Sound Waves: Characteristics and Applications Class 9 Notes and Numericals

What Sound Waves: Characteristics and Applications is about

Weightage: This chapter is part of Unit III, Motion, Force, Work and Sound, which carries 23 of the 80 theory marks in the school annual exam, shared with Describing Motion Around Us, How Forces Affect Motion, Work, Energy and and Simple Machines.Source: CBSE Science (086) syllabus, Class IX 2026-27, course structure, Unit III Motion, Force, Work and Sound

Key concepts

Production of sound by vibration

NCERT section 10.1 Production of Sound

Activity 10.1: a rubber band stretched over an open box makes sound only while it vibrates; when the vibration stops, the sound stops. Tightening it changes the sound.

Sound is produced by vibrating objects. A vibration is a repeated to-and-fro motion (oscillation). The object making the sound is the source.

• Strings (sitar, veena), membranes (tabla, dholak), air columns (bansuri), metal (bells, the *taal*, which shows sonority).
• Humans: the vocal cords in the larynx (voice box) vibrate; the tongue, lips, mouth and nose shape the sound into speech. Touch your throat while speaking and you feel the vibration.
• Insects: crickets and grasshoppers rub wings or legs.

Tuning fork (Activity 10.2): a U-shaped steel or aluminium bar with prongs and a stem. Struck on a rubber pad (never a hard surface), it hums; touching a prong to water makes ripples, proving the prongs vibrate. It gives an almost single-frequency sound.

In Kongthong, Meghalaya, each person has a whistled 'tune name' (*Jingrwai Iawbei*).

Worked example

A vibrating tuning fork that is too faint to see moving is touched to the surface of water in a bowl. What do you observe, and what does it prove? Why should you not strike it on a stone floor?

Try it first, then show the solutionSolution

1. Water splashes and ripples spread from the prong.
2. This shows the prongs are vibrating, so sound is produced by vibration.
3. Striking a hard surface can damage the prongs and change the frequency, and makes unwanted extra sounds.

Answer: Ripples form, proving the prongs vibrate; a rubber pad protects the fork and gives a clean note.

Sound needs a medium

NCERT section 10.2 Propagation of Sound

Activity 10.3: with your ear on a desk, you hear a friend's faint scratch clearly: sound travels through solids. Activity 10.4: spoons tapped under water can be heard: sound travels through liquids.

The material that carries sound is the medium (solid, liquid or gas). A space with no matter is a vacuum.

Bell-jar experiment: an electric bell rings inside a glass jar. As a vacuum pump removes the air, the sound gets fainter until almost nothing is heard, though the bell is still seen ringing. Letting air back in brings the sound back.

Conclusion: sound cannot travel through vacuum.

• Space is nearly a vacuum, so astronauts on a spacewalk cannot hear each other directly or hear metal clanking; they use radio sets in their suits.
• A movie showing a loud bang when a spacecraft explodes in space is wrong (the flash of light is fine, as light needs no medium).

Worked example

In the bell-jar experiment, why does the sound not become completely silent even when the pump runs for a long time? Why can we still see the bell?

Try it first, then show the solutionSolution

1. A school pump cannot remove all the air; a little remains, and the bell's support also touches the jar, so a faint sound still travels.
2. Light does not need a medium, so we see the bell through the near-vacuum.

Answer: Some air and the solid support still carry faint sound; light needs no medium.

Sound as a longitudinal wave: compressions and rarefactions

NCERT section 10.3 Sound Waves

Slinky (Activity 10.5): push and pull one end: regions where turns are close together and regions where they are spread out travel along the slinky. A marked turn only moves back and forth about its place.

Air in a tube with an oscillating piston:

• Piston moves forward → pushes air together → compression (C), a region of higher density (and pressure).
• Piston moves back → air spreads out → rarefaction (R), a region of lower density.
• Collisions pass these on, so a series of C and R travels away from the source.

A sound wave is a series of compressions and rarefactions moving through a medium without the medium itself flowing. The particles only vibrate about their mean positions; it is energy that travels. From a small source, sound spreads in all directions as spherical waves.

Types of waves:

• Longitudinal: particles vibrate parallel to the direction the wave travels (sound).
• Transverse: particles vibrate perpendicular to the direction of travel (ripples on water, a shaken rope; light is a transverse wave that needs no medium).
• Mechanical waves need a medium; sound is one.

