Full chapter
General properties of waves
All 6 topics and the revision summary on one page.
01
What travels in a wave?
A wave is a travelling disturbance that transfers energy. The material it passes through does not have to travel along with it.
An oscillation is a repeated motion about an equilibrium position. A vibrating part can move away from that position and return while passing a disturbance to its neighbours.
Follow a mark on a rope
Give one end of a stretched rope a brief up-and-down movement. A pulse travels along the rope. A small mark on the rope rises and falls as the pulse passes, then returns to its starting position.
The mark identifies the same piece of material throughout. It does not travel to the far end with the pulse. Yet the disturbance can make something at the far end move, showing that energy has been transferred.
Move the end repeatedly and a train of waves is produced. Each part oscillates about its local position as the pattern travels. This is what it means for a wave to transfer energy without transferring matter along with the disturbance. An independent bulk flow, such as a water current, is a separate effect.
A travelling disturbance and a local vibration
Green arrows show wave travel. Blue double arrows show the directions of local oscillation, not the instantaneous velocity at a turning point. Orange marks the same material point in each pair.
Transverse pulse on a rope
Longitudinal pulse in a spring
Compare transverse and longitudinal motion
- Transverse wave
- The material vibrates perpendicular to the direction of wave travel. A wave travelling horizontally along a rope while its material moves up and down is an example. A spring can also carry a transverse disturbance if its end is moved sideways relative to its length.
- Longitudinal wave
- The material vibrates parallel to the direction of wave travel. Push and pull the end of a stretched spring along its length: close and wide coil spacings travel along it while a marked coil moves to and fro locally. Sound in air is another example.
The spring's close-spaced region is a compression; its more widely spaced region is a rarefaction. Those regions travel, but a particular coil does not stay in a compression and travel all the way along the spring.
Worked classification
Two disturbances travel to the right
In the first, marked points move up and down. Their vibration is perpendicular to the rightward travel, so the wave is transverse.
In the second, marked points move left and right. Their vibration is parallel to the rightward travel, so the wave is longitudinal. Calling a wave "horizontal" is not enough: both patterns travel horizontally in these examples.
Wavefronts in a ripple tank
A ripple tank is a shallow transparent tank in which disturbances on the water surface can be observed. A vibrating straight dipper can produce approximately straight wavefronts. A small point source can produce circular wavefronts spreading outwards.
A wavefront joins points at the same stage of oscillation, such as points along one crest. In the uniform region shown, the direction of wave travel is perpendicular to the local wavefront. Straight fronts travel across the tank; circular fronts spread radially away from the source.
Ripple-tank wavefronts, seen from above
Each green line joins crests at one instant. Arrows show travel perpendicular to the local front in the uniform region shown. The lines are not paths of water particles.
A straight dipper gives straight fronts
A point source gives circular fronts
Water surface patterns help show wavefronts and crest spacing. The rope is the simpler model for purely up-and-down transverse motion; real water-particle motion need not be an exactly vertical line.
Separate observation from explanation
Watch a marked point on a rope or spring as the disturbance passes, and compare its position before and afterwards. Describe that local motion separately from the changing position of a crest or compression. In a ripple tank, observe successive fronts and their spacing; the front itself is a pattern rather than a permanent set of particles.
Compare two directions. Transverse and longitudinal describe vibration relative to propagation. They do not mean that the material traces the whole drawn wave shape or travels with the crest.
Optional check A pulse travels to the right along a rope. A small mark on the rope rises and falls, then returns to its starting position. What has travelled along the rope?
02
Wave quantities and graphs
A snapshot shows different points at one instant. A time trace shows one point at different instants. Their horizontal axes answer different questions.
Displacement describes a point's position relative to equilibrium, with a chosen positive direction. For the rope below, upward displacement is positive and downward displacement is negative. This is the local displacement of the rope, not the distance travelled by the wave.
