PhysicsCore20 min read

Sound

A longitudinal wave: pitch, loudness, echoes and ultrasound

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01

Sound is a longitudinal wave

Definition

Compression — A region of a longitudinal wave where the particles are pushed closer together and the pressure is higher than normal.

A sound wave is produced by something vibrating — a loudspeaker cone, a guitar string, your vocal cords. The vibration pushes on the air next to it, and that disturbance travels outwards.

As the cone moves forward it pushes the air particles closer together, creating a region of higher pressure called a compression. As it moves back, the particles spread apart, leaving a region of lower pressure called a rarefaction. A continuous vibration sends out a train of compressions and rarefactions.

The particles vibrate back and forth along the direction the wave travels, which is what makes sound longitudinal. They do not travel with the wave — each one oscillates about its own position while the disturbance moves on.

Because sound needs particles to pass the disturbance along, it cannot travel through a vacuum. Ring an electric bell inside a jar and pump the air out: the hammer can still be seen striking, but the sound fades to nothing. This is the standard demonstration and a standard exam question.

02

Speed of sound in different materials

Sound travels at about 330 to 340 m s⁻¹ in air, roughly 1500 m s⁻¹ in water and around 5000 m s⁻¹ in steel. The pattern is consistent: the closer the particles, the faster the sound.

The reason is straightforward. The wave travels by particles colliding with their neighbours and passing the disturbance on. In a solid the particles are packed tightly and strongly bonded, so a collision happens almost immediately and the disturbance is handed along very quickly. In a gas the particles are far apart and must travel some distance before colliding, so the wave is slow.

Sound in air also travels slightly faster when the air is warmer, because the particles are moving faster and collide more often. It does not depend on the loudness or the pitch — all sounds travel at the same speed in the same conditions, which is fortunate, or an orchestra would arrive at the back of a hall in the wrong order.

v = f λv = 2d / t(echo)in an echo the sound travels to the surface and back, so the distance is doubled
v
speed of soundm s⁻¹
d
distance to the reflectorm
t
time for the echos
Worked example 14 marks

A student stands 165 m from a cliff, claps once and hears the echo 1.0 s later. Calculate the speed of sound in air.

  1. The sound travels to the cliff and back: 2 × 165 = 330 m.Doubling is the mark most often missed.
  2. Use v = distance / time.
  3. v = 330 / 1.0.
  4. v = 330 m s⁻¹.A sensible value for air — if you get 165 you forgot the return trip.

330 m s⁻¹

Sound waves add exactly like this. Two loudspeakers producing the same note create places where crests meet crests and the sound is louder, and places where a crest meets a trough and it is quieter — interference you can walk through in a room.

03

Pitch, loudness and the limits of hearing

Two properties of a sound wave map onto two things you hear. Frequency determines pitch: a high-frequency wave is heard as a high note. Amplitude determines loudness: a larger amplitude carries more energy and is heard as a louder sound.

Keeping these separate matters, because a question will often change one and ask about the other. Turning up the volume increases the amplitude and leaves the frequency alone — the note is louder but not higher.

The human ear responds to frequencies between roughly 20 Hz and 20 000 Hz. This range narrows with age, mostly at the top end. Sound above 20 kHz is called ultrasound, and although we cannot hear it, it is extremely useful.

Property of the waveWhat you hearChange it by
Frequencypitch — how high or lowvibrating faster or slower
Amplitudeloudnessvibrating with a bigger swing
Wave speednothing — arrival time onlychanging the medium
04

Echoes and ultrasound

Sound reflects from hard surfaces, and the reflection heard afterwards is an echo. The delay exists because sound travels at a finite speed — a fact worth stating explicitly, since "it bounces back" alone does not explain the pause.

That delay can be measured and turned into a distance. Send a pulse, time its return, multiply by the speed, and halve — because the pulse made the journey twice. This is echo sounding, used to map the sea bed and to find shoals of fish.

The same principle with ultrasound is used for medical imaging. Ultrasound is sent into the body, reflects from the boundaries between different tissues, and the returning pulses are timed to build a picture. Because it is not ionising, it is safe enough to use on an unborn baby, which X-rays are not.

