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NCERT Solutions for Class 9 Science Chapter 10 Sound Waves Characteristics and Applications (PDF) – 2027

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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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

Think It Over
Q1Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each other and hear the sounds of metal clanking as they do on the Earth?
Answer

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.

Q2How do most bats use sound to locate their prey in the dark at night?
Answer

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

Q1Which form of energy gets converted to sound energy?
Answer

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.
Q2How is sound produced and how does it reach our ears?
Answer

Sound is produced by a vibrating object and reaches the ear as a travelling disturbance in the density of a medium.

  1. 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.
  2. 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.
  3. 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

Activity 10.1
Q1Holding the box steady with one hand, pluck the rubber band with a finger. Do you hear any sound?
Answer

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.

Q2Pluck the rubber band again and watch it carefully. Is it vibrating?
Answer

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.

vibration = periodic to-and-fro motion (oscillation) about a mean position
Q3Wait till the rubber band stops vibrating. Do you still hear the sound?
Answer

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.

Q4Change the tension in the rubber band by stretching it more or loosening it slightly and plucking it each time. Does the sound change? What changes do you notice?
Answer

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 doHow the band vibratesFrequency νTime period TWhat you hear
Stretch it more (higher tension)Snaps back fasterhighershortershriller, high pitch
Loosen it (lower tension)Returns sluggishlylowerlongerdeeper, low pitch
Pluck harderLarger swing about the mean positionunchangedunchangedlouder (larger amplitude)
Q5Remove the rubber band from the box. Stretch it between two fingers and pluck it near your ear. Is the sound still produced? Is it as loud as before?
Answer

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

Q1How do humans and animals create sound? While talking or singing, gently touch your throat. Do you feel vibrations anywhere?
Answer

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

Activity 10.2
Q1Strike one of the prongs of the tuning fork gently against the rubber pad (Fig. 10.4b) and bring it close to your ear. Do you hear a sound? (Take care not to strike the tuning fork against a hard surface).
Prongs Rubber pad Stem (a) A tuning fork and a rubber pad (b) Striking a prong against therubber pad
Fig. 10.4 (a) and (b), page 186 — redrawn sketch. The fork is held by its stem and one prong is struck gently on the soft rubber pad.
Answer

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.

Q2Now, gently touch a water surface with one of the vibrating prongs of the tuning fork. Do you see waves forming on the surface of water?
Answer

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.

Q3Repeat step 3 a few times while bringing the prongs of the tuning fork near your ear in different orientations. Do you hear the sound?
Answer

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

Pause and Ponder
Q1Explore various ways of producing sound.
Answer

Every method comes down to one thing — making something vibrate. The ways differ only in what is set vibrating.

Way of producing soundWhat vibratesEveryday example
PluckingA stretched stringSitar, veena, guitar, ektara, rubber band
Striking / hittingA stretched membrane or a metal bodyTabla, dholak, taal (manjira), temple bell, tuning fork
BlowingThe air column inside a pipeBansuri, shehnai, conch (shankh), whistle
BowingA string, rubbed by a bowSarangi, violin
Rubbing / scrapingThe two rubbing surfacesCrickets and grasshoppers rubbing wings or legs; chalk squeaking on a board
Forcing air past flapsVocal cordsSpeaking, singing, animal calls
Sudden expansion of hot gasThe air itselfFirecracker, thunder, bursting balloon, sonic boom
Electrical signal driving a coneSpeaker diaphragmMobile phone, radio, loudspeaker
Q2Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
Answer
InstrumentFamilyVibrating part that makes the soundHow it is set vibrating
Sitar, veena, sarod, ektaraString (tat vadya)Stretched metal strings; the hollow gourd/body then amplifiesPlucked
Sarangi, violinStringStretched stringsBowed (rubbed)
Tabla, dholak, mridangam, damruMembrane (avanaddha vadya)Stretched skin/membrane of the drum head, plus the air inside the shellStruck
Bansuri, shehnai, whistleWind (sushir vadya)The air column inside the hollow pipeBlown
HarmoniumWindThin metal reedsAir pushed past by the bellows
Manjira (taal), ghanti, ghatamSolid body (ghana vadya)The metal or clay body itselfStruck
Tuning forkSolid bodyThe two prongsStruck on a soft pad

In-text Questions — Page 18610.2 Propagation of Sound

Q1How does sound reach your ear from the source? Sound travels through air but does it also travel through solids and liquids?
Answer

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

Activity 10.3
Q1Now, place your ear against the desk, close your other ear and listen again, as shown in Fig. 10.5. Are you able to hear the sound through the table?
ear pressed on the desk other ear closed friend knocks the desk desk
Fig. 10.5, page 187 — redrawn sketch. One student presses an ear on the desk top and closes the other ear while the friend knocks on the same desk.
Answer

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

Activity 10.4
Q1Now, submerge the two metal spoons in water without touching the sides or bottom of the bucket and tap them against one another again (Fig. 10.6b). Do you again hear the sound produced?
(a) tapped in air water (b) tapped under water water
Fig. 10.6, page 187 — redrawn sketch. Two metal spoons are tapped together (a) in the air above the water and (b) fully submerged, without touching the sides or bottom.
Answer

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.

