NCERT Solutions for Class 9 Science
“Chapter 12- SOUND”
NCERT Solutions for Class 9
Science Chapter 12 “Sound”:
by M. Farrukh Asif
NCERT Solutions for Class 9 Science Chapter 12
“SOUND”, Updated.
5.
THE
FUNDAMENTAL UNIT OF LIFE
6.
TISSUES
7.
MOTION
9.
GRAVITATION
10.
WORK
AND ENERGY
11.
SOUND
12.
IMPROVEMENT
IN FOOD RESOURCES
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complex textbook content into straightforward solutions, students can easily
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Science.
NCERT SOLUTIONS FOR CLASS 9 SCIENCE
CHAPTER-12 SOUND
NCERT Solutions for class 9 Science chapter 12
SOUND have been prepared and uploaded for
reference by our team of academic experts at NCERT
Solution cbsencert21.blogspot.com. Access solutions to all chapters of
NCERT class 9 Science from our website for comprehensive guidance. Utilize the
following NCERT solutions of chapter 12, crafted by our team, as a valuable
reference. Take the time to review the chapter 12 on Sound before heading into the exam.
• Sound is produced due to vibration of different
objects.
• Sound travels as a longitudinal wave through a
material medium.
• Sound travels as successive compressions and
rarefactions in the medium.
• In sound propagation, it is the energy of the sound
that travels and not the particles of the medium.
• Sound cannot travel in vacuum.
• The change in density from one maximum value to the
minimum value and again to the maximum
value makes one complete oscillation.
• The distance between two consecutive compressions
or two consecutive rarefactions is called the
wavelength, λ.
• The time taken by the wave for one complete
oscillation of the density or pressure of the medium is
called the time period, T.
• The number of complete oscillations per unit time
is called the frequency (ν), 1 ν = T .
• The speed v, frequency ν, and wavelength λ, of
sound are related by the equation, v = λν.
• The speed of sound depends primarily on the nature
and the temperature of the transmitting medium. • The law of reflection of
sound states that the directions in which the sound is incident and reflected
make equal angles with the normal to
the reflecting surface at the point of incidence and the three lie
in the same plane.
• For hearing a distinct sound, the time interval
between the original sound and the reflected one must
be at least 0.1 s.
• The persistence of sound in an auditorium is the
result of repeated reflections of sound and is called
reverberation.
• Sound properties such as pitch, loudness and
quality are determined by the corresponding wave properties.
• Loudness is a physiological response of the ear to
the intensity of sound.
• The amount of sound energy passing each second
through unit area is called the intensity of sound.
• The audible range of hearing for average human
beings is in the frequency range of 20 Hz – 20 kHz.
• Sound waves with frequencies below the audible
range are termed “infrasonic” and those above the
audible range are termed “ultrasonic”.
• Ultrasound has many medical and industrial
applications.
• The SONAR technique is used to determine the depth
of the sea and to locate underwater hills,
valleys, submarines, icebergs, sunken
ships, etc.
|
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Study Reference for Class 9
Chapter 12 Sound |
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Revision Notes for Chapter 12 Sound ·
Extra Questions for Chapter
12 Sound ·
MCQ Questions for Chapter 12
Sound ·
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In Text Questions
Page No: 162
1. How does the sound produced by a vibrating object in a medium reach your
ear?
Page No: 163
1. Explain how sound is produced by your school bell.
Answer
When the bell continues to move forward and backward, it creates a series of
compressions and rarefactions making production of sound.
2. Why are sound waves called mechanical waves?
Answer
Sound waves need material medium to propagate therefore, they are called
mechnical waves. Sound waves propagate through a medium because of
theinteraction of the particles present in that medium.
3. Suppose you and your friend are on the moon. Will you be able to hear any
sound produced by your friend?
Answer
No, because sound waves needs a medium through which they can propagate. Since
there is no material medium on the moon due to absence of atmosphere, you
cannot hear any sound on the moon.