Five quantities to identify
- Wave speed, v
- The distance travelled by the wave pattern per unit time, commonly in m/s. Follow a recognisable feature such as a crest. This is different from the changing speed of an individual vibrating point.
- Frequency, f
- The number of complete oscillations per second, measured in hertz (Hz). A frequency of 2.5 Hz means 2.5 complete cycles each second.
- Wavelength, λ
- The distance between adjacent points at the same stage of the wave cycle, measured in a length unit such as m. Examples are crest to next crest, trough to next trough, or one compression centre to the next.
- Period, T
- The time taken for one complete oscillation, in seconds. A point returns to the same stage of its motion after one period.
- Amplitude, A
- The maximum displacement from equilibrium. Its unit is a length unit such as m or cm. Measure from the equilibrium line to a crest or trough, not from crest to trough.
A complete cycle links period and frequency
If a point completes five oscillations in 2.0 s, its period is 2.0/5 = 0.40 s, and its frequency is 5/2.0 = 2.5 Hz. Taking longer for each cycle means completing fewer cycles per second.
A point crosses equilibrium twice in a cycle, once in each direction. When timing from an equilibrium crossing, finish at an equivalent crossing moving in the same direction. For a longitudinal wave, a compression centre to the neighbouring rarefaction centre is half a wavelength.
Read the same rope wave in two ways
The following wave travels to the right. Its amplitude is 2.0 cm, wavelength is 0.80 m and period is 0.40 s. The first graph is a snapshot at t = 0. The second follows the marked material point at x = 0.20 m.
Two graphs describe the same right-moving rope wave
Amplitude is 2.0 cm, wavelength is 0.80 m and period is 0.40 s. The wave speed is 2.0 m/s. The orange point in the snapshot is the material point followed by the time graph.
Snapshot at t = 0 s
Different points on the rope, all at one instant: read wavelength from the position axis.
One point at x = 0.20 m
One point on the rope, followed through time: read period from the time axis.
Reading the snapshot
Displacement against distance
The crests are at x = 0.20 m and x = 1.00 m. Their separation is λ = 1.00 - 0.20 = 0.80 m. The troughs at 0.60 m and 1.40 m are also 0.80 m apart.
The maximum displacement from zero is A = 2.0 cm. From +2.0 cm at a crest to -2.0 cm at a trough is 4.0 cm, twice the amplitude.
The horizontal axis is position, so this graph alone gives no time for a cycle. Frequency cannot be found from its crest spacing alone.
Reading the time trace
Displacement against time
The marked point starts at +2.0 cm. It reaches equilibrium at 0.10 s, -2.0 cm at 0.20 s, equilibrium at 0.30 s, and +2.0 cm again at 0.40 s.
It has then completed one cycle: T = 0.40 s and f = 1/0.40 = 2.5 Hz. Its amplitude is still 2.0 cm.
This horizontal axis is time. The distance between two peaks along the drawn page represents a time interval, not a wavelength.
A graph of displacement is not a graph of wave speed
On the time trace, a rising section means that the marked point's displacement is increasing in the chosen positive direction. It does not mean that the travelling wave is speeding up. A negative displacement means the point is below equilibrium, not that the wave must travel to the left.
Read values from the axes and their scales. The graph's physical width on a phone or a printed page is not the represented wavelength or period.
Check the horizontal axis first. A distance snapshot supplies spacing. A fixed-point time trace supplies timing. They can describe the same wave while carrying different information.
Optional check A displacement-distance snapshot has crests at 0.20 m and 1.00 m, with maximum displacement +2.0 cm and minimum -2.0 cm. What can be read from this snapshot alone?
03
Determine wave speed
In one period, a repeating wave pattern advances by one wavelength. Its speed is wavelength divided by period.
Speed is distance divided by time. For a periodic wave, the distance is one wavelength λ and the time is one period T. Frequency f is 1/T.
Worked example
Use the rope graph values
The period is 0.40 s and the wavelength is 0.80 m.
- Frequency: f = 1/0.40 = 2.5 Hz.
