NCERT Solutions for Class 9 Science Chapter 10 Sound Waves Characteristics and Applications – step-by-step answers to every exercise and in-text question of the chapter, as per the latest NCERT syllabus for 2027. Read them online or download the free question-answer PDF.
A plucked rubber band, a struck tuning fork and a slinky pushed and pulled at one end all say the same thing: sound is a disturbance in the density of a medium, not a flow of the medium itself. This chapter builds that idea into measurable quantities — wavelength, frequency, time period, amplitude and speed — and then uses them on echoes, sonar and ultrasound.
NCERT Solutions for Class 9 Science Chapter 10 Sound Waves Characteristics and Applications PDF
All the answers are on this page below. The PDF also explains every answer – why it is right and how to check it – so download it for revision.
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|---|---|---|
| Solutions PDF | Sound Waves Characteristics and Applications | View PDF |
NCERT Solutions for Class 9 Science Chapter 10 Sound Waves Characteristics and Applications – Questions and Answers
All 63 questions of the chapter (textbook pages 184 – 207) are answered below, in the order of the book – in-text questions, activities and the end-of-chapter exercises. Each answer is written the way you would write it in the exam.
Think It Over — Page 184Chapter opener
No — not directly. Outer space is a near vacuum, and sound needs a material medium to travel. So the astronauts cannot hear each other speak or hear the clanking of metal as they would on the Earth. They talk through special radio devices fitted in their spacesuits.
Most bats send out short bursts of ultrasonic waves (frequency above 20 kHz), and then listen for the echoes that bounce back from objects and prey. This is called echolocation.
- The bat emits a pulse of ultrasound.
- The pulse is reflected by an insect, a wall or a branch.
- The bat’s ears pick up the returning echo.
- From the time delay the bat judges how far the object is; from the direction and strength of the echo it judges where the object is and how big it is.
In-text Questions — Page 184Chapter introduction
Mechanical energy — specifically the kinetic energy of a vibrating object — is converted into sound energy.
- When you pluck a rubber band or a sitar string, your muscles do work on it. That work becomes the kinetic energy of the vibrating string, which is then handed on to the air as sound.
- In a loudspeaker the chain is different: electrical energy → mechanical (vibration of the diaphragm) → sound.
- In a firecracker it is chemical energy → heat → rapid expansion of gas → sound.
Sound is produced by a vibrating object and reaches the ear as a travelling disturbance in the density of a medium.
- Production. A source (vocal cords, a tuning fork, a drum membrane, an air column in a bansuri) vibrates — moves to and fro about a mean position.
- Propagation. Each forward push of the source squeezes the nearby air particles together and makes a compression (density above average). Each backward movement leaves a rarefaction (density below average). Particles collide with their neighbours and pass the squeeze forward, so the compression travels on while each particle only oscillates about its own mean position.
- Detection. The arriving compressions and rarefactions push and release the eardrum, making it vibrate. Tiny bones amplify the vibration, the cochlea turns it into electrical signals, and the brain reads them as sound.
Activity 10.1 — Page 185Let us explore — 10.1 Production of Sound
Yes, a clear twang is heard.
Plucking pulls the stretched band away from its mean position. When you let go, the tension pulls it back, it overshoots, and it keeps moving to and fro — it vibrates. The vibrating band pushes the air next to it forwards and backwards, sending compressions and rarefactions towards your ear.
Yes. Just after plucking, the band looks blurred — you see a widened, fuzzy band instead of a sharp line, because it is moving to and fro faster than your eye can follow.
No. The moment the band comes to rest, the sound stops.
This one observation is the whole conclusion of the activity: sound is produced by vibrations; no vibration, no sound.
Yes, the sound changes. Stretch the band tighter and the note becomes shriller (higher pitch); loosen it and the note becomes deeper (lower pitch).
| What you do | How the band vibrates | Frequency ν | Time period T | What you hear |
|---|---|---|---|---|
| Stretch it more (higher tension) | Snaps back faster | higher | shorter | shriller, high pitch |
| Loosen it (lower tension) | Returns sluggishly | lower | longer | deeper, low pitch |
| Pluck harder | Larger swing about the mean position | unchanged | unchanged | louder (larger amplitude) |
Sound is still produced, but it is much fainter than before. The band still vibrates, so it still makes sound. But on its own it sets very little air vibrating; on the box, the box and the air inside it also vibrated, so much more air was disturbed and the sound was louder.