Page No: 166
1. Which wave property determines (a) loudness, (b) pitch?
Answer
(a) Amplitude
(b) Frequency
2. Guess which sound has a higher pitch: guitar or car horn?
Answer
Guitar has a higher pitch than car horn, because sound produced by the strings
of guitar has high frequency than that of car horn. High the frequency higher
is the pitch.
1. What are the wavelength, frequency, time period and
amplitude of a sound wave?
Answer
→ Wavelength: The distance between two consecutive compressions or two
consecutive rarefactions is known as the wavelength. Its SI unit is metre (m).
→ Frequency: The number of complete oscillations per second is known as the
frequency of a sound wave. It is measured in hertz (Hz).
→ Amplitude: The maximum height reached by the crest or trough of a sound wave
is called its amplitude.
Answer
Speed, wavelength, and frequency of a sound wave are related by the following
equation:
Speed (v) = Wavelength (λ) x Frequency (ν)
v = λ x ν
3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed
is 440 m/s in a given medium.
Answer
Frequency of the sound wave, ν= 220 Hz
Speed of the sound wave, v = 440 m s-1
For a sound wave,
Speed = Wavelength x Frequencyv = λ x ν
∴ λ= v / ν = 440 / 220 = 2m
Hence, the wavelength of the sound wave is 2 m.
4. A person is listening to a tone of 500 Hz sitting at a distance of 450 m
from the source of the sound. What is the time interval between successive
compressions from the source?
Answer
The time interval between two successive compressions is equal to the time
period of the wave. This time period is reciprocal of the frequency of the wave
and is given by the relation:T= 1 / Frequency = 1/ 500 = 0.002 s
1. Distinguish between loudness and intensity of sound.
Answer
The intensity of a sound wave is defined as the amount of sound energy passing
through a unit area per second. Loudness is a measure of the response of the
ear to the sound. The loudness of a sound is defined by its amplitude. The
amplitude of a sound decides its intensity, which in turn is perceived by the
ear as loudness.
Page No: 167
1. In which of the three media, air, water or iron, does sound travel the
fastest at a particular temperature?
Answer
The speed of sound depends on the nature of the medium. Sound travels the
fastest in solids. Its speed decreases in liquids and it is the slowest in
gases. Therefore, for a given temperature, sound travels fastest in iron.
Page No: 168
1. An echo returned in 3 s. What is the distance of the reflecting surface from
the source, given that the speed of sound is 342 m s−1?
Answer
Speed of sound, v = 342 m s−1
Echo returns in time, t = 3 s
Distance travelled by sound = v × t = 342 × 3 = 1026 m
In the given time interval, sound has to travel a distance that is twice the
distance of the reflecting surface and the source.
Hence, the distance of the reflecting surface from the source= 1026 / 2 m = 513
m.
Page No: 169
1. Why are the ceilings of concert halls curved?
Answer
Ceilings of concert halls are curved so that sound after reflection (from the
walls) spreads uniformly in all directions.
Page No: 170
1. What is the audible range of the average human ear?
Answer
The audible range of an average human ear lies between 20 Hz to 20,000 Hz.
2. What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer
(a) Infrasound has frequencies less than 20 Hz.
(b) Ultrasound has frequencies more than 20,000 Hz.
Page No: 172
1. A submarine emits a sonar pulse, which returns from an underwater cliff in
1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the
cliff?
Speed of sound in salt water, v = 1531 m s - 1
Distance of the cliff from the submarine = Speed of sound x Time taken
Distance of the cliff from the submarine = 1.02 x 1531 = 1561.62 m
Distance traveled by the sonar pulse during its
transmission and reception in water = 2 x Actual distance = 2d
Actual Distance, d= Distance of the cliff from the
submarine/2
= 1561/2
= 780.31 m
Page No: 174
1. What is sound and how is it produced?
Answer
Sound is a form of energy that gives the sensation of hearing. It is produced
by the vibrations caused in air by vibrating objects.
