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

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

CLASS 9 · SCIENCE

NCERT Solutions

Chapter 10: Sound Waves:
Characteristics and Applications

NCERT Textbook — Exploration

BOOK PAGES SECTIONS QUESTIONS MEDIUM

184 – 207 29 63 English

Solutions, notes, sample papers & more at 66 pages

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

CLASS 9 · SCIENCE · EXPLORATION

NCERT Solutions — Chapter 10: Sound Waves:
Characteristics and Applications
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.

TEXTBOOK BOOK PAGES

Exploration (Class 9) 184 – 207

SECTIONS QUESTIONS

29 63

MEDIUM

English

Think It Over — Page 184
Chapter opener

THINK IT OVER

Q1 Two 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?

No — not directly. Outer space is a near vacuum, and a sound wave has nothing to travel in
there.

Why it happens: sound is a series of compressions and rarefactions — regions
where the particles of a medium are pushed closer together and pulled further
apart. A particle can only pass the disturbance on by colliding with a neighbouring
particle. Where there are almost no particles, there are no collisions, so there is
nothing to carry the disturbance. This is exactly what the vacuum bell jar experiment
(Fig. 10.7) shows: as the air is pumped out, the bell is still seen ringing but the sound
fades to nothing.

On the Earth the same clank is heard easily because the air between the two astronauts is a
perfectly good medium.

Page 1 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Did you know? Astronauts talk over radio sets built into their suits. Radio waves are
electromagnetic waves, not mechanical waves, so they need no medium and cross
the vacuum happily. If two helmets are pressed together, the clank can be heard —
the solid helmet material then acts as the medium.

Q2 How do most bats use sound to locate their prey in the dark at night?

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.

distance to prey = v × t / 2 (the sound travels there and back)

For an echo returning in 0.02 s in air, distance = 340 m s⁻¹ × 0.02 s ÷ 2 = 3.4 m

Why ultrasound and not ordinary sound: a shorter wavelength reflects cleanly
from small objects. At 40 kHz in air, λ = 344 m s⁻¹ ÷ 40000 s⁻¹ ≈ 0.0086 m ≈ 0.9 cm —
small enough to bounce off a mosquito-sized insect. A 100 Hz sound has λ ≈ 3.4 m
and would simply flow around it.

Did you know? Dolphins, whales and some birds also echolocate, and the same
principle is used by humans in sonar and in ultrasonography.

In-text Questions — Page 184

Page 2 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Chapter introduction

Q1 Which form of energy gets converted to sound energy?

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.

Why it matters: energy can neither be created nor destroyed (you met this in
Chapter 7). So sound energy must always be paid for out of some other store. That is
also why a plucked band goes quiet after a while — its vibrational energy has been
drained away into the air as sound and into heat.

Q2 How is sound produced and how does it reach our ears?

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.

Page 3 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
e m.
m l as
.co g
C R C R C
m a
l a se
a g

co m
m . ag
la se
direction of propagation of the wave
a g
source vibrates → compressions and rarefactions travel → eardrum
vibrates

. c om
s e m out in a
glado not.
com rarefaction (R). The pattern moves to the right; the particles
Air particles between the source and the ear: crowded in a compression (C), spread

. a
as em
g l
a Tip: keep the two ideas apart — the disturbance travels, the particles do not.
m a s
agl
What actually arrives at your ear is energy, not the air that was next to the tuning

m.co
se
fork.

g l a
a

co m
.
Activity 10.1 — Page 185
e m
om as
Let us explore — 10.1 Production of Sound

. c10.1 a g l
s em
ACTIVITY

gl a
a
Q1 Holding the box steady with one hand, pluck the rubber band with a finger. Do you
se m
hear any sound?
com g l a
m . a
gl ase
a

Yes, a clear twang is heard.

c o m
.
Plucking pulls the stretched band away from its mean position. When you let go, the tension

s e m The vibrating band
pulls it back, it overshoots, and it keeps moving to and fro — it vibrates.

. c
pushes
a g la
omthe air next to it forwards and backwards, sending compressions and rarefactions

sem
towards your ear.
a
agl c
Why the box matters: a bare band moves very little air. Stretched across the open
m .
m a s e
co agl
box, it also sets the air inside the box and the cardboard itself vibrating, so far more

m .
e
air is disturbed and the sound is much louder. This is exactly why a sitar, veena or
as
guitar has a hollow body.
a g l

co m
m .
m ase
.co


a g l Page 4 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Q2 Pluck the rubber band again and watch it carefully. Is it vibrating?

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

the blur width across the band ≈ twice the amplitude of the vibration

Check it yourself: touch the plucked band very lightly with a fingertip. You feel the
buzz — and the sound dies at once, because your finger absorbs the vibrational
energy.

Q3 Wait till the rubber band stops vibrating. Do you still hear the sound?

No. The moment the band comes to rest, the sound stops.

Why it happens: the sound was being fed by the band's own vibrational energy.
Each swing hands a little energy to the air (and a little is lost as heat inside the
stretched rubber). Once that store is used up the band stops, no new compressions
are made, and there is nothing left to reach your ear.

This one observation is the whole conclusion of the activity: sound is produced by vibrations;
no vibration, no sound.

Q4 Change 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?

Yes, the sound changes. Stretch the band tighter and the note becomes shriller (higher pitch);
loosen it and the note becomes deeper (lower pitch).

Page 5 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

WHAT YOU DO HOW THE BAND FREQUENCY TIME WHAT YOU
VIBRATES Ν PERIOD T HEAR

Stretch it more Snaps back faster higher shorter shriller, high pitch
(higher tension)

Loosen it (lower Returns sluggishly lower longer deeper, low pitch
tension)

Pluck harder Larger swing about the unchanged unchanged louder (larger
mean position amplitude)

Why it happens: a tighter band is pulled back to its mean position by a bigger
restoring force, so it completes each to-and-fro trip in less time. Since ν = 1/T, a
shorter time period means a higher frequency, and higher frequency is heard as
higher pitch.

Tip: notice that tension changes the pitch while how hard you pluck changes the
loudness. That is the difference between frequency and amplitude, in your own
hands. It is also how a sitar or tanpura is tuned — by turning the pegs to change the
tension.

Q5 Remove 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?

Sound is still produced, but it is much fainter than before.

Why it happens: the source of the sound is the vibrating band, and it is still
vibrating — so sound must be produced. But a thin band by itself sweeps only a
small volume of air, so it transfers very little energy per second to the surrounding
air. On the box, the band's vibrations were passed to the box and to the air column
inside it, and that much larger vibrating surface set far more air moving. Larger
amplitude of the air's density oscillation means more energy carried, hence greater
loudness.

Page 6 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Try This: hold the plucked band against a wooden desk or a steel tumbler. The
sound jumps in loudness again — the desk or tumbler is acting as a sounding board,
exactly like the body of a violin.

In-text Questions — Page 185
10.1 Production of Sound

Q1 How do humans and animals create sound? While talking or singing, gently touch
your throat. Do you feel vibrations anywhere?

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.

Why the pitch of a voice differs: the vocal cords behave like the stretched rubber
band of Activity 10.1. During adolescence, boys' vocal cords lengthen and thicken, so
they vibrate fewer times per second — lower frequency, and the voice 'deepens'. The
exact tone of each person also depends on the shape of the throat, mouth and nasal
cavities, which is why you can recognise a friend's voice instantly.

Check it yourself: hum with your fingers on the throat, then whisper the same
words. The buzzing disappears when you whisper — in a whisper the vocal cords are
not made to vibrate.

