Sound

Class 10 Physics, unit 13, book pages 93–120

Sound is a mechanical, longitudinal wave that is produced by a vibrating source and needs a medium to travel. This unit describes how sound is produced and propagated, its terms (wavelength, amplitude, frequency and time period), the audible range with infrasound and ultrasound, and the characteristics pitch, loudness and quality. It then covers the speed of sound, noise, reflection, refraction and diffraction, echo and its calculations, acoustic protection in buildings and the working of the human hearing system.

Learning outcomes (20)

Describe the production of sound.
Describe the longitudinal nature of sound waves.
State the approximate range of frequencies audible to humans as $20 Hz$ to $20000 Hz$ .
Justify why sound waves cannot travel in a vacuum.
Describe how changes in amplitude and frequency affect the loudness and pitch of sound waves.
Describe how different sound sources produce sound waves with different timbres.
Describe an echo as the reflection of sound waves.
Justify simple experiments to show the reflection of sound waves.
Illustrate a method involving a measurement of distance and time for determining the speed of sound in air.
State that the speed of sound in air is approximately $330$ to $350 m s^{- 1}$ .
Describe that, in general, sound travels faster in solids than in liquids and faster in liquids than in gases.
Define ultrasound as sound with a frequency higher than $20 kHz$ .
Illustrate and analyse the uses of ultrasound (in cleaning, in prenatal and other medical scanning, and in sonar).
Illustrate the use of infrasound.
Analyse the effects of noise pollution on the environment.
Justify the importance of acoustic protection.
Describe how knowledge of the properties of sound waves is applied in the design of buildings with respect to acoustics.
Explain the use of soft materials to reduce echo sounding.
Explain, with examples, how sound can reflect, refract and diffract.
Explain how sound is converted by the eardrum and nerves into electrical signals that are then interpreted by the brain.

Introduction to sound

Sound is a mechanical wave that starts from the vibration of the particles of a medium such as air, water or a solid. These vibrations create longitudinal waves, in which the particles oscillate parallel to the direction in which the wave travels. This produces regions of compression and rarefaction along the wave.

The properties of a sound wave, namely its frequency, amplitude and speed, decide the characteristics of the sound that we hear, such as its pitch and loudness. Sound waves need a medium to travel, and their behaviour is affected by reflection, refraction, diffraction and interference. Sound is essential for communication, entertainment and many technological applications, which is why its study matters in so many fields (the photograph of sound devices in Figure 13.1 shows a studio mixing desk and microphone).

Sound: A disturbance of matter that is transmitted from its source outwards and that produces the sensation of hearing in the ear.

13.1Source of soundOften examined

Source of sound: The object which is basically the cause of vibration.

Examples of sources of sound (Figure 13.2) are our clapping hands, a chirping bird, thunder, a ringing cellphone, musical instruments, a school bell, footsteps and a car horn.

How sound is produced (13.1.1)

The production of sound depends entirely on the vibration of an object. A vibrating source changes its energy into vibration and compels the surrounding medium to vibrate, and through this medium the sound wave travels. The particles of the surrounding medium move to and fro. Sound waves are those waves which produce the sensation of hearing when they strike the eardrum. The pressure waves made by the vibration finally collide with the eardrum, and the auditory system lets us hear the sound. Sounds surround us, from machines and vehicles to birds and conversations.

Experiments show that vibrations are essential for the production of sound:

blowing across the mouth of a bottle,
plucking a stretched rubber band,
flapping cardboard or the wings of a bird,
striking a bell or a drumhead.

Activity: vibration is necessary for sound

The apparatus is a tuning fork, a rubber pad, a glass and water.

Procedure

Fill the glass with water.
Strike the tuning fork on the rubber pad and gently touch one prong of the vibrating fork to the surface of the water (Figure 13.3(a)).
Strike the fork again and dip it vertically into the glass of water (Figure 13.3(b)).

Figure 13.3: Production of sound. (a) A vibrating prong touching the water surface. (b) The vibrating fork dipped into the water. The coloured fork is the vibrating source.

In both positions the vibrating fork makes sound, because the fork sets the particles of the medium (the water molecules) vibrating. The experiment can also be done with an empty glass, and the sound of the fork can still be heard, but the effect is smaller than with water because the fork is then in contact only with air.

13.1.2Medium for propagation of soundOften examined

Sound waves are mechanical and longitudinal, so a medium is needed for them to travel from one point to another. Without a medium the waves may be produced by a vibrating object, but they cannot travel from one place to another. For example, if there were no air in a classroom, the students would not be able to hear the lecture of the teacher.

Experiment: the electric bell in a bell jar

Take an electric bell and an airtight glass bell jar. The bell is suspended inside the jar, which is connected to a vacuum pump (Figure 13.4).

Procedure and observation

Press the switch. The bell rings and you hear it.
Start the vacuum pump. As the air is pumped out of the jar gradually, the sound becomes fainter, although the same current is still passing through the bell.
When very little air is left in the jar, only a very feeble sound is heard.
When all the air has been pumped out, no sound of the bell can be heard.

Figure 13.4: Propagation of sound. The bell is hung inside an airtight bell jar connected to a vacuum pump.

The bell still vibrates when the air has gone, but there are no air particles to carry the vibrations to the glass and out to the ear. This shows that a medium is necessary for the propagation of sound waves.

13.2Nature of sound wavesOften examined

Sound waves are mechanical and longitudinal in nature. They are also called pressure waves. When an object (the source of sound) vibrates, the vibration makes the molecules of the medium (air, a fluid and so on) vibrate. This vibration is changed into a chain of vibrations as the molecules pass on their energy, until the sound wave has travelled through the medium.

When any object vibrates, the vibration shows that it has kinetic energy. This kinetic energy is passed to the molecules of the medium. These molecules transfer the kinetic energy they have received to the next molecules of the medium. Because of this kinetic energy the molecules start a vibrating movement in the forward direction (Figure 13.5). In this way the vibrational energy made by the sound source is carried from one point to another.

How a sound wave travels

The source vibrates and has vibrational kinetic energy.
It pushes and pulls the molecules of the medium next to it, passing on some of this energy.
Each molecule vibrates and passes the energy to the next molecule.
The energy therefore travels forward through the medium as a wave, although no molecule travels all the way from the source to the listener.

Figures 13.5 and 13.6: Nature of sound. The vibrating source makes regions of compression (particles close together, high pressure) and rarefaction (particles far apart, low pressure) that travel in the direction of the wave.

13.2.1Compression and rarefactionOften examined

Sound waves are the combination of compressions and rarefactions.

Compression: The high pressure portion within a sound wave, where the molecules of the medium are very close to each other. It is produced by the vibrational kinetic energy of the source of sound. The book gives the molecules a low velocity here.
Rarefaction: The low pressure portion within a sound wave, where the molecules of the medium are comparatively far apart. It is produced by the vibrational kinetic energy of the source of sound. The book gives the molecules a high velocity here.

Compression and rarefaction (Figures 13.5 and 13.6)
Region	Pressure	Distance between molecules	Velocity of molecules (book)
Compression	High	Very short	Low
Rarefaction	Low	Comparatively large	High

13.2.2Related terms of sound wavesOften examined

Figure 13.7: Properties of a sound wave. A loudspeaker makes compressions and rarefactions; the same wave is shown as a curve below, with the crest, trough, amplitude and wavelength marked.

A. Wavelength ( $λ$ ): The distance between two consecutive crests or troughs within a compression region or rarefaction region. It is represented by the Greek letter lambda, $λ$ . Its SI unit is the metre.
B. Amplitude: The maximum displacement covered by a particle disturbed when a sound wave travels through the medium. Amplitude tells about the energy level carried by the sound wave: high amplitude means high energy and low amplitude means low energy.
C. Frequency ( $f$ ): The number of sound waves produced in one second. Its SI unit is the hertz. One hertz means one sound wave is produced in one second.
D. Time period ( $T$ ): The time taken by a wave to complete one cycle. Its SI unit is the second.

Term	Symbol	SI unit	What it tells us
Wavelength	$λ$	metre ( $m$ )	Length of one complete wave
Amplitude	$A$	metre ( $m$ )	Energy carried (loudness)
Frequency	$f$	hertz ( $Hz$ )	Waves produced per second (pitch)
Time period	$T$	second ( $s$ )	Time for one wave

f = \frac{1}{T}, v = f λ

Worked example: wavelength and time period from frequency

A note of frequency

256 Hz

travels through air at

343 m s^{- 1}

. Find (a) its time period and (b) its wavelength. (c) The frequency is doubled to

512 Hz

in the same air. What happens to the speed, the wavelength and the time period?

Given

$f_{1} = 256 Hz$
$v = 343 m s^{- 1}$
$f_{2} = 512 Hz$

To find

$T_{1}$ , $λ_{1}$ , and the changes for $f_{2}$

(a) The time period is the time for one cycle, so it is the reciprocal of the frequency.
$T_{1} = \frac{1}{f _{1}} = \frac{1}{256} = 3.9 \times 1 0^{- 3} s = 3.9 ms$
(b) In one time period the wave travels one wavelength, so $λ = v T = v / f$ .
$λ_{1} = \frac{v}{f _{1}} = \frac{343}{256} = 1.34 m$
Check by working backwards: a wave of wavelength $1.34 m$ and frequency $256 Hz$ travels
$1.34 \times 256 = 343 m s^{- 1}$
(c) The speed depends on the medium, and the medium is the same air, so $v$ stays $343 m s^{- 1}$ . The wavelength becomes
$λ_{2} = \frac{343}{512} = 0.67 m (half of 1.34 m)$
The time period also halves.
$T_{2} = \frac{1}{512} = 1.95 \times 1 0^{- 3} s = 1.95 ms$

Same speed, double the frequency

0.67

1.34

2.68

−1

−0.5

0.5

distance (m)

displacement

256 Hz

512 Hz

The coloured wave has

f = 256 Hz

and wavelength

1.34 m

. The dark wave has

f = 512 Hz

: twice as many waves fit in the same distance, so its wavelength is half. Both are drawn for the same amplitude.

λ_{1} = 1.34 m, T_{1} = 3.9 ms; for 512 Hz : v unchanged, λ_{2} = 0.67 m, T_{2} = 1.95 ms

13.2.3Types of sound wavesOften examined

Sound waves have different names, which are taken as their types, and all these names depend on the listener: unpleasant or pleasant effect, soft or hard sound, musical sound and so on. In general, sound waves can be sorted into rhythmic and non-rhythmic sound waves (Figure 13.8).

Figure 13.8: Types of sound. A rhythmic wave repeats regularly and is heard as a musical sound. A non-rhythmic wave has no regular pattern and is heard as noise.

	Rhythmic sound wave	Non-rhythmic sound wave
Shape	Regular, repeating pattern	Irregular pattern
Heard as	Musical sound	Noise

Audible range, infrasound and ultrasound

The human ear has a limit to the sound waves it can detect. The audible range is between $20 Hz$ and $20000 Hz$ (Figure 13.9).

A wave with a frequency less than $20 Hz$ is not audible to our ear. Such waves are called infrasonic waves. They are used to detect map-rock, underground petroleum formations, earthquakes and so on. Many animals use infrasonic waves to communicate across impressive distances, for example whales, elephants and giraffes.
The human ear can hear sound with a frequency greater than $20 Hz$ and less than $20000 Hz$ .
Waves with a frequency greater than $20000 Hz$ are called ultrasonic waves. They are mostly used in the medical field to examine the internal organs of patients, and also for navigation.

Figure 13.9: Frequency of sound on a scale that grows by a factor of

10

at each mark. The audible range of humans (coloured) lies between infrasound and ultrasound. The marks below show the upper limit of hearing quoted in the book for other animals.

Type	Frequency	Heard by humans?
Infrasound	Less than $20 Hz$	No
Audible sound	$20 Hz$ to $20000 Hz$	Yes
Ultrasound	More than $20000 Hz$ ( $20 kHz$ )	No

13.3UltrasoundOften examined

Ultrasound: Sound waves that have frequencies above the upper limit of human hearing, which is around $20 kHz$ .

Humans can typically hear sounds in the range $20 Hz$ to $20 kHz$ . Sounds with frequencies higher than this range are considered ultrasonic.

Ultrasound has a wide range of applications. In medicine it is commonly used for diagnostic imaging, such as prenatal ultrasound scans to monitor the development of a fetus. Ultrasound waves can penetrate body tissues and produce images of internal structures without the need for invasive procedures. In industry, ultrasound is used for non-destructive testing of materials, detecting flaws and cleaning delicate objects. It is also used in applications like sonar, which uses sound waves to detect objects underwater.

Uses of ultrasound (13.3.1)

Figure 13.10 shows three applications: (a) ultrasonic cleaning, (b) medical scanning and (c) sonar.

