Topic 06 | Physical sciences

Waves & energy transfer through media

Year 9 (Levels 9-10 band): wave and particle models of energy transfer; transverse vs longitudinal waves; wave equation; the electromagnetic spectrum; reflection, refraction, and diffraction.

55-75 min Printable practice Answer key Challenge included
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Read the explanation, work through the examples, then complete the core practice before printing.

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What you will learn

Worked example 0 Real-world example: radio station tuning

A local radio station broadcasts at 101.1101.1 MHz. Radio waves travel at the speed of light, c=3.0×108c = 3.0 \times 10^{8} m/s. What is the wavelength?

  1. Convert frequency: f=101.1×106f = 101.1 \times 10^{6} Hz.
  2. Wave equation: λ=vf=3.0×1081.011×1082.97\lambda = \dfrac{v}{f} = \dfrac{3.0 \times 10^{8}}{1.011 \times 10^{8}} \approx 2.97 m.
  3. That is why an FM antenna is a metre or two long — it needs to match the wavelength.

Key idea: every wave technology is designed around the wavelength and frequency of the signal it uses.

1. Energy transfer: three modes

ModeHow energy movesNeeds matter?Example
Conductionthrough direct contact — particles vibrate and pass energy to neighboursyes (best in solids)metal spoon heating in soup
Convectionin fluids: hot fluid rises, cool fluid sinks, carrying energy with ityes (fluids only)boiling water, sea breezes
Radiationby electromagnetic waves; no medium needednosunlight reaching Earth

Waves of particles (e.g. moving air in sound) transfer kinetic energy; electromagnetic waves carry energy through the fields themselves.

2. Transverse vs longitudinal waves

Transverse wave: particles vibrate perpendicular to the direction of energy transfer. Examples: water ripples, light, EM waves, wave on a string.

Longitudinal wave: particles vibrate parallel to the direction of energy transfer — compressions and rarefactions. Example: sound, seismic P-waves.

Transverse:wavelength lambdaLongitudinal:compressionrarefaction
Top: a transverse wave with crests and troughs. Bottom: a longitudinal wave with alternating compressions and rarefactions. Wavelength is one full cycle in each.

3. Wave properties and the wave equation

Every periodic wave has:

Wave relations

Wave equation
v=fλv = f \lambda

Units: vv in m/s, ff in Hz, λ\lambda in m.

Period and frequency
T=1f,f=1T.T = \dfrac{1}{f}, \qquad f = \dfrac{1}{T}.
Worked example 1 Sound in air

A tuning fork vibrates at 440440 Hz. Sound speed in air is about 340340 m/s. Find the wavelength.

  1. λ=vf=3404400.77\lambda = \dfrac{v}{f} = \dfrac{340}{440} \approx 0.77 m.
  2. About 7777 cm — roughly an arm’s length.
Worked example 2 Ocean wave speed

Ocean waves reach a beach with crests 1212 m apart every 88 s. Find the speed.

  1. Wavelength λ=12\lambda = 12 m; period T=8T = 8 s; frequency f=1/8=0.125f = 1/8 = 0.125 Hz.
  2. v=fλ=0.125×12=1.5v = f\lambda = 0.125 \times 12 = 1.5 m/s.

4. The electromagnetic (EM) spectrum

All EM waves travel at c=3.0×108c = 3.0 \times 10^{8} m/s in vacuum. They differ in frequency and wavelength.

RegionTypical wavelengthExample use
Radiom to kmbroadcasting, phones, Wi-Fi
Microwavecmmicrowave ovens, radar, satellites
Infrared10μm\sim 10\,\mu mheat imaging, remote controls
Visible light400-700 nmvision, photography
Ultraviolet100\sim 100 nmsterilisation, suntan (DNA damage)
X-ray0.1\sim 0.1 nmmedical imaging
Gamma<0.01< 0.01 nmcancer treatment, sterilising food

As wavelength decreases, frequency and energy per photon increase. UV, X-rays and gamma are “ionising” — they can knock electrons from atoms and damage DNA.

Worked example 3 Photon wavelength

Visible green light has frequency 5.6×10145.6 \times 10^{14} Hz. Find the wavelength in nanometres.