Sudden loud sounds (crackers, thunder) come from air heated and expanding very fast; a supersonic aircraft makes a sonic boom.

Worked example

Assertion: compressions and rarefactions move through the air from a speaker to your ear. A student says this means air from the speaker reaches your ear. Correct the student.

Try it first, then show the solutionSolution

1. Each air particle only vibrates back and forth about its own mean position.
2. It bumps its neighbour, passing on the compression or rarefaction.
3. So the disturbance and its energy travel to your ear, not the air particles near the speaker.

Answer: The pattern of density and the energy travel; the air particles do not move along with the wave.

Sound carries energy; graph of a sound wave

NCERT section 10.5 Graphical Representation of a Sound Wave

Activity 10.6: grains sprinkled on a rubber sheet stretched over a bowl jump when a loud sound is made nearby, without touching the bowl. Louder sound makes them jump higher. So sound is a form of energy: the vibrating source passes energy to the medium, and particles pass it on by collisions.

• A microphone turns sound energy into an electrical signal (sound vibrates a thin diaphragm).
• A speaker does the reverse: an electrical signal vibrates a cone, which makes sound.

Graph of a sound wave: plot density of the medium (y-axis) against distance from the source (x-axis) at one instant. The curve goes above the average density (dashed line) at compressions and below it at rarefactions.

• Highest point (maximum density, centre of a compression): crest.
• Lowest point (minimum density, centre of a rarefaction): trough.

One can also plot density against time at a fixed place.

Worked example

Describe a density-distance graph for a sound wave of wavelength 6 cm and density amplitude 2 units, starting at average density at 0 cm and rising. Where are the first two crests and troughs?

Try it first, then show the solutionSolution

1. Draw a dashed horizontal line for average density; mark distance in cm on the x-axis.
2. Draw a smooth wave rising 2 units above and falling 2 units below the line, repeating every 6 cm.
3. A crest comes a quarter wavelength (1.5 cm) after the start, a trough half a wavelength later: crests at 1.5 and 7.5 cm, troughs at 4.5 and 10.5 cm.

Answer: A wave 2 units either side of average; crests at 1.5 and 7.5 cm, troughs at 4.5 and 10.5 cm (6 cm apart).

Wavelength, frequency and time period

NCERT section 10.6 Characteristics of a Sound Wave

• Wavelength (λ): distance between two consecutive crests (or two consecutive troughs), i.e. one compression + one rarefaction. SI unit: metre.
• One oscillation: at a fixed point, density goes from maximum to minimum and back to maximum.
• Frequency (ν): number of oscillations per second at a fixed point. SI unit: hertz (Hz) = s${}^{-1}$.
• Time period (T): time for one complete oscillation. SI unit: second.

$\nu =\frac{1}{T}$

A higher frequency means a shorter time period.

Activity 10.7: an app such as Phyphox shows the frequency of sung notes Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa rising step by step. Thinner or tighter rubber bands vibrate faster and give a higher frequency.

Most everyday sounds mix many frequencies; a tuning fork or whistle gives almost one frequency (a tone). A musical note has a lowest frequency (fundamental) plus overtones; their pattern gives each instrument its timbre (why a flute and a tabla sound different at the same pitch and loudness). An octave is a doubling of frequency.

Worked example

At a point, 30 density oscillations occur in 3 s. Find the frequency and time period. How many oscillations would a 25 Hz source complete in one minute?

Try it first, then show the solutionSolution

1. $\nu =\frac{30}{3}=10$ Hz.
2. $T=\frac{1}{\nu }=\frac{1}{10}=0.1$ s.
3. In 60 s at 25 Hz: $25\times 60=1500$ oscillations.

Answer: 10 Hz, 0.1 s; 1500 oscillations.

Amplitude, intensity and speed of sound

NCERT section 10.6 Characteristics of a Sound Wave

Amplitude: the largest change in density (in a compression or rarefaction) from the average density. A larger amplitude carries more energy (grains jump higher when the plate is struck harder).

Intensity: the sound energy passing through unit area (held perpendicular to the direction of travel) in unit time. As sound spreads over a bigger area away from the source, intensity decreases with distance.