Use	How it works	Applications	Advantages
A. Cleaning	High-frequency sound waves create tiny bubbles in a liquid. These bubbles implode, producing small shock waves that clean surfaces.	Cleaning jewellery, electronic parts and surgical tools	Non-damaging; reaches small and hard-to-clean areas; often needs only water or mild detergents
B. Medical scanning (including prenatal)	A device called a transducer sends high-frequency sound waves into the body. The waves bounce off tissues and return to the transducer, creating an image.	Monitoring fetal development during pregnancy; checking heart function (echocardiograms) and blood flow (Doppler ultrasound)	Non-invasive; safe (no harmful radiation); gives real-time images
C. Sonar (Sound Navigation and Ranging)	A sonar device sends out sound pulses underwater. The pulses hit objects and return as echoes. The time taken for the echoes to return is used to calculate distance.	Mapping the ocean floor; finding underwater objects such as shipwrecks; measuring water depth	Measures distances underwater

13.3.2Calculating depth or distanceOften examined

Ultrasound can be used for measuring depth or distance. It involves sending pulses from a ship to the ocean floor and measuring the time taken for the echoes to return. The depth is calculated using the formula below.

d = \frac{v \times t}{2}

13.1

$d$ : depth or distance; $v$ : speed of sound in the medium; $t$ : time for the pulse to go to the object and come back

Derivation Depth from the echo time

Aim. Find the depth $d$ from the time $t$ taken by a pulse to go down and return.

Sonar: a pulse goes down to the sea bed and the echo comes back. The depth

d

is half of the distance covered by the pulse.

The pulse travels from the ship to the sea bed and then back to the ship, so the total distance travelled is twice the depth.
$distance = 2 d$
Speed is distance divided by time ( $v = distance / t$ ), so the distance is speed times time.
$2 d = v \times t$
Divide both sides by $2$ .
$d = \frac{v \times t}{2}$

Result

d = \frac{v \times t}{2}

Always halve the product $v t$ , because $t$ is the time for the round trip. The book's check: if the sound takes $2 s$ to return in water where $v = 1500 m s^{- 1}$ , then $d = 1500 \times 2/2 = 1500 m$ .

Example 13.1: Depth of the Pacific Ocean

A ship sends a sound wave in the Pacific Ocean and it takes

14.7 s

to return to the ship. Find the depth of the Pacific Ocean (speed of sound in water

1500 m s^{- 1}

Given

$v = 1500 m s^{- 1}$
$t = 14.7 s$

To find

Depth of the Pacific Ocean $d$

The sound goes down to the sea floor and comes back, so the depth is half of the distance travelled.
$d = \frac{v \times t}{2}$
Put in the values.
$d = \frac{1500 \times 14.7}{2} = \frac{22 050}{2} = 11025 m$
Change to kilometres.
$11025 m = 11.025 km$

d = 11025 m = 11.025 km (approx.)

Depth against echo time in sea water

2000

4000

6000

8000

10000

12000

t (s)

d (m)

(14.7, 11025)

d = 750 t

For sea water (

v = 1500 m s^{- 1}

d = 750 t

, a straight line through the origin. Example 13.1 is the marked point:

14.7 s

gives

11025 m

Worked example: two echoes from one sonar pulse

A ship sends one ultrasound pulse straight down into the sea. Two echoes are received: the first after

0.24 s

(from a shoal of fish) and the second after

0.90 s

(from the sea bed). Take the speed of sound in sea water as

1500 m s^{- 1}

. Find the depth of the shoal, the depth of the sea bed and the distance between them.

Given

$v = 1500 m s^{- 1}$
$t_{1} = 0.24 s$
$t_{2} = 0.90 s$

To find

$d_{1}$ , $d_{2}$ and $d_{2} - d_{1}$

Each time given is for the whole trip down and back, so use $d = v t /2$ for each echo.
Shoal of fish.
$d_{1} = \frac{1500 \times 0.24}{2} = \frac{360}{2} = 180 m$
Sea bed.
$d_{2} = \frac{1500 \times 0.90}{2} = \frac{1350}{2} = 675 m$
Distance between the shoal and the bed.
$d_{2} - d_{1} = 675 - 180 = 495 m$
Check: the extra time for the second echo is $0.90 - 0.24 = 0.66 s$ , and $1500 \times 0.66/2 = 495 m$ , which agrees.

The pulse (dark arrow) goes down; the echoes (coloured arrows) return from the shoal after

0.24 s

and from the sea bed after

0.90 s

d_{1} = 180 m, d_{2} = 675 m, d_{2} - d_{1} = 495 m

13.4Infrasound

Infrasound: Sound waves with a frequency below the lower limit of human audibility, generally less than 20\,\mathrm{Hz}.

Infrasonics find the following applications in daily life.

Communication by animals (13.4.1)

Giraffes were once believed to be silent creatures. It was later found that they make murmuring sounds at a very low frequency, which makes them hard for humans to hear. Typically, larger animals produce sounds of lower frequencies and smaller animals produce higher frequencies. These sounds can travel far through the air and the ground, and the ability to hear them depends on the size and structure of the animal's ears.

Elephants use infrasound to pass messages to each other across vast distances (Figure 13.11(a)). The infrasound they produce can be detected only with complicated devices. They communicate warnings about possible threats and coordinate their movements. This communication lets them stay connected over long distances and in areas of limited visibility, such as dense vegetation.

Seismic activity (13.4.2)

Infrasound is generated by natural events like earthquakes, volcanic eruptions and explosions. Specialised instruments called infrasound sensors detect these low-frequency sound waves. Monitoring seismic activity helps to predict earthquakes and volcanic eruptions (Figure 13.11(b)). Studying the behaviour of natural disasters gives early warnings, and it also helps us understand the internal processes of the Earth. For example, infrasound sensors placed around a volcano can detect low-frequency rumblings that point to an imminent eruption, which allows nearby people to be evacuated in time.

Use of infrasound	Example
Communication by animals	Elephants, giraffes and whales use low-frequency sounds over long distances
Seismic monitoring	Infrasound sensors detect earthquakes and volcanic rumblings and give early warning

13.5.1Characteristics of sound: pitchOften examined

There are basically three main characteristics of a sound wave: pitch, loudness and quality (timbre).

Pitch: That characteristic of a sound wave by which we can tell a grave (deep) voice from a shrill voice. It is not a physical quantity, so it cannot be measured directly. It is the perception of frequency.

Pitch depends on the frequency of the wave (and, the book adds, also on its wavelength and speed). It cannot be detected directly and can be analysed only on the basis of the frequency of the sound wave. Pitch describes the number of vibrations produced by the vocal cords in a unit interval of time (one second), which is why pitch is directly related to the frequency of the sound wave.

\text{Pitch of a sound wave}\ \propto\ \text{Frequency of the sound wave}

A low pitch means a low frequency and a high pitch means a high frequency. For example, a bird produces a high-pitched sound whereas the roar of a lion is a low-pitched sound, and the voice of a woman has a higher pitch than that of a man.

Low pitch and high pitch

Figure 13.12: Pitch of sound. Both waves have the same amplitude. The high-pitched wave (coloured) has three times the frequency of the low-pitched wave, so it has three waves in the time the low-pitched one has one.

13.5.2Characteristics of sound: loudnessOften examined

Loudness: That characteristic of a sound wave by which we can tell a loud sound from a faint sound. It is also not a physical quantity, so it cannot be measured directly. It can be measured on the basis of the amplitude of the sound wave.

\text{Loudness of a sound wave}\ \propto\ (\text{Amplitude of the sound wave})^{2}

Loudness is the sensation of how strong a sound wave is. The amplitude of the wave is the best parameter for checking its strength. The loudness level is studied in terms of the decibel.

High amplitude and low amplitude

Figure 13.13: Loudness of sound. A wave of high amplitude is a loud sound; a wave of low amplitude is a faint sound. The frequency is the same.

Factors that affect loudness

Four factors directly affect the loudness of a sound.

Factor	Effect on loudness
A. Vibrational amplitude of the source	Direct relation: a greater vibration amplitude of the source gives a higher loudness
B. Surface area of the source	Directly proportional: a greater surface area of the source produces a higher loudness
C. Distance from the source	Inverse relation: a listener at a short distance feels more loudness than one at a long distance
D. Sensitivity of the ears of the listener	Direct relation: more sensitive ears feel more effect of loudness

Pitch and loudness compared

Table 13.1: Comparative sketch of loudness and pitch of a sound wave
Pitch	Loudness
The characteristic of sound used to tell a shrill sound from a flat sound	The characteristic of sound used to tell a loud sound from a faint sound
Independent of the energy received by the ear in unit time	Depends on the energy received by the ear in unit time
Depends on the change in frequency: as the frequency increases, the shrillness increases	Independent of the change in frequency

Figure 13.14: Pitch and loudness of waves, read from the screen
Change on the oscilloscope screen (Figure 13.14)	Effect
Taller wave (larger amplitude)	Loudness increases
Shorter wave (smaller amplitude)	Loudness decreases
Waves closer together (higher frequency)	Pitch increases
Waves further apart (lower frequency)	Pitch decreases

Worked example: striking a tuning fork harder

A tuning fork is struck gently and then struck harder, so that the amplitude of its vibration doubles. The fork and the air are the same. (a) Which of the amplitude, frequency, wavelength, speed of sound, pitch and loudness change? (b) By what factor does the loudness change if loudness is proportional to (amplitude)^2?

(a) Striking harder gives the fork more energy, so it swings through a larger distance: the amplitude increases.
The frequency is set by the fork itself, so the number of vibrations per second does not change. The pitch (which goes with frequency) is therefore unchanged.
The speed of sound depends on the medium, and the medium is the same air, so the speed is unchanged. With v=f\lambda and the same v and f, the wavelength is unchanged too.
The amplitude is larger, so the loudness increases.
(b) Loudness is proportional to the square of the amplitude, so with A_2=2A_1:
\dfrac{\text{loudness}_2}{\text{loudness}_1}=\left(\dfrac{A_2}{A_1}\right)^{2}=2^{2}=4
Doubling the amplitude makes the sound four times as loud. In the same way, halving the amplitude would make it \left(\tfrac12\right)^2=\tfrac14 as loud.

Same frequency, double amplitude

The coloured wave (struck harder) and the dark wave (struck gently) repeat after the same time T, so they have the same frequency and pitch. Only the height of the wave, the amplitude, differs.

\text{amplitude and loudness change (loudness}\times4);\ f,\ \lambda,\ v\ \text{and pitch stay the same}

13.5.3Quality of sound or timbreOften examined

Quality (timbre): That characteristic of sound which helps our ear to tell apart two sound waves that have the same loudness and the same pitch.

Timbre helps us to study the dynamic characteristics and the harmonic content of sound waves. The word harmonic content describes the way the molecules of the medium pass on vibrational energy while sound travels from one point to another. The quality of a sound depends on the relative intensities of its overtones. An overtone is any harmonic with a frequency greater than the fundamental frequency of the sound.

Figure 13.15: The three characteristics of sound and what each depends on: loudness on the amplitude of vibration, pitch on the frequency and quality (timbre) on the waveform.

Same pitch, different quality

A pure tone (dark) and a complex wave (coloured) with the same fundamental frequency. Both repeat after the same time T, so they have the same pitch, but the shapes differ, so they sound different. The complex wave contains overtones.

Characteristic	Depends on	Wave shape (Figure 13.15)
Loudness	Amplitude of vibration	Soft: small amplitude; loud: large amplitude
Pitch	Frequency	Low: waves far apart; high: waves close together
Quality or timbre	Waveform	Clearer: a smooth wave; mixed: an irregular wave with overtones

13.6Analysing sound waves

Sound waves are vibrations that travel through the air (or other media) and reach our ears. These vibrations are usually produced by an object such as a guitar string, vocal cords or a speaker. Timbre (pronounced "tam-ber") is the quality or colour of a sound that makes it unique. It is what lets us tell different instruments or voices apart, even when they play the same note. An oscilloscope is used to analyse sound waves.

Oscilloscope: A device that allows us to see sound waves. It shows a sound wave as a trace (a line) on a screen (Figure 13.16).
Waveform: The shape of the trace on the oscilloscope. It shows how different sound sources produce different shapes, which helps us understand the timbre of the sound.
Pure tone (sine wave): A smooth, regular sine wave, such as the sound of a tuning fork on an oscilloscope. It has a simple timbre.
Complex wave: The more complicated waveform of a note from a piano or a violin. These sounds have richer timbres because they include multiple frequencies and overtones.

Sound	Waveform on the oscilloscope	Timbre
Tuning fork	Smooth sine wave (pure tone)	Simple
Piano or violin	Complex wave	Richer: many frequencies and overtones
Plucked guitar string	Fundamental frequency plus various overtones, a unique waveform	Distinct
Flute	Vibrations of the air column inside the instrument give a different waveform	Distinctive timbre of the flute

Examples: plucking a string on a guitar produces a sound wave with a specific shape. The oscilloscope trace shows the fundamental frequency and various overtones, giving a unique waveform. Wind instruments are similar: blowing into a flute makes vibrations in the air column inside the instrument. This produces a different waveform on the oscilloscope and gives the flute its own timbre.