In-text Questions — Page 18510.1 Production of Sound
Yes — you feel a distinct buzzing in the front of the throat, at the voice box (larynx).
In humans and many animals, sound is produced by the vibration of the vocal cords — two tightly stretched muscular flaps inside the larynx. Air pushed up from the lungs forces its way between them and sets them vibrating, and those vibrations become the sound of your voice.
- The tongue, lips, mouth and nasal cavity then shape that raw buzz into speech or music.
- Animals do it in other ways too: grasshoppers and crickets rub their wings or legs together, so the rubbing surfaces are the vibrating source.
Activity 10.2 — Page 186Let us explore — 10.1.1 Tuning fork
Yes — a steady, pure humming note.
The blow sets the two prongs vibrating towards and away from each other. Each outward swing compresses the air near the prong and each inward swing rarefies it, so a train of compressions and rarefactions travels to your ear.
Yes — ripples spread out on the water from the touching prong (a fine spray may even be thrown up). The ripples show that the prongs are vibrating, which supports the idea that sound is produced by vibrating objects.
Yes — the sound is heard in every orientation, though its loudness changes a little as the fork is turned. This shows that sound spreads out from its source in many directions, not along one line only.
Pause and Ponder — Page 18610.1.1 Tuning fork
Every method comes down to one thing — making something vibrate. The ways differ only in what is set vibrating.
| Way of producing sound | What vibrates | Everyday example |
|---|---|---|
| Plucking | A stretched string | Sitar, veena, guitar, ektara, rubber band |
| Striking / hitting | A stretched membrane or a metal body | Tabla, dholak, taal (manjira), temple bell, tuning fork |
| Blowing | The air column inside a pipe | Bansuri, shehnai, conch (shankh), whistle |
| Bowing | A string, rubbed by a bow | Sarangi, violin |
| Rubbing / scraping | The two rubbing surfaces | Crickets and grasshoppers rubbing wings or legs; chalk squeaking on a board |
| Forcing air past flaps | Vocal cords | Speaking, singing, animal calls |
| Sudden expansion of hot gas | The air itself | Firecracker, thunder, bursting balloon, sonic boom |
| Electrical signal driving a cone | Speaker diaphragm | Mobile phone, radio, loudspeaker |
| Instrument | Family | Vibrating part that makes the sound | How it is set vibrating |
|---|---|---|---|
| Sitar, veena, sarod, ektara | String (tat vadya) | Stretched metal strings; the hollow gourd/body then amplifies | Plucked |
| Sarangi, violin | String | Stretched strings | Bowed (rubbed) |
| Tabla, dholak, mridangam, damru | Membrane (avanaddha vadya) | Stretched skin/membrane of the drum head, plus the air inside the shell | Struck |
| Bansuri, shehnai, whistle | Wind (sushir vadya) | The air column inside the hollow pipe | Blown |
| Harmonium | Wind | Thin metal reeds | Air pushed past by the bellows |
| Manjira (taal), ghanti, ghatam | Solid body (ghana vadya) | The metal or clay body itself | Struck |
| Tuning fork | Solid body | The two prongs | Struck on a soft pad |
In-text Questions — Page 18610.2 Propagation of Sound
Sound reaches the ear through a material medium, and yes — it travels through solids and liquids as well as gases.
The source sets the particles next to it oscillating. Those particles collide with their neighbours and pass on the compression; the neighbours pass it on further; and so the disturbance is relayed, particle to particle, all the way to the eardrum. The material that does this relaying is called the medium.
- Solid: Activity 10.3 — a scratch on a desk is heard much more clearly with the ear pressed on the desk than through the air.
- Liquid: Activity 10.4 — two spoons tapped under water are still heard.
- Gas: ordinary hearing through air.