Q2If sound did not travel through liquids, would you have heard this sound?
Answer

No. With the spoons held clear of the bucket, water is the only medium touching them.

Path taken by the sound: vibrating spoons → water → water surface → air → ear

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

Q1A space where there is no medium (matter) is referred to as vacuum. Would you hear sound in vacuum?
Answer

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):

  1. An electric bell inside a sealed bell jar is switched on — the ringing is loud.
  2. A vacuum pump slowly removes the air. The sound gets fainter and fainter.
  3. Near vacuum: the hammer can still be seen striking the gong, but almost no sound is heard.
  4. 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

Pause and Ponder
Q1Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out.
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.
Answer

(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

Activity 10.5
Q1Give the slinky at your end a sharp push towards your friend and then quickly pull it back again (Fig. 10.8). Do you observe a disturbance created in the slinky which moves towards your friend?
Turns are closer together Turns are more spread out
Fig. 10.8, page 188 — redrawn sketch. A sharp push and pull at one end sends a bunched-up region (the disturbance) travelling along the stretched slinky.
Answer

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.

Q2Now, push and pull the slinky end multiple times in quick succession (The pulling and pushing of the end of the slinky is similar to the sound being produced continuously). Are a series of disturbances produced in the slinky? Do these disturbances move across the length of slinky? Does the mark on the slinky move back and forth parallel to the direction of the disturbance?
Answer

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

Q1If there is no medium (i.e., no particles), is the propagation of sound waves possible?
Answer

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

Pause and Ponder
Q1Assertion (A): Compressions and rarefactions move through the medium.
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.
Answer

(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

Activity 10.6
Q1Produce a loud sound near the bowl without touching it. Observe the grains on the sheet. Does the sound have any effect on the grains?
Answer

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.

Q2Repeat step 4 with different sources of sounds and observe the effect on the grains. You can try increasing or reducing the volume of sound. Try with different grains.
Answer

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 changeWhat you observeReason
Strike the plate harder (louder sound)Grains jump much higherLarger amplitude → more energy carried by the wave → the sheet is displaced further
Move the source further awayGrains move lessThe 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 easilySame energy delivered to a smaller mass gives a larger displacement
Loosen the stretched sheetResponse becomes weaker and dullerA slack sheet does not follow the rapid pushes of the sound wave faithfully

Pause and Ponder — Page 19210.4 Energy of Sound Waves

Pause and Ponder
Q1When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?
(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
Answer

(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

Pause and Ponder
Q1The variation of density of the medium for two sound waves is shown in Fig. 10.17 (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in Fig. 10.17 (c) and (d), label the axes and draw the curves corresponding to Fig. 10.17 (a) and (b).
Figure
Fig. 10.17, page 193. (a) and (b) show the density of the medium for two sound waves; (c) and (d) are the empty graph grids to be labelled and filled in.
Answer

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.

Figure
Fig. 10.17 completed: C and R marked on (a) and (b); graph (c) drawn for (a) and graph (d) for (b).

Activity 10.7 — Page 194Let us experiment (demonstration activity) — 10.6.1 Wavelength, frequency and time period

Activity 10.7
Q1Compare the musical notes by taking the ratio of each frequency with respect to the ‘Sa’. Do you observe any pattern?
Answer

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.

NoteSample frequencyRatio to SaSimple fraction
Sa240 Hz1.001
Re270 Hz1.139/8
Ga300 Hz1.255/4
Ma320 Hz1.334/3
Pa360 Hz1.503/2
Dha400 Hz1.675/3
Ni450 Hz1.8815/8
Sa (upper)480 Hz2.002
Ratio for Pa = 360 Hz ÷ 240 Hz = 1.5 = 3/2
Ratio for upper Sa = 480 Hz ÷ 240 Hz = 2.0
Q2If both voice and mobile-generated notes are used, compare their frequencies for the same musical notes.
Answer

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

Pause and Ponder
Q1Conduct Activity 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber band?
Answer

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.

