K323 / 2027
General properties of waves overview

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

The rope pulse moves right while a marked part returns to its original heightTwo snapshots show one right-moving pulse. Earlier, the orange marker is at the pulse's crest. Later, the pulse is farther right and the marker has returned to the horizontal equilibrium line. The marker has the same horizontal position in both snapshots. A vertical blue double arrow indicates the directions in which the rope can vibrate, perpendicular to the horizontal green propagation arrow. The static snapshots are schematic and do not give a measured time interval.Pulse travels rightEarlierLaterThe marker stays at the same xLocal motion is up and down

Longitudinal pulse in a spring

A compression travels along a spring while a marked coil moves locally to and froTwo snapshots contain the same 51 schematic coils. The closely spaced compression moves from the left part of the spring to the right. The orange marked coil is slightly left of its dashed equilibrium reference in the earlier snapshot and back at that reference in the later snapshot. It does not follow the compression along the spring. A horizontal blue double arrow indicates local to-and-fro motion, parallel to the green propagation arrow. The models do not provide a numerical amplitude or speed.Compression travels rightEarlierLaterThe marked coil moves locallyLocal motion is along the spring
Compare the wave's direction of travel with the marked material's local motion. The double-headed arrows show the directions in which the material can oscillate, not an instantaneous velocity at a turning point. The rope pulse and the compressed spring pattern travel while their material moves locally.

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

Parallel straight crests travel away from a straight dipperIn a top view of a ripple tank, a straight dipper at the left produces parallel vertical crest lines. A horizontal green arrow points right, perpendicular to those lines. The dipper vibrates to produce the waves; the drawing does not show material travelling with a crest.DipperTop view: lines of crests

A point source gives circular fronts

Circular crests travel radially outwards from a point sourceA top view shows a small orange vibrating point source at the centre of three concentric circular crests. Outward radial arrows cross the fronts at right angles. Every circle is a wavefront, not the orbit of one water particle.Orange dot: vibrating point source
These are top views of successive crests in a ripple tank. The lines represent wavefronts, not rows of water particles travelling together. The travel arrows are perpendicular to the local wavefronts.

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?
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.

f = 1/T   and   T = 1/fUse T in seconds to obtain f in Hz. Count complete cycles, not every passage through equilibrium.

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

Snapshot at t = 0 s: signed displacement in centimetresHorizontal position ranges from zero to 1.60 metres. At time zero, crests are at 0.20 and 1.00 metres and troughs at 0.60 and 1.40 metres. A bracket between adjacent crests marks the 0.80 metre wavelength. The orange marker is at position 0.20 metres and displacement plus 2.0 centimetres. Both graphs use the same vertical scale from minus 2.0 to plus 2.0 centimetres. The bracket A on the right spans zero to plus 2.0 centimetres, so amplitude is 2.0 centimetres, not the 4.0 centimetre peak-to-peak span. These are different horizontal quantities, not two wave shapes at different instants.Displacement / cmWavelength = 0.80 m-20+20.00.40.81.21.6APosition along rope / m

Different points on the rope, all at one instant: read wavelength from the position axis.

One point at x = 0.20 m

One point at x = 0.20 m: signed displacement in centimetresTime ranges from zero to 0.80 seconds for the one material point at position 0.20 metres. Its displacement starts at plus 2.0 centimetres, reaches zero at 0.10 seconds, minus 2.0 at 0.20, zero at 0.30 and plus 2.0 again at 0.40. The cycle repeats. A bracket marks the 0.40 second period. Both graphs use the same vertical scale from minus 2.0 to plus 2.0 centimetres. The bracket A on the right spans zero to plus 2.0 centimetres, so amplitude is 2.0 centimetres, not the 4.0 centimetre peak-to-peak span. These are different horizontal quantities, not two wave shapes at different instants.Displacement / cmPeriod = 0.40 s-20+20.00.20.40.60.8ATime / s

One point on the rope, followed through time: read period from the time axis.

The distance graph compares different rope points at one instant. The time graph follows the single point at x = 0.20 m. Both use displacement vertically, but only the snapshot gives wavelength directly and only the time trace gives period directly.

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?
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.

v = λ/T = fλv is wave speed in m/s, f is frequency in Hz, and λ is wavelength in m. The equation concerns the travelling pattern, not the speed of a vibrating material point.