Activity 10.3 — Pages 186 – 187Let us investigate — 10.2 Propagation of Sound
Yes — and the knock or scratch sounds distinctly louder and sharper through the desk than it did through the air.
The conclusion: sound can propagate through solids. Closing the other ear matters — it shuts out the weaker airborne sound so that you are judging only the sound that came through the desk.
Activity 10.4 — Page 187Let us investigate — 10.2 Propagation of Sound
Yes, the clink is clearly heard — a little duller and different in quality from the clink in air, but unmistakably there.
The tap sets the spoons vibrating. The vibrating metal squeezes and releases the water around it, so compressions and rarefactions travel out through the water, cross the water surface into the air, and reach your ear.
No. With the spoons held clear of the bucket, water is the only medium touching them.
If sound could not propagate through liquids, the first link of that chain would be broken. The vibration would die out at the surface of the metal, no compression would ever leave the spoons, and you would hear nothing at all.
In-text Questions — Page 18710.2 Propagation of Sound
No. Sound cannot be heard in a vacuum, because a sound wave has no way to travel without particles.
The proof is the vacuum bell jar experiment (Fig. 10.7):
- An electric bell inside a sealed bell jar is switched on — the ringing is loud.
- A vacuum pump slowly removes the air. The sound gets fainter and fainter.
- Near vacuum: the hammer can still be seen striking the gong, but almost no sound is heard.
- Let the air back in and the sound returns, growing to its original loudness.
Pause and Ponder — Page 18810.2.1 Sound needs a medium to propagate
Reason (R): Sound requires a medium to travel.
Choose the correct statement:
(i) Both A and R are true, but R is not the correct explanation of A.
(ii) Both A and R are true, and R is the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
(ii) Both A and R are true, and R is the correct explanation of A.
A is true — the bell is seen ringing but is hardly heard. R is true — sound needs a medium to travel. And R explains A: pumping out the air removes the medium that carries the sound to the jar wall and on to our ears.
Activity 10.5 — Page 188Let us observe — 10.3 Sound Waves
Yes. A single bunched-up region — a place where the turns are pressed close together — races along the slinky towards your friend.
This is a perfect model of a single compression in a sound wave: a push at the source, a crowding relayed forward by collisions, and no bulk flow of the medium.
Yes to all three.
- A series of disturbances is produced. Regions where the turns are close together alternate with regions where they are more spread out.
- They travel along the slinky, one after another, from your end to your friend’s end.
- The marked turn does not travel. It only oscillates to and fro about its rest position, and it does so parallel to the direction in which the disturbance moves.
In-text Questions — Page 19010.3 Sound Waves
No. Without a material medium, sound waves cannot propagate.
A sound wave is a pattern of higher and lower particle density. If there are no particles, there is no density to raise or lower, and nothing to collide and hand the disturbance forward. Waves that need a material medium in this way are called mechanical waves, and sound is one of them.
Pause and Ponder — Page 19110.3 Sound Waves
Reason (R): Individual particles of the medium continuously move forward with the wave.
Choose the correct statement:
(i) Both A and R are true, but R is not the correct explanation of A.
(ii) Both A and R are true, and R is the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
(iii) A is true, but R is false.
- A is true. Compressions and rarefactions really do travel forward through the medium — that is what a sound wave is.
- R is false. The individual particles do not move forward with the wave. Each particle only oscillates about its own mean position, parallel to the direction of propagation, and returns there once the wave has passed.
Activity 10.6 — Pages 191 – 192Let us experiment — 10.4 Energy of Sound Waves
Yes — the grains jump, dance and shuffle across the stretched sheet, even though nothing has touched the container.
Since the grains have been lifted against gravity, work has been done on them. That work could only have come from the sound, so the activity proves that sound carries energy. When the source vibrates, it transfers energy to the medium; the vibrating particles of the medium collide with others, and the energy is relayed onward.