Speed (v): distance a crest moves per second. In one time period the wave moves one wavelength:

$v=\frac{\lambda }{T}=\lambda \nu$ (speed = wavelength × frequency)

Medium (at 15 °C)	Approximate speed
Steel	5000 m s${}^{-1}$
Water	1500 m s${}^{-1}$
Air	340 m s${}^{-1}$

• Fastest in solids, then liquids, slowest in gases (about 4 to 5 times faster in water and 15 to 20 times faster in solids than in air).
• In air, speed rises with temperature and humidity: about 331 m s${}^{-1}$ at 0 °C and 344 m s${}^{-1}$ at 22 °C.
• In a given medium, speed does not depend on frequency; a higher frequency simply has a shorter wavelength.

Lightning and thunder: light arrives almost at once, sound later. Distance = speed of sound × delay.

Worked example

A sound of 680 Hz travels in air at 340 m s${}^{-1}$. Find its wavelength. If it enters water (1500 m s${}^{-1}$), what are its frequency and wavelength there?

Try it first, then show the solutionSolution

1. $\lambda =\frac{v}{\nu }=\frac{340}{680}=0.5$ m.
2. Frequency is set by the source, so it stays 680 Hz in water.
3. ${\lambda }_{water}=\frac{1500}{680}\approx 2.2$ m.

Answer: 0.5 m in air; 680 Hz and about 2.2 m in water.

Pitch, loudness and the range of hearing

NCERT section 10.6 Characteristics of a Sound Wave

Physical properties (frequency, amplitude) can be measured; how we perceive them is personal.

• Pitch: how we perceive frequency. Shrill sounds (whistle, siren, a mosquito) have high pitch (high frequency); deep sounds (thunder, a lion's roar) have low pitch. Boys' voices deepen in adolescence as vocal cords lengthen and thicken and vibrate more slowly.
• Loudness: how we perceive amplitude (and intensity). Larger amplitude sounds louder; loudness falls with distance. Loudness is measured in decibels (dB): rustling leaves a few dB, conversation about 60 dB, crackers over 100 dB.

Audible range: about 20 Hz to 20 000 Hz (20 kHz); it narrows with age.

• Below 20 Hz: infrasonic (elephants hear it; earthquakes and volcanoes produce it).
• Above 20 kHz: ultrasonic (dogs, cats, bats and dolphins hear it).

Noise is unwanted or harmful sound. Long exposure above safe limits harms sleep, health and hearing; hearing is tested with audiograms, and hearing aids (microphone, amplifier, speaker) help.

The ear: sound vibrates the eardrum; tiny bones amplify the vibration; the cochlea turns it into electrical signals for the brain. Two ears let the brain judge direction from the tiny time difference.

Worked example

Using 340 m s${}^{-1}$, find the wavelengths in air of the lowest and highest frequencies humans can hear.

Try it first, then show the solutionSolution

1. 20 Hz: $\lambda =\frac{340}{20}=17$ m.
2. 20 000 Hz: $\lambda =\frac{340}{20000}=0.017$ m $=1.7$ cm.

Answer: 17 m and 1.7 cm.

Reflection of sound: echo and reverberation

NCERT section 10.7 Reflection of Sound

Sound reflects from solid or liquid surfaces and obeys the same laws as light: the incident and reflected directions make equal angles with the normal, and all three lie in one plane.

Echo: a reflected sound heard separately from the original.

• Our ear keeps a sound for about 0.1 s, so the echo must arrive at least 0.1 s later.
• At 340 m s${}^{-1}$, sound goes $340\times 0.1=34$ m in 0.1 s: there and back, so the reflector must be at least 17 m away.
• Distance to a reflector $=\frac{v\times t}{2}$ (sound goes there and back).
• Hard, smooth surfaces give strong echoes; soft surfaces (curtains) absorb sound; rough ones scatter it.

Reverberation: repeated reflections in a big hall that make sound linger after the source stops; reflections arrive within about 0.05 s of each other and blur together. Too much makes speech unclear. Halls use sound-absorbing panels, upholstered seats, curtains and carpets. Curved ceilings and stage walls direct sound to the audience.

The Whispering Gallery of Gol Gumbaz in Bijapur (Vijayapura), Karnataka, lets a whisper be heard many times across the dome.