Activity 10.2 — Page 186

Page 7 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Let us explore — 10.1.1 Tuning fork

ACTIVITY 10.2

Q1 Strike 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 the
rubber 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.

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.

Why a tuning fork is used in sound experiments: unlike a voice or a bell, it gives
almost a single frequency (one clean note). Its frequency is fixed by the length,
thickness and material of the prongs, so it does not change however hard you strike
it — striking harder only makes it louder, not higher.

Tip: strike it on a soft rubber pad, never on a hard desk or the edge of a table. A hard
surface can bend or chip the prongs and permanently change the fork's frequency.

Page 8 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
m.
Now, gently touch a water surface with one of the vibrating prongs of the tuning
e
Q2

m l as
.co
fork. Do you see waves forming on the surface of water?

a g
se m
g l a
a

Yes — ripples spread outward from the touching prong, and a fine spray may even be

m
thrown up.
. co ag
e m
g l as
Why this matters: you cannot see the prong vibrating, but you can see what it does.
a
Each time the prong moves down it pushes the water down; each time it moves up
the water rushes back. That repeated push and release is what launches the ripples.
co m
m.
So the ripples are visible evidence that the prongs really are vibrating — and

m as e
.co l
therefore that the sound you heard was produced by vibrations.
a g
a s em
a gl Check it yourself: touch the water with a fork that has not been struck. There are no
ripples. Strike it and touch again — ripples appear at once, and the humming note
m a s
.co agl
gets quieter because the water is draining energy from the prongs.

se m
g l a
a
Repeat step 3 a few times while bringing the prongs of the tuning fork near your
m
Q3
ear in different orientations. Do you hear the sound?
. co
se m
o m g l a
m .c a
se — the sound is heard in every orientation, though its loudness changes a little as you

g l aYes
a turn the fork.

se m
o m g l a
m
Why it happens: the vibrating prongs.c disturb the air all around them, not in one line
a
e
only. Each disturbed particle scollides
a
l
with neighbours in all directions, so
ag
compressions and rarefactions spread out from the source as roughly spherical
waves (Fig. 10.10). Wherever you put your ear, some part of that expanding wave

co m
reaches it.
m .
m as e
.co a g l
emdirections, so in a few special directions their disturbances partly cancel. The general lesson
The change in loudness with orientation happens because the two prongs move in opposite

a s
agl stands: sound propagates in many directions from a source at once.
.c
s e m
m a
m . co agl
ase 186
agl
Pause and Ponder — Page

com
m .
m ase
.co


a g l Page 9 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

10.1.1 Tuning fork

PAUSE AND PONDER

Q1 Explore various ways of producing sound.

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

WAY OF PRODUCING WHAT VIBRATES EVERYDAY EXAMPLE
SOUND

Plucking A stretched string Sitar, veena, guitar, ektara, rubber band

Striking / hitting A stretched membrane or Tabla, dholak, taal (manjira), temple bell, tuning
a metal body fork

Blowing The air column inside a Bansuri, shehnai, conch (shankh), whistle
pipe

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 Speaker diaphragm Mobile phone, radio, loudspeaker
cone

The common thread: in every row, some object is made to move to and fro. That
motion squeezes and stretches the neighbouring air, and the resulting density
disturbance is what we call sound.

Page 10 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Q2 Make a list of different types of musical instruments and identify their vibrating
parts which produce sound.

INSTRUMENT FAMILY VIBRATING PART THAT HOW IT IS SET
MAKES THE SOUND VIBRATING

Sitar, veena, sarod, String (tat vadya) Stretched metal strings; the hollow Plucked
ektara gourd/body then amplifies

Sarangi, violin String Stretched strings Bowed (rubbed)

Tabla, dholak, Membrane Stretched skin/membrane of the Struck
mridangam, damru (avanaddha drum head, plus the air inside the
vadya) shell

Bansuri, shehnai, Wind (sushir The air column inside the hollow Blown
whistle vadya) pipe

Harmonium Wind Thin metal reeds Air pushed past by
the bellows

Manjira (taal), ghanti, Solid body (ghana The metal or clay body itself Struck
ghatam vadya)

Tuning fork Solid body The two prongs Struck on a soft
pad

Why most instruments have more than one vibrating part: a bare string or reed
moves very little air, so it is quiet. The gourd of a veena, the shell of a tabla and the
body of a violin are made to vibrate along with it, and they set a large volume of air
moving — so more energy reaches the listener per second and the sound is loud.

Did you know? The black patch at the centre of a tabla head, the syaahi, changes
how the membrane vibrates. It is why an Indian drum can be tuned to a definite
note, something most drums in the world cannot do — a point Sir C. V. Raman
studied in detail.

In-text Questions — Page 186

Page 11 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

10.2 Propagation of Sound

Q1 How does sound reach your ear from the source? Sound travels through air but
does it also travel through solids and liquids?

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.

Why solids carry sound best: in a solid the particles are packed close together and
are strongly bonded, so a push is handed on to the next particle almost immediately.
In a gas the particles are far apart and must travel some distance before colliding, so
the disturbance creeps along. That is why sound is fastest in solids (steel ≈ 5000 m
s⁻¹), slower in liquids (water ≈ 1500 m s⁻¹) and slowest in gases (air ≈ 340 m s⁻¹).

Activity 10.3 — Pages 186 – 187

Page 12 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Let us investigate — 10.2 Propagation of Sound

ACTIVITY 10.3

Q1 Now, 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?

other ear closed friend knocks the desk

ear pressed on 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.

Yes — and the knock or scratch sounds distinctly louder and sharper through the desk
than it did through the air.

Why it happens: the knock sets the wood of the desk vibrating. Because the
particles of a solid are closely packed and tightly bound, each particle passes the
disturbance to the next almost at once and very little energy leaks sideways.
Through air the same energy spreads out in all directions and thins out rapidly. So
the desk delivers a much larger share of the original energy to your ear.

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.

Page 13 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
m.
Did you know? Sound travels roughly 15 – 20 times faster in solids than in air. This is

m l a se
why a railway track carries the sound of an approaching train long before you hear it
o
m .c air, and why Question 11 on page 198 finds a delay aofgnearly 0.93 s
through the

l a se the two paths along a 340 m steel fence.
ag
between

co m
e m . ag
as
Activity 10.4 — Page 187
Let us investigate — 10.2 Propagation of Sound
a g l

m
ACTIVITY 10.4

co
se m.
o m l a
g (Fig. 10.6b). Do you
.c of the bucket and tap them against one another again
Now, submerge the two metal spoons in water without touching the sides or
a
Q1

se m
bottom
a
agl
again hear the sound produced?

m a s
m.co agl
l a se
a g

com
m .
m as e
.co a g l
se m
g l a
a water water

se m
(a) tapped in air
com
(b) tapped under water
g l a
m . a
ase
agl
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.

co m
m .
m
o l a se
m .c the clink is clearly heard — a little duller and different aingquality from the clink in air, but
se unmistakably there.
Yes,

a
agl c
.
The tap sets the spoons vibrating. The vibrating metal squeezes and releases the water around

s e m
m a
it, so compressions and rarefactions travel out through the water, cross the water surface into
the air, and reach your ear.
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 14 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Why the spoons must not touch the bucket: if they touched the sides or bottom,
the vibration could travel to your ear through the solid bucket and the table. By
keeping them clear, the only path left is water → air, so the experiment really is a test
of whether liquids carry sound.

Tip: the sound changes in quality because water and air carry the different
frequencies in the note unequally, and some of the sound is reflected back at the
water surface instead of crossing into the air.

Q2 If sound did not travel through liquids, would you have heard this sound?

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.

Why the logic is sound: this is how a scientist tests a claim — remove every
alternative path, and see whether the effect survives. Because the clink is heard, the
only remaining explanation is that water carried it. Sound therefore propagates
through liquids as well as through solids and gases.

Did you know? Sound travels about 4 – 5 times faster in water than in air, and far
further before fading. That is exactly why sonar, and not light or radio, is used to
find submarines and map the ocean floor.

In-text Questions — Page 187

Page 15 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

10.2 Propagation of Sound

Q1 A space where there is no medium (matter) is referred to as vacuum. Would you
hear sound in vacuum?

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.

Why it happens: a compression is a region where particles have been crowded
together. Fewer particles in the jar means fewer collisions to hand the crowding
forward, so less and less energy is relayed to the jar walls and out to your ear. With
no particles at all, there is nothing to compress and nothing to relay — so no sound
wave can exist. Sound is a mechanical wave: it needs a material medium.

Tip: notice what stays visible while the sound dies — light reaches you from the bell
throughout. Light is not a mechanical wave, so it crosses vacuum easily. That is the
cleanest way to remember the difference.

Pause and Ponder — Page 188
10.2.1 Sound needs a medium to propagate

PAUSE AND PONDER

Q1 Assertion (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.

(ii) Both A and R are true, and R is the correct explanation of A.

Page 16 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

A is true: in the bell jar experiment, once most of the air has been pumped out the bell is
still seen ringing but is practically inaudible.
R is true: sound is a mechanical wave and needs a material medium to propagate.
R explains A: pumping out the air removes the very particles that were relaying the
compressions and rarefactions. With almost no medium left, the wave cannot reach the jar
wall and then your ear — so the sound is not heard. The reason is not merely a true fact
placed beside A; it is the cause of A.

How to attack an assertion–reason question: check three things in order — (1) Is
A true on its own? (2) Is R true on its own? (3) Does R answer the question 'why is A
true?' Only if all three are yes do you choose 'R is the correct explanation'.

Activity 10.5 — Page 188
Let us observe — 10.3 Sound Waves

ACTIVITY 10.5

Q1 Give 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.

Yes. A single bunched-up region — a place where the turns are pressed close together — races
along the slinky towards your friend.

Page 17 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Why it happens: your push squeezes the first few turns together. Each squeezed
turn presses on the turn ahead of it and then springs back, so the squeeze is handed
forward from turn to turn. What moves along the slinky is the pattern of crowding,
not the turns themselves — each turn simply moves forward a little and comes back
to where it started.

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.

Q2 Now, 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?

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 THE SLINKY IN A SOUND WAVE IN AIR

Turns close together Compression — density above average

Turns spread apart Rarefaction — density below average

The marked turn oscillating in place An air particle oscillating about its mean position

The bunching pattern moving forward The sound wave travelling forward

Why this makes sound a longitudinal wave: the particles vibrate along the same
line the wave travels along. A wave with that property is called a longitudinal wave.
Compare a wave you make by shaking a rope up and down: there the particles move
perpendicular to the travel direction, and that is a transverse wave (Fig. 10.13).

Page 18 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
m.
Tip: the marked turn is the whole point of the activity. It is the visible proof that

m ase
l
energy travels while matter stays put.

m .co a g
l a se
a g
In-text Questions — Page 190
co m
. ag
10.3 Sound Waves

se m
l a
g particles), is the propagation of sound waves
If there is no medium (i.e.,ano
Q1
possible?

co m
se m.
m l a
o
.c a material medium, sound waves cannot propagate.ag
m
se wave is a pattern of higher and lower particle density. If there are no particles, there is
No. Without

a
Alsound
agno density to raise or lower, and nothing to collide and hand the disturbance forward. Waves
m
that need a material medium in this way are called mechanical waves, and sound is one of
a s
them.
m.co agl
l a se
MECHANICALg
a WAVE NON-MECHANICAL WAVE

co m
Needs a Yes No
medium?
m .
m as e
.co a g l
m
Examples Sound; waves on a slinky; seismic waves Light

a s eCan
agl cross
vacuum?
No Yes — which is why sunlight and starlight reach
the Earth

se m
com g l a
. a
Type Sound is longitudinal; seismic waves can Light is transverse

m
ase
be either

agl
Did you know? Earthquakes send out both kinds of mechanical wave through the

. c om
Earth. The longitudinal seismic waves travel fastest and are the first to be picked up

s em earthquake
by a seismograph, which is why they are used to give the earliest
m a
m
o
.cwarnings. agl
l a se
ag
.c
s e m
Pause and Ponder — Page 191 m a
e m . co agl
g l as
a

co m
m .
m as e
.co


a g l Page 19 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

10.3 Sound Waves

PAUSE AND PONDER

Q1 Assertion (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.

(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.

Why the pattern can move while the particles do not: think of the marked turn
on the slinky in Activity 10.5. It goes forward a little, comes back, goes forward again
— yet the bunched-up region sweeps the whole length of the slinky. Each particle
pushes its neighbour and steps back; the neighbour pushes the next one and steps
back. The crowding is handed on, but no particle makes the journey. What is
transported from the source to your ear is energy, not matter.

Check it yourself: if R were true, standing in front of a loudspeaker would feel like
standing in a steady wind blowing out of it. It does not — proof that the air is not
flowing towards you.

Activity 10.6 — Pages 191 – 192

Page 20 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Let us experiment — 10.4 Energy of Sound Waves

ACTIVITY 10.6

Q1 Produce a loud sound near the bowl without touching it. Observe the grains on the
sheet. Does the sound have any effect on the grains?

Yes — the grains jump, dance and shuffle across the stretched sheet, even though nothing
has touched the container.

Why it happens, step by step:
1. The struck metal plate vibrates and sends compressions and rarefactions out
through the air.
2. These reach the tightly stretched sheet. A compression presses on it, a rarefaction
releases it — so the sheet is pushed and released many times a second and starts to
vibrate.
3. The vibrating sheet throws the grains resting on it up into the air.

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.

Tip: spread the grains evenly and do not let them clump — a clump is heavier and
needs more energy to lift, so the effect becomes hard to see.

Q2 Repeat 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.

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.

Page 21 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

WHAT YOU CHANGE WHAT YOU REASON
OBSERVE

Strike the plate harder (louder Grains jump much Larger amplitude → more energy carried
sound) higher 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 Lighter grains jump Same energy delivered to a smaller mass
powder) instead of heavy ones (salt, higher and more easily gives a larger displacement
rice)

Loosen the stretched sheet Response becomes A slack sheet does not follow the rapid
weaker and duller pushes of the sound wave faithfully

Why loudness and amplitude go together: striking the plate harder does not
change how many times per second it vibrates (the frequency is the same, so the
note sounds the same). It changes how far the plate swings, which makes the
density change in each compression bigger — a larger amplitude. A larger-
amplitude wave carries more energy, so more energy reaches the sheet each second
and the grains fly higher.

Pause and Ponder — Page 192
10.4 Energy of Sound Waves

PAUSE AND PONDER

Q1 When 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

(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.