The louder the sound, the higher and more violently the grains jump. With a soft sound they barely quiver; with a very loud one they leap right off the sheet.
| What you change | What you observe | Reason |
|---|---|---|
| Strike the plate harder (louder sound) | Grains jump much higher | Larger amplitude → more energy carried by the wave → the sheet is displaced further |
| Move the source further away | Grains move less | The same energy is spread over a larger area, so the intensity at the sheet falls |
| Use light grains (semolina, chalk powder) instead of heavy ones (salt, rice) | Lighter grains jump higher and more easily | Same energy delivered to a smaller mass gives a larger displacement |
| Loosen the stretched sheet | Response becomes weaker and duller | A slack sheet does not follow the rapid pushes of the sound wave faithfully |
Pause and Ponder — Page 19210.4 Energy of Sound Waves
(i) Air particles near the tuning fork
(ii) Energy carried by sound waves
(iii) The tuning fork material
(iv) A continuous stream of compressed air
(ii) Energy carried by sound waves.
The air particles that were beside the prongs stay beside the prongs. Each one merely oscillates about its mean position and nudges its neighbour; the neighbour nudges the next, and so on. What travels the whole way from the fork to your eardrum is the disturbance, and with it the energy.
Pause and Ponder — Page 19310.5 Graphical Representation of a Sound Wave
Labelling (a): the dots are crowded at the 1st, 3rd and 5th dashed lines — mark C there; they are spread out at the 2nd, 4th and 6th dashed lines — mark R there.
Labelling (b): the pattern is shifted by half a wavelength — R at the 1st, 3rd and 5th dashed lines and C at the 2nd, 4th and 6th.
Graphs (c) and (d): label the x-axis Distance and the y-axis Density; the dashed horizontal line is the average density. Draw a smooth curve with a crest at every C and a trough at every R, crossing the average-density line at every solid line.
Activity 10.7 — Page 194Let us experiment (demonstration activity) — 10.6.1 Wavelength, frequency and time period
Yes — the frequencies are not spaced equally, but their ratios to Sa are simple fractions, and the upper Sa comes out at exactly twice the lower Sa.
| Note | Sample frequency | Ratio to Sa | Simple fraction |
|---|---|---|---|
| Sa | 240 Hz | 1.00 | 1 |
| Re | 270 Hz | 1.13 | 9/8 |
| Ga | 300 Hz | 1.25 | 5/4 |
| Ma | 320 Hz | 1.33 | 4/3 |
| Pa | 360 Hz | 1.50 | 3/2 |
| Dha | 400 Hz | 1.67 | 5/3 |
| Ni | 450 Hz | 1.88 | 15/8 |
| Sa (upper) | 480 Hz | 2.00 | 2 |
Ratio for upper Sa = 480 Hz ÷ 240 Hz = 2.0
For the same named note the two frequencies come out close but usually not identical, and the ratios between notes still follow the same pattern.
- The app produces an almost pure single frequency, so its reading is steady and sharp on the spectrum.
- Your voice gives a fundamental frequency plus overtones, so the spectrum shows one strong low peak with several weaker higher peaks. The fundamental is the one to compare.
- The voice reading also wobbles by a few hertz because it is hard to hold a note perfectly steady.
Pause and Ponder — Page 19510.6.1 Wavelength, frequency and time period
Yes — for the same length and tension, the thin band vibrates faster than the thick one. So the sound of the thin band has a higher frequency and a shorter time period (T = 1/ν), and it sounds shriller.
| Band | Rate of vibration | Frequency ν | Time period T = 1/ν | Sound heard |
|---|---|---|---|---|
| Thin band | Faster | Higher | Shorter | Shriller — high pitch |
| Thick band | Slower | Lower | Longer | Deeper — low pitch |
1200 oscillations per minute.
∴ number of oscillations = ν × time
ν = 20 Hz = 20 s⁻¹, time = 1 minute = 60 s
number of oscillations = 20 s⁻¹ × 60 s = 1200
Half the wavelength is 1.5 cm (= 0.015 m). In Fig. 10.19 the curve completes one full density oscillation (one crest and one trough) between 0 and 3.0 cm.
From Fig. 10.19: λ = 3.0 cm
∴ λ/2 = 3.0 cm ÷ 2 = 1.5 cm = 0.015 m
What if … — Page 19610.6.3 Speed of Sound
No — music would fall apart as it travelled, and the further you sat from the stage the worse it would sound.
