven by $F(t)=2\cos(\omega t)\ \text{N}$ .

Step 1. Find $Q$ . $Q=\omega_0/2\gamma=10/(2\times0.5)=10$ . Amplitude falls to $1/e$ after roughly $Q/\pi\approx3$ free-decay cycles, and the fractional resonance width is $\Delta\omega/\omega_0\sim1/Q=0.1$ — a moderately sharp peak.

Step 2. Find the steady-state amplitude at exact resonance, $\omega=\omega_0$ . There $(\omega_0^2-\omega^2)=0$ in the amplitude formula $A(\omega)=\dfrac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(2\gamma\omega)^2}}$ , so it reduces to $A(\omega_0)=\dfrac{F_0/m}{2\gamma\omega_0}=\dfrac{2/0.2}{2\times0.5\times10}=1\ \text{m}$ (an idealized round number; the method matters more than the size).

Step 3. Now weakly couple a second, identical undriven oscillator via $k_c\ll k$ , giving normal-mode splitting $\Delta\omega=\omega_2-\omega_1$ much smaller than the resonance width $2\gamma=1\ \text{rad/s}=\omega_0/Q$ . Because $\Delta\omega\ll\omega_0/Q$ , the two would-be resonance peaks blur into a single broadened peak — damping hides the splitting. Two peaks resolve only once $\Delta\omega\gtrsim\omega_0/Q$ .

Answer: $Q=10$ , resonance amplitude $A(\omega_0)=1\ \text{m}$ , and a weak enough coupling shows one broadened peak, not two.

Competitive stretch

Reuse the coupled pair from the derivation above ( $m=0.5\ \text{kg}$ , $k=20\ \text{N/m}$ , $k_c=5\ \text{N/m}$ , mass 1 displaced $0.02\ \text{m}$ and released). The normal modes are $\omega_1=\sqrt{40}\approx6.325\ \text{rad/s}$ and $\omega_2=\sqrt{60}\approx7.746\ \text{rad/s}$ , so the beat period is $T_{\text{beat}}=2\pi/(\omega_2-\omega_1)\approx4.42\ \text{s}$ , and all the energy first reaches mass 2 at $t=\pi/(\omega_2-\omega_1)\approx2.21\ \text{s}$ — half the beat period, exactly as the envelope picture in Step 4 predicts.

Common traps

Forgetting damping shifts the free-oscillation frequency. A damped (undriven) oscillator free-oscillates at $\omega_1=\sqrt{\omega_0^2-\gamma^2}$ , not at $\omega_0$ itself.
Confusing beat frequency with envelope frequency by a factor of 2. Energy (intensity) beats at $|\omega_1-\omega_2|$ , but the amplitude envelope $\cos[(\omega_1-\omega_2)t/2]$ oscillates at half that rate.
Using the wrong resonance reference. Displacement amplitude peaks slightly below $\omega_0$ ; velocity amplitude peaks exactly at $\omega=\omega_0$ . Mixing the two is a classic slip.
Treating the wave equation as unrelated to coupled oscillators. It is literally the $N\to\infty$ limit of masses and springs; losing that link makes group velocity and dispersion harder to reason about from scratch.

Ask the tutor: If you are unsure which layer of the SHM-to-waves hierarchy a mixed problem belongs to, walk through a specific question with the AI tutor and have it point out the tell-tale clue.

Why does displacing just one mass in a coupled pair excite both normal modes at once?

Because the initial condition $x_1(0)\neq0,\ x_2(0)=0,\ \dot x_1(0)=\dot x_2(0)=0$ is not a pure normal mode — a pure symmetric mode needs $x_1=x_2$ and a pure antisymmetric mode needs $x_1=-x_2$ . Writing your actual initial state as $q_1=x_1+x_2$ and $q_2=x_1-x_2$ shows it is an equal mixture of both, so both frequencies appear and their interference produces the beat.

Is the beat frequency the same thing as the frequency at which the masses swap all their energy?

Not quite — they differ by a factor of two, and this is the single most common slip. The envelope of $x_1(t)$ or $x_2(t)$ goes to zero (all energy in the other mass) once every $\pi/(\omega_2-\omega_1)$ , but a full beat cycle (energy back in the original mass) takes twice that, $2\pi/(\omega_2-\omega_1)$ . Always ask which quantity — displacement envelope or energy transfer — the question wants.

If damping hides normal-mode splitting, does that mean the splitting is not really there?

It is still there mathematically — $\omega_1$ and $\omega_2$ remain distinct solutions of the coupled equations — but damping broadens each resonance peak by roughly $\omega_0/Q$ . If the splitting $\Delta\omega$ is smaller than that width, the two peaks overlap so much they look like one broad peak in a frequency-response measurement. You would need weaker damping (higher $Q$ ) or stronger coupling (larger $\Delta\omega$ ) to resolve them experimentally.

Consolidation tutorial — oscillations and waves

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads