CBSE Class 9 Science Sound Notes

CBSE Class 9 Science Sound Notes. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Class 9 Science, Chapter 12 Sound. These notes explain how sound is produced and travels as a longitudinal wave, the characteristics of sound (loudness, pitch and quality), wavelength, frequency and speed, reflection, echo and reverberation, ultrasound and SONAR, with intext questions, 20 solved numericals and all exercise answers.

CHAPTER - 12 SOUND

Sound is a mechanical wave and needs a material medium like air, water, steel etc. for its propagation.
A medium is the substance or material through which a pulse or a wave moves.

INTEXT QUESTIONS PAGE NO. 162

1. How does the sound produced by a vibrating object in a medium reach your ear?

Answer: When an object vibrates, it forces the neighbouring particles of the medium to vibrate. These vibrating particles then force the particles adjacent to them to vibrate. In this way, vibrations produced by an object are transferred from one particle to another till it reaches the ear.

Teacher's Note:
a) The particles only vibrate about their mean positions; the disturbance travels forward.
b) Without a medium the vibrations cannot be passed on.

LONGITUDINAL AND TRANSVERSE WAVE

A longitudinal wave is a wave where the particles in the medium move parallel to the direction of propagation of the wave.
"If the particles of a medium vibrate in a direction, parallel to or along the direction of propagation of wave, it is called longitudinal wave"
Sound waves are pressure waves caused by objects which are vibrating. Sound waves need a medium through which to travel.
A compression is a region in a longitudinal wave where the particles are closest together. A rarefaction is a region in a longitudinal wave where the particles are furthest apart.

[Figure: A longitudinal wave in a medium shown as alternating regions of compression (particles close together, high pressure) and rarefaction (particles far apart, low pressure) along the direction of propagation of the wave]

TRANSVERSE WAVE

"If the particles of the medium vibrate in a direction, perpendicular to the direction of propagation, the wave is called transverse wave."
Example: water waves, vibrations of stretched string.
Transverse waves propagate in a medium in the form of crests and troughs as shown in fig:

[Figure: A transverse wave drawn as a sine curve with crests above and troughs below the line of propagation]

Crest : The maximum displacement along the upward direction.
Trough: The maximum displacement along the downward direction.

INTEXT QUESTIONS PAGE NO. 163

1. Explain how sound is produced by your school bell.

Answer: When the school bell vibrates, it forces the adjacent particles in air to vibrate. This disturbance gives rise to a wave and when the bell moves forward, it pushes the air in front of it. This creates a region of high pressures known as compression. When the bell moves backwards, it creates a region of low pressure known as rarefaction. As the bell continues to move forward and backward, it produces a series of compressions and rarefactions. This makes the sound of a bell propagate through air.

Teacher's Note:
a) Compressions are regions of high pressure and rarefactions are regions of low pressure.
b) The wave travels outwards while the air particles only vibrate to and fro.

2. Why are sound waves called mechanical waves?

Answer: Sound waves force the medium particles to vibrate. Hence, these waves are known as mechanical waves. Sound waves propagate through a medium because of the interaction of the particles present in that medium.

Teacher's Note:
a) Mechanical waves need a material medium, unlike light.
b) Sound cannot travel in a vacuum.

3. Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?

Answer: Sound needs a medium to propagate. Since the moon is devoid of any atmosphere, you cannot hear any sound on the moon.

Teacher's Note:
a) There is no air on the Moon, so no medium exists for sound.
b) Astronauts talk through radio, which uses electromagnetic waves.

CHARACTERISTICS OF SOUND

Sound can be distinguished from each other by the following three characteristics:
1. Loudness or Intensity
2. Pitch or Frequency
3. Quality or Timbre

1. Loudness
The sensation produced in the ear, which enables us to distinguish between a loud, and a faint sound is called loudness.
The amplitude of the wave determines the loudness or softness of a sound basically.

Intensity
The intensity of sound is defined as the amount of energy passing per unit time per unit area in a direction perpendicular to the area.