A singer’s voice and each instrument give out many frequencies together — every note is a fundamental plus overtones. If each frequency travelled at a different speed, they would reach the listener at different times. The notes would get smeared, the singer and the instruments would drift out of step, and listeners at different distances would hear the performance differently. Music sounds pleasant because, in air, the speed of sound does not depend on frequency.
Pause and Ponder — Page 19710.6.3 Speed of Sound
| State | Substance/Medium | Approximate speed |
|---|---|---|
| Solid | Steel | 5000 m s⁻¹ |
| Liquid | Water | 1500 m s⁻¹ |
| Gas | Air | 340 m s⁻¹ |
(i) the speed of sound in water with respect to the speed in the air.
(ii) the speed of sound in steel with respect to the speed in the water.
(i) about 4.4 : 1 (ii) about 3.3 : 1
= 1500 m s⁻¹ ÷ 340 m s⁻¹
= 4.41 ≈ 4.4 → vwater : vair ≈ 4.4 : 1
(ii) ratio = vsteel / vwater
= 5000 m s⁻¹ ÷ 1500 m s⁻¹
= 3.33 ≈ 3.3 → vsteel : vwater ≈ 3.3 : 1
Pause and Ponder — Page 19810.6.3 Speed of Sound
| State | Substance/Medium | Approximate speed |
|---|---|---|
| Solid | Steel | 5000 m s⁻¹ |
| Liquid | Water | 1500 m s⁻¹ |
| Gas | Air | 340 m s⁻¹ |
The time difference is about 0.93 s, and yes — Gunjan can easily hear the two knocks separately.
Through steel (v = 5000 m s⁻¹):
tsteel = 340 m ÷ 5000 m s⁻¹ = 0.068 s
Through air (v = 340 m s⁻¹):
tair = 340 m ÷ 340 m s⁻¹ = 1.000 s
Time difference Δt = tair − tsteel
= 1.000 s − 0.068 s = 0.932 s ≈ 0.93 s
Can she tell them apart? Two sounds are heard separately only if they are at least 0.1 s apart. Here
Activity 10.8 — Page 198Let us experiment (demonstration activity) — 10.6.4 Human perception of sound
As the frequency rises from 100 Hz to 1000 Hz, the sound becomes steadily shriller — the pitch goes up — even though the volume setting has not been touched.
| Frequency set | How it sounds | Wavelength in air (v = 344 m s⁻¹) |
|---|---|---|
| 100 Hz | Deep, humming, like a distant engine | 344 ÷ 100 = 3.44 m |
| 300 Hz | Full, like a low male voice | 344 ÷ 300 = 1.15 m |
| 600 Hz | Bright, clear | 344 ÷ 600 = 0.57 m |
| 1000 Hz | Thin, piercing, whistle-like | 344 ÷ 1000 = 0.34 m |
As you go below about 50 Hz the note becomes a deep rumble, gets rapidly harder to hear, and near 20 Hz it fades away altogether — you may feel it as a vibration rather than hear it.
Below 20 Hz → infrasonic waves — produced, but not heard by us
Above 20 kHz → ultrasonic waves — also not heard by us
In-text Questions — Page 19910.6.4 Human perception of sound
No. Humans can hear only a limited band of frequencies — the audible range, from about 20 Hz to 20 000 Hz (20 kHz).
- Below 20 Hz the waves are called infrasonic waves. Elephants can detect them, and they are used to detect earthquakes, volcanic eruptions and severe storms.
- Above 20 kHz they are called ultrasonic waves. Dogs, cats, bats and dolphins can detect them, and we use them in sonar, ultrasonography and industrial flaw detection.
Pause and Ponder — Page 20110.7.1 Echo
The reflecting surface must be at least 34.3 m away.
The sound has to make a round trip: from the source to the wall and back to the listener. So the total path is twice the distance we are asked for.