2. Describe with the help of a diagram, how compressions and rarefactions are
produced in air near a source of sound.
Answer
When a vibrating body moves forward, it createsa region of high pressure in its
vicinity. This region of high pressure is known as compression. When it moves
backward, it creates a region of low pressure in its vicinity. This region is
known as a rarefaction. As the body continues to move forward and backward, it
produces a series of compressions and rarefactions. This is shown in the below
figure.
3. Cite an experiment to show that sound needs a material medium for its propagation.
Take an electric bell and an air tight glass bell
jar connected to a vacuum pump. Suspend the bell inside the jar, and press the
switch of the bell. You will be able to hear the bell ring. Now pump out the
air from the glass jar. The sound of the bell will become fainter and after
some time, the sound will not be heard. This is so because almost all air has
been pumped out. This shows that sound needs a material medium to travel.
4. Why is a sound wave called a longitudinal wave?
Answer
The quality or timber of sound enables us to identify our friend by his voice.
6. Flash and thunder are produced simultaneously. But thunder is heard a few
seconds after the flash is seen, why?
Answer
The speed of sound (344 m/s) is less than the speed of light(3 x
108 m/s). Sound of thunder takes more time to reach the Earth as compared
to light. Hence, a flash is seen before we hear a thunder.
7. A person has a hearing range from 20 Hz to 20 kHz. What are the typical
wavelengths of sound waves in air corresponding to these two frequencies? Take
the speed of sound in air as 344 m s−1.
Answer
For a sound wave,
Speed = Wavelength x Frequencyv = λ x ν
Speed of sound in air = 344 m/s (Given)
(i) For, ν= 20 Hz
λ1= v/ν = 344/20 = 17.2 m
(ii) For, ν= 20000 Hz
λ2= v/ν = 344/20000 = 0.172 m
Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.
8. Two children are at opposite ends of an
aluminium rod. One strikes the end of the rod with a stone. Find the ratio of
times taken by the sound wave in air and in aluminium to reach the second
child.
Answer
Velocity of sound in air= 346 m/s
Velocity of sound wwave in aluminium= 6420 m/s
Let length of rode be 1
Time taken for sound wave in air, t1= 1 / Velocity in air
Time taken for sound wave in Aluminium, t2= 1 / Velocity in aluminium
Therefore, t1 / t2 = Velocity in aluminium / Velocity in air = 6420 /
346 = 18.55 : 1
9. The frequency of a source of sound is 100 Hz. How many times does it vibrate
in a minute?
Answer
Frequency = 100 Hz (given)
This means the source of sound vibrates 100 times in one second.
Therefore, number of vibrations in 1 minute, i.e. in 60 seconds = 100 x 60 =
6000 times.
10. Does sound follow the same laws of reflection as light does? Explain.
Answer
Sound follows the same laws of reflection as light does. The incident sound
wave and the reflected sound wave make the same angle with the normal to the
surface at the point of incidence. Also, the incident sound wave, the reflected
sound wave, and the normal to the point of incidence all lie in the same plane.
11. When a sound is reflected from a distant object, an echo is produced. Let
the distance between the reflecting surface and the source of sound production
remains the same. Do you hear echo sound on a hotter day?
Answer
An echo is heard when the time for the reflected sound is heard after 0.1 s
Time Taken= Total Distance / Velocity
On a hotter day, the velocity of sound is more. If the time taken by echo is
less than 0.1 sec it will not be heard.
12. Give two practical applications of reflection of sound waves.
Answer
Two practical applications of reflection of sound waves are:
→ Reflection of sound is used to measure the distance and speed of underwater
objects. This method is known as SONAR.
→ Working of a stethoscope is also based on reflection of sound. In a
stethoscope, the sound of the patient's heartbeat reaches the doctor's ear by
multiple reflection of sound.