Intensity = \( \frac{\text{energy}}{\text{area} \times \text{time}} \) = \( \frac{\text{power}}{\text{area}} \)

The unit of intensity is watt/metre² (W/m²).

Difference between loudness and intensity

Pitch or Frequency
Pitch is that characteristics of sound which helps in differentiating between a shrill sound from a grave (flat or dull) sound.
High and low pitched sounds are called treble and bass respectively.

Quality or Timbre
The characteristic of a sound which distinguishes it from another of the same pitch and loudness is called quality or timbre.
A sound of single frequency is called a tone.
A sound which is produced due to a mixture of several frequencies is called a note.

MUSIC AND NOISE

The difference between music and noise is represented in below figure:

[Figure: Two waveforms. The first is a regular wave with a definite amplitude and represents a musical sound. The second is an irregular curve with no definite amplitude and represents noise]

First figure is regular and has a definite amplitude. It represents a musical sound.
A musical sound can be defined as a pleasant continuous and uniform sound produced by regular and periodic vibrations. Example: The pleasant sound produced by a guitar, piano, tuning fork etc.
The second curve represents a noise. The curve is irregular and has no definite amplitude.
Musical sound is pleasant to hear and is of rich quality. Noise on the other hand is unpleasant to hear.
Noise can be defined as an irregular succession of disturbances, which are discordant and unpleasant to the ear.

TERMS RELATED TO SOUND WAVES

The amplitude A is the maximum excursion of a particle of the medium from the particles undisturbed position.
The wavelength is the horizontal length of one cycle of the wave.
The period is the time required for one complete cycle.
The frequency is related to the period and has units of Hz, or s⁻¹. The number of such oscillations per unit time is the frequency of the sound wave.
The speed of sound is defined as the distance which a point on a wave, such as a compression or a rarefaction, travels per unit time.

We know, speed, v = \( \frac{\text{distance}}{\text{time}} \) = \( \frac{\lambda}{T} \)
Here λ is the wavelength of the sound wave. It is the distance travelled by the sound wave in one time period (T) of the wave. Thus, v = λ \( \frac{1}{T} \) = λν
That is, speed = wavelength × frequency.
The speed of sound remains almost the same for all frequencies in a given medium.

INTEXT QUESTIONS PAGE NO. 166

1. Which wave property determines (a) loudness, (b) pitch?

Answer:
1. (a) Amplitude (b) Frequency
2. The loudness of a sound depends on its amplitude. If the amplitude of a sound is large, then the sound produced will also be loud.
3. The pitch of a sound depends on its frequency. A sound will be considered a high pitched sound, if its frequency is high.

Teacher's Note:
a) Large amplitude gives a loud sound.
b) High frequency gives a shrill, high-pitched sound.

2. Guess which sound has a higher pitch: guitar or car horn?

Answer: The frequency of the vibration of a sound produced by a guitar is greater than that produced by a car horn. Since the pitch of a sound is proportional to its frequency, the guitar has a higher pitch than a car horn.

Teacher's Note:
a) Pitch depends on frequency, not on loudness.
b) A car horn is loud but has a low pitch.

INTEXT QUESTIONS PAGE NO. 166

1. What are wavelength, frequency, time period and amplitude of a sound wave?

Answer:
1. Wavelength: The distance between two consecutive compressions or two consecutive rarefactions is known as the wavelength. Its SI unit is metre (m).
2. Frequency: The number of complete oscillations per second is known as the frequency of a sound wave. It is measured in hertz (Hz).
3. Time period: The time taken for one complete oscillation is the time period. Its SI unit is second (s).
4. Amplitude: The maximum height reached by the crest or trough of a sound wave is called its amplitude.

Teacher's Note:
a) The printed answer leaves out the time period, which is the time for one complete oscillation, T = 1/ν.
b) Time period and frequency are reciprocals of each other.

2. How are the wavelength and frequency of a sound wave related to its speed?

Answer:
1. Speed, wavelength, and frequency of a sound wave are related by the following equation:
2. Speed (v) = Wavelength (λ) × Frequency (ν), that is, v = λν.