- Wave speed: v = 2.5 x 0.80 = 2.0 m/s.
- Check the meaning: in 0.40 s the pattern advances 0.80 m, giving the same speed 0.80/0.40 = 2.0 m/s.
Rearrange for the unknown
A wave with speed 3.6 m/s and frequency 12 Hz has wavelength 3.6/12 = 0.30 m. Dividing frequency by speed would not give a length.
Controlled comparison
Higher frequency at unchanged sound speed
Use a supplied sound speed of 340 m/s under unchanged propagation conditions.
- At 500 Hz, λ = 340/500 = 0.68 m.
- At 1000 Hz, λ = 340/1000 = 0.34 m.
Doubling frequency halves wavelength because the speed is held fixed. Do not apply that conclusion without checking the stated speed or conditions; not every possible water-wave comparison has frequency-independent speed.
Measure several wavelengths and several cycles
For a steady ripple-tank pattern at fixed water depth, measure the separation of equivalent crests in a correctly calibrated top view. Time complete source oscillations separately. The following are supplied model readings.
Six crests enclose five wavelengths
A supplied calibrated span is 40.0 cm from crest 1 to crest 6. Use that reading, not the displayed size of the picture.
One wavelength is 40.0/5 = 8.00 cm = 0.0800 m. Count the five gaps between the six crests rather than dividing by the number of crests.
| Trial | Time / s |
|---|---|
| 1 | 4.90 |
| 2 | 5.00 |
| 3 | 5.10 |
The mean time is (4.90 + 5.00 + 5.10)/3 = 5.00 s. Start and finish the count at the same stage of the source's motion.
T = 5.00/20 = 0.250 s
v = 4.00 x 0.0800 = 0.320 m/sThe 5.00 s interval is for 20 oscillations. It is not the time for one crest to travel from the first to the sixth position.
Choose improvements that address a cause
- Uncertain crest positions: measure a longer span containing several wavelengths and divide by the number of intervals. A similar endpoint-reading uncertainty then forms a smaller fraction of the full span.
- Start/stop reaction time: time several complete cycles. This reduces the fractional effect of reaction time compared with timing only one cycle.
- Incorrect length scale: calibrate the image or ruler in the observed plane and read without parallax. A projected image need not have the same physical scale as the water surface.
- Unclear pattern: keep the source steady and the water depth unchanged, and avoid an area where reflected waves overlap the incident pattern.
Repetition shows the variation in readings. It does not correct counting six intervals instead of five, using an incorrect scale, or changing the propagation conditions between the length and time measurements.
A different direct method follows the same pulse between two marked positions and uses distance divided by travel time. That travel time is not automatically the oscillation period. Identify which event starts and stops the timing.
Keep a spacing measurement separate from a cycle count. First find one wavelength and the number of cycles per second; then use v = fλ.
Optional check The span from the first to the sixth crest is 40.0 cm. The source completes 20 oscillations in a mean time of 5.00 s. What is the wave speed for these readings?
04
Sound, pitch and loudness
A vibrating source makes a disturbance in a material medium. In air, the disturbance travels as compressions and rarefactions.
Longitudinal motion is parallel to wave travel. Amplitude describes the size of an oscillation; frequency describes how often it repeats.
From a vibrating source to travelling sound
A tuning fork's prongs or a loudspeaker cone vibrate. As the source moves forwards, it pushes nearby air into a slightly smaller region. As it moves back, that region can become more spread out. Interactions between neighbouring regions pass the disturbance onwards.
- A compression has greater local pressure and density than the undisturbed air.
- A rarefaction has lower local pressure and density than the undisturbed air. It still contains particles; it is not a vacuum.
Sound in air is a travelling compression pattern
A vibrating speaker disturbs nearby air. The diagram shows a longitudinal model, with all distances and particle motions schematic.
The repeated back-and-forth motion along the direction of travel makes sound in air longitudinal. Sound in water can be described by the same compression-and-rarefaction model. The medium carries the disturbance and energy while its particles oscillate about local positions.