13.7Speed of soundOften examined

Speed of sound: The distance covered by a sound wave per unit time.

\text{speed}=\dfrac{\text{distance}}{\text{time}}\qquad\text{or}\qquad v=\dfrac{d}{t}

13.2

Speed of sound in air and temperature

The speed of sound in air is generally between 330\,\mathrm{m\,s^{-1}} and 350\,\mathrm{m\,s^{-1}}, depending on the temperature. When the air is warmer, sound travels faster because the air particles have more energy and move more rapidly. In colder air the particles move more slowly and the speed of sound decreases. For example, on a cold winter day sound might travel at about 330\,\mathrm{m\,s^{-1}} (1188\,\mathrm{km\,h^{-1}}), while on a hot summer day it could reach 350\,\mathrm{m\,s^{-1}} (1260\,\mathrm{km\,h^{-1}}).

During a thunderstorm you often see lightning before hearing thunder. This is because light travels almost instantly, while sound takes longer because its speed is much smaller. The difference in their speeds lets us estimate the distance of the storm by counting the seconds between seeing the lightning and hearing the thunder. When an object moves faster than the speed of sound, it is said to move at supersonic speed (Figure 13.17).

Speed of sound in solids, liquids and gases

The speed of sound is fastest in solids, slower in liquids and slowest in gases. This difference is due to how the particles are arranged in each state of matter (Figure 13.18). In general the speed of sound depends on particle spacing, elasticity and density.

Figure 13.18: Speed of sound in different media. The closer the particles, the more quickly a vibration passes from one to the next.

Medium	Particles	Speed of sound (book values)	Why
A. Solids	Tightly packed and held by strong forces	Fastest; over 5000\,\mathrm{m\,s^{-1}} in metals like steel	Vibrations pass efficiently from one particle to the next because they are so close
B. Liquids	Less tightly packed than solids, but closer than in gases	About 1500\,\mathrm{m\,s^{-1}} in water, about four times that in air	Faster than gases, not as fast as solids
C. Gases	Widely spaced	Slowest; about 340\,\mathrm{m\,s^{-1}} in air at room temperature	The large gaps between particles cause more delay in passing on the sound energy

Examples from the book: if you place your ear on a metal rod and tap one end, you hear the sound almost instantly. Marine animals such as dolphins and whales rely on sound for communication over long distances in water.

Example 13.2: Distance of lightning

A boy at night heard the sound of thunder with a gap of about 3.0\,\mathrm{s} after he saw lightning. How far away is the lightning, taking the speed of sound as 343\,\mathrm{m\,s^{-1}}?

Given

v = 343\,\mathrm{m\,s^{-1}}
t = 3.0\,\mathrm{s}

To find

Distance of the lightning d

Speed is distance divided by time, so distance is speed times time. The light reaches him almost at once, so the gap is the time taken by the sound.
v=\dfrac{d}{t}\quad\text{or}\quad d=v\times t
Put in the values.
d=343\times3.0=1029\,\mathrm{m}
Change to kilometres.
d=1.029\,\mathrm{km}\approx1\,\mathrm{km}

d=1029\,\mathrm{m}=1.029\,\mathrm{km}\approx1\,\mathrm{km}

Worked example: the same distance through three media

A bell rings at one end of a 1\,\mathrm{km} path. Find how long its sound takes to cover 1000\,\mathrm{m} through air (343\,\mathrm{m\,s^{-1}}), water (1500\,\mathrm{m\,s^{-1}}) and a steel rail (5000\,\mathrm{m\,s^{-1}}). Which medium is fastest, and why does the order come out as it does?

Given

d = 1000\,\mathrm{m}
v_{air} = 343\,\mathrm{m\,s^{-1}}
v_{water} = 1500\,\mathrm{m\,s^{-1}}
v_{steel} = 5000\,\mathrm{m\,s^{-1}}

To find

time t in each medium

The distance is the same, so use t=d/v for each medium.
Air.
t=\dfrac{1000}{343}=2.92\,\mathrm{s}
Water.
t=\dfrac{1000}{1500}=0.67\,\mathrm{s}
Steel.
t=\dfrac{1000}{5000}=0.20\,\mathrm{s}
Steel is fastest and air is slowest. The particles of a solid are packed tightly and held by strong forces, so they pass the vibration on quickly; the particles of a gas are far apart, so the vibration is passed on slowly. Compared with air, the sound is about 4 times faster in water (1500/343\approx4.4) and about 15 times faster in steel (5000/343\approx14.6).

Distance against time in three media

Each line is d=vt. The steeper the line, the greater the speed. Steel (steepest) reaches 1\,\mathrm{km} in 0.2\,\mathrm{s}, water in 0.67\,\mathrm{s} and air in 2.92\,\mathrm{s}.

t_{air}=2.92\,\mathrm{s},\quad t_{water}=0.67\,\mathrm{s},\quad t_{steel}=0.20\,\mathrm{s}

13.8NoiseOften examined

In general, the term noise means unpleasant sound waves experienced by the listener. In physics, noise has the following characteristics. These sound waves have:

non-rhythmic behaviour,
irregular time intervals for the completion of vibrations,
variable energy states,
irregular intensities.

Figure 13.19 compares (a) noise, a trace with no regular pattern, and (b) musical sound, a regular pattern (compare Figure 13.8 above).

Difference between noise and musical sound (Figure 13.19)
	Musical sound	Noise
Waveform	Regular, rhythmic (periodic)	Irregular, non-rhythmic (non-periodic)
Vibrations	Complete in regular time intervals	Irregular time intervals
Energy and intensity	Regular energies and intensities	Variable energy and irregular intensities
Effect on the listener	Pleasant	Unpleasant

Sound waves can be studied in two ways: as periodic sound waves and non-periodic sound waves. The structure of the human ear is designed for periodic sound waves. Our ears feel a pleasant effect from sound waves that have regular periodic vibrations, energies, frequencies and even intensity levels.

The range of periodic sound waves is about 15\,\mathrm{Hz} to 4000\,\mathrm{Hz}. Beyond this range, periodic sound waves are changed into non-periodic waves, which give a very unpleasant effect to the listener. Beyond this range, receiving sound signals is painful.

Noise pollution

Noise is also one of the main causes of psychological and physical stress in daily life. It also has a negative effect on the level of concentration needed to do daily activities. We are mostly affected by environmental noise sources (Figure 13.20), for example:

Industrial buildings: factories and machinery, air-conditioning systems.
Construction sites: site formation (for example excavation), piling, road work, demolition, renovation.
Transportation: aircraft, trains, road vehicles, vessels.
Public places: open markets, streets, parks.

Noise pollution can be reduced by:

using earplugs,
turning off appliances when not in use,
planting more trees,
lowering the volume,
regular maintenance of vehicles and machines.

By controlling noise we can control the negative health effects that noise pollution has on everyone.

13.9Reflection, refraction and diffraction of soundOften examined

Sound waves interact with their surroundings in different ways. They can reflect, refract and diffract depending on the surfaces or materials they meet. These behaviours help us understand how sound travels and why we hear it in various situations.

Figure 13.21: Sound waves in reflection, refraction and diffraction. (a) A wave bounces off a hard surface. (b) Sound bends as it passes through air of different temperatures. (c) Sound spreads out after passing through a small opening.

Reflection (13.9.1)

Sound can reflect when it hits a surface and bounces back, just like light reflecting from a mirror. Reflection of sound waves is used to find the depth of the ocean (Figure 13.21(a)). For example, when you clap your hands in a large room with tiled walls, the sound seems to "bounce" around the room. Similarly, in a small space like a bathroom, you might notice that your voice sounds louder because the sound waves reflect off the walls.

Refraction (13.9.2)

Sound waves change direction as they pass through different materials or layers of air with different temperatures. This causes the sound to bend as its speed changes. For example, distant sounds are clearer at night when the cooler ground air and warmer upper air cause sound waves to bend towards the ground (Figure 13.21(b)). Sound waves also bend when they move through water layers that have different temperatures and densities. This bending affects how submarines and marine animals, such as whales, communicate underwater.

Diffraction (13.9.3)

Sound waves bend around obstacles or spread out after passing through small openings. This allows sound to spread into areas where there is no direct line of sight. For example, you can hear music playing in another room even if the door is slightly closed, because the sound waves bend and spread (Figure 13.21(c)). Similarly, when someone calls your name from behind a corner, you can still hear them because the sound waves diffract around the wall.

Behaviour	What happens	Example from the book
Reflection	The wave bounces back from a surface	Clapping in a tiled room; a louder voice in a bathroom; depth of the ocean
Refraction	The wave changes direction because its speed changes in a different material or temperature layer	Distant sounds clearer at night; communication of submarines and whales
Diffraction	The wave bends around an obstacle or spreads after a small opening	Hearing music from another room; hearing someone call from behind a corner

13.10EchoOften examined

Echo: The behaviour of a sound wave by which it comes back to the listener after striking an obstacle, because of the reflection of the wave. A distinct sound wave that obeys reflection.

Like other waves, sound waves bounce back from any hard surface (obstacle), like a rubber ball bouncing from the ground or a wall. When a sound wave is produced by a source, its reflection from a surface produces another wave which is heard separately from the original sound produced by the source (Figure 13.22). The reflected sound is not as strong as the original sound, and it prolongs the sensation of hearing the original sound. With the help of an echo, the distance of any object under water can also be found.

Figures 13.22 and 13.23: Echo. The sound travels the distance d to the hard surface and the reflected sound (coloured) returns over the same distance d, so the total distance is 2d.

Necessary condition for human hearing of an echo (13.10.1)

The human ear can perceive sounds that are 0.1\,\mathrm{s} apart. Two sound waves are heard separately only if the time gap between the echo and the original sound is at least 0.1\,\mathrm{s} (the sensation of sound stays in the brain for about 0.1\,\mathrm{s}).

Derivation Minimum distance for an echo

Aim. Find the least distance d between the source and the reflecting surface for an echo to be heard.

Assumptions.

The speed of sound is constant.
The ear can separate two sounds only if they are 0.1\,\mathrm{s} apart.

The sound goes to the surface and comes back, so the distance travelled by the sound is
D=2d
The echo must take t=0.1\,\mathrm{s} and the speed of sound is v. From v=D/t we get D=v\times t.
D=v\times t
Put in D=2d and t=0.1\,\mathrm{s}.
2d=v\times0.1
Divide by 2.
d=v\times\dfrac{0.1}{2}=\dfrac{v}{20}

Resultd=\dfrac{v}{20}

The minimum distance for an echo is one twentieth of the speed of sound. If v=340\,\mathrm{m\,s^{-1}}, then the least distance is 340/20=17\,\mathrm{m}.

d=\dfrac{v}{20}

13.3

Applications of echo (13.10.2)

The speed of sound can be determined.
It is used in obstetric ultrasonography and so on.
Bats cannot see with their eyes, but by echolocation they can reach their destinations and even identify things that come in their way (Figure 13.24).
The depth of the sea or the presence of submarines can be found by echo techniques.

Example 13.3: Distance of a cliff from an echo

A teacher took students to a nearby hill for an echo experiment. Abdullah shouted, and the teacher and the students heard his echo after 1.2\,\mathrm{s} using a stop watch. If the speed of sound is 346\,\mathrm{m\,s^{-1}}, how far is the cliff from him?

Given

\Delta t = 1.2\,\mathrm{s}
v = 346\,\mathrm{m\,s^{-1}}

To find

Distance of the cliff d

The echo travels to the cliff and back, so use the equation for the echo.
d=\dfrac{v\times\Delta t}{2}
Put in the values.
d=\dfrac{346\times1.2}{2}=\dfrac{415.2}{2}=207.6\,\mathrm{m}

d=207.6\,\mathrm{m}

Worked example: will a wall give a separate echo?

A girl claps her hands standing 12\,\mathrm{m} from a wall. Her brother claps at a second wall 20\,\mathrm{m} away. Take the speed of sound as 340\,\mathrm{m\,s^{-1}}. Find the time for each echo to return and say who hears a separate echo. What is the least distance that gives an echo?