Teacher's Note:
a) At a fixed speed, a higher frequency means a shorter wavelength.
b) Use SI units: m, Hz and m/s.

3. Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.

Answer:
1. Frequency of the sound wave, ν = 220 Hz and speed of the sound wave, v = 440 m s⁻¹.
2. For a sound wave, speed = wavelength × frequency, so v = λν.
3. λ = \( \frac{v}{\nu} \) = \( \frac{440}{220} \) = 2 m.
4. Hence, the wavelength of the sound wave is 2 m.

Teacher's Note:
a) Rearrange v = λν to get λ = v ÷ ν.
b) The unit of wavelength is metre.

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:
1. The time interval between two successive compressions is equal to the time period of the wave.
2. The time period is the reciprocal of the frequency of the wave: T = \( \frac{1}{\text{frequency}} \) = \( \frac{1}{500} \) = 0.002 s.

Teacher's Note:
a) The distance of 450 m is extra data and is not needed.
b) Successive compressions are one wavelength apart, so the interval is one time period.

INTEXT QUESTIONS PAGE NO. 166

1. Distinguish between loudness and intensity of sound.

Answer: 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.

Teacher's Note:
a) Intensity is physical and measurable; loudness depends on the ear.
b) The unit of intensity is W/m².

SONIC BOOM

When the speed of any object exceeds the speed of sound it is said to be travelling at supersonic speed. Bullets, jet aircrafts etc. often travel at supersonic speeds. When a sound producing source moves with a speed higher than that of sound, it produces shock waves in air. These shock waves carry a large amount of energy. The air pressure variation associated with this type of shock waves produces a very sharp and loud sound called the "sonic boom". The shock waves produced by a supersonic aircraft have enough energy to shatter glass and even damage buildings.

INTEXT QUESTIONS 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.

Teacher's Note:
a) Order of speed: solids > liquids > gases.
b) Speed in a medium also rises with temperature.

REFLECTION OF SOUND

The reflection of sound follows the law "angle of incidence equals angle of reflection", sometimes called the law of reflection. The law states that:
1. The incident wave, the reflected wave and the normal (at the point of incidence), all lie in the same plane.
2. The angle of reflection is always equal to the angle of incidence. i.e. ∠i = ∠r

[Figure: Sound reflecting from a surface: an incident ray and a reflected ray make equal angles i and r with the normal drawn at the point of incidence]

ECHO

An echo is the phenomenon of repetition of sound of a source by reflection from an obstacle. The sensation of sound persists in our brain for about 0.1 s. This property is called persistence of hearing. Therefore, to hear a distinct echo the time interval between the original sound and the reflected one must be at least 0.1 s.
Let d = minimum distance of the obstacles from the source
v = speed of sound in air
t = total time taken by the sound to reach the listener after reflection.
Then by distance = speed × time formula we have, 2d = v × t
Substituting v = 344 m/s (speed of the sound in air at 20°C) and t = 0.1 s, we get
2d = 344 × 0.1 = 34.4 m
d = 17.2 m
Hence for hearing a distinct echo, the minimum distance of the obstacle from the source of sound should be 17.2 m.

d = \( \frac{v t}{2} \)
where d is the distance between the source of sound and the obstacle, t is the time taken by sound in going to the obstacle and coming back and v is the speed of sound.
d is measured in metre, v in m/s and t in second.

INTEXT QUESTIONS 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:
1. Speed of sound, v = 342 m s⁻¹ and the echo returns in time, t = 3 s.
2. Distance travelled by sound = v × t = 342 × 3 = 1026 m.
3. In the given time interval, sound has to travel a distance that is twice the distance of the reflecting surface and the source.
4. Hence, the distance of the reflecting surface from the source = \( \frac{1026}{2} \) = 513 m.

Teacher's Note:
a) The sound travels to the obstacle and back, so divide the total path by 2.
b) Use d = vt/2 for every echo problem.