13. A stone is dropped from the top of a tower 500 m high into a pond of water
at the base of the tower. When is the splash heard at the top? Given, g = 10 m
s−2 and speed of sound = 340 m s−1.
Answer
Height of the tower, s = 500 m
Velocity of sound, v = 340 m s−1
Acceleration due to gravity, g = 10 m s−2
Initial velocity of the stone, u = 0 (since the stone is initially at
rest)
Time taken by the stone to fall to the base of the tower, t1
According to the second equation of motion:
Now, time taken by the sound to reach the top from
the base of the tower, t2= 500 / 340 = 1.47 s
Therefore, the splash is heard at the top after time, t
Where, t= t1 + t2 = 10 + 1.47 = 11.47 s.
14. A sound wave travels at a speed of 339 m s−1. If its wavelength is 1.5 cm,
what is the frequency of the wave? Will it be audible?
Answer
Speed of sound, v= 339 m s - 1
Wavelength of sound, λ= 1.5 cm = 0.015 m
Speed of sound = Wavelength x Frequencyv= λ x v
∴ v= v / λ = 339 / 0.015 = 22600 Hz
The frequency range of audible sound for humans lies between 20 Hz to 20,000
Hz. Since the frequency of the given sound is more than 20,000 Hz, it is not
audible.
15. What is reverberation? How can it be reduced?
Answer
The repeated multiple reflections of sound in any big enclosed space is known
as reverberation.
The reverberation can be reduced by covering the ceiling and walls of the
enclosed space with sound absorbing materials, such as fibre board, loose woollens,
etc.
16. What is loudness of sound? What factors does it depend on?
Answer
The effect produced in the brain by the sound of different frequencies is
called loudness of sound.
Loudness depends on the amplitude of vibrations. In fact, loudness is proportional
to the square of the amplitude of vibrations.
17. Explain how bats use ultrasound to catch a prey.
Answer
Bats produce high-pitched ultrasonic squeaks. These high-pitched squeaks are
reflected by objects such as preys and returned to the bat's ear. This allows a
bat to know the distance of his prey.
18. How is ultrasound used for cleaning?
Answer
Objects to be cleansed are put in a cleaning solution and ultrasonic sound
waves are passed through that solution. The high frequency of these ultrasound
waves detaches the dirt from the objects.
19. Explain the working and application of a sonar.
Answer
SONAR is an acronym for Sound Navigation And Ranging. It is an acoustic device
used to measure the depth, direction, and speed of under-water objects such as
submarines and ship wrecks with the help of ultrasounds. It is also used to
measure the depth of seas and oceans.
A beam of ultrasonic sound is produced and
transmitted by the transducer (it is a device that produces ultrasonic sound)
of the SONAR, which travels through sea water. The echo produced by the
reflection of this ultrasonic sound is detected and recorded by the detector,
which is converted into electrical signals. The distance (d) of the under-water
object is calculated from the time (t) taken by the echo to return with speed
(v) is given by 2d = v × t. This method of measuring
distance is also known as ‘echo-ranging’.
20. A sonar device on a submarine sends out a signal and receives an echo 5 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625 m.
Answer
Time taken to hear the echo, t= 5 s
Distance of the object from the submarine, d= 3625 m
Total distance travelled by the sonar waves during the transmission and reception in water= 2d
Velocity of sound in water, v=
2d / t = 2 x 3625 / 5 = 1450 ms-1.
21. Explain how defects in a metal block can be detected using ultrasound.
Answer
Defects in metal blocks do not allow ultrasound to pass through them and they
are reflected back. This fact is used to detect defects in metal blocks.
Ultrasound is passed through one end of a metal block and detectors are placed
on the other end. The defective part of the metal block does not allow
ultrasound to pass through it. As a result, it will not be detected by the
detector. Hence, defects in metal blocks can be detected using ultrasound.
→ Outer ear: This is also called ‘pinna’. It
collects the sound from the surrounding and directs it towards the auditory canal.