Page 22 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

OPTION VERDICT WHY

(i) Air particles near the tuning Wrong Those particles only vibrate about their mean positions;
fork they never make the journey

(ii) Energy carried by sound Correct Energy is transferred from particle to particle by collisions
waves

(iii) The tuning fork material Wrong The steel stays in your hand; nothing leaves it

(iv) A continuous stream of Wrong Compressions and rarefactions alternate; there is no one-
compressed air way flow of air

Tip: a microphone makes this concrete. Sound energy arriving at its diaphragm is
converted into electrical energy; a speaker runs the same chain backwards. Energy is
the thing being passed around at every stage.

Pause and Ponder — Page 193

Page 23 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

10.5 Graphical Representation of a Sound Wave
co m
em.
m as
PAUSE AND PONDER

.co a g l
a s emvariation of density of the medium for two sound waves is shown in Fig. 10.17
gl
The
a
Q1
(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

co m
. ag
(a) and (b).
e m
g l as
a

co m
e m.
m l as
m .co a g
l a se (a) (b)
a g
m a s
m .co agl
l a se
a g

co m
.
a s em
com l
(c) (d)

e m . ag
a s
agl
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.

se m
com g l a
m . a
ase

agl
Labelling (a) and (b). Wherever the dots are crowded together the density is above average —
mark that C. Wherever the dots are thinned out the density is below average — mark that R. In

co m
.
the printed figure, guide lines are drawn through the centres of these regions.

se m
m l a
In (a) the first dashed guide line — the one at the left-hand edge — runs through a dense
o agR alternately at every dashed line
.cband, so mark C there, R at the next dashed line, and C and
se m after that. The solid lines midway between them mark the places where the density is exactly
g l a
a average.
c
m .
In (b) the guide lines stand in exactly the same places, but the dot pattern is shifted by half a
s e
. c om
wavelength: the first dashed line now runs through a thinned-out band, so it is R there and C
a g la
s e
at the next dashed line. A placemthat is a compression in (a) is a rarefaction in (b). Both strips
a
agl
have the same wavelength.

com
m .
m ase
.co


a g l Page 24 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Labelling the axes of (c) and (d). Along the horizontal axis write Distance; along the vertical
axis write Density; and draw a horizontal dashed line for the average density.
Drawing the curves. Every C becomes a crest (maximum density) and every R becomes a
trough (minimum density). So:

(c), matching (a): the curve starts at a crest above the average-density line, at the first
dashed guide line on the vertical axis.
(d), matching (b): the same curve shifted by half a wavelength — it starts at a trough where
(c) had a crest.

(a) / (c)

C R C R C R C

Density avg.

Distance →

(b) / (d)

Density

R C R C
Top: the dot pattern of (a) with C and R marked, and below it the matching density–distance graph (c)
— a crest at every C, a trough at every R. Bottom: graph (d) for strip (b), the same wave shifted by half
a wavelength.

Why the graph looks like a smooth wave when the picture is only dots: the
graph does not plot particles — it plots the density of the medium at each distance,
at one instant. Density rises smoothly to a maximum in the middle of a compression
and falls smoothly to a minimum in the middle of a rarefaction, so the plot comes
out as a smooth curve about the average-density line.

Tip: keep the vocabulary straight — crest is the highest point of the graph
(maximum density, i.e. a compression), trough is the lowest point (minimum density,
i.e. a rarefaction). A sound wave is longitudinal even though its graph looks like a
wiggle.

Page 25 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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

ACTIVITY 10.7

Q1 Compare the musical notes by taking the ratio of each frequency with respect to
the 'Sa'. Do you observe any pattern?

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 TYPICAL MEASURED 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 Pa = 360 Hz ÷ 240 Hz = 1.5 = 3/2

Ratio for upper Sa = 480 Hz ÷ 240 Hz = 2.0

The pattern: frequency rises from Sa to the upper Sa, and each step is a simple
whole-number ratio. Because the upper Sa is exactly double the lower Sa, the two
ends of the scale form an octave — an interval between two notes where one has
twice the fundamental frequency of the other (for example 200 Hz and 400 Hz). This
is why the two Sa's sound like 'the same note, higher up'.

Page 26 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Tip: your own numbers will not match the table exactly, because the pitch you start
on depends on your voice or on the app's setting. What must come out right are the
ratios, especially the 2 : 1 for the octave.

Q2 If both voice and mobile-generated notes are used, compare their frequencies for
the same musical notes.

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.

Why the two can differ and still be 'the same note': what our ear judges as a note
is fixed mainly by the fundamental frequency. If you sing Sa at 245 Hz while the app
plays 240 Hz, the note is recognisably the same, and Pa will still be about 1.5 times
your own Sa. What differs is the timbre — the pattern and strength of the overtones
— which is exactly why a flute, an ektara and a human voice playing the same note
at the same loudness still sound completely different.

Pause and Ponder — Page 195
10.6.1 Wavelength, frequency and time period

PAUSE AND PONDER

Q1 Conduct 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?

Yes — for the same length and the same tension, the thin band vibrates faster than the
thick one.

Page 27 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

BAND RATE OF FREQUENCY TIME PERIOD T = SOUND HEARD
VIBRATION Ν 1/Ν

Thin band Faster Higher Shorter Shriller — high
pitch

Thick Slower Lower Longer Deeper — low
band pitch

ν = 1/T (Eq. 10.1)

If the thin band completes 400 oscillations in 1 s, ν = 400 Hz and T = 1/400 s = 0.0025 s

If the thick band completes 200 oscillations in 1 s, ν = 200 Hz and T = 1/200 s = 0.005 s

Why thickness matters: a thicker band carries more mass on every centimetre of
its length. The same restoring force must now accelerate more mass, so the band is
slower to swing back and forth — each complete oscillation takes longer. A longer
time period means a lower frequency, because ν and T are inverses of each other.

Did you know? This is why a sitar or veena carries strings of different thicknesses
side by side: the thick strings give the low notes and the thin ones the high notes, all
at roughly the same tension.

Q2 If 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?

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

Page 28 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
m.
Why the units work out: hertz means 'per second', so 20 Hz literally reads '20

m l a se
oscillations every second'. Multiplying by 60 s cancels the seconds and leaves a pure
o
.c a count of oscillations, which is exactly what was asked
a gfor. Always
m
number —

l a se the minute to seconds before multiplying; that single step is where most
g
convert
amistakes happen.

o m
Tip: 20 Hz sits right at the lower edge m
e
. c ag
s
of the human audible range (20 Hz – 20 kHz).
a produce infrasound, which we cannot hear
Anything slower than this piston lwould
g
at all. a

co m
se m.
o m l a
Forcthe sound wave represented by the graph shown in Fig.g10.19, what is half of its
m . a
e
Q3

g l as wavelength?

a
m a s
m.co agl
l a se
a g

co m
Density

m .
m as e
.co a g l
se m
g l a
a
0 1.5 3.0 4.5 Distance (cm)
se m
com g l a
m . a
ase
Fig. 10.19, page 195 — graphical representation of a sound wave (density against

agl
distance).

co m
m .
m as e
.co
Half the wavelength is 1.5 cm (= 0.015 m).
a g l
a s emRead the graph first. The curve starts on the average-density line at 0, rises to a crest, comes
agl back down through the line, dips to a trough, and returns to the line at the mark labelled 3.0
.c
cm. That completes one full density oscillation — so one whole wavelength is measured on the
s e m
m a
. co agl
printed scale as

e m
g l as
a

co m
m .
m ase
.co


a g l Page 29 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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

λ = 3.0 cm
λ/2 = 1.5 cm

Density

0 1.5 Distance (cm) 3.0 4.5

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.