MULTIPLE ECHOES

When sound is repeatedly reflected from a number of obstacles at suitable distances, many echoes are heard one after the other. This constitutes multiple echoes.

USES OF MULTIPLE REFLECTION OF SOUND

The phenomenon of multiple reflection of sound are given below:
1. Megaphones or Speaking tubes.
Megaphones or loudhailers, horns, musical instruments are all designed to send sound in a particular direction without spreading it in all directions. In these instruments, a tube followed by a conical opening reflects sounds successively to guide most of the sound waves from the source in the forward direction.

[Figure: A megaphone or horn: a tube followed by a conical opening, with sound waves from the source reflected successively along the walls and sent forward]

2. Stethoscope
Stethoscope is a medical instrument used for listening to sounds produced within the body, chiefly in the heart or lungs. In stethoscopes the sound of the patient's heartbeat reaches the doctor's ears by multiple reflection of sound, as shown in below figure.

[Figure: A stethoscope in which sound from the patient's chest piece is reflected repeatedly along the tube to the doctor's ears]

3. Design of concerts Halls, Cinema Halls and Conference Halls
The ceilings of these halls are curved. This enables the sound to reach all corners of the hall after reflection from the ceiling as shown in below left figure.

A sound board, which is a curved sound reflecting surface, is placed behind the stage. The source is located at the focus of this reflecting surface. Sound waves coming from the source become parallel after reflection from the sound board and spread evenly throughout the width of the hall as shown in above right sided figure.

[Figure: Left: a hall with a curved ceiling reflecting sound from the stage to all corners. Right: a curved sound board behind the stage with the source at its focus, sending parallel sound waves through the width of the hall]

REVERBERATION

A sound created in a big hall will persist by repeated reflection from the walls until it is reduced to a value where it is no longer audible. The repeated reflection that results in this persistence of sound is called reverberation.

INTEXT QUESTIONS 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.

Teacher's Note:
a) A curved ceiling sends the reflected sound to every part of the hall.
b) Sound boards at the stage work on the same idea.

RANGE OF FREQUENCIES

There are three categories of longitudinal mechanical waves which cover different range of frequencies:
1. Sound waves or Audible waves
These waves have frequencies which lie between 20 Hz to 20 kHz.
2. Infrasonic waves or Infrasound
Those longitudinal mechanical waves whose frequencies are below 20 Hz are called infrasonic waves.
3. Ultrasonic waves or Ultrasound
Those longitudinal mechanical waves whose frequencies lie above 20 kHz are called ultrasonic waves.

APPLICATIONS OF ULTRASOUND

INDUSTRIAL USES OF ULTRASOUND
1. Cleaning instruments and electronics components
Ultrasound is generally used to clean parts located in hard-to-reach places, for example, spiral tube, odd shaped parts, electronic components etc. Objects to be cleaned are placed in a cleaning solution and ultrasonic waves are sent into the solution. Due to the high frequency, the particles of dust, grease and dirt get detached and drop out. The objects thus get thoroughly cleaned.
2. Plastic welding
Application of small pressure and ultrasonic vibration to two similar surface produce sufficient thermal energy to bond the surfaces together.
3. Detecting flaws and cracks in metal blocks.
Ultrasounds can be used to detect cracks and flaws in metal blocks. Metallic components are generally used in construction of big structures like buildings, bridges, machines and also scientific equipment. The cracks or holes inside the metal blocks, which are invisible from outside reduces the strength of the structure. Ultrasonic waves are allowed to pass through the metal block and detectors are used to detect the transmitted waves. If there is even a small defect, the ultrasound gets reflected back indicating the presence of the flaw or defect, as shown in figure.