→ Middle ear: The sound reaches the end of the
auditory canal where there is a thin membrane called eardrum or tympanic
membrane. The sound waves set this membrane to vibrate. These vibrations are
amplified by three small bones- hammer, anvil and stirrup.
→ Inner ear: These vibration reach the cochlea in
the inner ear and are converted into electrical signals which are sent to the
brain by the auditory nerve, and the brain interprets them as sound.
=====================
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Notes:
Ch-12 Sound | Class 9th Science
Topics in the Chapter
Introduction
• Production of Sound
• Propogation of Sound
• Sound waves as Longitudinal waves
• Characteristics of Sound wave
• Wavelength
• Frequency
• Time Period
• Amplitude
Ø
Pitch
Ø
Loudness
Quality
or Timbre
• Velocity
• Speed of sound in various medium
• Sonic Boom
• Reflection of Sound
• Echo
• Reverberation
• Range of Hearing
• Hearing Aid
• Application of Ultrasound
• SONAR
• Structure of Human Ear
Ø
Working of Human Ear
§
The sensation felt by our ears is called sound.
§
Sound is a form of energy which makes us hear.
§
Law of conservation of energy is also applicable to sound
§
Sound travels in form of wave.
Ø
Production of Sound
o
Sound is produced when object vibrates or sound
is produced by vibrating objects.
o
The
energy required to make an object vibrate and produce sound is provided by some
outside source (like our hand, wind etc.).
o
Example: Sound of our voice is produced by
vibration of two vocal cords in our throat.
Ø
Sound of a drum or tabla is produced by
vibration of its membrane
when struck.
o
In laboratory experiments, sound is produced by
vibrating tuning fork. The vibrations of tuning fork can be shown by touching a
small suspended pith ball (cork ball) with a prong of the sounding tuning fork.
The pith ball is pushed away with a great force.
Sound can be produced by following methods:
(i) By vibrating string (sitar)
(ii) By vibrating air (flute)
(iii) By vibrating membrane (table, drum)
(iv) By vibrating plates (bicycle bell)
(v) By friction in objects
(vi) By scratching or scrubbing the objects etc.
Propogation of Sound
o The
substance through which sound travels is called a medium.
o The
medium may be solid, liquid or gas.
o When
an object vibrates, then the air particles around it also start vibrating in
exactly the same way and displaced from their stable position.
o These vibrating air particles exert a force on
nearby air particles so they are also displaced from their rest position and
start to vibrate.
o This
process is continued in the medium till sound reaches our ears.
o The
disturbance produced by sound travels through the medium (not the particles of
the medium).
A wave is a disturbance that travels through a
medium and carries energy.
o So
sound travels in wave form known as mechanical waves.
• When a body vibrates then it compresses the air
surrounding it and form a area of high density called compression (C).
o Compression
is the part of wave in which particles of the medium are closer to one another
forming high pressure.
o →
This compression move away from the vibrating body.
• When vibrating body vibrates back a area of low
pressure is formed called rarefaction (R).
o →
Rarefaction is the area of wave in which particles of the medium are further
apart from one another forming a low pressure or low density area.
o →
When body vibrates back and forth, a series of compression and rarefaction is
formed in air resulting in sound wave.
o →
Propagation of sound wave is propagation of density change.
Sound needs Medium for Propogation
• Sound waves are mechanical waves.
o →
It needs material medium for propogation like air, water, steel etc.
o →
It cannot travel in vaccum.
o →
An electric bell is suspended in airtight bell jar connected with vacuum
pump.
o →
When bell jar is full of air, we hear the sound but when air is pumped out from
the bell jar by vacuum pump and we ring the bell, no sound is heard.
o →
So, medium is necessary for propagation of sound.