= 343 m s⁻¹ × 0.2 s
= 68.6 m
This is the to-and-fro distance, so
distance to the reflecting surface = 68.6 m ÷ 2
= 34.3 m
What if … — Page 20210.8 Ultrasonic and Infrasonic Waves, and their Applications
We would gain a great deal of extra information about the world, but we would also be flooded with noise we currently do not notice at all.
| Advantages | Disadvantages |
|---|---|
| We could hear bats and dolphins hunting, and follow animal communication that is silent to us today. | Constant noise from ultrasonic pest repellers, some machinery, vehicle parking sensors and cleaning equipment. |
| We could tell by ear whether ultrasonic devices — parking sensors, cleaners, pest repellers — are working. | Ultrasonic welding and industrial cleaning plants would become unbearable workplaces and would need heavy shielding. |
| We would get early warning of some machine faults, since worn or leaking machine parts often give out ultrasound. | Long hours near machines giving out strong ultrasound could tire or harm our hearing, so such workplaces would need ear protection. |
| Better echolocation of our own: short-wavelength sound reflects from small objects, so we could judge small obstacles in the dark. | Speech and music would be buried under a permanent hiss, making concentration and conversation harder. |
Pause and Ponder — Page 20310.8.1 Echolocation
The ocean is 3000 m (3 km) deep at that place.
total distance = speed × time = 1500 m s⁻¹ × 4 s = 6000 m
depth = total distance ÷ 2 = 6000 m ÷ 2 = 3000 m = 3 km
Revise, Reflect, Refine — Pages 204 – 206End-of-chapter questions
(i) Sound shows reflection
(ii) Sound needs a medium to propagate
(iii) Sound has frequency
(iv) Sound carries energy
(ii) Sound needs a medium to propagate.
A mechanical wave is defined as a wave that requires a material medium to propagate. So the observation that pins sound down as mechanical is precisely the one that shows it dies out without a medium — the vacuum bell jar experiment.
(i) wavelength
(ii) speed
(iii) number of compressions per second
(iv) time period
(iii) number of compressions per second.
Frequency is the number of complete density oscillations passing a fixed point each second, and every complete oscillation brings one compression with it. So raising the frequency directly raises the number of compressions arriving per second. (The speed stays the same, while the wavelength and the time period decrease.)
(i) 80 Hz
(ii) 5 Hz
(iii) 10 Hz
(iv) 0.2 Hz
(ii) 5 Hz.
= 20 ÷ 4 s
= 5 s⁻¹ = 5 Hz
Each compression that sweeps past the point marks one complete density oscillation there, so counting compressions is the same as counting oscillations.
It will produce reverberation, not an echo.
Minimum gap needed to hear two sounds separately = 0.1 s
0.05 s < 0.1 s → the brain cannot separate them
Because the reflected sound arrives while the original is still being registered, the two merge. The sound seems to persist or linger a little after the source has stopped — and that persistence caused by reflections is exactly what reverberation means.
(i) Wave (a) has the greater wavelength. (ii) Wave (a) also has the smaller amplitude. Over the same distance, (a) shows 3 complete waves while (b) shows 6, so the wavelength of (a) is about twice that of (b). The crests of (a) rise less above the average-density line, so its amplitude is smaller.
The green curve is A, the red curve is B and the blue curve is C.
All three curves are drawn over the same distance axis, so the one that fits in the most cycles has the shortest wavelength and therefore the highest frequency.
| Curve in Fig. 10.31 | Complete cycles across the axis | Wavelength | Frequency ν = v/λ | Label |
|---|---|---|---|---|
| Green (tallest, most closely spaced) | 4 | Shortest | Maximum | A |
| Red (middle) | 3 | Middle | Middle | B |
| Blue (widest, smallest crests) | 2 | Longest | Minimum | C |
Plot density on the y-axis and distance on the x-axis, draw a horizontal dashed line for the average density, and then draw a smooth wave that
- rises 3 units above the average-density line at each crest and falls 3 units below it at each trough (that is the density amplitude), and
- repeats every 4 cm along the distance axis — crest to crest, or trough to trough, is 4 cm.
There are two errors.
