3 Waves

Exam analysis

Most tested topics

Waves units

A wave measurement is complete only when its unit identifies the quantity being measured. The required unit symbols are case-sensitive and are written after the numerical value.

Wave quantity or context	Unit	Unit symbol	Example
angle	degree	$^\circ$	$35^\circ$
frequency	hertz	Hz	50 Hz
wavelength or distance	metre	m	0.80 m
wave speed	metre per second	m/s	340 m/s
time or period	second	s	0.020 s

Hertz is a rate unit: 1 Hz means one cycle per second. Metres measure length, while metres per second measure speed. Seconds measure a time interval, including the time for one cycle when a period is stated.

Do not confuse a quantity symbol with its unit: frequency may be represented by $f$ but is measured in Hz; wave speed may be represented by $v$ but is measured in m/s. Write Hz with a capital H, but m and s in lower case.

(b) Properties of waves

Wave type is determined by comparing the direction of the oscillations with the direction in which the wave travels and transfers energy. It is not determined by the shape of a drawn line alone.

Feature	Transverse wave	Longitudinal wave
Direction of oscillations	Perpendicular (at right angles) to wave travel	Parallel to wave travel
Repeating pattern	Crests and troughs	Compressions and rarefactions
Example	Light and surface water waves	Sound waves in air

In a transverse water wave, the surface moves mainly up and down while the disturbance travels across the water. In a longitudinal sound wave, air particles move backwards and forwards along the same line as the travelling disturbance, producing alternating compressions and rarefactions.

The particles or fields oscillate; the wave pattern travels. Saying only that one wave is ‘up and down’ and another is ‘side to side’ is incomplete unless the direction is compared with the direction of wave travel.

A wave can be described by how far it oscillates, how its repeating pattern is spaced, and how quickly each cycle occurs.

Quantity	Definition	How to identify or measure it	Unit
amplitude	maximum displacement from the equilibrium position	equilibrium to a crest or trough	m
wavefront	line or surface joining points at the same stage of an oscillation	for example, one crest line	-
frequency, $f$	number of complete cycles passing a point each second	cycles divided by time	Hz
wavelength, $\lambda$	shortest distance between two neighbouring points in the same phase	crest to next crest, or compression to next compression	m
period, $T$	time taken for one complete cycle	total time divided by number of cycles	s

Amplitude is measured from equilibrium, not from crest to trough; crest-to-trough height is twice the amplitude. A wavelength must join equivalent points on consecutive cycles, not just any two nearby points.

A wave transfers energy from one place to another without carrying matter along with the travelling wave pattern. Changes in a wave can also carry information.

In a material medium, each particle oscillates about its equilibrium position and passes the disturbance to neighbouring particles. Energy therefore moves through the medium even though the particles have no overall journey with the wave.

A floating cork rises and falls as a water wave passes rather than travelling all the way with a crest. In sound, air particles vibrate locally while energy reaches a listener. In a communication signal, controlled changes in the wave represent information sent to a receiver.

Matter may oscillate while a mechanical wave passes, but oscillation is not the same as net transfer of that matter from source to receiver.

During one period, a wave travels one wavelength. Therefore its speed equals the number of wavelengths produced each second multiplied by the length of each wavelength.

v=f\lambda

$v$ is wave speed in m/s, $f$ is frequency in Hz, and $\lambda$ is wavelength in m. Rearrangements are $f=v/\lambda$ and $\lambda=v/f$ .

For sound travelling at 340 m/s with frequency 1.7 kHz, first convert $1.7\text{ kHz}=1700\text{ Hz}$ . Then $\lambda=v/f=340/1700=0.20\text{ m}$ .

Convert all values to compatible units before substituting. A frequency in kHz or a wavelength in cm cannot be used directly with a speed in m/s.

Frequency counts cycles per second, while period is the time for one cycle. They are reciprocals: more cycles each second means less time for each cycle.

f=\frac{1}{T}\qquad T=\frac{1}{f}

$f$ must be in hertz and $T$ must be in seconds when these equations are used with SI units.