BandRate of vibrationFrequency νTime period T = 1/νSound heard
Thin bandFasterHigherShorterShriller — high pitch
Thick bandSlowerLowerLongerDeeper — low pitch
Q2If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?
Answer

1200 oscillations per minute.

Frequency ν = number of oscillations ÷ time taken
∴ 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
Q3For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?
0 1.5 3.0 4.5 Distance (cm) Density
Fig. 10.19, page 195 — graphical representation of a sound wave (density against distance).
Answer

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.

λ = distance between two consecutive crests (or two consecutive troughs)
From Fig. 10.19: λ = 3.0 cm
∴ λ/2 = 3.0 cm ÷ 2 = 1.5 cm = 0.015 m
0 1.5 3.0 4.5 Distance (cm) Density λ = 3.0 cm λ/2 = 1.5 cm
Reading Fig. 10.19: one complete density oscillation spans 0 to 3.0 cm, so λ = 3.0 cm and half the wavelength is 1.5 cm.

What if … — Page 19610.6.3 Speed of Sound

What if …
Q1What if … the speed of sound in air depended on its frequency? Would music still sound pleasant when a singer performs with instruments? Why or why not?
Answer

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

Pause and Ponder
Q1Table 10.1 shows the speed of sound in a few media at atmospheric pressure.
Table 10.1: Speed of sound in different media at 15 °C
StateSubstance/MediumApproximate speed
SolidSteel5000 m s⁻¹
LiquidWater1500 m s⁻¹
GasAir340 m s⁻¹
Compare the speeds in different media by finding the ratio of
(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.
Answer

(i) about 4.4 : 1   (ii) about 3.3 : 1

(i) ratio = vwater / vair
= 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

Pause and Ponder
Q1Two friends are standing along a steel fence at a distance of 340 m from each other (Fig. 10.23). Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table 10.1, calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least 0.1 s to be heard separately.)
steel fence friend knocks with a metal object Gunjan’s ear on the fence 340 m
Fig. 10.23, page 198 — redrawn sketch. The knock reaches Gunjan by two paths at once: along the steel fence and through the air.
Table 10.1: Speed of sound in different media at 15 °C
StateSubstance/MediumApproximate speed
SolidSteel5000 m s⁻¹
LiquidWater1500 m s⁻¹
GasAir340 m s⁻¹
Answer

The time difference is about 0.93 s, and yes — Gunjan can easily hear the two knocks separately.

time = distance ÷ speed

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

0.932 s  >  0.1 s  →  yes, she hears two distinct knocks

Activity 10.8 — Page 198Let us experiment (demonstration activity) — 10.6.4 Human perception of sound

Activity 10.8
Q1Increase the frequency in steps of 100 Hz up to 1000 Hz and describe how the sound changes.
Answer

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 setHow it soundsWavelength in air (v = 344 m s⁻¹)
100 HzDeep, humming, like a distant engine344 ÷ 100 = 3.44 m
300 HzFull, like a low male voice344 ÷ 300 = 1.15 m
600 HzBright, clear344 ÷ 600 = 0.57 m
1000 HzThin, piercing, whistle-like344 ÷ 1000 = 0.34 m
Q2Next, set the frequency to 50 Hz. Reduce the frequency till about 20 Hz or the point where you cannot hear the sound anymore.
Answer

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.

Human audible range = 20 Hz to 20 000 Hz (20 kHz)
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

Q1Can humans hear all sounds?
Answer

No. Humans can hear only a limited band of frequencies — the audible range, from about 20 Hz to 20 000 Hz (20 kHz).

Infrasonic Audible range Ultrasonic 20 Hz 20 kHz frequency elephants can detect dogs, cats, bats, dolphins
The human audible range is only the middle strip. Sound exists on both sides of it, but our ears do not respond to it.
  • 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

Pause and Ponder
Q1An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 m s–1.
Answer

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.

total distance travelled = speed × time
= 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

What if …
Q1What if … humans could detect ultrasonic waves like dogs can? What would be the advantages and disadvantages?
Answer

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.

AdvantagesDisadvantages
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

Pause and Ponder
Q1Sound travels much farther in water than light, and thus, is used for various underwater applications. A sonar signal sent to find the depth of ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 m s–1?
Answer

The ocean is 3000 m (3 km) deep at that place.

The 4 s is the time for the signal to go down and come back.