Worked example

You clap in front of a cliff and hear the echo after 0.8 s (speed of sound 340 m s${}^{-1}$). How far away is the cliff? What is the least distance at which you could hear an echo?

Try it first, then show the solutionSolution

1. Distance $=\frac{vt}{2}=\frac{340\times 0.8}{2}=136$ m.
2. Least distance $=\frac{340\times 0.1}{2}=17$ m.

Answer: 136 m; 17 m.

Ultrasound, infrasound, echolocation and sonar

NCERT section 10.8 Ultrasonic and Infrasonic Waves, and their Applications

Uses of ultrasonic waves (above 20 kHz):

• Ultrasonography: imaging organs and a baby in the womb without surgery.
• Breaking kidney stones into small pieces.
• Cleaning delicate machine parts and ultrasonic welding.
• Finding cracks and defects inside metal blocks.
• Locating objects by reflection.

Uses of infrasonic waves (below 20 Hz): detecting earthquakes, volcanic eruptions and severe storms, because they travel far through air and the ground.

Echolocation: locating objects by reflected sound. Most bats send out short bursts of ultrasound and use the echoes from prey and obstacles; dolphins, whales and some birds do this too.

SONAR (sound navigation and ranging): a ship sends ultrasonic pulses into water and times the echo to find the distance, direction and speed of submarines, shoals of fish, shipwrecks or the sea bed:

$\text{depth}=\frac{v\times t}{2}$

Car parking sensors use the same idea in air (about 40 kHz). Audio surveillance detects drones from their low humming. Sir C. V. Raman also studied the acoustics of the tabla and mridangam; the black *syaahi* patch on the drum head gives them their rich tones.

Worked example

A ship's sonar pulse returns from the sea bed after 3 s. If sound travels at 1500 m s${}^{-1}$ in seawater, how deep is the sea there?

Try it first, then show the solutionSolution

1. The pulse goes down and comes back, so one-way time $=\frac{3}{2}=1.5$ s.
2. Depth $=1500\times 1.5=2250$ m.

Answer: 2250 m.

Science in real life: Sound Waves: Characteristics and Applications

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Echo, Reverberation, Sonar and Ultrasound
• Wavelength, Frequency, Speed, Pitch and Loudness of Sound

Sound Waves: Characteristics and Applications: formulas and key facts

Frequency and time period

$\nu =\frac{1}{T}$; $\nu =\frac{\text{number of oscillations}}{\text{time}}$

Wave equation

$v=\lambda \nu =\frac{\lambda }{T}$

Echo or sonar distance

$d=\frac{vt}{2}$ (t is the time for the round trip)

Minimum echo distance

$d_{min}=\frac{v\times 0.1}{2}$ ≈ 17 m in air at 340 m s⁻¹

Lightning distance

$Distance=speedofsound\times timebetweenflashandthunder.$

Speeds of sound

In air	about 340 m s⁻¹ (331 at 0 °C, 344 at 22 °C)
In water	about 1500 m s⁻¹
In steel	about 5000 m s⁻¹

Hearing range

20 Hz to 20 kHz; below 20 Hz infrasonic; above 20 kHz ultrasonic.

Perception

Pitch ↔ frequency; loudness ↔ amplitude; timbre ↔ pattern of overtones.

Units

Wavelength m; frequency Hz (s⁻¹); time period s; speed m s⁻¹; loudness dB.

Reverberation

Reflections arriving within about 0.05 s blend; echo needs at least 0.1 s.

Sound Waves: Characteristics and Applications: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Sound Waves: Characteristics and Applications

The question patterns to expect, and the method that works for each.

Tricks to remember Sound Waves: Characteristics and Applications

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Vacuum is voiceless

  No particles, no sound; light still gets through.

  Use it when: Medium questions.

• Very Lazy Nightingale

  v = λ × ν: Velocity = Lambda × Nu.

  Use it when: The wave equation.

• Echo halves

  Sound goes there and back, so halve the time (or the distance).

  Use it when: Echo and sonar numericals.

• Pitch–frequency, Loud–amplitude

  P and F are 'Pitch Follows Frequency'; L and A are 'Loudness Arises from Amplitude'.

  Use it when: Perception questions.

• Steel, Water, Air: 5000, 1500, 340

  Solids are fastest, gases slowest.

  Use it when: Speed comparisons.