Why half a wavelength is worth naming: it is exactly the distance from the centre
of a compression to the centre of the next rarefaction. Half a wavelength away from
a crest, the density has swung from its maximum right down to its minimum.

Tip: a common slip is to measure from a crest to the next trough and call that the
wavelength. That gap is only λ/2. Always measure crest-to-crest or trough-to-trough.

What if … — Page 196

Page 30 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

10.6.3 Speed of Sound

WHAT IF …

Q1 What 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?

No — music would fall apart as it travelled, and the further you sat from the stage the
worse it would sound.
A sung note or an instrument's note is never a single frequency: it is a fundamental together
with several overtones. All of them leave the stage at the same instant.

Suppose the fundamental (200 Hz) travelled at 340 m s⁻¹ and an overtone (400 Hz) at 350 m
s⁻¹.

Time for the fundamental to cross a 34 m hall = 34 m ÷ 340 m s⁻¹ = 0.100 s

Time for the overtone = 34 m ÷ 350 m s⁻¹ = 0.097 s

Gap on arrival = 0.100 s − 0.097 s = 0.003 s, and it grows with distance

The overtones of a single note would arrive before or after the fundamental, so the note
would smear out and its timbre would change with distance — a tabla would not sound like
a tabla at the back of the hall.
The singer's high notes and the accompanying instrument's low notes would arrive at
different times, so the two would drift out of rhythm even though they were played
together.
Different listeners at different distances would hear different rhythms and different tone
colours from the same performance.

Why the real world is kind to music: in air, the speed of sound depends only on the
medium — on temperature and humidity — and not on the source or its frequency.
So every frequency in a chord travels at the same 344 m s⁻¹ and arrives together,
keeping the note whole. If the frequency changes, it is the wavelength that adjusts
(λ = v/ν), never the speed.

Did you know? Some engineered materials, such as porous foams and specially
designed structures, do make sound speed depend on frequency. Acoustic
engineers use them precisely because of that unusual behaviour.

Page 31 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Pause and Ponder — Page 197
10.6.3 Speed of Sound

PAUSE AND PONDER

Q1 Table 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

STATE SUBSTANCE/MEDIUM APPROXIMATE SPEED

Solid Steel 5000 m s⁻¹

Liquid Water 1500 m s⁻¹

Gas Air 340 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.

(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

Notice that the ratios have no unit — m s⁻¹ divided by m s⁻¹ cancels. And combining the two,
vsteel : vair = 5000 : 340 ≈ 14.7 : 1, which agrees with the chapter's statement that sound travels
typically 15 – 20 times faster in solids than in air.

Page 32 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Why solid > liquid > gas: a sound wave moves by particles bumping into their
neighbours. In steel the atoms are locked in place by strong bonds and are very
close together, so a push is passed on almost instantly. In water the molecules are
close but free to slide, so the push is passed on more slowly. In air the molecules are
far apart and weakly interacting, so each one must travel a comparatively long way
before it collides — the disturbance creeps along. Closer packing and stiffer bonding
therefore mean a higher speed of sound.

Tip: the table is quoted at 15 ºC for a reason. Speed in a gas depends noticeably on
temperature — air carries sound at 331 m s⁻¹ at 0 ºC but 344 m s⁻¹ at 22 ºC — so a
speed for a gas is meaningless without a temperature.

Pause and Ponder — Page 198

Page 33 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

10.6.3 Speed of Sound
co m
e m.
m as
PAUSE AND PONDER

.co a g l
a s emfriends are standing along a steel fence at a distance of 340 m from each other
gl
Two
a
Q1
(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

. com
Table 10.1, calculate the time difference between the sound that reached Gunjan
ag
a s em it have been possible for her to distinguish
through the air and the steel. Would

agl time interval between two sounds must be at least
between the two sounds? (The
0.1 s to be heard separately.)

co m
se m.
m l a
m .cofriend knocks with a metal object ag ear on the fence
se
Gunjan’s

g l a
a
m a s
m .co steel fence agl
l a se
a g 340 m

co m
m .
o m l a seby two paths at once:
m .c ag
Fig. 10.23, page 198 — redrawn sketch. The knock reaches Gunjan

l a se along the steel fence and through the air.

ag
Table 10.1: Speed of sound in different media at 15 °C
se m
com g l a
STATE
.
SUBSTANCE/MEDIUM APPROXIMATE SPEED
a
a s em
agl
Solid Steel 5000 m s⁻¹

m
Liquid Water 1500 m s⁻¹

340 m s⁻¹ m.
co
se
Gas Air

co m l a
m . ag
l a se
ag ANSWER
c
m .
The time difference is about 0.93 s, and yes — Gunjan can easily hear the two knocks
s e
separately.
. com a gla
a
One knock sends sound along two
s empaths at once: through the steel fence, and through the air.
The distance is the same, g
a 340l m; only the speed differs.

com
m .
m ase
.co


a g l Page 34 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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

The one through the steel arrives first, then almost a second later the same knock arrives
through the air.

steel fence, 340 m

friend knocks Gunjan listens

through steel: 5000 m s⁻¹ → arrives in 0.068 s

through air: 340 m s⁻¹ → arrives in 1.000 s

gap = 0.93 s, well above the 0.1 s the ear needs
The same knock, two paths. Steel carries it about 14.7 times faster than air, so the two arrivals are
nearly a second apart.

Page 35 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Why the ear needs 0.1 s: the brain merges two sounds arriving closer together than
about 0.1 s into a single sensation, which is also why an echo needs a reflecting
surface at least 17 m away. Here the gap is nine times larger than that limit, so there
is no difficulty at all.

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

ACTIVITY 10.8

Q1 Increase the frequency in steps of 100 Hz up to 1000 Hz and describe how the sound
changes.

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

Why the pitch rises: pitch is how our brain interprets frequency. A higher frequency
means the density of the air at your eardrum swings from maximum to minimum
and back more times each second, and the ear reads that as a shriller sound. In
general high pitch goes with high frequency and low pitch with low frequency,
although the exact relation is not a simple one. Note that the speed of sound stays
344 m s⁻¹ throughout — it is the wavelength that shrinks, since λ = v/ν.

Tip: loudness may seem to change a little as you sweep the frequency. That is not
the app — the human ear is naturally most sensitive around 1000 – 4000 Hz, so
equal-energy sounds do not seem equally loud.

Page 36 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Q2 Next, set the frequency to 50 Hz. Reduce the frequency till about 20 Hz or the point
where you cannot hear the sound anymore.

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

Why the sound disappears: the wave is still being produced and is still travelling
through the air. What fails is the detector. Below about 20 Hz the eardrum and
cochlea no longer respond well enough to send a usable signal to the brain, so we
register nothing. Nothing has happened to the physics of the wave — only to our
ability to sense it.

The exact cut-off differs from person to person and narrows with age, so the point at which the
sound vanishes will not be the same for everyone in the class. Many students hear it down to
about 20 Hz; some lose it at 30 Hz.

Did you know? Elephants communicate over kilometres using infrasound below 20
Hz, and dogs, cats, bats and dolphins hear ultrasound above 20 kHz. Our range is
just one window on a much wider spectrum.

In-text Questions — Page 199
10.6.4 Human perception of sound

Q1 Can humans hear all sounds?

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

Page 37 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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.