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

Revise, Reflect, Refine
Q1Which observation best supports the idea that sound is a mechanical wave?
(i) Sound shows reflection
(ii) Sound needs a medium to propagate
(iii) Sound has frequency
(iv) Sound carries energy
Answer

(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.

Q2For a sound wave propagating in a medium, increasing its frequency will increase its
(i) wavelength
(ii) speed
(iii) number of compressions per second
(iv) time period
Answer

(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.)

Q3If 20 compressions pass a point in 4 seconds, the frequency is
(i) 80 Hz
(ii) 5 Hz
(iii) 10 Hz
(iv) 0.2 Hz
Answer

(ii) 5 Hz.

frequency ν = number of oscillations ÷ time taken
= 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.

Q4In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.
Answer

It will produce reverberation, not an echo.

Time gap between the direct sound and the reflected sound = 0.05 s
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.

Q5Graphs representing two sound waves are given in Fig. 10.30. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has (i) greater wavelength, and (ii) smaller amplitude?
Distance Density (a) Distance Density (b)
Fig. 10.30, page 205 — density–distance graphs of two sound waves, drawn to the same scale on both axes.
Answer

(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.

Distance Density (a) 3 waves: longer λ lower crests: smaller amplitude Distance Density (b) 6 waves: shorter λ taller crests: larger amplitude
Both graphs cover the same span of distance and use the same scales, so cycles can be counted directly and crest heights compared directly.
Q6The sound waves emitted by three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.
Distance Density
Fig. 10.31, page 205 — the sound waves emitted by the three sources, drawn on one pair of axes and deliberately left unlabelled.
Answer

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. 10.31Complete cycles across the axisWavelengthFrequency ν = v/λLabel
Green (tallest, most closely spaced)4ShortestMaximumA
Red (middle)3MiddleMiddleB
Blue (widest, smallest crests)2LongestMinimumC
Distance Density A B C
Fig. 10.31 marked: A = green curve (4 waves), B = red curve (3 waves), C = blue curve (2 waves).
Q7Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
Answer

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.
+3 −3 0 C C C R R R 0 2 4 6 8 10 12 Distance (cm) Density λ = 4 cm amplitude = 3 units average density
Density–distance graph with a density amplitude of 3 units and a wavelength of 4 cm. Each crest is a compression, each trough a rarefaction.
Q8In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?
Answer

There are two errors.

  1. 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.)
  2. 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.
Q9A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 m s–1 find its time period.
Answer

The time period is 0.01 s.

Step 1 — find the frequency.   v = ν × λ  (Eq. 10.2)
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
Q10A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 m s–1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
Answer

The wreck lies about 3812.5 m (roughly 3.8 km) below the ship.

The 5 s is the time for the pulse to go down to the wreck and back.

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
Q11A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s–1.
Answer

About 6.96 × 10⁻³ s, that is roughly 7 milliseconds.

The pulse goes to the obstacle and returns, so it covers twice 1.2 m.

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
Q12The speed of sound in air is about 331 m s–1 at 0 ºC and nearly 344 m s–1 at 22 ºC. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 ºC to 0 ºC? Assume that all other conditions remain unchanged.
Answer

About 0.2 s of extra time.

time = distance ÷ speed

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
Q13The variation of density of medium for a sound wave propagating with a speed of 340 m s–1 is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
8 cm
Fig. 10.32, page 206 — variation of the density of the medium along the direction of travel; the marked span is 8 cm.
Answer

Wavelength λ = 4 cm = 0.04 m, and frequency ν = 8500 Hz.

The 8 cm arrow runs from one compression to the third compression, i.e. over 2 wavelengths.
8 cm = 2λ  →  λ = 8 cm ÷ 2 = 4 cm = 0.04 m
Figure
Reading Fig. 10.32: the 8 cm arrow spans two whole repeats of the compression pattern, so one wavelength is 4 cm.
Now the frequency.   v = ν × λ  (Eq. 10.2)
ν = v / λ
= 340 m s⁻¹ ÷ 0.04 m
= 8500 s⁻¹ = 8500 Hz
Q14The graphical representation of two sound waves A and B propagating at the same speed of 345 m s–1 is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.
0 2.5 5.0 Distance (cm) Density A B
Fig. 10.33, page 206 — graphical representation of the two sound waves A and B.
Answer

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 A: 2 wavelengths in 5.0 cm → λA = 5.0 cm ÷ 2 = 2.5 cm = 0.025 m
Wave B: 1 wavelength in 5.0 cm → λB = 5.0 cm = 0.050 m
Frequencies, using ν = v / λ with v = 345 m s⁻¹ for both:

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
Q15Two identical sound sources are placed at A and B — one in air and one submerged in water (Fig. 10.34). Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times than that of B, what is the ratio between the speeds of sound in air and water?
A B cliff source A in air source B under water sea wall
Fig. 10.34, page 206 — redrawn sketch. Source A stands in the air above the water and source B is submerged, both the same horizontal distance from the vertical rock face.
Answer

vair : vwater = 1 : 4.5 = 2 : 9, i.e. about 0.22 : 1.