- There should be no sound at all. Outer space is a near vacuum. Sound is a mechanical wave and needs a material medium; with almost no particles there is nothing to compress and nothing to pass the disturbance on. The explosion would be completely silent to a distant observer. (Light is not a mechanical wave, so the flash would be seen — which is why sunlight reaches the Earth across empty space.)
- Even in a medium, the two could not arrive together. Light travels enormously faster than sound, so the flash always arrives first and the sound follows after a delay.
The time period is 0.01 s.
ν = v / λ = 344 m s⁻¹ ÷ 3.44 m
= 100 s⁻¹ = 100 Hz
Step 2 — find the time period. ν = 1/T (Eq. 10.1)
T = 1 / ν = 1 ÷ 100 Hz
= 0.01 s
The wreck lies about 3812.5 m (roughly 3.8 km) below the ship.
total distance = speed × time
= 1525 m s⁻¹ × 5 s
= 7625 m
depth of the wreck = total distance ÷ 2
= 7625 m ÷ 2
= 3812.5 m ≈ 3.8 km
About 6.96 × 10⁻³ s, that is roughly 7 milliseconds.
total distance = 2 × 1.2 m = 2.4 m
time = distance ÷ speed
= 2.4 m ÷ 345 m s⁻¹
= 0.00696 s ≈ 6.96 × 10⁻³ s ≈ 7.0 ms
About 0.2 s of extra time.
At 22 ºC (v = 344 m s⁻¹):
t₁ = 1720 m ÷ 344 m s⁻¹ = 5.000 s
At 0 ºC (v = 331 m s⁻¹):
t₂ = 1720 m ÷ 331 m s⁻¹ = 5.196 s
Extra time = t₂ − t₁
= 5.196 s − 5.000 s = 0.196 s ≈ 0.2 s
Wavelength λ = 4 cm = 0.04 m, and frequency ν = 8500 Hz.
8 cm = 2λ → λ = 8 cm ÷ 2 = 4 cm = 0.04 m
ν = v / λ
= 340 m s⁻¹ ÷ 0.04 m
= 8500 s⁻¹ = 8500 Hz
λA = 2.5 cm and νA = 13 800 Hz; λB = 5.0 cm and νB = 6900 Hz.
Read the wavelengths off the distance axis of Fig. 10.33. Between 0 and 5.0 cm, curve A completes two full cycles while curve B completes just one.
Wave B: 1 wavelength in 5.0 cm → λB = 5.0 cm = 0.050 m
νA = 345 m s⁻¹ ÷ 0.025 m = 13 800 Hz = 13.8 kHz
νB = 345 m s⁻¹ ÷ 0.050 m = 6900 Hz = 6.9 kHz
vair : vwater = 1 : 4.5 = 2 : 9, i.e. about 0.22 : 1.
time = distance ÷ speed
For A (in air): tA = 2d / vair
For B (in water): tB = 2d / vwater
Given: tA = 4.5 × tB
∴ 2d / vair = 4.5 × (2d / vwater)
The 2d cancels from both sides:
1 / vair = 4.5 / vwater
∴ vwater = 4.5 × vair
vair : vwater = 1 : 4.5 = 2 : 9 ≈ 0.22
The Journey Beyond — Page 207Project work
| Grade of hearing loss | Hearing threshold (better ear) | What it means day to day |
|---|---|---|
| Normal | Below 20 dB | No difficulty |
| Mild | about 20 – 35 dB | Trouble following soft speech or conversation in a noisy place |
| Moderate | about 35 – 50 dB | Difficulty with ordinary conversation; a hearing aid usually helps |
| Moderately severe | about 50 – 65 dB | Ordinary speech hard to follow even in a quiet room; needs a hearing aid |
| Severe | about 65 – 80 dB | Normal speech not heard; needs a powerful aid or an implant |
| Profound | about 80 dB and above | Even loud sounds hardly heard; a cochlear implant may be needed |
Sample observation: with the cone over the phone, the music sounds clearly louder in front of the cone’s open end and quieter at the sides and behind it. A narrow cone makes the sound louder straight ahead but more directional; a wide cone spreads it more. The cone does not create energy — it guides the sound into a narrower beam, so more sound energy passes through each unit area in front of it (higher intensity).
| Arrangement | Sample dB reading at 1 m | What you hear |
|---|---|---|
| Phone alone | about 62 dB | Thin, spreads in all directions |
| Narrow cone | about 68 dB in front, less at the sides | Louder ahead, sharply directional |
| Wide cone | about 66 dB in front, spread wider | Louder but less sharply aimed |
| Behind the cone | lower than the phone alone | Noticeably quieter |
A curved surface behind and above the stage reflects sound outward and downward towards the audience, so more of the performer’s energy reaches the listeners and it reaches all seats more evenly.