If $T=2.5\text{ ms}$ , convert first: $2.5\text{ ms}=2.5\times10^{-3}\text{ s}$ . Then $f=1/T=1/(2.5\times10^{-3})=400\text{ Hz}$ .

A period of 2.5 ms is not 2.5 s. Missing the milli conversion changes the answer by a factor of 1000.

The same wave relationships apply to sound and electromagnetic waves. Identify the wave and medium, select the speed stated or appropriate to that context, convert prefixes, and then rearrange before substituting.

Context	Speed to use	Common conversion	Example result
sound in air	use the value given for that air condition	kHz to Hz: multiply by $10^3$	$v=330\text{ m/s}$ and $f=660\text{ Hz}$ give $\lambda=0.50\text{ m}$
electromagnetic wave in free space	$3.0\times10^8\text{ m/s}$	MHz to Hz: multiply by $10^6$	$f=100\text{ MHz}$ gives $\lambda=3.0\text{ m}$

If a wave enters a region where its speed changes, the source still fixes the frequency. From $v=f\lambda$ , the wavelength changes in the same ratio as the speed.

Do not use $3.0\times10^8\text{ m/s}$ for sound, and do not assume every sound-speed value is identical; use the speed for the named wave and medium.

The Doppler effect is the change in observed frequency and wavelength caused by relative motion between a wave source and an observer.

A moving source emits each new wavefront from a different position. Ahead of an approaching source, successive wavefronts are closer together; behind a receding source, they are farther apart. The wave speed in the same medium remains constant, so $v=f\lambda$ links the changed wavelength to the changed observed frequency.

Relative motion	Observed wavelength	Observed frequency or pitch
source approaching observer	shorter	higher
source receding from observer	longer	lower

The source does not need to emit at a different frequency. Motion changes the spacing and arrival rate of wavefronts at the observer; with no relative motion towards or away from the observer, there is no Doppler shift.

All types of wave can be reflected and refracted. These behaviours occur at boundaries, but they describe different changes to the wave.

Behaviour	What happens	Cause	Examples
reflection	the wave returns into the original region	interaction with a boundary	a sound echo; light from a mirror
refraction	wave speed and wavelength change as the wave enters a different medium or region; its direction can also change	wave speed differs across the boundary	light entering glass; water waves entering shallower water

During refraction the frequency remains fixed by the source. Since $v=f\lambda$ , a change in speed produces a change in wavelength.

Refraction does not always mean bending: a wave crossing the boundary along the normal changes speed and wavelength without changing direction. Detailed ray laws are taught later.

(c) The electromagnetic spectrum

Visible light is one small region of a continuous electromagnetic spectrum. The seven named regions are radio waves, microwaves, infrared, visible light, ultraviolet, x-rays and gamma rays.

Continuous means that electromagnetic wavelengths form an unbroken range. The named regions are useful bands within that range, not separate kinds of disturbance with gaps between them.

Every electromagnetic wave travels at the same speed in free space: $3.0\times10^8\text{ m/s}$ . A radio wave and a gamma ray can have very different frequencies and wavelengths while sharing this free-space speed.

The common speed applies in free space. Electromagnetic waves can travel at different, lower speeds in materials, so do not apply the free-space statement to every medium.

Across the electromagnetic spectrum, decreasing wavelength means increasing frequency because all regions share the same free-space speed and $v=f\lambda$ .

Direction	Ordered spectrum
wavelength decreases; frequency increases	radio → microwave → infrared → visible → ultraviolet → x-ray → gamma
wavelength increases; frequency decreases	gamma → x-ray → ultraviolet → visible → infrared → microwave → radio

Within visible light, the order from longer wavelength and lower frequency to shorter wavelength and higher frequency is red → orange → yellow → green → blue → indigo → violet.

Ultraviolet is beyond violet but is not a visible colour; infrared is beyond red but is not visible. ‘Higher in the spectrum’ is ambiguous, so always state whether frequency or wavelength is increasing.

An electromagnetic wave is chosen for a use because its interaction with matter, transmission or detection makes that job possible.