Worked example

Use the rope graph values

The period is 0.40 s and the wavelength is 0.80 m.

  1. Frequency: f = 1/0.40 = 2.5 Hz.
  2. Wave speed: v = 2.5 x 0.80 = 2.0 m/s.
  3. 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

λ = v/f   and   f = v/λConvert lengths to metres and frequency to hertz before using speed in m/s. For example, 1 kHz = 1000 Hz and 1 cm = 0.01 m.

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.

A forty-centimetre span across six crests contains five wavelengthsSix equally spaced straight crests are numbered one to six in a top view. Each of the five intervening gaps is labelled lambda for one wavelength. A total bracket from the first crest to the sixth is labelled 40.0 centimetres. Therefore one wavelength is 8.00 centimetres, or 0.0800 metres. These are supplied model readings, not a screen ruler measurement.Top view of six crests123456λλλλλ40.0 cm
The span from the first to the sixth crest contains five wavelengths. Its stated length is 40.0 cm; use the supplied scale, not the diagram's physical size on the screen.

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.

Repeated times for 20 complete source oscillations
TrialTime / s
14.90
25.00
35.10

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.

f = 20/5.00 = 4.00 Hz
T = 5.00/20 = 0.250 s
v = 4.00 x 0.0800 = 0.320 m/s
The 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?
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.

A speaker produces compressions and rarefactions while air particles oscillate locallyA speaker at the left is the vibrating source; a short blue double arrow below it indicates its vibration. Rows of air-particle symbols show two crowded compression regions and a more spread-out rarefaction between them. The rarefaction still contains air particles. A green arrow shows the sound pattern travelling right. A blue horizontal double arrow below the orange marked particle shows the directions of its local oscillation, parallel to propagation. Adjacent compression centres are one wavelength apart. Particle rows are a simplified representation and not a solid lattice. The air does not stream from the speaker to a listener with the wave.Sound travels rightCompressionCompressionRarefactionParticle moves to and froSpeakerOne wavelength
The source's vibration creates travelling compressions and rarefactions. A marked air particle oscillates locally, parallel to the wave's travel. Adjacent compression centres are one wavelength apart; the particle does not travel from the speaker to the listener.

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

Sound A: amplitude 0.20 micrometres and period 4.0 millisecondsThe vertical axis is particle displacement in micrometres, with the same scale from minus 0.40 to plus 0.40 for all three graphs. The horizontal axis runs from zero to eight milliseconds on the same scale. This sound has amplitude 0.20 micrometres, period 4.0 milliseconds and frequency 250 hertz. It completes two cycles in eight milliseconds.Particle displacement / micrometres-0.4-0.20.0+0.2+0.402468Time / ms

Amplitude 0.20 micrometres; period 4.0 ms; frequency 250 Hz.

B: larger amplitude

Sound B: amplitude 0.40 micrometres and period 4.0 millisecondsThe vertical axis is particle displacement in micrometres, with the same scale from minus 0.40 to plus 0.40 for all three graphs. The horizontal axis runs from zero to eight milliseconds on the same scale. This sound has amplitude 0.40 micrometres, period 4.0 milliseconds and frequency 250 hertz. It has twice the displacement amplitude of A and the same frequency.Particle displacement / micrometres-0.4-0.20.0+0.2+0.402468Time / ms

Amplitude 0.40 micrometres; period 4.0 ms; frequency 250 Hz.

C: shorter period

Sound C: amplitude 0.20 micrometres and period 2.0 millisecondsThe vertical axis is particle displacement in micrometres, with the same scale from minus 0.40 to plus 0.40 for all three graphs. The horizontal axis runs from zero to eight milliseconds on the same scale. This sound has amplitude 0.20 micrometres, period 2.0 milliseconds and frequency 500 hertz. It completes four cycles in eight milliseconds, twice as many as A, so its frequency is double. Equal displacement amplitude at a different frequency does not establish equal perceived loudness.Particle displacement / micrometres-0.4-0.20.0+0.2+0.402468Time / ms

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.

A and B have the same period but different amplitudes. C has a shorter period than A. These are particle-displacement traces, not curved paths along which air particles travel.
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?
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.

2d = vt   so   d = vt/2t is the delay from emission to reception of the echo. v is the sound speed. d is the one-way distance from the source/receiver to the reflector.