  1. λ=cf=3.0×1085.6×1014=5.36×107\lambda = \dfrac{c}{f} = \dfrac{3.0 \times 10^{8}}{5.6 \times 10^{14}} = 5.36 \times 10^{-7} m.
  2. =536= 536 nm — green.

5. Wave behaviour: reflection, refraction, diffraction

6. Sound in different media

Sound is a longitudinal wave that needs particles. Its speed depends on how tightly particles are bound.

MediumSpeed of sound (approx.)
Air (20 C^{\circ}\text{C})343 m/s
Water1480 m/s
Steel5000 m/s

Sound cannot travel through a vacuum — no particles to vibrate.


Practice: Year 9

Fluency

Types and properties

    1. Classify as transverse or longitudinal: (a) light, (b) sound, (c) water surface waves, (d) wave on a rope.
    2. Define amplitude, frequency, wavelength, and period.
    3. State the wave equation in symbols and in words.
    4. List the regions of the EM spectrum in order of increasing frequency.
    5. State the three modes of energy transfer and which require matter.
    6. Why can sound not travel through a vacuum?
Fluency

Wave equation calculations

    1. A wave has frequency 5050 Hz and wavelength 44 m. Find the speed.
    2. Find the wavelength of a 2.42.4 GHz Wi-Fi signal (c=3×108c = 3 \times 10^{8} m/s).
    3. A radio station broadcasts at wavelength 33 m. What is the frequency?
    4. A wave on a string travels at 1212 m/s with wavelength 0.40.4 m. Find the frequency and period.
    5. Sound travels at 14801480 m/s in water. A whale’s call has frequency 2020 Hz. Wavelength?
    6. A red laser has wavelength 650650 nm. Find the frequency.
Reasoning

Apply the ideas

    1. Explain why a low bass note reaches your ear from around a corner but a high treble note does not.
    2. Give one everyday use of each of: microwaves, infrared, X-rays.
    3. Why is UV light harmful to skin while visible light is not?
    4. Describe what happens to the speed, wavelength, and frequency of a light wave when it passes from air into water.
    5. A student claims sound travels faster in air than in steel because steel is heavier. Evaluate this claim.
Problem solving

Wave scenarios

    1. You see lightning and hear thunder 6 seconds later. Estimate how far away the strike was (sound speed 343343 m/s).
    2. A bat’s ultrasonic call is 4040 kHz and travels at 343343 m/s. Calculate the wavelength, and suggest why bats use such a high frequency for echolocation.
    3. A swimming-pool lane rope is shaken to produce a wave at 22 Hz with wavelength 1.51.5 m. Find the wave speed. How long to travel the 2525 m pool?
    4. An AM radio station at 1.0×1061.0 \times 10^{6} Hz diffracts well around hills; an FM station at 1.0×1081.0 \times 10^{8} Hz does not. Explain using wavelength.

Challenge

Reasoning

Harder reasoning

    1. Earthquakes produce both P-waves (longitudinal) and S-waves (transverse). S-waves cannot travel through the outer core. Explain how this is evidence that Earth’s outer core is liquid.
    2. Ultrasound imaging sends pulses into the body at 3\sim 3 MHz and detects echoes. Sound speed in soft tissue is about 15401540 m/s. Calculate the wavelength and comment on the smallest feature that can reasonably be resolved.
    3. Two loudspeakers emit the same frequency sound waves. At certain points listeners hear a loud sound; at others almost silence. Explain using superposition (constructive and destructive interference).
    4. A radar station sends a pulse at 1010 GHz; the echo returns 1.2×1041.2 \times 10^{-4} s later. Find the distance to the target. State the two wave ideas you used (speed of EM waves, and distance-time).
Answers

Answer key

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Year 9 answers

Fluency

Types and properties

    1. (a) transverse, (b) longitudinal, (c) mostly transverse (water surface moves up/down), (d) transverse.
    2. Amplitude: max displacement from equilibrium. Frequency: cycles per second (Hz). Wavelength: distance between consecutive crests (m). Period: time for one full cycle (s).
    3. v=fλv = f \lambda: the speed of a wave equals its frequency multiplied by its wavelength.
    4. Radio -> microwave -> infrared -> visible -> ultraviolet -> X-ray -> gamma.
    5. Conduction (needs matter, best in solids), convection (needs fluid), radiation (no medium needed).
    6. Sound is a mechanical wave needing particles to vibrate. Vacuum has (almost) no particles, so no vibration can propagate.
Fluency