Sound needs a medium
Sound can travel through gases, liquids and solids because their particles interact. It cannot propagate through an ideal vacuum, where there are no particles to pass on this mechanical disturbance.
Observation and explanation
A vibrating source in a bell jar
A sounding source is placed in a bell jar. As air is removed, it becomes much less audible outside even though the source can still be seen vibrating.
The reduced sound supports the need for a material transmission medium. The observation does not mean that vibration has stopped. In a real apparatus, solid supports and remaining gas can still transmit some sound, so the demonstration is not a perfect removal of every pathway.
Change amplitude and frequency separately
Under comparable conditions, a sound wave of greater amplitude is louder. A sound wave of greater frequency has a higher pitch. Increasing amplitude does not by itself require a higher frequency.
The following graphs represent the displacement of an air particle against time at the same observation position for three separate sounds. Their vertical units are micrometres (µm), where 1 µm = 0.000001 m, not microphone voltage.
Compare amplitude and period separately
These are air-particle displacement-time graphs at one observation point, with the same axis scales and comparable conditions. They are not paths through the air or microphone voltages.
A: reference sound
Amplitude 0.20 micrometres; period 4.0 ms; frequency 250 Hz.
B: larger amplitude
Amplitude 0.40 micrometres; period 4.0 ms; frequency 250 Hz.
C: shorter period
Amplitude 0.20 micrometres; period 2.0 ms; frequency 500 Hz.
B is louder than A under the comparable conditions and has the same pitch. C has a higher pitch than A; its unchanged displacement amplitude alone does not establish equal perceived loudness.
- Sound A: 0.20 µm amplitude, 4.0 ms period
- 4.0 ms = 0.0040 s, so f = 1/0.0040 = 250 Hz.
- Sound B: 0.40 µm amplitude, 4.0 ms period
- Its frequency is also 250 Hz, so it has the same pitch as A. Its larger amplitude makes it louder under the stated comparable conditions. Doubling amplitude does not mean that perceived loudness is exactly doubled.
- Sound C: 0.20 µm amplitude, 2.0 ms period
- 2.0 ms = 0.0020 s, so f = 500 Hz. Its higher frequency gives a higher pitch than A. Equal displacement amplitude at a different frequency does not establish an exact equality of perceived loudness.
Read a microphone trace with its own units
A microphone and recording system can display a sound signal against time. That display usually represents the microphone's electrical output, rather than air-particle displacement measured in metres. Use the labelled vertical quantity.
To compare signal amplitudes, keep the microphone, recording gain, source distance and relevant surroundings the same. A taller trace caused by turning up the recording gain is not evidence that the source became louder. The time between repeated equivalent peaks still gives the period when the time scale is known.
Amplitude and frequency answer different questions. Taller oscillations concern amplitude; more cycles in the same time concern frequency. Neither means that the air itself travels all the way from source to receiver.
Optional check At the same observation point in comparable conditions, sound B has twice the particle-displacement amplitude of sound A, but both have period 4.0 ms. How do they compare?
05
Echoes and distance
An echo is reflected sound received after travelling to a surface and back. The measured delay includes the complete sound path.
For approximately constant speed, distance travelled = speed x time. Before calculating, identify where the sound is emitted, where it reflects and where it is received.
Draw the outward and return paths
Place a sound source and receiver together, a distance d from a reflecting surface. A short pulse travels distance d to the surface and distance d back. If the sound speed is approximately uniform and the paths match, the total distance is 2d.
Count both parts of the echo's path
First example: the source and receiver are together. Sound travels at the supplied 340 m/s and returns after 0.40 s.
Separate recorded example: subtract the timestamps
Worked example
An echo returns after 0.40 s
Use the supplied sound speed 340 m/s. The source and receiver are together, and the reflector remains stationary.
- Total sound path: vt = 340 x 0.40 = 136 m.
- One-way reflector distance: d = 136/2 = 68 m.