Sound Waves: Characteristics and Applications: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Sound Waves: Characteristics and Applications

The slips examiners see most often, and what to write instead.

• ✗Mistake: Forgetting to halve the echo time

  Students use the full time as one-way travel.

  Wrong working

  Echo after 0.8 s: $d=340\times 0.8=272$ m.

  Right working

  $d=\frac{340\times 0.8}{2}=136$ m.

  ✓Do this: Round trip → divide by 2.

  Usually costs: The answer mark.

• ✗Mistake: Saying air particles travel from the source to the ear

  We 'feel' sound coming at us.

  Wrong working

  Sound reaches us because air near the tuning fork moves to our ear.

  Right working

  Particles only vibrate about mean positions; the compressions and rarefactions and the energy travel.

  ✓Do this: Remember the marked turn on the slinky.

  Usually costs: 1 mark.

• ✗Mistake: Thinking frequency changes speed

  Higher notes 'sound faster'.

  Wrong working

  A 1000 Hz sound travels faster in air than a 500 Hz sound.

  Right working

  In the same medium both travel at the same speed; the 1000 Hz sound has half the wavelength.

  ✓Do this: Speed depends on the medium (and temperature), not the frequency.

  Usually costs: 1 mark in MCQs.

• ✗Mistake: Confusing pitch with loudness

  Both are 'how a sound feels'.

  Wrong working

  A loud sound has a high pitch.

  Right working

  Loudness depends on amplitude; pitch depends on frequency. A lion's roar is loud but low-pitched.

  ✓Do this: Pitch–frequency, loudness–amplitude.

  Usually costs: 1 mark.

• ✗Mistake: Measuring amplitude from trough to crest

  It looks like the 'size' of the wave.

  Wrong working

  Amplitude = vertical distance from trough to crest.

  Right working

  Amplitude = maximum change from the average density (half the trough-to-crest height).

  ✓Do this: Draw the average line first.

  Usually costs: 1 mark in graph questions.

NCERT exercise guide: Sound Waves: Characteristics and Applications

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Sound Waves: Characteristics and Applications

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   Sound cannot travel through
   (a) steel
   (b) water
   (c) vacuum
   (d) air

   Hint

   No particles.

   Answer

   (c) vacuum

2. Q2 · 1 mark · MCQExam level

   Halving the frequency of a sound in the same medium will
   (a) halve its speed
   (b) double its wavelength
   (c) double its speed
   (d) halve its time period

   Hint

   $v=\lambda \nu$ with $v$ fixed.

   Answer

   (b) double its wavelength (speed is unchanged; the time period doubles).

3. Q3 · 1 mark · Assertion-reasonExam level

   Assertion (A): Astronauts on a spacewalk talk to each other by radio.
   Reason (R): Radio waves, like light, do not need a material medium.
   (a) A and R are both true, and R explains A correctly.
   (b) A and R are both true, but R does not explain A.
   (c) A is true, R is false.
   (d) A is false, R is true.

   Hint

   Which waves cross a vacuum?

   Answer

   (a)

4. Q4 · 2 marks · Very short answerExam level

   Define wavelength and time period of a sound wave and give their SI units.

   Hint

   Crest to crest; one oscillation.

   Answer

   Wavelength: distance between two consecutive crests (or troughs), in metres. Time period: time for one complete density oscillation at a point, in seconds.

5. Q5 · 2 marks · Very short answerHarder exam level

   A tuning fork of 512 Hz is sounded in air at 344 m s⁻¹. Find the wavelength and the time period.

   Hint

   $\lambda =\frac{v}{\nu }$.

   Answer

   $\lambda =\frac{344}{512}\approx 0.67$ m; $T=\frac{1}{512}\approx 0.00195$ s.

6. Q6 · 2 marks · Very short answerHarder exam level

   In a cartoon, an astronaut on the Moon hears a boulder crash on the far side of a crater. Is this possible? Could he sense the crash if he were kneeling with his hands on the ground?

   Hint

   Medium: none above the ground, but the ground is solid.

   Answer

   He cannot hear it through the Moon's surroundings, which are a near vacuum with no air to carry sound. Through the solid ground, however, the vibrations could travel and be felt (or reach his suit through contact), because solids carry sound.

7. Q7 · 3 marks · Short answerHarder exam level

   Distinguish between echo and reverberation. How are reverberations reduced in a cinema hall?