Why there is a limit at all: hearing is a mechanical process. The eardrum and the
tiny bones behind it must be able to follow the incoming density changes, and the
hair cells in the cochlea must convert them into electrical signals. Both fail outside a
certain band of frequencies. This is a limit of the detector, not of the sound — the
wave is perfectly real either way.

Did you know? The upper limit falls with age, and it also falls with damage from
loud sound. A teenager may hear up to 20 kHz while an adult may stop at 14 – 16
kHz, which is why the range is quoted as varying 'from person to person'.

Pause and Ponder — Page 201
10.7.1 Echo

PAUSE AND PONDER

Q1 An 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.

The reflecting surface must be at least 34.3 m away.

Page 38 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

The sound has to make a round trip: from the source to the wall and back to the listener. So the
co m
total path is twice the distance we are asked for.
e m.
m l as
m .co a g
l a se
total distance travelled = speed × time
g
a= 343 m s⁻¹ × 0.2 s

co m
ag
= 68.6 m
m .
as e
This is the to-and-fro distance, soa g l
distance to the reflecting surface = 68.6 m ÷ 2
co m
em.
m as
= 34.3 m

.co a g l
se m
g l a
Written as a single formula, d = v t / 2 = (343 m s⁻¹ × 0.2 s) / 2 = 34.3 m. Any wall closer than this
a returns the echo too soon for the experiment.
m a s
.co
Why the factor of 2 is not optional: forgetting it doubles the answer to 68.6 m.
em agl
a s
gl is always the round-trip time, so always halve either
Every echo, sonar and ultrasonic-sensor calculation in this chapter has the same
structure — the measuredatime
the time or the distance, never neither and never both.

co m
m .
o m l a se
.c at all is set by the ear's 0.1 s limit: d = (340 m s⁻¹ × 0.1ags) / 2 = 17 m. This
Tip: compare this with the book's own case. The minimum distance for hearing any

m
aseexperiment simply demands twice the delay, so it needs roughly twice the distance.
echo

agl
se m
com g l a
m . a
What if … — Page 202
ase
agl
10.8 Ultrasonic and Infrasonic Waves, and their Applications

co m
WHAT IF …

m .
o m l a secan? What would be the
.c ag
What if … humans could detect ultrasonic waves like dogs

em
Q1

l a s advantages and disadvantages?

ag c
m .
e

om a s
. c agl
We would gain a great deal of extra information about the world, but we would also be

s e
flooded with noise we currently mdo not notice at all.
a
agl

com
m .
m ase
.co


a g l Page 39 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

ADVANTAGES DISADVANTAGES

We could hear bats and dolphins hunting, and follow Constant noise from ultrasonic pest repellers, some
animal communication that is silent to us today. machinery, vehicle parking sensors and cleaning
equipment.

A doctor or technician could hear a flaw in a metal block Ultrasonic welding and industrial cleaning plants would
or a blocked pipe directly, without instruments. become unbearable workplaces and would need heavy
shielding.

Ultrasonic reversing sensors, security alarms and range Higher-frequency sound is also higher-energy for the
finders could be checked by ear. same amplitude, so long exposure could damage hearing
faster.

Better echolocation of our own: short-wavelength sound Speech and music would be buried under a permanent
reflects from small objects, so we could judge small hiss, making concentration and conversation harder.
obstacles in the dark.

Why ultrasound is so useful for locating things: at 40 kHz in air, λ = 344 m s⁻¹ ÷ 40
000 s⁻¹ ≈ 0.0086 m, less than a centimetre. A wave reflects sharply only from objects
that are not much smaller than its wavelength, so ultrasound bounces cleanly off
small targets that ordinary sound would simply flow around. That is exactly why it
can image a baby's organs, break a kidney stone, or find a crack inside a metal block.

Tip: when a question asks for advantages and disadvantages, give the physical
reason for each side rather than a list of opinions. Here both sides come from the
same fact — short wavelength and high energy.

Pause and Ponder — Page 203
10.8.1 Echolocation

PAUSE AND PONDER

Q1 Sound 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?

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

Page 40 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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

Or in one step: d = v t / 2 = (1500 m s⁻¹ × 4 s) / 2 = 3000 m.

Why sonar and not light or radar: water absorbs and scatters light within a few
tens of metres, and radio waves are absorbed almost immediately. Sound, on the
other hand, travels about 4 – 5 times faster in water than in air and keeps going for
kilometres before its intensity falls away. So sound is the only practical probe for the
deep sea — which is why ships map the sea floor and find submarines and wrecks
with sonar (sound navigation and ranging).

Tip: if a question ever gives you the one-way time instead of the return time, do not
halve it. Read carefully whether the time quoted is 'to return', 'for the echo', or 'to
reach'.

Revise, Reflect, Refine — Pages 204 – 206
End-of-chapter questions

REVISE, REFLECT, REFINE

Q1 Which 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

(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.

Page 41 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

OPTION IS IT TRUE OF SOUND? DOES IT PROVE 'MECHANICAL'?

(i) Shows reflection Yes No — light reflects too, and light is not mechanical

(ii) Needs a medium Yes Yes — this is the definition

(iii) Has frequency Yes No — every wave has a frequency

(iv) Carries energy Yes No — every wave carries energy

Why the other three fail: they are all true statements about sound, but they are
true of all waves. An observation only counts as evidence for a classification if it
distinguishes that class from the others.

Q2 For 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

(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.

Speed: v depends only on the medium → unchanged

Wavelength: from v = ν × λ, λ = v/ν → decreases when ν increases

Time period: T = 1/ν → decreases when ν increases

Check with numbers (air, v = 344 m s⁻¹):

ν = 200 Hz → λ = 344 ÷ 200 = 1.72 m, T = 0.005 s

ν = 400 Hz → λ = 344 ÷ 400 = 0.86 m, T = 0.0025 s

Page 42 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Why the speed does not change: the speed of sound is fixed by the medium —
how closely packed and how tightly bound its particles are, plus temperature and
humidity. The source can only decide how often it sends compressions out. The
medium then decides how fast each one travels, and the wavelength adjusts itself to
fit.

Q3 If 20 compressions pass a point in 4 seconds, the frequency is (i) 80 Hz (ii) 5 Hz (iii) 10
Hz (iv) 0.2 Hz

(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.

Time period T = 1/ν = 1 ÷ 5 Hz = 0.2 s — one compression every 0.2 s, which checks out: 20

× 0.2 s = 4 s ✓

Why 0.2 Hz is the trap: option (iv) is 4 ÷ 20, which is the time period in seconds, not
the frequency. Watch the units — hertz means 'per second', so the count must go on
top and the time underneath.

Q4 In a room, the reflected sound reaches the ear 0.05 s after its production. Will it
produce an echo or reverberation? Justify your answer.

It will produce reverberation, not an echo.