Given

v = 340\,\mathrm{m\,s^{-1}}
d_1 = 12\,\mathrm{m}
d_2 = 20\,\mathrm{m}
t_{min} = 0.1\,\mathrm{s}

To find

t_1, t_2 and the least distance

The sound goes to the wall and back, so the distance travelled is 2d and the time is t=2d/v.
First wall.
t_1=\dfrac{2\times12}{340}=\dfrac{24}{340}=0.071\,\mathrm{s}
Second wall.
t_2=\dfrac{2\times20}{340}=\dfrac{40}{340}=0.118\,\mathrm{s}
The ear can tell two sounds apart only if they are at least 0.1\,\mathrm{s} apart. t_1=0.071\,\mathrm{s} is less than 0.1\,\mathrm{s}, so the echo merges with the original sound (the girl hears no separate echo). t_2=0.118\,\mathrm{s} is more than 0.1\,\mathrm{s}, so the brother hears a clear echo.
Least distance for an echo.
d=\dfrac{v}{20}=\dfrac{340}{20}=17\,\mathrm{m}
This agrees: 12\,\mathrm{m}<17\,\mathrm{m}<20\,\mathrm{m}.

The sound (dark arrow) goes to the wall and the echo (coloured arrow) returns. Only the longer path takes more than 0.1\,\mathrm{s}.

t_1=0.071\,\mathrm{s}\ (\text{no echo}),\ t_2=0.118\,\mathrm{s}\ (\text{echo}),\ d_{min}=17\,\mathrm{m}

Echo and reverberation

	Echo	Reverberation
What it is	Reflected sound that can be distinguished from the original sound	Reflected sound that cannot be distinguished from the original sound; the overlapping of sound caused by multiple reflections
Time	Heard 0.1\,\mathrm{s} or more after the original sound	Heard before 0.1\,\mathrm{s} of the original sound

13.11Acoustics and acoustic protectionOften examined

Acoustics: That branch of physics which deals with the study of sound: its characteristics, production, transmission, effects and so on.
Environmental noise: Undesirable or destructive open-air sound created by human activity, such as noise from road traffic, industrial activity, rail traffic and air traffic.

There are many types of acoustics, such as environmental noise, musical acoustics, ultrasounds, infrasounds, vibration and dynamics.

Acoustic protection (13.11.1)

Acoustic protection: The technique or method used to absorb undesirable sounds by soft and porous surfaces.

Reflection of sound is more prominent if the surface is rigid and smooth, and less if the surface is soft and irregular. Soft porous materials, such as draperies and rugs, absorb a large amount of sound energy and so quieten echoes and soften noises. By using such materials in noisy places we can reduce the level of noise pollution. However, if the surfaces of classrooms or public halls are too absorbent, the sound level may be too low for the audience.

Surface	Reflection of sound	Effect
Rigid and smooth (tiles, concrete, glass)	More	Echoes and louder sound
Soft, porous and irregular (draperies, rugs)	Less (more absorbed)	Quieter echoes and softer noise

13.11.2Importance of acoustic protection

Acoustic protection is crucial for several key reasons.

Area	Benefit	How
A. Health and well-being	Hearing protection	Prevents hearing loss from prolonged noise exposure
A. Health and well-being	Mental health	Reduces stress, anxiety and sleep disorders by minimising noise pollution
B. Productivity and performance	Workplace efficiency	Enhances concentration, reduces errors and prevents accidents by limiting distractions
B. Productivity and performance	Learning environments	Improves the ability of students to focus and retain information
C. Environmental and community impact	Wildlife preservation	Protects animals that rely on sound for communication and navigation
C. Environmental and community impact	Quality of life	Ensures quieter, more pleasant living environments for communities
D. Legal and regulatory compliance	Occupational safety	Helps organisations meet noise exposure regulations
D. Legal and regulatory compliance	Building standards	Ensures buildings provide adequate noise control as per codes
E. Technological and industrial development	Innovation	Drives the creation of new materials and quieter machinery, benefiting various industries

13.11.3Applying acoustics in building design

Understanding the properties of sound waves is crucial for designing buildings with good acoustics.

Design idea	What is done
A. Sound absorption	Materials like carpets, acoustic panels and curtains are used to absorb sound and reduce echoes
B. Sound reflection	Walls and ceilings are designed to reflect sound where it is needed; materials like glass and concrete are used for this purpose
C. Sound transmission	Thick walls, double-glazed windows and insulated doors are used to prevent sound from travelling between spaces; mostly dense or heavy materials like concrete blocks are used
D. Sound diffusion	Elements that scatter sound evenly are installed, like irregular surfaces or special panels
E. Resonance control	The size and design of the room are chosen to avoid amplification of certain sounds; bass traps are used in studios to control low-frequency sounds
F. Noise control	Gaps are sealed so that doors and windows block noise; barrier walls, fences and landscaping reduce outside noise
G. Vibration control	Rubber mounts and floating floors are used for damping, and noisy equipment is isolated so that vibrations do not spread

Practical examples: concert halls combine absorbing, reflecting and diffusing materials for clear sound; offices use absorbing materials to reduce noise and raise productivity; homes use soundproofing for privacy and quiet. Using the properties of sound waves helps design buildings that sound good and are comfortable.

Soft materials on classroom walls

Soft materials are used on the walls of classrooms and public buildings to reduce echo, which has the following benefits.

Sound absorption: soft materials like carpets, curtains and acoustic panels absorb sound waves, reducing reflections and echoes. This helps students to hear and understand the teacher, leading to better learning.
Reducing reverberation time: soft materials shorten the time it takes for sound to decay, making speech and other sounds clearer. This suits concentration and productivity in libraries and lecture halls. A quieter environment reduces stress and tiredness and adds comfort in places like hospitals and auditoriums.

13.12Human hearing systemOften examined

In physics, the process of changing sound into electrical signals that are interpreted by the brain involves several steps that can be described through the principles of wave mechanics, mechanical vibrations and signal transduction. The ear has three main parts: the outer ear, the middle ear and the inner ear (Figure 13.26).

Figure 13.26: Human hearing system. Sound enters the outer ear and makes the eardrum vibrate; the three small bones (coloured) of the middle ear pass the vibrations on to the cochlea of the inner ear; the cochlear nerve then carries the signals to the brain.

From the outer ear to the brain

Sound waves enter the ear. Sound waves are vibrations in the air that travel into the ear canal and hit the eardrum.
Eardrum vibration. The eardrum vibrates in response to these sound waves.
Ossicles amplify the vibrations. The vibrations are passed to three tiny bones in the middle ear called the ossicles (malleus, incus and stapes), which amplify the vibrations.
Vibrations enter the cochlea. The stapes (one of the ossicles) sends the vibrations into the cochlea, a fluid-filled spiral structure in the inner ear.
Fluid waves in the cochlea. The vibrations create waves in the fluid inside the cochlea.
Hair cells detect vibrations. These fluid waves move the basilar membrane inside the cochlea, causing the hair cells on the membrane to bend.
Conversion to electrical signals. When the hair cells bend, they generate electrical signals.
Electrical signals travel to the brain. These electrical signals are sent through the auditory nerve to the brain.
The brain interprets the signals. The brain processes these signals and interprets them as sound, such as music or speech.

Part of the ear	Parts named in the book	Job
Outer ear	Ear canal	Collects the sound waves and leads them to the eardrum
Middle ear	Eardrum, ossicles (malleus, incus, stapes)	The eardrum vibrates; the ossicles amplify the vibrations
Inner ear	Cochlea (basilar membrane, hair cells), auditory nerve	Changes the fluid waves into electrical signals for the brain

Formula sheet

Depth or distance from an echo

d=\dfrac{v\times t}{2}

d: depth or distance; v: speed of sound in the medium; t: time for the pulse or echo to go to the object and return (13.1)

Unit: \mathrm{m}

Speed of sound

v=\dfrac{d}{t}

d: distance covered by the sound; t: time taken (13.2)

Unit: \mathrm{m\,s^{-1}}

Least distance for an echo

d=\dfrac{v}{20}

v: speed of sound; the echo must return at least 0.1\,\mathrm{s} after the original sound (13.3)

Unit: \mathrm{m}

Frequency and time period

f=\dfrac{1}{T}

f: frequency (waves per second); T: time for one wave. From Unit 12 (Waves); used in the numerical problems of this unit

Units: \mathrm{Hz} and \mathrm{s}

Speed, frequency and wavelength

v=f\lambda

v: speed of the wave; f: frequency; \lambda: wavelength. From Unit 12 (Waves); used in the numerical problems of this unit

Unit: \mathrm{m\,s^{-1}}

Pitch

\text{Pitch}\propto f

f: frequency of the sound wave

Loudness

\text{Loudness}\propto A^{2}

A: amplitude of the sound wave

Key terms

Sound: A disturbance of matter that is transmitted from its source outwards and that produces the sensation of hearing in the ear.
Source of sound: The object which is basically the cause of vibration.
Medium: The material (air, water or a solid) through which a sound wave travels. Sound cannot travel in a vacuum.
Longitudinal wave: A wave in which the particles oscillate parallel to the direction in which the wave travels. Sound waves are longitudinal.
Compression: The high pressure portion of a sound wave, where the particles of the medium are very close together.
Rarefaction: The low pressure portion of a sound wave, where the particles of the medium are comparatively far apart.
Wavelength: The distance between two consecutive crests or troughs; symbol \lambda, SI unit metre.
Amplitude: The maximum displacement covered by a particle disturbed when a sound wave travels through the medium; it tells the energy carried by the wave.
Frequency: The number of sound waves produced in one second; symbol f, SI unit hertz.
Time period: The time taken by a wave to complete one cycle; symbol T, SI unit second.
Rhythmic and non-rhythmic waves: Regular, repeating sound waves (musical sound) and irregular sound waves (noise).
Audible frequency range: The range of frequency from 20\,\mathrm{Hz} to 20\,000\,\mathrm{Hz} that humans can hear.
Infrasound: Sound waves with a frequency less than 20\,\mathrm{Hz}, which humans cannot hear.
Ultrasound: Sound waves with a frequency greater than 20\,\mathrm{kHz}, which humans cannot hear.
Sonar: Sound Navigation and Ranging: a device that sends sound pulses underwater and uses the time for the echoes to return to find distance.
Pitch: The perception of the frequency of sound; the greater the frequency, the greater the pitch and the shriller the sound.
Loudness: The perception of the intensity of sound; it depends on the amplitude, and the greater the amplitude, the louder the sound. It is measured in decibels.
Quality (timbre): The characteristic that tells two sounds of the same loudness and pitch apart; it depends on the number and relative intensity of the frequencies (overtones) present, that is on the waveform.
Overtone: Any harmonic with a frequency greater than the fundamental frequency of a sound.
Oscilloscope: A device that shows a sound wave as a trace (a line) on a screen.
Pure tone and complex wave: A pure tone is a smooth sine wave (for example a tuning fork); a complex wave contains several frequencies and overtones (for example a piano or violin).
Speed of sound: The distance covered by a sound wave per unit time; about 330 to 350\,\mathrm{m\,s^{-1}} in air.
Supersonic speed: The speed of an object that moves faster than the speed of sound.
Noise: Any loud, discordant or disagreeable sound; in physics, sound with non-rhythmic behaviour, irregular time intervals and irregular intensities.
Musical sound: A sound that is pleasant and harmonious.
Reflection of sound: The bouncing back of sound when it hits a surface.
Refraction of sound: The bending of sound as it passes through materials or layers of air of different temperature, because its speed changes.
Diffraction of sound: The bending of sound around obstacles or its spreading out after passing through small openings.
Echo: Reflected sound that can be distinguished from the original sound (heard 0.1\,\mathrm{s} or more after it).
Reverberation: The overlapping of sound caused by multiple reflections; reflected sound that cannot be distinguished from the original sound (heard before 0.1\,\mathrm{s}).
Acoustics: The branch of physics that deals with the study of sound: its characteristics, production, transmission and effects.
Environmental noise: Undesirable or destructive open-air sound created by human activity, such as road traffic, industry, rail and air traffic.
Acoustic protection: The application of soft and porous material to absorb undesirable sounds and protect people against noise.
Eardrum: The membrane in the ear that vibrates in response to sound waves.
Ossicles: The three tiny bones of the middle ear (malleus, incus and stapes) that amplify the vibrations of the eardrum.
Cochlea: The fluid-filled spiral structure of the inner ear in which the vibrations make hair cells bend and produce electrical signals.
Auditory nerve: The nerve that carries the electrical signals from the cochlea to the brain.

Multiple choice questionsp. 117–118

What are the three main parts of the human ear?

AOuter ear, Middle ear, Inner earAnswer
BInner ear, Cochlea, Eardrum
CMiddle ear, Eardrum, Auditory nerve
DOuter ear, Auditory nerve, Cochlea

Why

The human ear has three main regions: the outer ear (pinna and ear canal), the middle ear (eardrum and the three ossicles) and the inner ear (cochlea).

Options B, C and D each mix parts of the ear with the auditory nerve or the cochlea, which belong to the inner ear and the hearing pathway, not to the three regions.

Sound produced by piano and violin gives ______ waveform on oscilloscope.

ASimple
BHarmonic
CComplexAnswer
DSymmetric

Why

A piano or violin note contains a fundamental frequency together with several overtones. The combined vibrations give a complex waveform on the oscilloscope.

A tuning fork gives a pure tone, which is a simple sine wave (A). Harmonic (B) names an overtone, not the shape of the trace, and symmetric (D) is not a type of waveform.

Which is NOT true for sound?

ASound is produced by vibrating body
BSound travels due to variation of pressure in air
CSound transfer energy along with matter of mediumAnswer
DSound is compressional and longitudinal wave

Why

Sound transfers energy through a medium, but the particles of the medium only oscillate about their rest positions. The matter of the medium does not travel along with the wave, so statement C is false.

Statements A, B and D are true: sound comes from a vibrating body (A), it travels as pressure variations in air (B), and it is a compressional longitudinal wave (D).

4FBISE SSC-II 2026 Annual I, Set D, MCQ 12

Sound travels faster in:

Arubber
Bair
Cwater
DsteelAnswer

Why

Sound travels fastest in solids, where the particles are tightly packed and pass the vibrations on quickly. Steel is a solid and gives a speed above 5000 m/s.

Air (B) and water (C) are slower because their particles are farther apart (gas) or less tightly bound (liquid). Rubber (A) is a soft solid, so its speed is far below that of steel.

Speed of sound in air is 332\,\mathrm{m\,s^{-1}}. What is its speed in vacuum?

AEqual to 332\,\mathrm{m\,s^{-1}}
BGreater than 332\,\mathrm{m\,s^{-1}}
CLess than 332\,\mathrm{m\,s^{-1}}
DZeroAnswer

Why

Sound is a mechanical wave and needs particles of a medium to carry the vibrations. A vacuum has no particles, so sound cannot travel in it and its speed there is zero.

The value 332\,\mathrm{m\,s^{-1}} belongs to air only. Options A, B and C wrongly assume that sound can travel through vacuum at some non-zero speed.

The sound travels from water to an iron rod and then into air and back into water. The speed of sound will successively:

Aincrease, decrease, increaseAnswer
Bdecrease, increase, decrease
Cincrease, increase, increase
Ddecrease, decrease, decrease

Why

Water to iron rod: the speed rises, because the iron is a solid and sound travels faster in solids than in liquids (about 5000\,\mathrm{m\,s^{-1}} against 1500\,\mathrm{m\,s^{-1}} in water). The speed increases.

Iron to air: the speed falls, since air is a gas and is much slower. Air back into water: the speed rises again. The order is increase, decrease, increase, which is option A.

What will happen to speed of sound, if frequency is doubled? It will become:

Ahalf
Bdouble
Cfour times
Dremain sameAnswer

Why

The speed of sound depends on the medium and its temperature, not on the frequency. Doubling the frequency halves the wavelength, because v=f\lambda, but the speed stays the same.

Options A, B and C each make the speed depend on frequency, which is wrong.

Infrasonic waves have frequency:

Agreater than 20 Hz
Bless than 20 HzAnswer
Cof 20 Hz
Dof 20 kHz

Why

Infrasonic waves are sound waves below the audible range of humans, so their frequency is less than 20 Hz.

Frequencies greater than 20 Hz (A) are in the audible or ultrasonic range. Options C and D give only the boundary values, not the frequency range of infrasound.

For echo, the minimum distance of a person from obstacle is:

A17 mAnswer
B34 m
C0.1 m
Dany distance above 50 m

Why

For an echo to be heard separately, the reflected sound must return at least 0.1 s after the original. Then d=\dfrac{v}{20} (equation 13.3), so d=\dfrac{343}{20}\approx17\,\mathrm{m} for air.

Option B is double this distance, which is more than the minimum. Option C (0.1 m) is far too close for an echo, and option D is not the minimum at all.

The property of waves which is directly related with loudness of sound is:

AFrequency
BWavelength
CSpeed
DAmplitudeAnswer

Why

Loudness is the sensation of the strength of a sound, and it depends on the amplitude of the wave: loudness is proportional to (\text{amplitude})^{2}.

Frequency (A) decides the pitch, and wavelength (B) and speed (C) do not decide loudness.

Sound entered in ear is amplified by:

AOssiclesAnswer
BCochlea
CEardrum
DBrain

Why

The three small bones of the middle ear, the malleus, incus and stapes (the ossicles), amplify the vibrations of the eardrum before they reach the cochlea.

The cochlea (B) turns the fluid waves into nerve signals, the eardrum (C) only receives and vibrates, and the brain (D) interprets the signals; none of them amplifies the sound.

Silent whistle is used to train dog. When trainer blows the whistle, human beings do not hear it but dog do listen. Its possible frequency is:

A> 20\,\mathrm{Hz}
B< 20\,\mathrm{Hz}
C< 20000\,\mathrm{Hz}
D> 20000\,\mathrm{Hz}Answer

Why

Humans cannot hear sounds above about 20\,000\,\mathrm{Hz} (20 kHz), so a silent whistle must be ultrasonic, with frequency greater than 20000\,\mathrm{Hz}. A dog can hear up to about 35000\,\mathrm{Hz} (book information box).

Option A is wrong because any sound above 20 Hz includes the audible range that humans hear. Options B and C include audible or infrasonic frequencies, so a human would hear them.

The pitch of sound depends upon:

AfrequencyAnswer
Bamplitude
Cquality
Ddisplacement

Why

Pitch is the sensation of how high or low a sound is. It depends on the frequency: a higher frequency gives a higher pitch (Section 13.5.1).

Amplitude (B) controls loudness, quality (C) depends on the waveform, and displacement (D) is not a characteristic of sound.

Two sound waves having same loudness and pitch can be distinguished by one of the characteristics of sound called:

Aloudness
Bpitch
CqualityAnswer
Dintensity

Why

Two sounds with the same loudness and pitch can still differ in their quality (timbre), because quality depends on the waveform, that is, on the mix of overtones.

Loudness (A), pitch (B) and intensity (D) are the very properties that are the same in the two sounds, so they cannot tell them apart.

The sensation of sound persists in our brain for about:

A10 s
B1 s
C0.1 sAnswer
D0.01 s

Why

The ear and brain keep the sensation of a sound for about 0.1 s. This is the same 0.1 s limit that the book uses for echo: a reflection arriving within 0.1 s overlaps with the original sound.

Ten seconds (A) and one second (B) are far too long, and 0.01 s (D) is too short for the sensation to persist.

Constructed response questionsp. 118

Mark the frequency region for Infrasound, Human Audible Frequency and Ultrasound.

Mark each frequency in the empty row.
0\,\mathrm{Hz}	10\,\mathrm{Hz}	20\,\mathrm{Hz}	1000\,\mathrm{Hz}	3500\,\mathrm{Hz}	13000\,\mathrm{Hz}	20000\,\mathrm{Hz}	40000\,\mathrm{Hz}	55000\,\mathrm{Hz}	80000\,\mathrm{Hz}

Mark each frequency in the empty row.
0\,\mathrm{Hz}
10\,\mathrm{Hz}
20\,\mathrm{Hz}
1000\,\mathrm{Hz}
3500\,\mathrm{Hz}
13000\,\mathrm{Hz}
20000\,\mathrm{Hz}
40000\,\mathrm{Hz}
55000\,\mathrm{Hz}
80000\,\mathrm{Hz}

Solution

Infrasound is sound below about 20 Hz, the audible range for humans is from 20 Hz to 20000 Hz, and ultrasound is sound above 20 kHz. Each frequency is placed in the region that contains it.

Region of each frequency.
0\,\mathrm{Hz}	10\,\mathrm{Hz}	20\,\mathrm{Hz}	1000\,\mathrm{Hz}	3500\,\mathrm{Hz}	13000\,\mathrm{Hz}	20000\,\mathrm{Hz}	40000\,\mathrm{Hz}	55000\,\mathrm{Hz}	80000\,\mathrm{Hz}
Infrasound	Infrasound	Audible	Audible	Audible	Audible	Audible (upper limit)	Ultrasound	Ultrasound	Ultrasound

Region of each frequency.
0\,\mathrm{Hz}	Infrasound
10\,\mathrm{Hz}	Infrasound
20\,\mathrm{Hz}	Audible
1000\,\mathrm{Hz}	Audible
3500\,\mathrm{Hz}	Audible
13000\,\mathrm{Hz}	Audible
20000\,\mathrm{Hz}	Audible (upper limit)
40000\,\mathrm{Hz}	Ultrasound
55000\,\mathrm{Hz}	Ultrasound
80000\,\mathrm{Hz}	Ultrasound

The value 20000 Hz is the upper limit of human hearing, so it is audible. The 0 Hz value is taken as the lower end of infrasound. The values 40000 Hz, 55000 Hz and 80000 Hz lie above 20 kHz and are ultrasound.

Note. The book prints the question as a table with an empty row to mark. The answer row above gives the region of each frequency.

Label the major parts of human hearing system.

Solution

The book gives an outline of the ear (Figure 13.26). Label its parts in the order in which sound passes through them, from the outer ear to the brain.

The labelled parts of the human hearing system, in the order sound passes through them.

Part	Region	Role in hearing
Pinna and ear canal	Outer ear	Collect the sound and direct it to the eardrum
Eardrum	Middle ear	Vibrates in response to the sound waves
Ossicles (malleus, incus, stapes)	Middle ear	Amplify the vibrations
Cochlea	Inner ear	Fluid-filled spiral; hair cells on its membrane bend and generate signals
Auditory (cochlear) nerve	Inner ear to brain	Carries the electrical signals to the brain
Brain	Centre	Interprets the signals as sound, such as music or speech

Short response questionsp. 118–119

Analyses the process of sound transmission from outer ear to brain.

Solution

Sound is a pressure wave in air. It is collected by the outer ear and carried along the ear canal to the eardrum, so the sound travels through this canal before it reaches the middle ear.

The sound waves hit the eardrum, which vibrates in response to them.
The vibrations pass to the three ossicles (malleus, incus and stapes) in the middle ear, which amplify them.
The stapes sends the vibrations into the cochlea, the fluid-filled spiral structure in the inner ear. The vibrations create waves in the fluid.
The fluid waves move the basilar membrane, which bends the hair cells on it. When the hair cells bend they generate electrical signals.
The signals travel along the auditory nerve to the brain, which interprets them as sound, such as music or speech.

So the sound is changed from air pressure waves to mechanical vibrations, then to fluid waves, then to electrical signals, and the brain completes the process by interpreting them.

What are common sources of harmful noise, assess their impact on human health?

Solution

Noise is harmful when it is loud, irregular and prolonged. The common sources, and their examples, are:

Source	Examples
Industrial buildings	factories, machinery, air-conditioning systems
Construction sites	excavation, piling, road work, demolition, renovation
Transportation	aircraft, trains, road vehicles, vessels
Public places	open markets, streets, parks
Personal listening	earphones and headphones used for long periods at high volume

Impact on hearing. Long exposure to high sound levels causes noise-induced hearing loss and tinnitus (ringing in the ears).

Impact on mind and body. Noise causes psychological and physical stress, and disturbs sleep. It also reduces concentration, so people make more errors and accidents are more likely.

Control. Noise can be reduced by using earplugs, turning off appliances when not in use, planting more trees, lowering the volume of vehicles and regular maintenance of machines.

How do animals use infrasound for communication? Compare this with human communication methods.

Solution

Many animals use infrasound, which is below 20 Hz, to communicate over long distances. Elephants send messages to each other across vast distances, warn of threats and coordinate their movements. Whales and giraffes also use low-frequency sounds. Infrasound travels far through air and ground, so the animal can be heard by its group even when it is far away or hidden in dense vegetation.

Feature	Animals using infrasound	Human communication
Frequency	below 20 Hz	audible sound, from 20 Hz to 20000 Hz
Distance	very long, through air and ground	usually short, for speech and conversation
Hearing	the animal hears it; humans need complicated devices to detect it	humans hear speech directly
Purpose	warnings, coordination and contact with the group	information, ideas and conversation

Humans cannot hear infrasound, so they must rely on audible sound, speech and other signs. Animals depend on their own hearing and ear structure, which decides how well they can receive these low frequencies.

Evaluate the advantages and limitations of ultrasonic cleaning.

Solution

Ultrasonic cleaning uses high-frequency sound waves in a liquid. The waves create tiny bubbles that implode and give small shock waves, and these clean the surface of the object.

Advantage: it is non-damaging and reaches small and hard-to-clean areas.
Advantage: it often needs only water or a mild detergent.
Advantage: it is used for cleaning jewellery, electronic parts and surgical tools.

Limitation: the method depends on a liquid, so the object must be placed in the liquid, and it is best suited to small and delicate items that can be put into a cleaning bath.

Note. The book lists only the advantages of ultrasonic cleaning. The limitation given here follows from the method and is not stated in the book.

5FBISE SSC-II 2026 Annual I, Set B, MCQ 12

Why does sound travel faster in solids than liquids and gases?