Let d = distance of A and of B from the cliff (the same for both); each sound travels 2d.
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

The Journey Beyond
Q1Many people use earphones extensively these days. Find out the research studies that might have been done to understand the impact of excessive use of earphones on hearing (if any). Also, find out how hearing is tested and what are the decibel ranges for defining mild, moderate and severe hearing loss. What are the government schemes for purchasing or fitting of aids or appliances and free cochlear implants? Write an article on your findings.
Answer
Sample answer (short article): Earphones and our hearing. Earphones send sound straight into the ear canal, so the sound reaching the eardrum can be very loud. The WHO warns that listening above about 80 dB for long hours can damage hearing, and studies of young people who use earphones for many hours a day report early hearing loss. The damage is to the hair cells of the cochlea, which do not grow back. Hearing is tested with an audiometer: tones of different frequencies are played at increasing loudness, and the softest level heard at each is plotted as an audiogram. Loss is graded by that level, as in the table. Under the Government of India’s ADIP scheme, hearing aids are given free or at low cost, and eligible young children can get free cochlear implants.
Grade of hearing lossHearing threshold (better ear)What it means day to day
NormalBelow 20 dBNo difficulty
Mildabout 20 – 35 dBTrouble following soft speech or conversation in a noisy place
Moderateabout 35 – 50 dBDifficulty with ordinary conversation; a hearing aid usually helps
Moderately severeabout 50 – 65 dBOrdinary speech hard to follow even in a quiet room; needs a hearing aid
Severeabout 65 – 80 dBNormal speech not heard; needs a powerful aid or an implant
Profoundabout 80 dB and aboveEven loud sounds hardly heard; a cochlear implant may be needed
Q2Make a cone using a poster paper or cardboard and adhesive tape. Cover a mobile phone that is playing music with the cone. Compare the loudness of the sound with and without the cone. You can also use another mobile phone with an app to measure the characteristics of the sound in both cases. Try experimenting with different shapes and record your observations. (This activity is to be facilitated by the teacher.)
Answer

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).

ArrangementSample dB reading at 1 mWhat you hear
Phone aloneabout 62 dBThin, spreads in all directions
Narrow coneabout 68 dB in front, less at the sidesLouder ahead, sharply directional
Wide coneabout 66 dB in front, spread widerLouder but less sharply aimed
Behind the conelower than the phone aloneNoticeably quieter
Q3How does the curved design of ceilings and walls behind the stage in concert and conference halls improve the quality of sound for the audience compared to flat surfaces? You may consult an architect or search it on the internet.
Answer

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.

Q4Carry out a simple activity to measure the speed of sound, along with a friend in a large open ground of size 200 m or more. (This activity is to be facilitated by the teacher.)
(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?
Answer

(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:

Trial12345
Time (s)0.750.700.780.720.75
average time = (0.75 + 0.70 + 0.78 + 0.72 + 0.75) s ÷ 5
= 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.

Q5Explore the internet resources to explore the effect of humidity and temperature on the speed of sound. Some such resources are:
(i) https://phet.colorado.edu/en/simulations/sound-waves/
(ii) https://musiclab.chromeexperiments.com/Experiments
(iii) https://phyphox.org/experiments
Answer

Both temperature and humidity increase the speed of sound in air, and temperature has by far the larger effect.

ConditionApproximate speed in airSource
Dry air at 0 ºC331 m s⁻¹Section 10.6.3
Air at 15 ºC340 m s⁻¹Table 10.1
Air at 22 ºC344 m s⁻¹Section 10.6.3
Air at 25 ºC346 m s⁻¹The Journey Beyond, page 207
Humid air at the same temperaturea little higher than dry airSection 10.6.3

Chapter 10 of 12 – Science

Also for This Chapter

NCERT Solutions Class 9 Science Chapter 10 Sound Waves Characteristics and Applications – Overview

ClassClass 9
SubjectScience
Chapter10 – Sound Waves Characteristics and Applications
Textbook pages184 – 207
Questions solved63
Session2027
FormatFull answers on this page + PDF

NCERT Solutions Class 9 Science Chapter 10 – FAQs

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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.

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