Reflection at fixed and free ends
A wave pulse carries energy along a rope. At the end of the rope it has to go somewhere, so it reflects.
- Fixed end (tied to a wall): the end cannot move. The rope pulls the wall up, so the wall pulls the rope down (Newton's third law). The pulse returns inverted. We say it has a phase change of π (half a cycle).
- Free end (a ring that slides on a smooth pole): the end swings up even higher and then pulls the rope back. The pulse returns upright, with no phase change.
The same rule works for sound in pipes: a closed end acts like a fixed end, an open end like a free end.
Transmission and reflection between two media
When a wave meets a new medium, it splits: some energy is transmitted, some is reflected.
- Frequency does not change: the boundary vibrates at the rate the wave arrives.
- Speed changes, so wavelength changes (λ = v/f). In a slower medium the pulse is shorter.
- Light (fast) → heavy (slow) rope: reflected pulse inverted, like a fixed end.
- Heavy (slow) → light (fast) rope: reflected pulse upright, like a free end.
- The transmitted pulse is always upright.
For a string the amounts are r = (v₂ − v₁)/(v₁ + v₂) and t = 2v₂/(v₁ + v₂) (as fractions of the incoming height). A big difference between the media means more reflection. This is why ultrasound gel is used: without it, almost all the sound would reflect at the air gap between probe and skin.
Polarisation
In a transverse wave the vibration is at right angles to the direction of travel. It can be up-down, side-to-side or any angle in between.
Unpolarised light (sunlight, a bulb) has vibrations in all these directions mixed. A polariser lets through only one direction. The light is now plane-polarised and has half the intensity.
Malus's law: a second polariser (analyser) at angle θ to the first passes
I = I₀ cos²θ
At θ = 0 all passes; at 45° half; at 90° nothing ("crossed polarisers").
Sound cannot be polarised: it is longitudinal, so its vibration is along the direction of travel. Polarisation proves light is a transverse wave.
Uses: glare-free sunglasses, LCD screens, 3D cinema glasses, stress testing of plastic parts.
The Doppler effect
When a source of waves moves, each new crest is sent from a point closer to the listener ahead. The crests bunch up ahead and spread out behind.
- Source or observer approaching: higher observed frequency (higher pitch, bluer light).
- Moving apart: lower observed frequency (lower pitch, redder light).
For sound (speed v in still air):
f' = f × (v + vo) / (v − vs) when they approach each other;
f' = f × (v − vo) / (v + vs) when they move apart.
For light and slow speeds: Δf / f ≈ v / c. Astronomers use the redshift of galaxies to show the universe is expanding.
Try it: ask a friend to swing a phone playing a steady beep in a circle on a string. You will hear the pitch rise and fall each turn.
Wave technologies
Devices use wave behaviour to carry, capture or store information and energy:
- Ultrasound scans and sonar: time the echo of a pulse that reflects at a boundary; distance = speed × time ÷ 2.
- Radar speed guns and Doppler scans: measure the frequency shift of the echo to find speed (cars, blood flow, storms).
- Optical fibres: total internal reflection keeps light inside glass for internet signals.
- Digital signals: information sent as on/off pulses. They resist noise better than analogue signals and can be copied and stored without loss.
- Solar cells, cameras and phones: capture electromagnetic waves and turn them into electric signals.
Key formulas and definitions
- v = f λ (f unchanged at a boundary)
- Fixed end / slow medium: reflection inverted (phase change π)
- String: r = (v₂ − v₁)/(v₁ + v₂), t = 2v₂/(v₁ + v₂)
- Malus's law: I = I₀ cos²θ
- Unpolarised through one polariser: I = I₀ / 2
- Doppler (sound): f' = f (v ± v_o)/(v ∓ v_s), upper signs when approaching
- Light, v ≪ c: Δf / f ≈ v / c
- Echo distance d = v t / 2
Worked examples
1. A wave of frequency 50 Hz moves from a rope where v = 10 m/s into one where v = 4 m/s. Find the wavelengths.
Frequency stays 50 Hz. λ₁ = 10/50 = 0.20 m. λ₂ = 4/50 = 0.08 m.
2. Unpolarised light of intensity 80 W/m² passes through two polarisers at 60° to each other. Find the final intensity.
After the first: 80/2 = 40 W/m². After the second: 40 × cos²60° = 40 × 0.25 = 10 W/m².
3. A siren of 700 Hz moves towards you at 30 m/s. Speed of sound 340 m/s. What frequency do you hear?
f' = 700 × 340/(340 − 30) = 700 × 1.097 ≈ 768 Hz.
4. The same siren moves away at 30 m/s. What do you hear?
f' = 700 × 340/(340 + 30) = 700 × 0.919 ≈ 643 Hz.
5. A cyclist rides at 10 m/s towards a still 500 Hz horn. v = 340 m/s. What does she hear?
Observer moving: f' = 500 × (340 + 10)/340 ≈ 515 Hz.
6. A ship's sonar pulse returns from the sea bed after 0.80 s. Sound in sea water travels at 1,500 m/s. How deep is the sea?
d = vt/2 = 1500 × 0.80 ÷ 2 = 600 m.
Common mistakes
- Thinking the frequency changes when a wave enters a new medium. Frequency stays; speed and wavelength change.
- Saying sound can be polarised. Only transverse waves can be polarised.
- Forgetting to halve the intensity at the first polariser for unpolarised light.
- Thinking the Doppler effect means the sound gets louder. It is about frequency (pitch), not loudness.