The 136 m is not the reflector distance. It is the sum of the outward and return distances.
A reception time is not automatically a delay
In a separate recording, a pulse is emitted at 20 ms and its echo is received at 220 ms. The timing origin occurred before emission.
Reading recorded times
Subtract, convert, then use the full path
- Delay: t = 220 - 20 = 200 ms.
- Convert: 200 ms = 0.200 s, since 1 ms = 0.001 s.
- Total path: 340 x 0.200 = 68.0 m.
- Reflector distance: d = 68.0/2 = 34.0 m.
If the reflector distance were independently measured as 34.0 m, the same record could instead determine speed: v = 2d/t = 68.0/0.200 = 340 m/s.
Measure the intended echo
A microphone with electronic timing or a recorded signal is useful for a short interval. Identify the emission event and the echo of that same pulse. Use a time scale with enough resolution, and account for any stated delay introduced by the instrument.
Several surfaces can produce different echoes. Select the return from the intended reflector rather than pairing the emission with an unrelated later pulse. If determining speed, measure the distance to the actual reflecting surface.
Repeating a human stopwatch measurement does not remove reaction-time limitations or guarantee that a very short echo interval has been resolved. An electronic record addresses timing; a clearer reflector arrangement addresses competing echoes. Choose the improvement that matches the problem.
The path determines whether to divide by two. A one-way transmission from a source to a separate receiver uses its one-way path. Do not divide every sound distance by two merely because sound is involved.
Ultrasound uses the same pulse-echo reasoning to measure underwater distances and locate tissue boundaries.
Optional check A source and receiver are together. A pulse is emitted at 20 ms and its echo arrives at 220 ms. With sound speed 340 m/s and the same outward and return path, how far away is the reflector?
06 / Pure
Using ultrasound
Ultrasound is sound above the usual upper limit of human hearing, conventionally above about 20 kHz. Its echoes can locate reflecting boundaries.
1 kHz = 1000 Hz, so 20 kHz is 20000 Hz. Ultrasound is still a mechanical sound wave and needs a material medium. The high frequency does not make it electromagnetic radiation.
Underwater echo sounding
An active echo-sounding system can emit a short ultrasound pulse into water. Its transducer converts an electrical signal into sound and converts a received echo back into an electrical signal. The time between emission and reception indicates the distance along the beam when the sound speed is known.
For a vertical beam reflected from the seabed, the pulse travels down from the transducer and back up. Using an approximately uniform water sound speed, depth below the transducer = vt/2.
Sonar calculation
A vertical return after 0.060 s
Use the supplied speed 1500 m/s. The round-trip interval is 0.060 s.
d = 1500 x 0.060 / 2 = 45 m.
This is the seabed distance below the transducer. It is not automatically the depth below the sea surface if the transducer is submerged. A sloping beam would measure distance along its path rather than the vertical depth.
This example uses an active sonar system with ultrasound. Sonar more broadly uses sound: some systems listen without emitting a ranging pulse, and not every sonar frequency is ultrasonic.
Use reflected pulses to locate a boundary
Active echo sounding with ultrasound
Supplied sound speed 1500 m/s; return delay 0.060 s. A vertical beam locates the seabed 45 m below the transducer.
A probe receives echoes from tissue boundaries
This supplied uniform-speed model uses 1540 m/s along the path. Depths are measured from the probe face; the coupling layer is included in this simplified path.
A: 3.08 cm, echo after 40 microseconds. B: 6.16 cm, echo after 80 microseconds.
Locate soft-tissue boundaries
A scanning probe sends ultrasound pulses into the body and receives returning echoes. At a boundary between tissues, some of a pulse can reflect while some continues farther. Later echoes can therefore return from deeper boundaries.
- Emit a pulse: record its emission time.
- Receive echoes: identify the delays of the returned signals.
- Estimate depth: combine each round-trip delay with an appropriate assumed sound speed.
- Build an image: combine information from many beam positions. Echo strength as well as timing contributes to the displayed image.