The curved surfaces also send the reflected sound to the listeners soon after the direct sound (within about 0.05 s), so it strengthens the sound instead of forming echoes. Flat, parallel walls bounce sound back and forth many times, causing too much reverberation and a garbled sound.
(i) Your friend stands at one end of the open ground with the balloons, while you stand at the other end with the stopwatch.
(ii) Signal your friend to burst one balloon. When you see the balloon burst, start the stopwatch. As soon as you hear the ‘pop’ sound of the bursting balloon, stop the timer and note down the reading.
(iii) Repeat this experiment multiple times and take the average value of the times noted.
(iv) Note the approximate distance between you and your friend using a map application on a mobile phone.
(v) Divide the distance measured with the average time to get the average speed of sound. What value of speed did you get from the experiment? Compare it with the speed of sound in air, which is typically about 346 m s–1 at 25 °C.
(vi) Why did you measure the time between ‘seeing’ and ‘hearing’ the balloon burst?
(v) The experiment should give a speed of roughly 330 – 360 m s⁻¹, close to the accepted 346 m s⁻¹ at 25 ºC.
Sample readings for a measured distance of 250 m:
| Trial | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Time (s) | 0.75 | 0.70 | 0.78 | 0.72 | 0.75 |
= 3.70 s ÷ 5 = 0.74 s
speed of sound = distance ÷ average time
= 250 m ÷ 0.74 s
= 338 m s⁻¹
(vi) Why measure between ‘seeing’ and ‘hearing’: because light travels so much faster than sound that the flash of the burst can be treated as arriving instantly. So the time between seeing the burst and hearing the pop is the time taken by the sound alone to travel the distance.
(i) https://phet.colorado.edu/en/simulations/sound-waves/
(ii) https://musiclab.chromeexperiments.com/Experiments
(iii) https://phyphox.org/experiments
Both temperature and humidity increase the speed of sound in air, and temperature has by far the larger effect.
| Condition | Approximate speed in air | Source |
|---|---|---|
| Dry air at 0 ºC | 331 m s⁻¹ | Section 10.6.3 |
| Air at 15 ºC | 340 m s⁻¹ | Table 10.1 |
| Air at 22 ºC | 344 m s⁻¹ | Section 10.6.3 |
| Air at 25 ºC | 346 m s⁻¹ | The Journey Beyond, page 207 |
| Humid air at the same temperature | a little higher than dry air | Section 10.6.3 |
Chapter 10 of 12 – Science
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Also for This Chapter
NCERT Solutions Class 9 Science Chapter 10 Sound Waves Characteristics and Applications – Overview
| Class | Class 9 |
| Subject | Science |
| Chapter | 10 – Sound Waves Characteristics and Applications |
| Textbook pages | 184 – 207 |
| Questions solved | 63 |
| Session | 2027 |
| Format | Full answers on this page + PDF |
NCERT Solutions Class 9 Science Chapter 10 – FAQs
Where can I download NCERT Solutions Class 9 Science Chapter 10 Sound Waves Characteristics and Applications PDF?
Read every answer on this page, or download the free PDF from the table at the top.
Is this for the 2027 session?
Yes, it follows the latest NCERT Class 9 Science book for 2027.
How many questions are solved in NCERT Class 9 Science Chapter 10?
All 63 questions of the chapter (textbook pages 184 – 207) – the in-text questions, activities and the end-of-chapter exercises.
How should I use these solutions?
Read the chapter in the NCERT book first, attempt the questions yourself, then check your answers here.