Waves & Oscillations · Chapter 8

Wave Properties

Waves carry energy without carrying matter. Understanding their properties unlocks optics, acoustics, electromagnetism, and quantum mechanics.

PrerequisitesSimpleharmonicmotionTrigonometryBasiccalculusSimple harmonic motion \cdot Trigonometry \cdot Basic calculus
Learning Goals
  • Identifyamplitude,wavelength,frequency,wavenumber,andwavespeedfromtheequationy=entify amplitude, wavelength, frequency, wave number, and wave speed from the equation y = sin(kx\omegat+ϕsin(kx - \omegat + \phi
  • Verify that a sinusoidal wave satisfies the wave equation and relate wave speed to medium properties.
  • Derive the standing wave pattern from superposition of two counter-propagating waves.
  • Calculate resonant harmonic frequencies for a string fixed at both ends.
  • Apply the conditions for constructive and destructive interference to two-source problems.

8.1 What is a Wave?

A wave is a disturbance that propagates through space and time, transferring energy without permanently displacing matter. The medium oscillates locally while the pattern moves. There are two fundamental types:

Definition 8.1Wave Parameters
A sinusoidal traveling wave is described by:y(x,t)=Asin(kx\omegat+ϕy(x, t) = A sin(kx - \omegat + \phiwhere:
  • A — amplitude (maximum displacement, meters)
  • k=2π/λk = 2\pi/\lambda — wave number (radians per meter)
  • ω=2\pif=2π/T\omega = 2\pif = 2\pi/T — angular frequency (radians per second)
  • φ — initial phase (radians)
  • v=ω/k=fλv = \omega/k = f\lambda — wave speed (meters per second)

8.2 The Wave Equation

All waves satisfy the wave equation — a second-order partial differential equation that relates the spatial and temporal second derivatives of displacement:

2yt2=v22yx2\frac{\partial^2 y}{\partial t^2}=v^2\frac{\partial^2 y}{\partial x^2}(8.1)

You can verify that y = A sin(kx − ωt) satisfies (8.1) with v = ω/k. This equation arises from Newton's second law applied to an elastic medium. The wave speed v depends on the medium:

vstring=FTμvsound=Bρvlight=c=3.00×108m/sv_\mathrm{string}=\sqrt{\frac{F_T}{\mu}} \qquad v_\mathrm{sound}=\sqrt{\frac{B}{\rho}} \qquad v_\mathrm{light}=c=3.00\times10^8\,\mathrm{m/s}(8.2)

where F_T is string tension, μ is linear mass density, B is bulk modulus, and ρ is density. Note: wave speed in a medium is a property of that medium, not of frequency.

8.3 Standing Waves

When two identical waves travel in opposite directions, their superposition creates a standing wave — a pattern that oscillates in place with fixed nodes and antinodes.

Theorem 8.1Standing Wave Formation
Adding two traveling waves of equal amplitude moving in opposite directions:y=Asin(kx\omegat)+Asin(kx+\omegat)=2Asin(kx)cos(\omegaty = A sin(kx - \omegat) + A sin(kx + \omegat) = 2A sin(kx) cos(\omegatThisfactorsintoaspatialpart2Asin(kx)andatemporalpartcos(\omegat).Thenodes(zerosThis factors into a spatial part 2A sin(kx) and a temporal part cos(\omegat). The nodes (zeros arefixedatkx=nπx=nλ/2.Theantinodes(maxima)areatx=(2n+1)λ/4are fixed at kx = n\pi \to x = n\lambda/2. The antinodes (maxima) are at x = (2n+1)\lambda/4

Standing waves on a string fixed at both ends satisfy the boundary condition: nodes at x = 0 and x = L. This forces the allowed wavelengths:

λn=2Lnfn=nv2Ln=1,2,3,\lambda_n=\frac{2L}{n} \qquad f_n=\frac{nv}{2L} \qquad n=1,2,3,\ldots(8.3)

These are the harmonics (or overtones). n = 1 is the fundamental frequency; n = 2 is the first overtone, and so on. This is why guitar strings produce musical notes: the string length forces specific resonant frequencies.

Loading 3D simulation…
Figure 8.1. 3D wave surface simulation. Switch between traveling, standing, and circular (point source) modes. Note how the standing wave has fixed nodes — points that never move. Drag to rotate, scroll to zoom.
Example 8.1Guitar String Harmonics

A guitar string is 65 cm long. The wave speed on this string is 400 m/s. Find the first three harmonic frequencies.