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.

An echo travels 68 metres out and 68 metres backA combined sound source and receiver is on the left and a reflector is 68 metres to its right. The outward arrow points right and the returning echo arrow points left. They are vertically separated only for readability: the example uses the same outward and return path. The total path is 136 metres in 0.40 seconds at 340 metres per second. The reflector distance is half the total path.SourceandreceiverReflectorOutward: 68 mReturn: 68 mTotal path: 136 m in 0.40 s

Separate recorded example: subtract the timestamps

Emission at 20 milliseconds and return at 220 milliseconds give a 200 millisecond intervalThis separate example is a timeline, not the 0.40 second path example above. An emission event is at 20 milliseconds and the corresponding echo arrives at 220 milliseconds. The elapsed time is 200 milliseconds, or 0.200 seconds. At 340 metres per second this interval gives a reflector distance of 34.0 metres. Vertical event marks identify timestamps; their heights do not represent a sound amplitude.Emitted20 msEcho received220 ms060120180240Recorded time / ms220 - 20 = 200 ms
The sound path goes to the reflector and back. The 0.40 s path example and the separate timestamp example both use a round-trip delay, but the timestamp example first requires subtraction.

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.

  1. Total sound path: vt = 340 x 0.40 = 136 m.
  2. 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

  1. Delay: t = 220 - 20 = 200 ms.
  2. Convert: 200 ms = 0.200 s, since 1 ms = 0.001 s.
  3. Total path: 340 x 0.200 = 68.0 m.
  4. 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?
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 sonar pulse travels down from the transducer and returns from the seabedA submerged transducer sends an ultrasound pulse vertically down to a level seabed. A second upward arrow shows the returning echo. The paths are offset to distinguish directions. The supplied 0.060 second delay and uniform speed of 1500 metres per second give a one-way distance of 45 metres. A depth bracket starts at the transducer face, below the water surface, and ends at the seabed. The drawing is schematic and does not assign a depth below the water surface.Water surfaceTransducerOutBack45 mSeabed

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 probe sends ultrasound through gel and receives separate echoes from nearer and farther boundariesA probe face contacts a thin gel layer above the skin. It emits a pulse downwards. Part reflects from boundary A, 3.08 centimetres from the probe face; part continues and reflects from boundary B, 6.16 centimetres away. Upward arrows show separate echoes from the two boundaries. The directions are separated sideways for clarity, not literal bent sound paths. In the supplied uniform-speed model, A returns an echo after 40 microseconds and B after 80 microseconds. Their depths have a two-to-one ratio. Many beam measurements, using echo times and strengths, are needed to build an image.ProbeGelSkinAB6.16cmDown: pulse. Up: returning echoes.

A: 3.08 cm, echo after 40 microseconds. B: 6.16 cm, echo after 80 microseconds.

Echo times distinguish the two boundariesA schematic pulse trace has time in microseconds on its horizontal axis and signal in arbitrary units on its vertical axis. The emission reference is at zero. Echo A is at 40 microseconds and echo B at 80 microseconds. The later echo corresponds to the farther boundary under the stated uniform-speed model. Pulse heights and widths are schematic and do not supply a numerical reflectivity, tissue identity or depth scale.Signal / arbitrary unitsEmittedAB020406080100Time / microsecondsPulse heights are schematic
The sonar model measures a return from the seabed. The soft-tissue model receives separate echoes from two boundaries. Under its uniform-speed assumption, the later echo comes from the farther boundary. Echo heights are schematic and do not measure depth or identify tissue.

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.

  1. Emit a pulse: record its emission time.
  2. Receive echoes: identify the delays of the returned signals.
  3. Estimate depth: combine each round-trip delay with an appropriate assumed sound speed.
  4. 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.

  1. Complete path: vt = 1540 x 0.0000800 = 0.1232 m.
  2. 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?
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

Read the axis before interpreting the shape
RepresentationWhat it gives directly
Displacement-distance snapshotDifferent points at one instant. Read wavelength from equivalent adjacent points and amplitude from equilibrium to maximum displacement.
Displacement-time traceOne point at different instants. Read period from a complete cycle and amplitude from maximum displacement; use f = 1/T.
Ripple-tank wavefrontsEach 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 traceUse 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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