Page 43 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
em.
Time gap between the direct sound and the reflected sound = 0.05 s
m l as
.co
Minimum gap needed to hear two sounds separately = 0.1 s
m a g
l a se
g
a0.05 s < 0.1 s → the brain cannot separate them

. c om ag
s
Because the reflected sound arrives while the
e m original is still being registered, the two merge.
a g
The sound seems to persist or linger alalittle after the source has stopped — and that
persistence caused by reflections is exactly what reverberation means.
We can also check how big such a room is:
co m
e m.
m l as
.co
d = v t / 2 = (340 m s⁻¹ × 0.05 s) ÷ 2 = 17 m ÷ 2 = 8.5 m
m a g
l a se
g
a A wall 8.5 m away is an ordinary large room — and indeed the minimum distance for a true
s
echo is 17 m, twice as far.
m a
m.co agl
ECHO
l a se REVERBERATION

Time gap At least 0.1 sa g Less than 0.1 s (typically within 0.05 s)

co m
.
Reflections Usually a single reflection from a distant Many reflections from the walls of a

se m
om a
involved surface hall

What.c
l
agsound prolonged, blurred into
m
ase
you hear The sound repeated, clearly separate The

agl
itself

Where Hills, cliffs, long empty corridors Halls, auditoriums, empty rooms
se m
com g l a
m . a
ase
agl
Tip: auditoriums are designed with sound-absorbing panels, upholstered chairs and
curtains so that reverberation is controlled. Too little and the hall sounds dead; too
much and speech turns into a garble.

co m
m .
m as e
.co a g l
se m
g l a
a c
m .
m a s e
e m . co agl
g l as
a

com
m .
m ase
.co


a g l Page 44 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Q5 Graphs 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?

Density

Density
Distance Distance
(a) (b)

Fig. 10.30, page 205 — density–distance graphs of two sound waves, drawn to the same
scale on both axes.

(i) Wave (a) has the greater wavelength. (ii) Wave (a) also has the smaller amplitude.
Read the printed graphs across the same length of the distance axis:

WAVE (A) WAVE (B)

Complete cycles shown over the same distance 3 6

Wavelength (crest to crest) Larger — about twice that of (b) Smaller

Height of a crest above the average-density line Smaller Larger — roughly 1.5 times

So Greater λ, smaller amplitude Shorter λ, larger amplitude

Page 45 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Density

(a) Distance (b) Distance
3 cycles → longer λ 6 cycles → shorter λ
shorter crests → smaller amplitude 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.

Why the scales had to be the same: wavelength is read off the horizontal axis and
amplitude off the vertical axis. If the two graphs used different scales, a wave could
look longer or taller without actually being so. Once the scales match, counting
cycles and comparing crest heights is a fair comparison. In sound terms, (b) has the
higher frequency (ν = v/λ) and, having the larger amplitude, also carries more energy
— so it would be heard as a shriller and louder sound.

Page 46 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Q6 The 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.

Density

Distance

Fig. 10.31, page 205 — the sound waves emitted by the three sources, drawn on one pair
of axes and deliberately left unlabelled.

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. COMPLETE CYCLES WAVELENGTH FREQUENCY Ν LABEL
10.31 ACROSS THE AXIS = V/Λ

Green (tallest, most 4 Shortest Maximum A
closely spaced)

Red (middle) 3 Middle Middle B

Blue (widest, 2 Longest Minimum C
smallest crests)

Page 47 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

All three travel in the same medium, so v is the same for all three.

From v = ν × λ, ν ∝ 1/λ — shorter wavelength means higher frequency.

Since λgreen < λred < λblue, νgreen > νred > νblue

∴ green = A, red = B, blue = C

Why you must not use the crest height: the tallest curve here also happens to be
the one with the highest frequency, but that is a coincidence of the drawing. Crest
height is amplitude, which tells you about loudness and energy, not about
frequency. Frequency is read from how tightly packed the cycles are along the
distance axis.

Tip: count from crest to crest, not from the start of the graph. If a curve begins mid-
cycle, counting crests is far more reliable than trying to count whole cycles by eye.

Q7 Draw a graph to represent a sound wave for which the density amplitude is 3 units
and wavelength is 4 cm.

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.

Page 48 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
λ = 4 cm
em.
m l as
+3
m .co a g
l a se
a g amplitude = 3 units

co m
Density
e m . aver ag
g l as
a

co m
e m.
m l as
.co a g
−3

se m
g l a
a
0 4 8 12 Distance (cm) 16
Density–distance graph with a density amplitude of 3 units and a wavelength of 4 cm. Each crest is a

m a s
.co agl
compression, each trough a rarefaction.

a s em
agl (1) crest-to-crest and trough-to-trough must both
How to check your own drawing:
measure 4 cm; (2) the crest must be exactly as far above the dashed line as the

. com
trough is below it, namely 3 units; (3) the curve must cross the average-density line
m and R at each
at every quarter-wavelength, i.e. every 1 cm. Also mark C at eachecrest
s
m a
gl and rarefactions.
co — that reminds you the graph is describing compressions
m .
trough
a
ase
agl
m
Tip: if a speed were also given, you could add the frequency. For example in air at

a se
com l
344 m s⁻¹, ν = v/λ = 344 m s⁻¹ ÷ 0.04 m = 8600 Hz.
. a g
m
ase
agl
Q8 In a movie, while showing the explosion of a spacecraft in space, a flash of light is

co m
.
shown along with sound at the same time. What are the errors in this depiction?

em
m l as
.co a g
emThere are two errors, and the first one is fatal.

a s
agl 1. There should be no sound at all. Outer space is a near vacuum. Sound is a mechanical
.c
s e m
m gl a
wave and needs a material medium; with almost no particles there is nothing to compress

. co a
m
and nothing to pass the disturbance on. The explosion would be completely silent to a

l a se a mechanical wave, so the flash would be seen — which is why
distant observer. (Light is not
sunlight reaches thea g across empty space.)
Earth

com
m .
m ase
.co


a g l Page 49 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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.

speed of light = 3 × 10⁵ m s⁻¹ speed of sound in air = 340 m s⁻¹

ratio = 3 × 10⁵ ÷ 340 ≈ 880 000 times faster

For a source 1700 m away in air:
time for the sound = 1700 m ÷ 340 m s⁻¹ = 5 s

time for the light = 1700 m ÷ (3 × 10⁵ m s⁻¹) ≈ 0.0000057 s — effectively instant

The everyday version of the same physics: during a thunderstorm you see the
lightning first and hear the thunder seconds later, even though both were produced
at the same instant. Counting that delay and multiplying by 340 m s⁻¹ tells you how
far away the strike was.

Tip: a fair correction for the film would be a bright silent flash, followed by nothing
— or, if the camera is meant to be inside a spacecraft, the sound arriving through the
ship's own metal hull, which is a medium.

Q9 A 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.

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

Page 50 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Or in a single step, since v = λ/T: T = λ/v = 3.44 m ÷ 344 m s⁻¹ = 0.01 s.

What the answer means physically: T = 0.01 s is the time the wave takes to move
forward by exactly one wavelength, and equally the time a single air particle takes to
complete one full oscillation about its mean position. At 100 Hz this is a low,
humming note, comfortably inside the audible range of 20 Hz – 20 kHz.

Tip: watch the units. m ÷ (m s⁻¹) = s, so dividing a wavelength by a speed always
gives a time. If your answer comes out in the wrong unit, you have divided the
wrong way round.

Q10 A 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?

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

Why ultrasound is used here: ultrasonic waves have very short wavelengths, so
they reflect sharply from an object as small as a ship's hull instead of spreading
around it, and they can be sent out as a narrow beam that gives a definite direction.
Sound also travels far in water without dying out, unlike light. That combination —
direction, sharp reflection and long reach — is what makes sonar work.

Page 51 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Tip: the number 1525 m s⁻¹ is close to the 1500 m s⁻¹ in Table 10.1. Speed in
seawater varies a little with temperature, depth and salinity, so a question will
always tell you which value to use.

Q11 A 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.