Region	Required uses	Why it is useful
radio waves	broadcasting and communications	signals can carry information over large distances
microwaves	cooking; satellite transmissions	absorbed energy heats food; directed signals can pass through the atmosphere to and from satellites
infrared	heaters; night vision	readily transfers thermal energy; warm objects emit infrared that detectors can sense
visible light	optical fibres; photography	can be guided along transparent fibres; cameras detect it to form photographs
ultraviolet	fluorescent lamps	fluorescent materials absorb ultraviolet and emit visible light
x-rays	seeing internal structures, including medical imaging	pass through some materials and soft tissue more readily than denser material such as bone
gamma rays	sterilising food and medical equipment	penetrating radiation kills microorganisms

A region may have other valid uses, but these are the pairings required here. The use is not explained by naming the wave alone: connect the wave's behaviour to what the device or process needs.

Excessive exposure to electromagnetic radiation can harm the body. Risk is reduced by limiting exposure time, increasing distance where practical, and placing suitable shielding between the source and people.

Radiation	Required harmful effect	Simple protective measures
microwaves	internal heating of body tissue	metal shielding and safety interlocks; keep away from a leaking source
infrared	skin burns	heat-resistant shielding or clothing; increase distance and limit exposure time
ultraviolet	damage to surface cells and blindness	cover skin, use sunscreen and UV-blocking eye protection; limit time in strong sunlight
gamma rays	cell mutation and cancer	dense shielding, remote handling, greater distance and the shortest practical exposure time

Protection must interrupt the exposure route: shielding absorbs or blocks radiation, distance reduces the radiation reaching the body, and shorter exposure reduces the total received.

Do not swap the named hazards: microwaves cause internal heating, infrared causes surface burns, ultraviolet damages surface cells and eyes, and gamma radiation can cause mutation and cancer.

(d) Light and sound

Light is a transverse electromagnetic wave: its oscillations are perpendicular to the direction in which the wave travels and transfers energy.

Behaviour	What happens to light
reflection	light returns into the original medium at a boundary
refraction	light changes speed and usually changes direction when it crosses into a different medium

Reflection and refraction describe what happens at a boundary; transverse describes the direction of oscillation. These are independent properties, so a reflected or refracted light wave remains transverse.

For reflection from a surface, the angle of incidence equals the angle of reflection.

i=r

Draw a normal at 90° to the surface where the incident ray meets it. Measure $i$ between the incident ray and the normal, and $r$ between the reflected ray and the normal. For example, $i=38^\circ$ gives $r=38^\circ$ .

Both angles are measured from the normal, not from the reflecting surface. A ray making 30° with the surface makes 60° with the normal.

A ray diagram uses straight lines with arrowheads to show the direction of light. Every boundary construction begins with a normal drawn at 90° through the point of incidence.

Draw the boundary and normal.
Draw the incident ray ending at the normal and add an arrow towards the boundary.
For reflection, draw the reflected ray on the original side with $r=i$ .
For refraction into a higher-refractive-index medium, draw the refracted ray closer to the normal; into a lower-index medium, draw it farther from the normal.
Add an arrow showing the continuing direction.

At normal incidence, $i=0^\circ$ , so the ray changes speed but not direction. Through a rectangular block with parallel faces, the emerging ray is parallel to the incident ray but laterally displaced.

Do not measure an angle from the surface or draw a curved ray inside one uniform medium. Direction changes at a boundary; each ray segment is straight.

Investigate refraction by tracing a narrow light ray through a transparent block and measuring angles from the normal.

Place the block on paper and draw around it.
Direct a single narrow ray at one face and mark two points on the incident ray and two on the emerging ray.
Remove the block, join the marks with a ruler, and draw the normal at each boundary.
Measure incidence and refraction angles with a protractor.
Repeat for a range of incidence angles.

Shape	What it lets the investigation show
rectangular block	refraction at two parallel faces and lateral displacement
semicircular block	a ray aimed through the centre meets the curved face normally, isolating refraction at the flat face
triangular prism	successive non-parallel faces produce an overall change in direction

Keep the block fixed while marking each path and use a thin ray; a wide ray or moving outline makes the measured angles uncertain.