[Figure: A metal block with an ultrasound source on one side and a detector on the other; a crack or hole inside the block reflects the ultrasonic waves so they do not reach the detector]

Medical uses of Ultrasound
1. Echocardiography
Ultrasonic waves are made to reflect from various parts of the heart and form the image of the heart. This technique is called 'echocardiography'.
2. Ultrasonography
Ultrasound scanner is an instrument which uses ultrasonic waves for getting images of internal organs of the human body. A doctor may image the patient's organs such as the liver, gall bladder, uterus, kidney, etc. It helps the doctor to detect abnormalities, such as stones in the gall bladder and kidney or tumours in different organs. In this technique the ultrasonic waves travel through the tissues of the body and get reflected from a region where there is a change of tissue density. These waves are then converted into electrical signals that are used to generate images of the organ. These images are then displayed on a monitor or printed on a film. This technique is called 'ultrasonography'. Ultrasonography is also used for examination of the foetus during pregnancy to detect congenital defects and growth abnormalities.
3. Surgical uses.
Ultrasound is used for bloodless brain surgery as well as painless extraction of teeth etc. Ultrasound may be employed to break small 'stones' formed in the kidneys into fine grains. These grains later get flushed out with urine.
4. Therapeutic uses.
Ultrasound is used for treatment of neuralgic and rheumatic pains.

COMMUNICATION (SONAR)

It is an acronym which means SOund Navigation And Ranging.
A sonar is a device which measures the distance, direction and speed of objects lying under water using ultrasonic waves.

How does the sonar work?
The transmitter produces and transmits ultrasonic waves. These waves travel through water and after striking the object on the seabed, get reflected back and are sensed by the detector. The detector converts the ultrasonic waves into electrical signals which are appropriately interpreted. The distance of the object that reflected the sound wave can be calculated by knowing the speed of sound in water and the time interval between transmission and reception of the ultrasound.
Let the time interval between transmission and reception of ultrasound signal be t and the speed of sound through seawater be v. The total distance, 2d travelled by the ultrasound is then, 2d = v × t. The above method is called echo-ranging.
The sonar technique is used to determine the depth of the sea and to locate underwater hills, valleys, submarine, icebergs, sunken ship etc.

[Figure: A ship with a SONAR transmitter and detector sending ultrasonic waves down to the seabed or a submarine and receiving the reflected echo]

INTEXT QUESTIONS 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. Humans cannot hear sounds having frequency less than 20 Hz and greater than 20,000 Hz.

Teacher's Note:
a) 20,000 Hz is the same as 20 kHz.
b) The upper limit of hearing falls with age.

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.

Teacher's Note:
a) Infrasound lies below the audible range and ultrasound above it.
b) Elephants use infrasound; bats use ultrasound.

INTEXT QUESTIONS 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?

Answer:
1. Time taken by the sonar pulse to return, t = 1.02 s and speed of sound in salt water, v = 1531 m s⁻¹.
2. Distance travelled by the pulse = speed × time = 1.02 × 1531 = 1561.62 m.
3. This distance is twice the distance of the cliff, 2d.
4. d = \( \frac{1561.62}{2} \) = 780.81 m (the PDF prints 780.31 m).

Teacher's Note:
a) The printed answer 780.31 m is a division slip; 1561.62 ÷ 2 = 780.81 m.
b) The pulse goes to the cliff and comes back, so halve the total path.

NUMERICALS

1. A boy hears an echo of his own voice from a distant hill after 1 s. the speed of sound is 340m/s. What is the distance of the hill from the boy?

Answer:
1. 2d = v × t = 340 × 1 = 340 m.
2. d = \( \frac{340}{2} \) = 170 m.

Teacher's Note:
a) The sound covers the distance twice.
b) The hill is 170 m away.

2. A boy is standing in front of wall at a distance of 85m produces 2 claps per second. He notices that the sound of his clapping coincides with the echo. The echo is heard only once when clapping is stopped. Calculate the speed of sound.

Answer:
1. Time between claps = \( \frac{1}{2} \) = 0.5 s, and the echo takes this time to return.
2. Distance travelled = 2 × 85 = 170 m.
3. v = \( \frac{170}{0.5} \) = 340 m/s.

Teacher's Note:
a) The echo of one clap coincides with the next clap, so the echo time is 0.5 s.
b) The speed agrees with the usual value in air.