Solution

The speed of sound depends on how closely the particles of the medium are packed and how strongly they are held together. Sound passes from one particle to the next, so the closer and more firmly bound the particles, the faster the vibrations pass on.

Solids: particles are tightly packed and held together by strong forces, so the vibrations transfer efficiently from one particle to the next. In steel the speed is over 5000\,\mathrm{m\,s^{-1}}.
Liquids: particles are less tightly packed than in solids but still closer together than in gases. Sound travels at about 1500\,\mathrm{m\,s^{-1}} in water.
Gases: particles are widely spaced, so the transfer of vibrations takes longer. Sound travels at about 340\,\mathrm{m\,s^{-1}} in air.

Therefore sound travels fastest in solids, slower in liquids and slowest in gases.

How two plastic glasses with a string stretched between them could be better way to communicate than merely shouting through the air?

Solution

The plastic cups and the string form a simple string telephone. When one person speaks into an open cup, the sound makes the cup's base vibrate, and the vibration passes into the tight string.

A string is a solid, and sound travels faster and more efficiently in solids than in air. The vibrations move along the string from one cup to the other, and the sound is confined to the string instead of spreading out in all directions. The listener at the other cup hears the speech clearly, even at a distance where shouting would not be heard. The string must be kept tight so that the vibrations pass along it.

How can we distinguish between two sounds having same loudness?

Solution

Two sounds with the same loudness can be distinguished by their pitch and their quality (timbre).

Pitch: if the frequencies differ, one sound is higher (shriller) than the other. A bird gives a high-pitched sound and a lion a low-pitched roar.
Quality: if the pitch is also the same, the sounds still differ in their waveform, because the overtones present differ. On an oscilloscope a piano note and a violin note of the same pitch and loudness show different shapes, so the ear tells them apart.

During a match in cricket stadium, you see a batsman striking the ball but we hear stroke sound slightly later. Explain this time difference?

Solution

Light from the bat travels almost instantly to the eye, but sound travels much more slowly, at about 340\,\mathrm{m\,s^{-1}} in air (between 330 and 350 \mathrm{m\,s^{-1}}). The sound from the stroke needs time to cover the distance from the bat to the listener.

The time taken is t=\dfrac{d}{v}. For example, if the spectator is 100\,\mathrm{m} away, t=\dfrac{100}{340}\approx0.3\,\mathrm{s}, so the sound arrives about a third of a second after the stroke is seen.

When a pendulum vibrates, we do not hear its sound. Why?

Solution

A pendulum swings slowly, so its frequency is very low, well below 20 Hz. A sound with frequency below 20 Hz is infrasound, and humans cannot hear it. For example, a pendulum that completes one vibration in 2 s has a frequency of 0.5\,\mathrm{Hz}.

The pendulum also produces only a very small pressure variation in the air, so the sound it makes is weak. Both reasons together mean that the sound cannot be heard by the human ear.

Two students are talking in the corridor of your school, you can hear them in your class room but you cannot see them. How?

Solution

Sound waves can bend around obstacles and spread through openings, a process called diffraction. The students' voices pass through the open door and around the edge of the wall, so they reach the classroom and are heard.

Light travels in straight lines and does not bend around the wall or door in the same way, so the students cannot be seen from the classroom. This is also why you can hear music from another room even when the door is slightly closed.

What steps would you take to stop echo and reverberation effects in a large room?

Solution

Echo and reverberation are caused by reflection of sound from hard, smooth surfaces. The steps are to absorb the reflected sound, scatter it, and control the size and shape of the room.

Absorb the sound: fix soft and porous materials such as carpets, curtains, rugs and acoustic panels to the walls and ceiling. They absorb sound and reduce reflections and echoes.
Shorten the reverberation: soft materials shorten the time for the sound to decay, so that the overlapping reflections become faint.
Scatter the sound: install irregular surfaces or special panels that spread the sound evenly instead of reflecting it back in one direction.
Control the room: choose the size and design of the room so that certain sounds are not amplified, and use bass traps for low-frequency sound.

Long response questionsp. 119–120

Explain the production of sound waves with examples.

Solution

Production of sound

Every sound is produced by a vibrating object. The object pushes on the particles of the surrounding medium, and each particle passes the vibration on to its neighbour. The result is a pressure wave that travels outwards from the source.

The vibrating source transfers its energy to the medium, which is why a source must be vibrating before any sound is heard. A motionless object gives no sound.

Examples

Tuning fork: strike it on a rubber pad so that it vibrates. Touch one prong to the surface of water and the water splashes, which shows that the vibration of the fork is producing sound (book Figure 13.3).
Stretched rubber band: pluck it and it vibrates; the air around it carries the sound to the ear.
Blowing across a bottle, flapping cardboard or bird wings: the air is set into vibration, and the vibrating air reaches the ear.
Striking a bell or a drumhead: the object vibrates and the surrounding air carries the sound away.
Clapping hands, a chirping bird, the vocal cords and a car horn all work in the same way: each is a vibrating source.

Justify why sound waves cannot travel in a vacuum. Design an experiment that demonstrates this principle.

Solution

Why sound cannot travel in a vacuum

Sound is a mechanical wave. It needs the particles of a medium, which pass the vibrations from one particle to the next. A vacuum contains no particles, so there is nothing to pass the vibrations on, and the energy cannot be carried from the source to the ear. For this reason the speed of sound in vacuum is zero.

Experiment: electric bell in a bell jar

Apparatus: an electric bell suspended from a cork in an airtight glass bell jar, a switch, and a vacuum pump connected to the base of the jar (book Figure 13.4).

Figure 13.4: Electric bell in an airtight bell jar, connected to a vacuum pump. The bell keeps working while the air is removed.

Method and observation

Press the switch with the jar full of air. The bell is heard clearly.
Start the vacuum pump. Air is removed gradually, and the sound becomes fainter, although the same current still passes through the bell.
When very little air is left, only a very feeble sound is heard. When the air is pumped out completely, no sound can be heard, although the bell is still vibrating.

Conclusion: the sound fades as the medium is removed and is absent in a vacuum. A medium is therefore necessary for the propagation of sound, and the bell vibrating in vacuum shows that the source still vibrates, which is the reason the sound is not heard.

What is nature of sound waves? How is sound propagated? Explain.

Solution

Nature of sound waves

Sound waves are mechanical and longitudinal. They are mechanical because they need a medium. They are longitudinal because the particles of the medium vibrate parallel to the direction in which the wave travels. Sound waves are also called pressure waves, because they consist of regions of high pressure and low pressure.

Compression: a region of high pressure, where the particles are pushed close together.
Rarefaction: a region of low pressure, where the particles are spread farther apart.

Pressure variation in a sound wave

Pressure variation along a sound wave of wavelength 2 m. Compressions are at x=0, 2 and 4 m, and rarefactions at x=1 and 3 m.

How sound is propagated

The vibrating source pushes the nearest particles of the medium forward, so they are compressed.
Those particles push their neighbours, so the compression moves outwards from the source.
As each particle moves forward, it leaves behind a region where the particles are farther apart, which is a rarefaction.
Each particle only oscillates back and forth about its own rest position. The energy, not the matter, travels outwards.

So the vibrational energy of the source is transferred from one point to another through the medium, and the wave does not carry the particles with it.

4FBISE SSC-II 2026 Annual I, Set B, Q.4FBISE SSC-II 2026 Annual I, Set D, Q.4FBISE SSC-II 2026 Annual I, Set M, Q.4

Discuss how changes in amplitude and frequency affect the loudness and pitch of sound waves.

Solution

Amplitude and loudness

The amplitude of a wave is the maximum displacement of its particles from their rest position, and it measures the energy carried by the wave. Loudness is the sensation of the strength of a sound. Loudness depends on the amplitude, and it is proportional to the square of the amplitude:

\text{loudness}\propto A^{2}

Same frequency, different amplitude

Two waves of the same frequency. The louder sound has the larger amplitude.

Doubling the amplitude gives four times the loudness, because (2A)^{2}=4A^{2}. The amplitude does not change the pitch of the sound.

Frequency and pitch

Pitch is the sensation of how high or low a sound is. It depends on the frequency: more vibrations each second give a higher pitch. A bird produces a high-pitched sound, and a lion's roar is low-pitched.

Same amplitude, different frequency

Two waves of the same amplitude. The higher-frequency wave has the higher pitch.

Summary

Change in the wave	Effect on loudness	Effect on pitch
Amplitude increases	loudness increases	no change
Frequency increases	no change	pitch increases

Thus loudness is controlled by the amplitude and pitch by the frequency. Changing one does not change the other.

What is quality of sound? How do the shape and material of a sound source influence the waveform of the sound it produces?

Solution

Quality of sound

Quality, also called timbre, is the characteristic that lets the ear tell apart two sounds of the same loudness and the same pitch. It depends on the waveform, that is, on the number and relative strengths of the overtones (harmonics) present along with the fundamental frequency.

A pure tone, such as the note from a tuning fork, has a single frequency and gives a smooth sine wave with a simple timbre. A note from a piano or a violin contains several overtones, so its waveform is complex and its timbre is richer.

Pure tone

A pure tone from a tuning fork: a smooth sine wave.

Complex tone

A complex tone, such as a piano or violin note: the fundamental plus an overtone with three times its frequency.

Effect of the shape and material of the source

The shape of a sound source and the material it is made of decide which overtones it produces and how strongly they are present. Each source therefore gives a waveform with its own shape on the oscilloscope, even when the pitch and loudness are the same.

Sound source	Waveform on the oscilloscope
Tuning fork	smooth sine wave, simple timbre
Piano or violin	complex waveform, rich timbre
Plucked guitar string	a specific shape set by the string
Wind instrument (flute)	a different waveform, from the vibration of the air column
Drumhead (animal skin or plastic)	a waveform set by the material and the way it is struck

Because quality depends on the waveform, the same note played on two different instruments is still recognised as different.

Explain the speed of sound in different media. How does the speed of sound differ in solids, liquids, and gases? Explain why, and discuss real-world implications.

Solution

Speed of sound in different media

The speed of sound is the distance covered by the sound wave per unit time, v=\dfrac{d}{t}. It depends on the medium: sound is fastest in solids, slower in liquids and slowest in gases.

Medium	Typical speed	Example
Solid	above 5000\,\mathrm{m\,s^{-1}}	steel
Liquid	about 1500\,\mathrm{m\,s^{-1}}	water (about four times the speed in air)
Gas	about 340\,\mathrm{m\,s^{-1}} (330 to 350 \mathrm{m\,s^{-1}} in air)	air at room temperature

Why the speed differs

Solids: the particles are tightly packed and held by strong forces, so each particle passes the vibration on to the next very quickly.
Liquids: the particles are less tightly packed than in solids, but closer together than in gases, so the speed is intermediate.
Gases: the particles are widely spaced, so the transfer of the vibration from one particle to the next takes longer and the speed is lowest.
Temperature: in warmer air the particles have more energy and move more rapidly, so sound travels faster. Sound is nearer 330\,\mathrm{m\,s^{-1}} on a cold day and nearer 350\,\mathrm{m\,s^{-1}} on a hot day.

Real-world implications

Thunder and lightning: light arrives almost at once, but thunder arrives later. The time gap gives the distance to the storm.
Sound in metals: a tap at one end of a metal rod or rail is heard almost at once at the other end, through the metal, while the same sound through air arrives much later.
Marine life: sound travels about four times faster in water than in air, so dolphins and whales depend on sound to communicate over long distances in water.
Distance measurement: sonar and echo methods use the speed of sound in water, about 1500\,\mathrm{m\,s^{-1}}, to find depths and distances.

What is reflection of sound? Differentiate between echo and reverberation.

Solution

Reflection of sound

Reflection is the bouncing back of sound when it strikes a surface, just as light is reflected by a mirror. Reflection is stronger from a rigid and smooth surface, and weaker from a soft and irregular surface, which absorbs more of the sound.

Echo and reverberation

Both are reflections of sound. They differ in the time gap between the reflected sound and the original sound, and this decides whether the reflection is heard as a separate sound.

Feature	Echo	Reverberation
Time gap	0.1 s or more after the original sound	less than 0.1 s after the original sound
What is heard	a distinct repeat of the original sound	a prolonged, overlapping and muddled sound
Cause	a single reflection from a distant obstacle, such as a hill or a cliff	many reflections from walls, ceiling and floor of a room
Example	shouting towards a hill and hearing the repeat	clapping in a large hall with tiled walls

Echo is a distinct reflected sound that can be told apart from the original. Reverberation is the reflected sound that cannot be distinguished from the original, because the reflections overlap before the original has finished.

Explain the phenomenon of an echo as the reflection of sound waves. Design an experiment to measure the time delay of an echo, and discuss how this can be used to determine distances in various applications, such as sonar.