Ultrasound is also used industrially to detect cracks inside metal castings without cutting them open, and to clean delicate objects such as jewellery and surgical instruments by shaking dirt loose in a liquid bath.

Key points

  1. Sound is longitudinal: compressions and rarefactions, vibration along the direction of travel.
  2. It cannot travel through a vacuum.
  3. Faster in solids than liquids, and slowest in gases.
  4. Frequency gives pitch; amplitude gives loudness.
  5. Echo problems: the sound goes there and back, so halve the distance.

Practice questions

6 questions · 26 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

2 · 4 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Describe how a compression and a rarefaction are produced by a vibrating loudspeaker cone.
Model answer

As the cone moves forward it pushes air particles together, forming a compression. As it moves back the particles spread apart, forming a rarefaction.

Examiner tip. Link each to the direction of the cone's motion — that pairing is what earns both marks.

SQ2[2 marks]
Explain why an echo is heard some time after the original sound.
Model answer

The sound reflects from a distant hard surface and travels back to the listener. Because sound travels at a finite speed, the reflected sound arrives later than the direct one.

Examiner tip. "Finite speed" is the physics. Saying only "it bounces back" misses the reason for the delay.

Solved numericals

1 · 4 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
A ship sends an ultrasound pulse vertically downward and receives the reflection from the seabed 0.40 s later. The speed of sound in water is 1500 m s⁻¹. Calculate the depth of the water.

Given. v = 1500 m s⁻¹, t = 0.40 s

Full working
  1. Total distance travelled = v × t = 1500 × 0.40 = 600 m[1]
  2. Recognises this is the distance down and backthe mark the question exists to test[1]
  3. Depth = 600 / 2[1]
  4. = 300 m[1]

300 m

Examiner tip. Any question involving an echo, sonar or an ultrasound scan needs the factor of two. Write "there and back" beside your first line so you cannot forget it.

Long questions

1 · 9 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[9 marks]
A student stands between two parallel cliffs and claps once. She hears the first echo after 0.60 s and a second echo after 1.0 s. The speed of sound in air is 340 m s⁻¹.
  1. Calculate her distance from the nearer cliff. [3]
  2. Calculate her distance from the further cliff. [3]
  3. Explain why sound travels faster in water than in air. [3]
Mark scheme
  1. Distance travelled by the first echo = 340 × 0.60 = 204 m[1]
  2. Recognises this is there and back[1]
  3. Distance to the nearer cliff = 102 m[1]
  4. Distance travelled by the second echo = 340 × 1.0 = 340 m[1]
  5. Halves it[1]
  6. Distance to the further cliff = 170 m[1]
  7. Particles in water are much closer together than in air[1]
  8. So collisions between them are more frequent[1]
  9. And the vibration is passed on more quickly[1]

(a) 102 m (b) 170 m

Examiner tip. Two echoes, two independent calculations — do not try to combine them. Each one needs its own halving.

Exam questions

2 · 9 marks

Multi-part questions with a full mark scheme.

Q1[5 marks]
An electric bell is suspended inside a sealed glass jar. The air is slowly pumped out.
  1. State what is observed.
  2. Explain the observation.
  3. State what this shows about sound.
Mark scheme
  1. The sound gets quieter and eventually cannot be heard[1]
  2. The hammer is still seen to strike, so the bell is still vibratingthis detail is a mark — it rules out the bell simply stopping[1]
  3. There are fewer air particles to be compressed and to pass the vibration on[1]
  4. Sound requires a medium to travel through[1]
  5. It cannot travel through a vacuum[1]

Examiner tip. Mentioning that the hammer is still visibly moving is worth a mark on its own, because it shows the sound stopped for the right reason.

Q2[4 marks]
Two notes are displayed on an oscilloscope. Note B shows waves that are taller and closer together than note A.
  1. Compare the loudness of the two notes and justify your answer.
  2. Compare the pitch of the two notes and justify your answer.
Mark scheme
  1. B is louder than A[1]
  2. Because it has a larger amplitude[1]
  3. B has a higher pitch than A[1]
  4. Because waves closer together means a higher frequency[1]

Examiner tip. Each comparison needs its wave property named. "B is louder and higher" with no reasons scores two of four.