59–64, 70WavesFoundation

Wave Properties

A wave transfers energy from one place to another without transferring the medium itself. A cork on a pond bobs up and down as ripples pass, but ends up where it started. Describing that motion precisely — its speed, frequency, wavelength and phase — is the foundation for everything else in this topic.

30 min Topic 5 · Waves and the Particle Nature of Light

What you'll be able to do

  • Distinguish longitudinal from transverse waves with examples
  • Define amplitude, wavelength, frequency, period and wave speed
  • Use v=fλv = f\lambda and f=1/Tf = 1/T
  • Express phase and phase difference in degrees and radians
1

Transverse and longitudinal

In a transverse\textbf{transverse} wave the particles of the medium oscillate perpendicular to the direction of energy transfer — waves on a string, ripples on water, and all electromagnetic waves.

In a longitudinal\textbf{longitudinal} wave the oscillation is parallel to the direction of travel, producing regions of compression\textbf{compression} and rarefaction\textbf{rarefaction} — sound in air, and the P-waves of an earthquake.

Only transverse waves can be polarised\textbf{polarised}, since a longitudinal oscillation has only one possible direction. That difference is how polarisation proves light is transverse.

2

Describing a wave

Amplitude\textbf{Amplitude} AA is the maximum displacement from equilibrium. Wavelength\textbf{Wavelength} λ\lambda is the distance between two consecutive points in phase, such as crest to crest. Period\textbf{Period} TT is the time for one complete oscillation, and frequency\textbf{frequency} ff is the number of oscillations per second, so f=1/Tf = 1/T.

In one period the wave advances one wavelength, so its speed is v=λ/T=fλv = \lambda/T = f\lambda — the wave equation.

The speed is set by the medium, not the source. When a wave crosses into a new medium its frequency stays fixed (it is set by the source) and the wavelength changes to match the new speed.

v=fλv = f\lambda
The wave equation.
f=1Tf = \dfrac{1}{T}
Frequency and period.
Example 1A sound wave of frequency 440 Hz travels at 340 m s−1340\ \mathrm{m\,s^{-1}}. Find its wavelength and period.
1λ=vf=340440=0.773\lambda = \dfrac{v}{f} = \dfrac{340}{440} = 0.773 m.
2T=1440=2.27×10−3T = \dfrac{1}{440} = 2.27\times10^{-3} s.
Answerλ=0.77 m\lambda = 0.77\ \mathrm{m}; T=2.3 msT = 2.3\ \mathrm{ms}.

Tip — On a displacement–distance\textit{distance} graph you read wavelength; on a displacement–time\textit{time} graph you read period. Check the horizontal axis before measuring.

3

Phase and phase difference

Phase\textbf{Phase} describes the position of a point within its cycle. One whole cycle is 360∘360^{\circ} or 2π2\pi radians.

Two points one wavelength apart are in phase\textbf{in phase} (difference 2π2\pi, equivalent to zero). Two points half a wavelength apart are in antiphase\textbf{antiphase} — difference π\pi — moving in opposite directions at every instant.

For points separated by a distance xx along the wave, the phase difference is the fraction of a wavelength, times 2π2\pi.

Δϕ=2πxλ\Delta\phi = \dfrac{2\pi x}{\lambda}
Phase difference from separation.
Example 1Two points on a wave of wavelength 0.60 m are 0.15 m apart. Find their phase difference.
1Fraction of a wavelength: 0.150.60=14\dfrac{0.15}{0.60} = \tfrac14.
2Δϕ=14×2π\Delta\phi = \tfrac14 \times 2\pi.
Answerπ2\dfrac{\pi}{2} rad, or 90∘90^{\circ}.
4

Energy, intensity and the oscilloscope

Intensity\textbf{Intensity} is power per unit area, I=P/AI = P/A, in W m−2\mathrm{W\,m^{-2}}. For a wave, intensity is proportional to amplitude squared — doubling the amplitude quadruples the energy delivered.

For a point source radiating equally in all directions, the power spreads over a sphere of area 4πr24\pi r^{2}, giving the inverse-square law.

An oscilloscope displays a signal’s displacement against time. Frequency is found from the period read off the screen using the time-base setting, and amplitude from the vertical divisions using the volts-per-division setting.

I=PAI = \dfrac{P}{A}
Intensity.
I∝A2I \propto A^{2}
Intensity scales with amplitude squared.
I=P4πr2I = \dfrac{P}{4\pi r^{2}}
Point source.
Example 1On an oscilloscope set to 2.0 ms div−12.0\ \mathrm{ms\,div^{-1}}, one full wave spans 4.0 divisions. Find the frequency.
1T=4.0×2.0=8.0T = 4.0 \times 2.0 = 8.0 ms.
2f=18.0×10−3f = \dfrac{1}{8.0\times10^{-3}}.
Answer125 Hz125\ \mathrm{Hz}

Tip — Measure across several complete waves and divide, rather than reading one cycle. It reduces the percentage uncertainty in the period.

Equation recap

v=fλv = f\lambda
Wave equation.
f=1/Tf = 1/T
Frequency and period.
Δϕ=2πx/λ\Delta\phi = 2\pi x/\lambda
Phase difference.
I∝A2I \propto A^{2}
Intensity and amplitude.

Common mistakes to avoid

Measuring wavelength from a displacement–time graph.
That graph gives the period. Wavelength comes from displacement–distance.
Saying frequency changes when a wave enters a new medium.
Frequency is set by the source; speed and wavelength change.
Taking amplitude as crest-to-trough distance.
Amplitude is equilibrium to maximum displacement — half that.
Claiming the medium travels with the wave.
Energy is transferred; particles oscillate about fixed positions.

Key takeaways

  • Waves transfer energy without transferring matter.
  • Transverse oscillations are perpendicular to travel; longitudinal are parallel.
  • v=fλv = f\lambda and f=1/Tf = 1/T; frequency is fixed by the source.
  • Phase difference is 2πx/λ2\pi x/\lambda; in phase and antiphase are the key cases.
  • Intensity is power per area, proportional to amplitude squared.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 11 lessons