   Hint

   Time gap; soft materials.

   Answer

   Echo: a single reflection heard separately, at least 0.1 s after the original (reflector at least about 17 m away). Reverberation: many reflections arriving close together (within about 0.05 s), making sound linger. Reduced by sound-absorbing panels, curtains, carpets and upholstered seats, and by shaping walls and ceilings.

8. Q8 · 3 marks · Short answerHarder exam level

   Explain how a bat catches a moth in total darkness. Name the waves used and the property of sound involved.

   Hint

   Echolocation.

   Answer

   The bat sends short bursts of ultrasonic waves. They reflect from the moth and return; from the time delay and direction of the echoes the bat judges the moth's position and closes in. This is echolocation, using reflection of sound.

9. Q9 · 3 marks · Short answerHarder exam level

   Compare the speeds of sound in steel, water and air by finding (i) water : air (ii) steel : water, using 5000, 1500 and 340 m s⁻¹. Why is sound faster in solids?

   Hint

   Divide.

   Answer

   (i) $\frac{1500}{340}\approx 4.4$ (ii) $\frac{5000}{1500}\approx 3.3$. In solids, particles are packed closely and strongly linked, so a disturbance is passed on very quickly.

10. Q10 · 3 marks · Short answerHarder exam level

    List four uses of ultrasound and one use of infrasound.

    Hint

    Medicine, industry, sea.

    Answer

    Ultrasound: imaging organs (ultrasonography), breaking kidney stones, cleaning delicate parts, detecting cracks in metal, sonar, parking sensors. Infrasound: detecting earthquakes, volcanic eruptions or severe storms.

11. Q11 · 3 marks · Short answerHOTS / competency

    A child shouts towards a cliff 85 m away on a day when sound travels at 340 m s⁻¹. After 3 s the child shouts again. Does the second shout overlap the first echo? Explain with calculation.

    Hint

    Echo time.

    Answer

    Echo time $=\frac{2\times 85}{340}=0.5$ s, so the echo of the first shout returns at 0.5 s, well before the second shout at 3 s. No overlap; the echo is heard clearly.

12. Q12 · 3 marks · Short answerHOTS / competency

    Why can the same note on a flute and a tabla sound different even at the same loudness? Explain with the terms fundamental, overtones and timbre.

    Hint

    Mixture of frequencies.

    Answer

    Each instrument's note has the same lowest frequency (fundamental) but a different mix and strength of higher frequencies (overtones), set by its shape, material and construction. This pattern is the timbre, which lets us tell the instruments apart.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What produces sound?Show answer

   Vibrating objects.

2. 2.Which part of the human body produces the voice?Show answer

   The vocal cords in the larynx (voice box).

3. 3.Define a compression and a rarefaction.Show answer

   Compression: region of higher density than average; rarefaction: region of lower density than average.

4. 4.What is a longitudinal wave?Show answer

   A wave in which particles of the medium vibrate parallel to the direction in which the wave travels.

5. 5.What is the SI unit of frequency?Show answer

   The hertz (Hz), equal to one oscillation per second.

6. 6.Write the relation between speed, wavelength and frequency.Show answer

   $v=\lambda \nu$.

7. 7.In which medium is sound fastest: steel, water or air?Show answer

   Steel.

8. 8.What is the audible range of humans?Show answer

   About 20 Hz to 20 000 Hz.

9. 9.What minimum time gap is needed to hear an echo?Show answer

   About 0.1 s.

10. 10.What does SONAR stand for?Show answer

    Sound navigation and ranging.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you describe two activities showing sound comes from vibration?
• ☐Can you explain the bell-jar experiment and the astronaut puzzle?
• ☐Can you draw compressions and rarefactions and label λ?
• ☐Can you explain the difference between longitudinal and transverse waves?
• ☐Can you mark crest, trough, λ and amplitude on a density graph?
• ☐Can you use $v=\lambda \nu$ and $T=\frac{1}{\nu }$?
• ☐Can you rank speeds in steel, water and air and say how temperature matters?
• ☐Can you link pitch and loudness to physical properties?
• ☐Can you solve echo and sonar problems with the factor of 2?
• ☐Can you list three uses each of ultrasound and infrasound?

Sound Waves: Characteristics and Applications: frequently asked questions