# 3.5 Wave Phenomena and Stationary Waves

F.5 · Physics · Wave Motion · Lesson 2

Use wavefronts and path difference to explain observations, then calculate refraction, interference and stationary-wave spacing.

## Figure

Each stationary-wave loop spans half a wavelength; nodes stay at rest.

## Key points

- **Wavefronts** join adjacent particles in phase; rays are perpendicular to them. Consecutive crest fronts are one wavelength apart.
- **Reflection** returns waves into the original medium; angles are measured from the normal, with $i=r$.
- **Refraction (折射)** changes wave speed; at a stationary boundary, frequency stays unchanged. Normal incidence gives no bending.
- **Shallower water** gives smaller speed and wavelength; oblique rays bend towards the normal.
- **Diffraction (衍射)** spreads waves through gaps or around edges; longer wavelengths or narrower gaps give more spreading.
- **Superposition** adds displacements algebraically; cancelling displacement does not destroy energy.
- **Coherent sources (相干波源)** have the same frequency and a constant phase difference; they need not be in phase.
- **Stationary waves** superpose equal-amplitude waves of equal frequency and wavelength travelling in opposite directions.
- **Nodes (波節)** have zero amplitude; antinodes have the largest. Same-loop particles are in phase; adjacent loops are in antiphase.
- **Energy transfer** has zero net time average along an ideal stationary wave; a travelling wave carries energy along the medium.

## Must know

| Rule | Condition / result |
|---|---|
| Wave speed | $v=f\lambda$ |
| Refraction | $\dfrac{\sin i}{\sin r}=\dfrac{v_1}{v_2}=\dfrac{\lambda_1}{\lambda_2}$; stationary boundary, oblique incidence |
| Path difference | $\Delta=\lvert S_1P-S_2P\rvert$; cannot exceed source separation |
| Constructive interference | In-phase sources: $\Delta=n\lambda$ |
| Destructive interference | In-phase sources: $\Delta=(n+\tfrac12)\lambda$; zero amplitude needs equal arriving amplitudes |
| Stationary-wave spacing | Adjacent nodes: $\lambda/2$; node to nearest antinode: $\lambda/4$ |

Here $n=0,1,2,\ldots$. Antiphase sources reverse the constructive/destructive conditions. Reflection at a stationary barrier in unchanged depth, and diffraction in a uniform medium, preserve $v$, $f$ and $\lambda$.

## Common mistakes

- ✗ Coherent means in phase. ✓ Same frequency and constant phase difference.
- ✗ Refraction lowers frequency. ✓ The source fixes frequency; at a stationary boundary it is unchanged. Use $v=f\lambda$ to link the new speed and wavelength.
- ✗ Destructive interference always gives zero amplitude. ✓ Complete cancellation needs equal arriving amplitudes.
- ✗ One stationary-wave loop is one wavelength. ✓ Each loop spans $\lambda/2$.
- ✗ A displacement–time graph gives wavelength. ✓ It gives particle motion and period; use a displacement–distance graph for wavelength.

## Quick check

1. Water waves cross a stationary boundary into shallower water: $v_1=0.40\,\mathrm{m/s}$, $v_2=0.20\,\mathrm{m/s}$ and $f=2.0\,\mathrm{Hz}$. Find both wavelengths and $r$ if $i=30^\circ$.
2. At fixed wavelength in a uniform medium, how can a gap give more diffraction? Do $v$, $f$ or $\lambda$ change?
3. Two in-phase sources of wavelength $0.20\,\mathrm{m}$ are $1.10\,\mathrm{m}$ and $1.40\,\mathrm{m}$ from P. Find the path difference and interference type. When is amplitude zero?
4. A string has three loops over $0.60\,\mathrm{m}$ and vibrates at $25\,\mathrm{Hz}$. Find its wavelength and travelling-wave speed.
5. For the string in Q4, find adjacent-node spacing and node-to-nearest-antinode spacing. Compare the phase of particles in adjacent loops.

## Answers

1. $\lambda_1=0.40/2=0.20\,\mathrm{m}$; $\lambda_2=0.10\,\mathrm{m}$. $\sin r=(0.20/0.40)\sin30^\circ=0.25$; $r\approx14.5^\circ$. Frequency unchanged.
2. Narrow the gap; $v$, $f$ and $\lambda$ are unchanged.
3. $\Delta=1.40-1.10=0.30\,\mathrm{m}=1.5\lambda$: destructive; zero amplitude requires equal arriving amplitudes.
4. $\lambda=2(0.60)/3=0.40\,\mathrm{m}$; $v=f\lambda=25(0.40)=10\,\mathrm{m/s}$.
5. $\lambda/2=0.20\,\mathrm{m}$; $\lambda/4=0.10\,\mathrm{m}$; adjacent loops are in antiphase.

## Did you know?

A plucked guitar string can support several stationary-wave modes at once. Their different frequencies help give the note its characteristic sound.