Fundamental (n=1):f1=v/(2L)=400/(2×0.65)=f_{1} = v/(2L) = 400/(2 \times 0.65) = 307.7 Hz E4note\approx E_{4} note
Second harmonic (n=2):f2=2f1=f_{2} = 2f_{1} = 615 Hz E5\approx E_{5}
Third harmonic (n=3):f3=3f1=f_{3} = 3f_{1} = 923 Hz
Note:The harmonics are integer multiples of the fundamental — this is what gives musical instruments their timbre.

8.4 Wave Interference

When two or more waves overlap in the same medium, the resulting displacement is the sum of the individual displacements. This is the superposition principle.

Wavelength λ60 px
Source spacing5 × λ
Phase difference Δφ0°
Interference condition
Constructive (Δφ = 0)

Constructive — waves in phase (Δφ = 0, 2π, …) → bright bands

Destructive — waves out of phase (Δφ = π, 3π, …) → dark bands

Blue = positive amplitude, Red = negative amplitude.

The pattern is the 2D superposition: y = y₁ + y₂

Figure 8.2. Twosourceinterferencepattern.Blue=constructive(wavesinphase),Red=destructive(Two-source interference pattern. Blue = constructive (waves in phase), Red = destructive (waves out of phase). Adjust phase difference to see the pattern invert.
Example 8.2Double-Slit Fringe Spacing

InYoungsdoubleslitexperiment:slitseparationd=0.2mm,screendistanceL=2m,λIn Young's double-slit experiment: slit separation d = 0.2 mm, screen distance L = 2 m, \lambda = 550 nm. Find fringe spacing.

Formula:\Deltay=\lambdaL/d\Deltay = \lambdaL/d
Calculate:\Deltay=(550×109×2)/(0.2×103)=\Deltay = (550\times10^{-9} \times 2) / (0.2\times10^{-3}) = 5.5 mm
Interpretation:Brightfringesappearevery5.5mm.Shorterwavelengthcloserfringes;largerslitseparBright fringes appear every 5.5 mm. Shorter wavelength \to closer fringes; larger slit separationcloserfringesation \to closer fringes.
Definition 8.2Common Traps
  • Amplitude is not wave speed: larger amplitude carries more energy, but the speed is set by the medium.
  • Frequency and wavelength trade off in one medium: ifvisfixed,increasingfdecreasesλif v is fixed, increasing f decreases \lambda
  • Standing waves do not transport energy along the string on average: the pattern stores energy locally between nodes.
  • Nodes are fixed points: antinodes have maximum displacement amplitude, not maximum displacement at every instant.
Exercises — 8.1–8.4 Waves
1.
Asoundwaveinair(v=340m/s)haswavelengthλ=68cm.Finditsfrequency,period,anA sound wave in air (v = 340 m/s) has wavelength \lambda = 68 cm. Find its frequency, period, and wave number.
Hz
Straightforward
2.
Write the equation for a sound wave with frequency 440 Hz (concert A), amplitude 0.5 mm, traveling in the +x direction at 340 m/s.
m
Straightforward
3.A guitar string has tension 80 N and linear density 5 g/m. What length string produces 440 Hz as its fundamental? As its third harmonic?
Intermediate
4.Twocoherentsourcesare4λapart(samefrequency,inphase).FindtheangleofthefirstTwo coherent sources are 4\lambda apart (same frequency, in phase). Find the angle of the first dark fringe.
Intermediate
5.Analyze the energy distribution in a standing wave. Where are the nodes and antinodes of displacement? Of velocity? At what phase in the oscillation does a standing wave have maximum kinetic energy vs maximum potential energy?
Challenging
Key Takeaways
  • Allsinusoidalwavessatisfyy=Asin(kx\omegat+ϕ);speedv=ω/k=fλAll sinusoidal waves satisfy y = A sin(kx - \omegat + \phi); speed v = \omega/k = f\lambda.
  • Wave speed depends on the medium (tension, density, elasticity) — not on frequency.
  • Standing waves arise from superposition of counter-propagating waves; they have fixed nodes.
  • Allowedharmonicsonafixedstring:fn=nv/(2L)thebasisofstringedinstrumentphysAllowed harmonics on a fixed string: f_{n} = nv/(2L) — the basis of stringed instrument physics.
  • Interferenceisfundamental:constructivewhere\Deltar=mλ,destructivewhere\Deltar=(m+12)λInterference is fundamental: constructive where \Deltar = m\lambda, destructive where \Deltar = (m+\frac{1}{2})\lambda.