Experiment to show that sound cannot travel through
vacuum
Sound Waves as Longitudinal Waves
• A wave in which the particles of the medium vibrate
back and forth in the same direction in which the wave is moving, is called a longitudinal
wave.
o When we push and pull the slinky compression
(number of turns are more or closer) and rarefaction (number of turns are less
or farther) are formed.
o When
a wave travels along with slinky, its each turn moves back and forth by only a
small distance in the direction of wave. So the wave is longitudinal.
o The
direction of vibrations of the particles is parallel to the direction of wave.
• When one end of a slinky is moved up and down
rapidly whose other end is fixed, it produces transverse wave.
o This
wave possess along the slinky in horizontal direction, while turns of slinky
(particles) vibrate up and down at right angle to the direction of wave.
o Thus
in transverse wave particles of the medium vibrate up and down at right angles
to the direction of wave.
o Light waves are transverse waves but they
don’t need a material
medium for propagation.
Characteristics of Sound Wave
• The characteristics of sound waves are :
wavelength, frequency, amplitude, time period and velocity.
o When a wave travel in air the density and
pressure of air changes from their mean position.
o Compression is shown by crest while
rarefaction is shown by trough.
o Compression
is the region of maximum density or pressure.
o Rarefaction
is the region of minimum density or pressure.
Wavelength
o In
sound waves the combined length of a compression and an adjacent rarefaction is
called its wavelength.
o The
distance between the centres of two consecutive compressions or two consecutive
rarefactions is also called its wavelength.
o It
is denoted by the Greek letter lamda (λ). Its SI unit is metre.
Frequency
o
No. of complete waves produced in one second or
number of vibrations per second is called frequency.
o
Number of compressions or rarefactions passed in
one second is also frequency.
o
Frequency of wave is same as the frequency of
the vibrating body which produces the wave.
• The SI unit of frequency is hertz (Hz). The symbol
of frequency is v (nu).
• 1 Hertz: One Hz is equal to 1 vibration
per second.
• Bigger unit of frequency is kilohertz kHz = 1000
Hz.
Time Period
o Time
taken to complete one vibration is called time period.
o Time
required to pass two consecutive compressions or rarefactions through a point
is called time period.
• SI unit of time period is second (s). Time period
is denoted by T.
• The frequency of a wave is the reciprocal of the
time period.
• v = 1/T
Amplitude
o The
maximum displacement of the particle of the medium from their original
undisturbed position is called amplitude of the wave.
• Amplitude is denoted by A and its SI unit is metre
(m).
o Sound
have characteristics like pitch loudness and timbre.
• Pitch: The pitch of sound depends on the frequency
of sound (vibration).
o t
is directly proportional to its frequency. Greater the frequency, higher is the
pitch and lesser the frequency, lower is the pitch.
o A
woman’s voice is shrill having a high pitch while a man’s voice is flat having
low pitch.
o High
pitch sound has large number of compressions and rarefactions passing a fixed
point per unit time.
• Loudness: The loudness depends on the amplitude of
the sound wave.
o Loudness
is the measure of the sound energy reaching the ear per sec.
o Greater
the amplitude of sound wave, greater is the energy, louder the sound; short is
the amplitude, less is the energy, soft is the sound.
o Loudness
is measured in decibel ‘dB’.
• Quality or Timbre: The timbre of a sound depends on
the shape of sound wave produced by it. It is the characteristic of musical
sound.
o It
helps us to distinguish between two sounds of same pitch & loudness.
• Sound of single (same) frequency is called tone while
a mixture of different frequencies is called note.
o Noise
is unpleasant to hear while music is pleasant to hear and it is of good
quality.
Velocity
• The distance travelled by a wave in one second is
called velocity of the wave.
• Its SI unit is metre per second (ms-1).
Velocity = Distance travelled/Time taken
⇒ v = λ/T
(λ is the wavelength of the waves travelled in one time time period T)
v = λv (1/T = v)
So, Velocity = Wavelength × Frequency
This is the wave equation.
Example: What is the frequency of sound wave
whose time period is 0.05 second ?