Solution

Echo as reflection of sound

An echo is a reflected sound that reaches the listener after the original sound, with a time gap of 0.1 s or more. Sound that strikes a hard obstacle is reflected back to the listener. Because the sound travels to the obstacle and back, the echo covers twice the distance between the listener and the obstacle.

Derivation Distance from the time delay of an echo

Aim. Find the distance d to the reflecting surface from the time delay \Delta t of the echo.

Assumptions.

Sound travels at a constant speed v in the air.
The echo is heard a time \Delta t after the original sound.

The sound travels from the observer to the wall (d) and back again (d).

The sound travels from the observer to the wall and back, so the total distance covered is 2d. The time taken is \Delta t.
2d=v\,\Delta t
Write the speed as distance divided by time, v=\dfrac{\text{total distance}}{\text{time}}.
v=\dfrac{2d}{\Delta t}
Make d the subject of the equation.
d=\dfrac{v\,\Delta t}{2}

Resultd=\dfrac{v\,\Delta t}{2}

The distance to the reflecting surface is half the total distance covered by the sound in the time delay. For the echo to be heard separately, \Delta t\geq0.1\,\mathrm{s}, which gives d\geq\dfrac{v}{20} (equation 13.3).

Experiment to measure the time delay of an echo

Method

Stand facing a large, hard surface such as a wall or the side of a hill, at a convenient distance. Use a stopwatch.
Shout or clap, and start the stopwatch at the instant the sound is made.
Stop the stopwatch at the instant the echo is heard. Repeat the measurement several times and take the mean of the times, to reduce the error in reaction time. The mean is \Delta t.
Calculate the distance using d=\dfrac{v\,\Delta t}{2} with the speed of sound in air, v=340\,\mathrm{m\,s^{-1}}. For example, \Delta t=1.2\,\mathrm{s} gives d=\dfrac{340\times1.2}{2}=204\,\mathrm{m}.

The experiment works only when the echo is heard after the original sound, which requires an obstacle at least 17\,\mathrm{m} away.

Using echoes to find distances

Sonar: a ship sends ultrasound pulses to the seabed, and the echo returns after a time t. The depth is d=\dfrac{vt}{2}. For example, an echo taking 14.7 s in water at 1500\,\mathrm{m\,s^{-1}} gives a depth of 11025\,\mathrm{m} (book Example 13.1).
Underwater objects: the depth of the sea, shipwrecks and submarines can be located by echo techniques.
Medical ultrasound: echoes from internal organs are used to form images, for example in prenatal scanning.
Bats: they use echolocation to reach their destinations and identify objects on their way.

Differentiate between noise and music? Explain that how is noise nuisance?

Solution

Noise and music

Musical sound is pleasant and harmonious. Noise is any loud, discordant or disagreeable sound. Physically, the two differ in the regularity of their waveforms (book Figure 13.19 and Figure 13.8).

Feature	Musical sound	Noise
Waveform	rhythmic and periodic, with a regular repeated pattern	non-rhythmic, with no regular pattern
Time intervals	regular time intervals for each vibration	irregular time intervals for the vibrations
Energy and intensity	steady energy and regular intensities	variable energy and irregular intensities
Effect on the listener	pleasant and harmonious	unpleasant and disagreeable

Why noise is a nuisance

Noise is a nuisance when it is unwanted and disturbs people. The ear is designed for periodic sound waves, which give a pleasant effect. Non-periodic sounds give an unpleasant effect, and sounds of high intensity are painful. Noise is therefore a nuisance because:

it causes psychological and physical stress in daily life;
it reduces concentration, so tasks are done less well and errors become more likely;
it disturbs sleep, and so causes fatigue;
at high levels it can cause hearing loss and tinnitus.

Justify the importance of acoustic protection in environments such as schools and hospitals. How can design elements and materials contribute to creating sound-friendly spaces?

Solution

What acoustic protection is

Acoustic protection is the use of soft and porous materials to absorb unwanted sounds. Soft materials reflect less sound and absorb more of its energy, so echoes and noise are reduced.

Why it is important in schools and hospitals

Learning: students can hear and understand the teacher, and focus and retain information better, when the reflections and background noise are reduced.
Health: protection from noise prevents hearing loss and reduces stress, anxiety and sleep disorders.
Recovery: a quieter environment reduces stress and fatigue, which helps patients in hospitals to recover.

Design elements and materials

Element or material	How it helps
Carpets, curtains and rugs	absorb sound and reduce echoes
Acoustic panels	absorb sound and reduce reverberation
Irregular surfaces or special panels	scatter sound evenly (diffusion)
Thick walls, double-glazed windows and insulated doors	stop sound passing between spaces
Rubber mounts and floating floors	damp vibrations and stop them spreading
Room size and design, bass traps	avoid amplification of certain sounds (resonance control)

The design must balance absorption and reflection. If the classroom or hall is too absorbent, the sound level may be too low for the audience, so the walls and ceiling are designed to reflect where needed and absorb where needed.

Analyze the effects of noise pollution on the environment and human health. What strategies can be implemented to mitigate these effects, and how can communities balance development with acoustic quality?

Solution

Effects of noise pollution

Noise pollution is undesirable or destructive sound created by human activity, such as traffic, industry and construction.

Human health: hearing loss from prolonged exposure, psychological and physical stress, and sleep disorders.
Productivity and safety: lower concentration, more errors and a greater risk of accidents at work and in study.
Environment and wildlife: noise disturbs animals that rely on sound for communication and navigation, such as the infrasound-using elephants and whales.
Community: a poorer quality of life in homes and public places.

Strategies to reduce the effects

Personal protection: use earplugs and limit the volume of earphones and headphones.
Reduce noise at source: turn off appliances when not in use, lower the volume of vehicles, and keep machines and vehicles in regular maintenance.
Absorb and block sound: use soft materials, seal gaps around doors and windows, and use barrier walls, fences and landscaping.
Control vibration: use rubber mounts and floating floors, and isolate noisy equipment.
Plant trees: more trees help to reduce the noise level in an area.

Balancing development with acoustic quality

Communities can continue to develop while keeping good acoustic quality by applying building standards that require adequate noise control, enforcing occupational noise limits in industry, and encouraging quieter machinery and materials. Acoustic design is then part of planning from the start, not an afterthought.

Humans can hear frequencies from approximately 20 Hz to 20,000 Hz. Analyze how sounds outside this range (infrasound and ultrasound) are utilized in nature and technology. What are the implications of these frequencies for human hearing?

Solution

The audible range and the two outside ranges

Humans hear sounds from about 20 Hz to 20 kHz. Sound below 20 Hz is infrasound, and sound above 20 kHz is ultrasound. People cannot hear either, but both are used in nature and in technology.

Uses in nature and technology

Type	Frequency	Uses in nature and technology
Infrasound	below 20 Hz	communication by elephants, whales and giraffes over long distances; detecting earthquakes and volcanic eruptions; locating underground rock and petroleum formations
Ultrasound	above 20 kHz	cleaning delicate objects; medical scanning, including prenatal scanning and heart checks; sonar for finding depths and objects underwater; echolocation by bats

Implications for human hearing

Humans are not able to hear infrasound, so warnings from natural events such as volcanoes and earthquakes must be detected with special sensors.
Ultrasound is also inaudible, yet it can be used safely for scanning, because it is non-invasive. Its use depends on the echoes it produces, not on hearing.
The size of an animal decides the frequencies it produces: larger animals produce lower frequencies, and the ability to hear them depends on the size and structure of the animal's ears.
A silent whistle works because it produces ultrasound, which humans cannot hear but dogs can. This shows that the human audible range is not the full range of sound.

Explain how sound waves are converted into electrical signals by the eardrum and auditory nerves.

Solution

Conversion of sound into electrical signals

The ear converts sound into electrical signals in several steps, each passing the vibration on in a different form: sound waves, mechanical vibrations, fluid waves, and finally electrical signals sent to the brain.

Sound waves travel into the ear canal and strike the eardrum, which vibrates in response to them.
The vibrations are passed to the three ossicles (malleus, incus and stapes) in the middle ear, which amplify them.
The stapes sends the amplified vibrations into the cochlea, a fluid-filled spiral structure in the inner ear. The vibrations create waves in the fluid.
The fluid waves move the basilar membrane inside the cochlea, which bends the hair cells on the membrane.
When the hair cells bend, they generate electrical signals.
The electrical signals are sent through the auditory nerve to the brain, which processes and interprets them as sound, such as music or speech.

So the eardrum and the ossicles convert the sound into mechanical vibrations, the cochlea and hair cells convert these into electrical signals, and the auditory nerve carries the signals to the brain for interpretation.

Note. The book's long questions are numbered 1 to 12 and then 15; questions 13 and 14 are not printed. The numbering is kept as printed.

Numerical response questionsp. 120

A tourist clapped his hands near a cliff and heard the echo after 5 s. What is the distance of the cliff from the student if the speed of the sound is taken as 344 m/s?

Solution

Given

v = 344\,\mathrm{m\,s^{-1}} speed of sound
t = 5\,\mathrm{s} time for the echo to return

To find

d, distance of the cliff

The sound travels to the cliff and back, so the total distance covered is 2d. Use v=\dfrac{2d}{t}, which gives d=\dfrac{vt}{2}.
d=\dfrac{vt}{2}
Substitute the values.
d=\dfrac{344\times5}{2}
Evaluate.
d=\dfrac{1720}{2}=860\,\mathrm{m}

d=860\,\mathrm{m}

Book answer. 860 m

Note. The book says 'the student' where the person who clapped is meant; the distance is from the clapping person to the cliff.

Speed of sound in seawater is 1500 m/s. If its wavelength is 45 cm, find the frequency of these sound waves.

Solution

Given

v = 1500\,\mathrm{m\,s^{-1}} speed of sound in seawater
\lambda = 45\,\mathrm{cm}=0.45\,\mathrm{m} wavelength

To find

f, frequency

The speed, frequency and wavelength are related by v=f\lambda. Make f the subject.
f=\dfrac{v}{\lambda}
Substitute the values, with the wavelength in metres.
f=\dfrac{1500}{0.45}
Evaluate.
f=3333.3\,\mathrm{Hz}

f\approx3333.3\,\mathrm{Hz}\approx3.33\,\mathrm{kHz}

Book answer. 3,333.3 Hz

A sound wave in seawater travels at speed 1500 m/s, it is sent and detected back after reflection from seabed in 8.5 s. Find the distance from surface to the bottom of the sea.

Solution

Given

v = 1500\,\mathrm{m\,s^{-1}} speed of sound in seawater
t = 8.5\,\mathrm{s} time for the wave to go down and come back

To find

d, depth of the sea

The wave travels down to the seabed and back up, so the depth is half the total distance: d=\dfrac{vt}{2}.
d=\dfrac{vt}{2}
Substitute the values.
d=\dfrac{1500\times8.5}{2}
Evaluate.
d=\dfrac{12750}{2}=6375\,\mathrm{m}

d=6375\,\mathrm{m}

Book answer. 6,375 m

A physician counts 69 heartbeats in 1 min. Calculate the frequency and period of the heartbeats.

Solution

Given

N = 69 number of heartbeats
t = 1\,\mathrm{min}=60\,\mathrm{s} time

To find

f, frequency
T, period

Frequency is the number of vibrations in one second, so divide the number of beats by the time in seconds.
f=\dfrac{N}{t}=\dfrac{69}{60}
Evaluate the frequency.
f=1.15\,\mathrm{Hz}
The period is the reciprocal of the frequency, T=\dfrac{1}{f}.
T=\dfrac{1}{1.15}
Evaluate the period.
T=0.8696\,\mathrm{s}

f=1.15\,\mathrm{Hz},\qquad T=0.8696\,\mathrm{s}

Book answer. 1.15 Hz, 0.8696 s

Find the range of wavelengths at speed 343 m/s for audible frequency range (20 Hz to 20 kHz).

Solution

Given

v = 343\,\mathrm{m\,s^{-1}} speed of sound
f_{\min} = 20\,\mathrm{Hz} lowest audible frequency
f_{\max} = 20\,\mathrm{kHz}=2\times10^{4}\,\mathrm{Hz} highest audible frequency

To find

\lambda_{\max} and \lambda_{\min}

The wavelength is \lambda=\dfrac{v}{f}. The longest wavelength is at the lowest frequency.
\lambda_{\max}=\dfrac{343}{20}
Evaluate the longest wavelength.
\lambda_{\max}=17.15\,\mathrm{m}
The shortest wavelength is at the highest frequency.
\lambda_{\min}=\dfrac{343}{2\times10^{4}}
Evaluate the shortest wavelength.
\lambda_{\min}=0.01715\,\mathrm{m}

\lambda_{\min}=0.01715\,\mathrm{m}\ \text{to}\ \lambda_{\max}=17.15\,\mathrm{m}

Book answer. 17.15 m, 0.01715 m

During thunder storm, thunder sound is heard after 3 second of lightning flash. Find the distance of clouds from ground. (Speed of sound = 340 m/s)

Solution

Given

v = 340\,\mathrm{m\,s^{-1}} speed of sound
t = 3\,\mathrm{s} time between the flash and the thunder

To find

d, distance of the clouds

The light from the flash reaches us almost instantly, so the time is taken by the sound alone, travelling one way from the cloud to the ground. Use d=vt (not d=\dfrac{vt}{2}, since there is no return journey).
d=vt
Substitute the values.
d=340\times3
Evaluate.
d=1020\,\mathrm{m}

d=1020\,\mathrm{m}\approx1.02\,\mathrm{km}

Book answer. 1020 m

SONAR (sound navigation and ranging system) sends ultrasound signal towards sea bed. It is received back after 5.3 s. If speed of sound in sea water is 1550 m/s. Find the depth of sea there.