Wave equation calculations

    1. v=50×4=200v = 50 \times 4 = 200 m/s.
    2. λ=3×1082.4×109=0.125\lambda = \dfrac{3 \times 10^{8}}{2.4 \times 10^{9}} = 0.125 m =12.5= 12.5 cm.
    3. f=3×1083=1×108f = \dfrac{3 \times 10^{8}}{3} = 1 \times 10^{8} Hz = 100 MHz.
    4. f=v/λ=12/0.4=30f = v/\lambda = 12/0.4 = 30 Hz. Period T=1/300.033T = 1/30 \approx 0.033 s.
    5. λ=1480/20=74\lambda = 1480/20 = 74 m.
    6. f=3×108/(650×109)4.6×1014f = 3 \times 10^{8} / (650 \times 10^{-9}) \approx 4.6 \times 10^{14} Hz.
Reasoning

Apply the ideas

    1. Low frequency = long wavelength. Diffraction is strongest when wavelength is similar to or larger than the obstacle/gap. Bass has wavelengths of metres, comparable to door widths, so it diffracts around corners. Treble has wavelengths of centimetres and diffracts much less.
    2. Microwaves: mobile phone signals, microwave ovens, radar. Infrared: TV remotes, thermal imaging. X-rays: medical imaging, airport security.
    3. UV photons carry more energy per photon and can ionise atoms or damage DNA directly, causing sunburn and skin cancer. Visible photons have less energy and are mostly absorbed harmlessly.
    4. Speed decreases (light is slower in water); wavelength decreases proportionally; frequency stays the same (set by the source).
    5. Incorrect. Sound travels faster in steel (rigid bonds transmit vibration quickly) than in air; “heavier” is the wrong factor — stiffness matters more than density.
Problem solving

Wave scenarios

    1. Distance =343×62060= 343 \times 6 \approx 2060 m or about 22 km.
    2. λ=343/(40000)=0.00858\lambda = 343 / (40\,000) = 0.00858 m 8.6\approx 8.6 mm. Short wavelength lets bats resolve small objects (insects) and focus the beam more tightly.
    3. v=fλ=2×1.5=3v = f\lambda = 2 \times 1.5 = 3 m/s. Time =25/38.3= 25 / 3 \approx 8.3 s.
    4. AM wavelength =3×108/106=300= 3 \times 10^{8} / 10^{6} = 300 m, comparable to hills — diffracts well. FM wavelength =3×108/108=3= 3 \times 10^{8} / 10^{8} = 3 m, much smaller than hills — little diffraction, so FM needs line of sight.
Reasoning

Challenge

    1. Transverse waves cannot propagate through fluids because fluids have no shear rigidity. Observations show a “shadow zone” where S-waves do not arrive, meaning they pass through a liquid layer. P-waves still arrive (longitudinal, slower in liquid). This is evidence that the outer core is liquid.
    2. λ=1540/(3×106)5.1×104\lambda = 1540 / (3 \times 10^{6}) \approx 5.1 \times 10^{-4} m =0.51= 0.51 mm. Smallest resolvable feature is roughly the wavelength, so about 0.50.5 mm — enough to see an unborn baby’s bones and organs.
    3. Where waves from the two speakers arrive in phase (crests coincide), amplitudes add — loud (constructive interference). Where they arrive out of phase (crest meets trough), they cancel — quiet (destructive interference). The pattern depends on the path-length difference relative to wavelength.
    4. Pulse travels to target and back, total distance =c×t=3×108×1.2×104=3.6×104= c \times t = 3 \times 10^{8} \times 1.2 \times 10^{-4} = 3.6 \times 10^{4} m. Distance to target =1.8×104= 1.8 \times 10^{4} m =18= 18 km. Ideas used: EM waves travel at cc; distance = speed times time.

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