Solution
Frequency, v = 1/T
Given T = 0.05 s
v = 1/0.005 = 100/5 = 20Hz
∴ Frequency = 20 Hz.
Speed of Sound in Various Mediums
o Speed
of sound depends on the nature of material through which it travels. It is
slowest in gases, faster in liquids and fastest in solids.
o Speed
of sound increases with the rise in temperature.
o Speed of sound increases as humidity of air increases.
o Speed
of light is faster than speed of sound.
o In
air, speed of sound is 344 ms-1 at 22ºC.
Sonic Boom
o
Some aircrafts, bullets, rockets etc. have
‘supersonic speed’.
• Supersonic refers to the speed of an object which
is greater than the speed of sound and it produces extremely loud sound waves
called ‘shock waves’ in air.
o
Sonic boom is an explosive noise caused by shock waves.
o
It emits tremendous sound energy which can shatter the glass panes of windows.
o
Reflection of Sound
o
Like light, sound also bounces back when it falls
on a hard surface. It is called reflection of sound.
• The laws of reflection of light are obeyed during
reflection of sound.
(i) The incident sound wave, the reflected sound wave and normal at the point
of incidence lie in the same plane.
(ii) Angle of reflection of sound is always equal to
the angle of incidence of sound.
Echo
• The repetition of sound caused by the reflection of
sound waves is called an echo.
o We
can hear echo when there is a time gap of 0.1 second in original sound and echo
(reflected sound).
o Echo is produced when sound reflected from a hard surface (i.e. brick wall,
mountain etc.) as soft surface tends to absorb sound.
Minimum distance to hear an echo
Speed = Distance/Time
Here, Speed of sound in air = 344 ms-1 at 22ºC
Time = 0.1 second
344 = Distance/0.1 sec
⇒ Distance = 344 × 0.1 = 34.4 m
So, distance between reflecting surface and audience
= 34.4/2 = 17.2 (at 22ºC).
o
Rolling of thunder is due to multiple reflection
of sound of thunder from a number of reflectingsurfaces such as clouds and the
earth.
Reverberation
• The persistence of sound in a big hall due to
repeated reflection of sound from the walls, ceiling and floor of the hall is
called reverberation.
o
If reverberation is too long, sound becomes
blurred, distorted and confusing due to overlapping of different sound.
Methods to reduce reverberation in big halls or auditoriums
o
Panels made of felt or compressed fibre board are put on walls and ceiling to
absorb sound.
o
Heavy curtains are put on doors and windows.
o
Carpets are put on the floor.
o
Seats are made of material having sound
absorbing properties.
Applications of Reflection of Sound
(i) Megaphone, loudspeakers, bulb horns and trumpets,
shehnai etc. are designed to send sound in a particular direction without
spreading all around.
o
All these instruments have funnel tube which
reflects sound waves repeatedly towards audience. In this amplitude of sound
waves adds up to increase loudness of sound.
(ii) Stethoscope: It is a medical instrument used for
listening the sounds produced in human body mainly in heart and lungs. The
sound of the heartbeats reaches the doctor’s ears by the multiple reflection of
the sound waves in the rubber tube of stethoscope.
(iii) Sound Board: In big halls or auditoriums sound
is absorbed by walls, ceiling, seats etc. So a curved board (sound board) is
placed behind the speakers so that his speech can be heard easily by audiences.
The soundboard works on the multiple reflection of sound.
(iv) The ceiling of concert halls are made curved, so
that sound after reflection from ceiling, reaches all the parts of the hall.
Range of Hearing
(i) Range of hearing in human is 20 Hz to 20000 Hz.
o
Children younger than 5 years and dogs can hear
upto 25 KHz.
(ii) The sounds of frequencies lower than 20 Hz are known as ‘infrasonic sounds’.
o
A vibrating simple pendulum produces infrasonic
sounds.
o
Rhinoceroses communicate each other using
frequencies as low as 5 Hz.
o
Elephants and whales produces infrasonic waves.
o
Earthquakes produces infrasonic waves (before
shock waves)
which some animals can hear and get disturbed.