Solution

Given

v = 1550\,\mathrm{m\,s^{-1}} speed of sound in sea water
t = 5.3\,\mathrm{s} time for the signal to return

To find

d, depth of the sea

The signal goes down to the sea bed and comes back, so the depth is d=\dfrac{vt}{2}.
d=\dfrac{vt}{2}
Substitute the values.
d=\dfrac{1550\times5.3}{2}
Evaluate.
d=\dfrac{8215}{2}=4107.5\,\mathrm{m}

d=4107.5\,\mathrm{m}

Book answer. 4107.5 m

FBISE 2026 questions

Questions from the FBISE SSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

FBISE SSC-II 2026 Annual I, Set B, MCQ 121 mark

Sound travels faster in solids than in gases mainly because solids have:

ALess pressure
BHigher densityAnswer
CLower density
DLess particles

Why

In a solid the particles are tightly packed (closely spaced and strongly bound), so the solid is much denser than a gas. Each particle is close to its neighbour and passes the vibration on quickly, so sound travels fastest in solids. Of the four options only higher density matches this.

Options C and D describe a gas, where the particles are far apart and the vibration is passed on slowly. Option A is wrong because the speed of sound in a solid does not come from low pressure.

FBISE SSC-II 2026 Annual I, Set B, Q.2 (vii)1+2 = 3 marks

What is echo? How is it produced?

Solution

Echo (1 mark). An echo is a distinct sound that comes back to the listener after the sound wave strikes an obstacle and is reflected from it. It is heard separately from the original sound.

How it is produced (2 marks). When a source makes a sound, the wave travels to a hard, rigid surface such as a cliff, a hill or a wall, and bounces back by reflection, like a rubber ball bouncing from a wall. The reflected wave is heard as a second, weaker sound.

The sound travels to the obstacle and back, so the total distance covered is twice the distance d of the obstacle. The ear keeps the sensation of a sound for about 0.1\,\mathrm{s}, so the reflected wave is heard separately only if it returns at least 0.1\,\mathrm{s} after the original sound:

2d=v\times0.1\quad\Rightarrow\quad d=\dfrac{v}{20}

With v=340\,\mathrm{m\,s^{-1}} the obstacle must be at least 17\,\mathrm{m} away. If the time gap is less than 0.1\,\mathrm{s}, the reflected sound overlaps the original and no separate echo is heard (reverberation).

FBISE SSC-II 2026 Annual I, Set B, Q.43+2 = 5 marks

a. Describe the physical factors affecting loudness and pitch of sound. b. How the amplitude of a sound wave affects its loudness. Give suitable examples.

Solution

(a) Factors affecting loudness and pitch

Loudness is the characteristic of a sound by which we tell a loud sound from a faint sound. It depends on the amplitude of the wave, and it is proportional to the square of the amplitude:

\text{loudness}\propto A^{2}

The physical factors affecting loudness are:

Vibrational amplitude of the source: a larger amplitude of vibration gives a louder sound.
Surface area of the source: a larger vibrating surface gives a louder sound.
Distance from the source: the farther the listener, the fainter the sound; a listener near the source hears it louder.
Sensitivity of the listener's ears: more sensitive ears feel a greater loudness.

Pitch is the characteristic by which we tell a shrill (high) sound from a grave (deep) sound. It depends on the frequency of the wave (the number of vibrations per second):

\text{pitch}\propto f

A larger frequency gives a higher pitch and a smaller frequency gives a lower pitch. Pitch does not depend on the amplitude. A bird gives a high-pitched sound and the roar of a lion is low-pitched.

(b) Effect of the amplitude on loudness

The amplitude is the maximum displacement of the particles of the medium, and it measures the energy carried by the wave. A larger amplitude means more energy reaches the ear every second, so the sound is louder. Doubling the amplitude gives (2A)^{2}=4A^{2}, that is four times the loudness. The frequency stays the same, so the pitch does not change.

Same frequency, different amplitude

Two waves of the same frequency. The wave with the larger amplitude (A=2) is louder.

Examples: a drum struck hard vibrates with a larger amplitude and sounds louder than the same drum struck gently; turning up the volume of a loudspeaker makes its cone vibrate with a larger amplitude; a speaker who shouts produces a larger amplitude than one who whispers.

FBISE SSC-II 2026 Annual I, Set D, MCQ 121 mark

In which medium does sound travel the fastest?

AVacuum
BGases
CLiquids
DSolidsAnswer

Why

Sound travels fastest in solids, slower in liquids and slowest in gases. The particles of a solid are tightly packed, so they pass the vibrations on most quickly (over 5000\,\mathrm{m\,s^{-1}} in steel).

Liquids (C) come second, about 1500\,\mathrm{m\,s^{-1}} in water, and gases (B) are slowest, about 340\,\mathrm{m\,s^{-1}} in air. In a vacuum (A) there are no particles, so sound cannot travel at all.

FBISE SSC-II 2026 Annual I, Set D, Q.2 (vii)1+2 = 3 marks

What is the minimum time required to produce echo? Why is an echo not heard in a small room? Justify.

Solution

Minimum time (1 mark). The sensation of a sound stays in the brain for about 0.1\,\mathrm{s}, so the echo must reach the ear at least 0.1\,\mathrm{s} after the original sound. The minimum time is 0.1\,\mathrm{s}.

Small room (2 marks). The sound travels to the wall and back, a distance 2d, in the time t. For an echo t\ge0.1\,\mathrm{s}, which needs the wall to be at least

d=\dfrac{v}{20}=\dfrac{340}{20}=17\,\mathrm{m}

away. In a small room the walls are nearer than 17\,\mathrm{m}. Take a wall at 5\,\mathrm{m}: the sound returns after t=2d/v=(2\times5)/340\approx0.03\,\mathrm{s}, which is less than 0.1\,\mathrm{s}. The reflected sound therefore reaches the ear before the sensation of the original sound has faded, and the two cannot be told apart. They overlap (reverberation) and make the sound seem louder and prolonged, but no separate echo is heard.

FBISE SSC-II 2026 Annual I, Set D, Q.43+2 = 5 marks

a. Explain the effect of changing amplitude and frequency of sound waves. Also draw diagram. b. Differentiate between loudness and pitch of sound draw diagrams as well.

Solution

(a) Effect of changing amplitude and frequency

Amplitude. The amplitude is the maximum displacement of the particles of the medium and it measures the energy of the wave. Increasing the amplitude makes the sound louder, and decreasing it makes the sound fainter. The loudness is proportional to the square of the amplitude, and the pitch does not change.

Same frequency, different amplitude

Two waves of the same frequency. The wave with the larger amplitude (A=2) is louder.

Frequency. The frequency is the number of waves produced per second. Increasing the frequency makes the waves come closer together and raises the pitch (a shriller sound); decreasing it lowers the pitch. The loudness does not change.

Same amplitude, different frequency

Two waves of the same amplitude. The wave with twice the frequency (2f) has the higher pitch.

Change in the wave	Effect on loudness	Effect on pitch
Amplitude increases	loudness increases	no change
Frequency increases	no change	pitch increases

(b) Loudness and pitch compared

Feature	Loudness	Pitch
Meaning	tells a loud sound from a faint sound	tells a shrill (high) sound from a grave (low) sound
Depends on	amplitude of the wave, loudness \propto A^{2}	frequency of the wave, pitch \propto f
Energy received by the ear in unit time	depends on it	does not depend on it
Change of frequency	not affected	increases as the frequency increases
Unit	decibel	no unit; judged from the frequency in hertz
In the diagram	taller wave means louder sound	waves closer together means higher pitch

Diagrams: the first graph above shows loudness (same frequency, larger amplitude is louder). The second graph shows pitch (same amplitude, the wave with more cycles in the same distance has the higher pitch).

FBISE SSC-II 2026 Annual I, Set M, MCQ 121 mark

Sound cannot travel through:

AVacuumAnswer
BAir
CWater
DSteel

Why

Sound is a mechanical wave, so it needs particles of a medium to pass the vibrations on. A vacuum has no particles, so sound cannot travel through it. This is why the bell in a bell jar cannot be heard when the air is pumped out.

Air (B), water (C) and steel (D) all have particles, so they carry sound. Steel is in fact the fastest of the three.

FBISE SSC-II 2026 Annual I, Set M, Q.2 (vii)3 marks

Differentiate between pitch and loudness. (Any three)

Solution

Pitch	Loudness
The characteristic used to tell a shrill sound from a grave (flat) sound	The characteristic used to tell a loud sound from a faint sound
Depends on the frequency of the wave, pitch \propto f	Depends on the amplitude of the wave, loudness \propto A^{2}
Independent of the energy received by the ear in unit time	Depends on the energy received by the ear in unit time
Increases when the frequency increases	Not changed by a change in frequency

Any three of these differences carry full marks. Example: a bird has a high pitch and a lion a low pitch, while a shout is louder than a whisper.

FBISE SSC-II 2026 Annual I, Set M, Q.4(2+1)+2 = 5 marks

a. Explain the effect of changing amplitude and frequency on sound waves. Also draw diagram. b. How the frequency of a sound wave determines its pitch?

Solution

(a) Effect of changing amplitude and frequency

Amplitude (2 marks). The amplitude is the maximum displacement of the particles of the medium and it shows the energy carried by the wave. A larger amplitude gives a louder sound (loudness \propto A^{2}), and the pitch stays the same.

Same frequency, different amplitude

Two waves of the same frequency. The wave with the larger amplitude (A=2) is louder.

Frequency (1 mark). The frequency is the number of waves per second. A larger frequency brings the waves closer together and gives a higher pitch, while the loudness stays the same.

Same amplitude, different frequency

Two waves of the same amplitude. The wave with twice the frequency (2f) has the higher pitch.

(b) Frequency and pitch (2 marks)

Pitch is the sensation by which we tell a shrill sound from a deep sound. It is not a physical quantity and cannot be measured directly, but it can be found from the frequency of the wave, because it describes the number of vibrations produced in one second. Hence pitch is directly proportional to frequency:

\text{pitch}\propto f

A high frequency gives a high pitch (shrill) and a low frequency gives a low pitch (deep). Examples: a bird gives a high-pitched sound and the roar of a lion is low-pitched; the voice of a woman has a higher pitch than that of a man.

Exam focus

Explain how amplitude and frequency change loudness and pitch, with diagrams; give the factors affecting each. (long)
Q.4 (5 marks) in all three 2026 sets (B, D and M), with a different part (b) each time; also comprehensive question 4 of the book.
Go to answer
Differentiate between pitch and loudness (three points). (short)
Set M Q.2 (vii) and part (b) of Set D Q.4; learn the table of meaning, dependence and energy.
Go to answer
What is an echo? How is it produced? Give the minimum distance d=v/20. (short)
Set B Q.2 (vii); Set D Q.2 (vii) asked the minimum time and why a small room has no echo; also MCQ 9 and comprehensive questions 7 and 8.
Go to answer
Why is no echo heard in a small room? (Echo against reverberation.) (short)
Set D Q.2 (vii) asked this with a justification; the book also asks echo against reverberation in comprehensive question 7.
Go to answer
Speed of sound in solids, liquids and gases, and why solids are fastest. (mcq)
MCQ 12 in Set B (why solids) and Set D (fastest medium); also MCQ 4, short question 5 and comprehensive question 6.
Go to answer
Sound needs a medium: it cannot travel through a vacuum (bell jar experiment). (mcq)
Set M MCQ 12; the book asks it as MCQ 5 and as comprehensive question 2 with an experiment.
Go to answer
Echo numericals: d=vt/2 for the cliff, seabed and SONAR. (numerical)
No numerical from this unit appeared in the 2026 papers, but numerical problems 1, 3, 6 and 7 use the same echo method and the unit has seven numericals.
Go to answer
Quality (timbre): how two sounds of the same loudness and pitch are told apart. (short)
Not asked in 2026, but short question 7 and comprehensive question 5 of the book test it, and it completes the three characteristics of sound.
Go to answer