One echo does not give a complete photograph of an organ. The system assembles information from many reflections. In the simple model here, timing locates a boundary along the beam; a taller signal does not by itself mean a deeper boundary.
Why the probe uses coupling gel
Gel fills gaps between the probe and skin so that sound can pass into the body more effectively. An intervening air layer would cause strong reflection at the boundaries and reduce transmission. This does not mean that air cannot carry sound; the change between materials affects how much is transmitted or reflected.
Soft-tissue calculation
An echo after 80.0 microseconds
Use the supplied uniform sound speed 1540 m/s. One microsecond (µs) is 0.000001 s, so 80.0 µs = 0.0000800 s.
- Complete path: vt = 1540 x 0.0000800 = 0.1232 m.
- Boundary depth: d = 0.1232/2 = 0.0616 m = 6.16 cm.
Under the same model, a 40 µs echo comes from a nearer boundary at 3.08 cm. These depths are measured from the probe along the beam. The supplied speed is a model value, not a claim that every tissue has exactly the same sound speed.
Convert the time and retain the return path. Microseconds are millionths of a second. Multiplying speed by an echo delay gives the complete path; the boundary depth is half of it in this model.
Optional check An ultrasound pulse returns from one tissue boundary after 80.0 microseconds. Using the supplied uniform speed 1540 m/s, what depth does this echo indicate below the probe?
Revision summary
- Wave motion
- A travelling disturbance transfers energy without matter travelling with it. Compare material vibration with propagation: perpendicular for transverse, parallel for longitudinal.
- Frequency and period
- f = 1/T. Frequency is complete cycles per second, in Hz. Period is time per complete cycle, in s. A complete cycle contains two equilibrium crossings, one in each direction.
- Wave speed
- v = fλ = λ/T. Use f in Hz and λ in m for v in m/s. Wave speed is the pattern's propagation speed.
- Echo distance
- d = vt/2 when the source and receiver are together and the outward and return paths match. Subtract emission time from reception time before converting units and calculating.
Choose the graph for the quantity
| Representation | What it gives directly |
|---|---|
| Displacement-distance snapshot | Different points at one instant. Read wavelength from equivalent adjacent points and amplitude from equilibrium to maximum displacement. |
| Displacement-time trace | One point at different instants. Read period from a complete cycle and amplitude from maximum displacement; use f = 1/T. |
| Ripple-tank wavefronts | Each front joins points at the same stage, such as one crest. Travel is perpendicular to the local front in the uniform region shown. |
| Sound or microphone trace | Use the actual vertical quantity and units. Greater amplitude relates to louder sound under comparable conditions; greater frequency gives higher pitch. |
Keep the physical explanation
- A marked part of a rope or spring oscillates locally; a crest or compression travels.
- Sound comes from a vibrating source and needs a material medium. Compressions and rarefactions are regions of higher and lower local pressure/density, not air particles travelling from source to listener.
- A wavelength is one full spatial cycle; compression to neighbouring rarefaction is half of one. Amplitude is half the crest-to-trough displacement.
- Six consecutive crests span five wavelengths. Timing several cycles is different from timing a pulse's travel between two positions.
- A sound echo follows an outward and return path. Select the intended return, the correct time interval and the stated sound speed.
Ultrasound
Ultrasound is sound above about 20 kHz and still needs a medium. Active echo sounding locates an underwater reflector; soft-tissue scanning uses echo delays and strengths from many beam positions. For the simple uniform-speed pulse-echo model, depth = vt/2. A gel layer improves transmission between probe and skin.
Units and measurement reminders
1 cm = 0.01 m; 1 ms = 0.001 s; 1 kHz = 1000 Hz; 1 µs = 0.000001 s. Match the units before substitution. Use a calibrated scale, count full wavelength intervals and complete cycles, keep propagation conditions steady, and choose timing resolution that can resolve the event. Repetition does not correct an incorrect scale or the wrong echo.
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