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

We can also confirm that 40 kHz really is ultrasonic and see why it suits the job:

λ = v / ν = 345 m s⁻¹ ÷ 40 000 s⁻¹ = 0.0086 m ≈ 0.9 cm

40 kHz > 20 kHz → ultrasonic, inaudible to the driver

Why the system works so well: the round trip takes only about 7 ms, so the sensor
can fire many pulses every second and update the beep almost continuously as the
car creeps back. The wavelength of under a centimetre means the pulse reflects
cleanly off a low bollard or a kerb. And because 40 kHz is above the audible range,
the driver hears only the electronic beep, never the ultrasound itself.

Page 52 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Tip: milliseconds are natural units here. 0.00696 s = 6.96 ms, since 1 ms = 10⁻³ s.

Q12 The 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.

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

Why cold air slows sound down: in a gas, sound travels by molecules colliding with
their neighbours. Temperature is a measure of how fast those molecules are already
moving about. At 22 ºC the molecules are moving faster than at 0 ºC, so they meet
their neighbours sooner and pass the compression on more quickly. Cool the air and
every collision is slightly delayed, so the whole disturbance creeps forward more
slowly. Humidity works the same way — moister air carries sound slightly faster.

Tip: notice how small the effect is: a 22 ºC drop changes a 5 s journey by only about
4%. That is why the rough rule for a thunderstorm — 'count the seconds and
multiply by about 340 m s⁻¹' — works well enough on a winter night as on a summer
one.

Page 53 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
m.
The variation of density of medium for a sound wave propagating with a speed of
e
Q13

m
cowave.
340 m s–1 is shown in Fig. 10.32. Calculate the wavelength and frequency of the
g l as
. a
em
sound

l a s
ag
m
8 cm
.co ag
a sem
agl

co m
e m.
m l as
m .co a g
l a se Fig. 10.32, page 206 — variation of the density of the medium along the direction of

a g travel; the marked span is 8 cm.

m a s
em
.co agl
a s
gl frequency ν = 8500 Hz.
Wavelength λ = 4 cm = 0.04 m,aand
First read the wavelength off the figure. The dot pattern shows alternating dense bands

. c om
(compressions) and thin bands (rarefactions), with faint guide lines drawn through their centres.

s e m — that is, from
The arrow marked 8 cm stretches across two complete repeats of the pattern

. com
one compression to the third compression. Therefore
a gla
a s em
a gl 8 cm = 2λ → λ = 8 cm ÷ 2 = 4 cm = 0.04 m

se m
com g l a
m . a
e
as8 cm
agl

co m
m .
m as e
.co C a g l
m
ase
C C

agl
λ = 4 cm λ = 4 cm
.c
s e m
comwavelength is 4 cm. a
agl
Reading Fig. 10.32: the 8 cm arrow spans two whole repeats of the compression pattern, so one

m .
as e
a g l

com
m .
m ase
.co


a g l Page 54 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Now the frequency. v = ν × λ (Eq. 10.2)

ν=v/λ

= 340 m s⁻¹ ÷ 0.04 m

= 8500 s⁻¹ = 8500 Hz

Why the conversion to metres matters: the speed is given in m s⁻¹, so the
wavelength must be in metres before dividing — 4 cm = 4/100 m = 0.04 m. Working
in centimetres would give 340 ÷ 4 = 85, which is not a frequency in hertz at all.
Always bring every quantity to SI units before substituting.

Check it yourself: 8500 Hz lies between 20 Hz and 20 kHz, so this is an audible
sound — a high, shrill note. The time period is T = 1/ν = 1/8500 ≈ 1.18 × 10⁻⁴ s.

Q14 The 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.

A
B
Density

0 2.5 5.0 Distance (cm)

Fig. 10.33, page 206 — graphical representation of the two sound waves A and B.

λA = 2.5 cm and νA = 13 800 Hz; λB = 5.0 cm and νB = 6900 Hz.

Page 55 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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

WAVE CYCLES IN 5.0 WAVELENGTH FREQUENCY Ν = V/ TIME PERIOD T =
CM Λ Λ 1/Ν

A 2 2.5 cm = 0.025 m 13 800 Hz 7.25 × 10⁻ ⁵ s

B 1 5.0 cm = 0.050 m 6900 Hz 1.45 × 10⁻ ⁴ s

Why the frequencies come out in the ratio 2 : 1: both waves are in the same
medium, so both travel at 345 m s⁻¹. With v fixed, ν is inversely proportional to λ.
Wave A's wavelength is exactly half of B's, so its frequency is exactly double. In
musical language A is one octave above B — an octave is precisely a doubling of
frequency.

Tip: the two curves also differ in amplitude (A's crests are taller), which affects
loudness. The question asks only about wavelength and frequency, so amplitude
plays no part in the calculation.

Page 56 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

Q15 Two 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?

cliff

source A in air
A

sea wall
source B under water
B

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.

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

Source A is in air and source B is submerged in water, but both are the same horizontal distance
from the cliff face, so both pulses cover the same round-trip distance. Call that distance 2d for
each.

Page 57 of 66

Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

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

cliff

A (in air)

same horizontal distance d each way

B (in water)

t₀ = 4.5 t₋ with the same 2d → v₋ = 4.5 v₀
Both pulses cover the same round trip 2d to the cliff and back. Only the speed of the medium differs,
so the times are in inverse ratio to the speeds.

Page 58 of 66

as e
a g l
Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications AglaSem · NCERT Solutions

co m
m.
Why the answer is physically sensible: the chapter states that sound travels about

m l a se
4 – 5 times faster in water than in air, and this problem gives exactly 4.5. Using the
o
values in .Table a g Water's
c 10.1, 1500 m s⁻¹ ÷ 340 m s⁻¹ = 4.4 — a very close match.
m
se are far more closely packed than air's, so a compression is passed on
l a
g more quickly.
molecules
amuch
o m
e
. c
m is asked. The question asks for air : water, ag
s
Tip: be careful which way round the ratio
a : 4.5). If it had asked for water : air, the answer
which is the smaller number firstl(1
g
would be 4.5 : 1. a

. c om
m a
Note on the printed figure: in the English edition Fig. 10.34 is printed
s em with only the
.
labels A l which shows
coand B and no picture. The Hindi edition prints the fullagscene,
a s em A mounted in the air on a pier and source B hanging below the water surface,
gl both facing the vertical rock face across the water. The physics is unaffected: A and B
source
a
s
are at the same horizontal distance from the cliff.
m a
m.co agl
l a se
a 207
The Journey Beyond — Page
g
Project work
com
m .
m
THE JOURNEY BEYOND
as e
.co a g l
a s em Many people use earphones extensively these days. Find out the research studies
a gl Q1
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
se m
com g l a
.
ranges for defining mild, moderate and severe hearing loss. What are the
m a
ase
government schemes for purchasing or fitting of aids or appliances and free

agl
cochlear implants? Write an article on your findings.

. c om
m a s em organise, then write.
Method. This is a small research project, so work in three stages: gather,
o gl ICMR and AIIMS studies, and
1..cGather. Use reliable sources — WHO reports on safe listening,
a
a s em
agl
the Ministry of Social Justice and Empowerment website. Note the source and year against
every fact you copy.
.c
s e m
com gla
2. Understand how hearing is tested. An audiometer plays tones of known frequency at

.
slowly increasing loudness; the softest level the listener can just hear is the hearing
m threshold against frequency gives an audiogram.
threshold at that frequency.ePlotting
a
a s
a
3. Write a 400 – 600 word glarticle: what earphones do to the ear, what the studies found, how
loss is graded, and what help is available.

co m
m .
m ase
.co


a g l Page 59 of 66