(iii) The sounds of frequencies higher than 20 KHz
are known as ‘ultrasonic waves’.
o Dogs,
parpoises, dolphins, bats and rats can hear ultrasonic sounds.
o Bats
and rats can produce ultrasonic sounds.
Hearing Aid
o It
is battery operated electronic device used by persons who are hard of hearing.
o Microphone
convert sound into electrical signals, than those are amplified by amplifier.
Amplified signals are send to the speaker of hearing aid. The speaker converts the
amplified signal to sound and sends to ear for clear hearing.
Applications of Ultrasound
(i) It is used to detect cracks in metal blocks in
industries without damaging them.
(ii) It is used in industries to clean ‘hard to
reach’ parts of objects such as spiral tubes, odd shaped machines etc.
(iii) It is used to investigate the internal organs
of human body such as liver, gall bladder, kidneys, uterus and heart.
(iv) Ecocardiography: These waves are used to reflect the action of heart
and its images are formed. This technique is called echocardiography.
(v) Ultrasonography: The technique of obtaining
pictures of internal organs of the body by using echoes of ultrasound waves is
called ultrasonography.
(vi) Ultrasound is used to split tiny stones in
kidneys into fine grains.
SONAR
o The
word ‘SONAR’ stands for ‘Sound Navigation And Ranging’.
SONAR is a device which is used to find distance,
direction and speed of underwater objects.
o SONAR
consists of a transmitter and a receptor or detector and installed at the
bottom of a ship.
o The
transmitter produces and transmits ultrasonic waves.
o These
waves travel through water and after striking the objects on the bottom of sea,
are reflected back and received by detector.
o These
reflected waves are converted into electric signals by detector.
o The
sonar device measures the time taken by ultrasound waves to travel from ship to
bottom of sea and back to ship.
o Half
of this time gives the time taken by the ultrasound waves from ship to bottom.
• Let the time interval between transmission and
reception of ultrasound signal is t.
Speed of sound through sea water is v
Total distance travelled by waves = 2d.
Then, 2d = v × t.
This is called echo ranging.
o The
sonar is used to find the depth of sea, to locate underwater hills, valleys,
submarines, icebergs and sunken ships etc.
o Bats
fly in the dark night by emitting high pitched ultrasound waves which are reflected
from the obstacle or prey and returned to bats ear.
o The
nature of reflection tells the bat where the obstacle or prey is and what it is
like.
Structure of Human Ear
o The
ear consists of three parts: outer ear, middle ear and inner ear.
o The
ears are the sense organs which help us in hearing sound.
o The
outer ear is called pinna. It collects the sound from surroundings.
o This
sound passes through the auditory canal.
o The
middle ear contains of three bones: hammer, anvil and stirrup linked with one
another. Free end of hammer touches ear drum and that of stirrup linked with
membrane of oval window of inner ear.
o The
lower part of middle ear has a narrow ‘Eustachian tube’.
o The
inner ear has a coiled tube called cochlea, which is connected with oval
window. Cochlea is filled with a liquid containing nerve cells.
o Other
side of cochlea is connected to auditory nerve which goes to brain.
Working of Human ear
Pinna → Ear canal → Ear drum → Hammer → Anvil →
Stirrup → Oval window → Cochlea → Auditory nerve → Brain
→ When compression of sound wave strikes the ear
drum, the pressure on the outside of ear drum increases and pushes the ear drum
inwards.
→ While during rarefaction ear drum moves outwards.
Thus, ear drum starts vibrating back and forth.
→ These vibrations are increased by three bones and
middle ear transmits these amplified pressure variations received from sound waves
to inner ear.
→ In the inner ear the pressure variations are turned
into electric signals by the cochlea.
→ These electric signals are sent to the brain via
auditory nerve and the brain interprets them as sound.
===========================
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