For light entering a material from air, refractive index compares the sine of the incidence angle with the sine of the refraction angle.

n=\frac{\sin i}{\sin r}

$n$ has no unit; $i$ and $r$ are measured from the normal. Use degree mode on the calculator.

If $i=45^\circ$ and $r=28^\circ$ , then $n=\sin45^\circ/\sin28^\circ=0.7071/0.4695=1.51$ .

Do not calculate $i/r$ . The equation uses the sine of each angle, and reversing the numerator and denominator gives the wrong refractive index.

A glass block investigation determines refractive index from repeated measurements of incidence and refraction angles.

Draw around a rectangular glass block on paper.
Shine a narrow ray from air into the block and mark the incident and refracted paths.
Remove the block, draw the normal, and measure $i$ and $r$ .
Repeat for several incidence angles, avoiding very small angles that give large percentage uncertainty.
Calculate $n=\sin i/\sin r$ for each pair and take the mean after checking anomalies.

A stronger analysis plots $\sin i$ on the vertical axis against $\sin r$ on the horizontal axis. Since $\sin i=n\sin r$ , the gradient of a best-fit line through the origin is $n$ .

Repeat angle pairs, not just the final arithmetic. A mean cannot reveal a systematic error such as measuring from the surface instead of the normal.

Total internal reflection occurs when light travels from a higher-refractive-index medium towards a lower-index medium and the incidence angle is greater than the critical angle. No refracted ray then leaves through that boundary.

Device	Role of total internal reflection
optical fibre	repeated internal reflections keep light pulses inside the core so encoded information travels along the fibre
prism	internal reflection redirects a beam through a chosen angle in devices such as binoculars and periscopes

Both conditions are required: travel from higher to lower refractive index, and $i>c$ . If either condition fails, some light is refracted through the boundary.

The critical angle $c$ is the incidence angle in the higher-refractive-index medium for which the refracted ray in the lower-index medium is at $90^\circ$ to the normal and travels along the boundary.

Incidence angle in the higher-index medium	Outcome
$i<c$	light refracts out, bending away from the normal
$i=c$	refracted ray travels along the boundary
$i>c$	total internal reflection

The critical angle is not the first angle at which any reflection occurs: partial reflection can occur below $c$ . It is the threshold that separates refraction out from total internal reflection.

For a material-to-air boundary, critical angle and refractive index are related by:

\sin c=\frac{1}{n}

To find the angle, use $c=\sin^{-1}(1/n)$ . To find refractive index, use $n=1/\sin c$ . The angle $c$ is in degrees and $n$ has no unit.

For glass with $n=1.50$ , $c=\sin^{-1}(1/1.50)=41.8^\circ$ . A larger refractive index gives a smaller critical angle.

Use inverse sine when finding an angle; $c=1/n$ is incorrect. This syllabus equation applies to the material-air boundary described here.

Sound is a longitudinal mechanical wave. Particles of the medium oscillate parallel to the direction of wave travel, forming compressions and rarefactions.

Behaviour	What happens to sound	Example
reflection	sound returns from a boundary	an echo
refraction	sound changes speed and direction across regions where its speed differs	sound bending through air at different temperatures

Sound requires particles to pass on the vibration, so it cannot travel through a vacuum. The particles oscillate locally; they do not travel from the source to the listener.

The syllabus frequency range for human hearing is 20 Hz to 20 000 Hz, which is also 20 Hz to 20 kHz.

Frequency	Classification relative to human hearing
below 20 Hz	below the hearing range (infrasound)
20 Hz to 20 000 Hz inclusive	within the stated hearing range
above 20 000 Hz	above the hearing range (ultrasound)

$20\text{ kHz}=20 000\text{ Hz}$ , not 20 Hz. A vibrating source can produce sound outside the human hearing range even though it is still oscillating.

An electronic two-microphone method measures sound travel time over a known distance without relying on human reaction time.

Place two microphones a measured distance $d$ apart and connect them to an oscilloscope or data logger.
Make a sharp sound close to the first microphone.
Measure the delay $\Delta t$ between the two recorded signals.
Calculate $v=d/\Delta t$ .
Repeat at several large separations, identify anomalies, and average consistent values or find the gradient of a distance-time graph.

v=\frac{d}{\Delta t}

Measure distance between the microphone positions and keep air temperature as constant as practical because sound speed changes with temperature. A larger separation makes the delay a larger fraction of the measured time and reduces percentage uncertainty.

For an echo method, sound travels to the wall and back, so the distance is $2d$ . A handheld stopwatch is unsuitable for short separations because the travel time is comparable with reaction time.

A microphone converts sound-pressure variations into a changing electrical signal; an oscilloscope displays that signal as voltage against time.

Screen direction	Represents	Setting
horizontal ( $x$ )	time	timebase: time per division
vertical ( $y$ )	signal voltage, related to sound-wave amplitude	gain: volts per division

Connect the microphone, produce the sound, then adjust the timebase until at least one complete cycle is visible and the trace is steady. Adjust vertical gain so the trace is large enough to measure without leaving the screen.

The screen is not a picture of air particles moving through space. Its horizontal axis is time, and its vertical displacement represents the microphone's electrical signal.

Find sound frequency by measuring the period of a steady oscilloscope trace and using $f=1/T$ .

Connect a microphone and obtain a steady trace with several complete cycles.
Count the horizontal divisions across $N$ complete cycles.
Multiply divisions by the timebase to find their total time.
Divide by $N$ to obtain one period $T$ in seconds.
Calculate $f=1/T$ and repeat the measurement across another set of cycles.

T=\frac{(\text{divisions})(\text{time per division})}{N},\qquad f=\frac{1}{T}

If 8 divisions contain 4 cycles and the timebase is 0.50 ms/division, total time is 4.0 ms. Thus $T=4.0/4=1.0\text{ ms}=1.0\times10^{-3}\text{ s}$ and $f=1000\text{ Hz}$ .

Use the horizontal timebase, not the vertical gain. Convert milliseconds or microseconds to seconds before using $f=1/T$ .

Pitch is the perception linked to the frequency at which the source vibrates: increasing source frequency produces a higher-pitch sound, while decreasing it produces a lower-pitch sound.

Source change	Wave change	Heard change
vibrates more times each second	higher frequency, shorter period	higher pitch
vibrates fewer times each second	lower frequency, longer period	lower pitch

Amplitude does not determine pitch. Two sounds can have the same frequency and pitch but different amplitudes and loudnesses.

For the same source and listening conditions, a larger vibration amplitude produces a larger-amplitude sound wave and a louder sound; a smaller amplitude produces a quieter sound.

Source vibration	Oscilloscope trace	Heard sound
larger amplitude	taller trace from equilibrium	louder
smaller amplitude	shorter trace from equilibrium	quieter

Changing amplitude does not change frequency or pitch. Loudness at a listener can also change with distance, so the amplitude relationship must compare otherwise similar conditions.

3 Waves

Exam analysis

Most tested topics

Waves units

Choose the correct unit for waves

(b) Properties of waves

Distinguish transverse and longitudinal waves

Read the five quantities that describe a wave

Waves carry energy and information, not matter

Use wave speed, frequency and wavelength

Convert between frequency and period

Apply wave equations in sound and electromagnetic contexts

Explain the Doppler effect from moving wavefronts

Recognise reflection and refraction in every kind of wave

(c) The electromagnetic spectrum

Meet the electromagnetic spectrum

Order the electromagnetic spectrum

Connect electromagnetic waves to their uses

Match electromagnetic hazards to protection

(d) Light and sound

Describe light as a transverse wave

Apply the law of reflection

Construct reflection and refraction ray diagrams

Investigate refraction with three block shapes

Calculate refractive index from two angles

Determine the refractive index of glass

Use total internal reflection in fibres and prisms

Define the critical angle

Calculate critical angle and refractive index

Describe sound as a longitudinal wave

Use the human hearing range

Measure the speed of sound in air

Display sound with a microphone and oscilloscope

Measure sound frequency with an oscilloscope

Relate pitch to source frequency

Relate loudness to source amplitude