3. A man stationed between two parallel cliffs fires a gun. He hears the first echo after 1.5 s and the next after 2.5 s. What is the distance between the cliffs and when does he hear the third echo? Take the speed of sound in air is 340 m/s.

Answer:
1. Distance to the first cliff = \( \frac{340 \times 1.5}{2} \) = 255 m.
2. Distance to the second cliff = \( \frac{340 \times 2.5}{2} \) = 425 m.
3. Distance between the cliffs = 255 + 425 = 680 m.
4. The third echo comes from the sound that goes to one cliff, then to the other and back: path = 2 × 680 = 1360 m, so t = \( \frac{1360}{340} \) = 4 s.

Teacher's Note:
a) The third echo is heard 4 s after firing.
b) The sum of the first two echo times is also 4 s.

4. A man fires a shot and hears an echo from a cliff after 2s. He walks 85m towards the cliff and the echo of a second shot is now heard after 1.4s What is the velocity of sound and how far was the man from the cliff when he first heard the echo?

Answer:
1. Let the first distance be d and the speed of sound v. Then 2d = 2v and 2(d - 85) = 1.4v.
2. Subtracting, 2 × 85 = 0.6v, so v = \( \frac{170}{0.6} \) = 283.3 m/s.
3. d = v = 283.3 m.

Teacher's Note:
a) The speed from the given data is 283.3 m/s, lower than the usual 340 m/s because the data are not exact for air.
b) Use two equations in the two unknowns d and v.

5. A boy hears an echo of his own voice from a distant hill after 2.5 s. the speed of sound is 340m/s. Calculate the distance of the hill from the boy.

Answer:
1. d = \( \frac{v t}{2} \) = \( \frac{340 \times 2.5}{2} \) = 425 m.

Teacher's Note:
a) The hill is 425 m away.

6. A person clapped his hands near a cliff and heard the echo after 5 s. What is the distance of the cliff from the person if the speed of the sound, v is taken as 346 m/s?

Answer:
1. d = \( \frac{346 \times 5}{2} \) = 865 m.

Teacher's Note:
a) Halve the total distance travelled by the sound.

7. A child hears an echo from a cliff after the sound of powerful cracker is produced. How far away is the cliff from the child? (Take speed of the sound in air as 340 m/s).

Answer:
1. The time of the echo is not printed in the question, so use the general formula d = \( \frac{v t}{2} \).
2. With v = 340 m/s, d = 170 t metres, where t is the time of the echo in seconds.
3. For example, if the echo is heard after 2 s, d = 170 × 2 = 340 m.

Teacher's Note:
a) The time of the echo is missing in the printed question, so the general result is given with an example.
b) A distinct echo needs t of at least 0.1 s.

8. An observer standing between two cliffs fires a gun. He hears one echo after 1.5 s and another after 3.5 s. If the speed of sound is 340 m/s, find (a) the distance of the observer from the first cliff and (b) distance between the cliffs.

Answer:
1. (a) d₁ = \( \frac{340 \times 1.5}{2} \) = 255 m.
2. (b) d₂ = \( \frac{340 \times 3.5}{2} \) = 595 m.
3. Distance between the cliffs = 255 + 595 = 850 m.

Teacher's Note:
a) Each echo time belongs to one cliff.
b) The cliffs are on opposite sides of the observer, so add the distances.

9. A boy stands 60m in front of a tall wall and claps. The boy continues to clap every time an echo is heard. Another boy finds that the time between the first (1st) and the fifty first (51st) clap is 18s. Calculate the speed of the sound.

Answer:
1. From the 1st to the 51st clap there are 50 intervals, so each interval = \( \frac{18}{50} \) = 0.36 s.
2. Each interval is the time for the sound to go to the wall and back: 2 × 60 = 120 m.
3. v = \( \frac{120}{0.36} \) = 333.3 m/s.

Teacher's Note:
a) Count intervals, not claps: 51 claps make 50 intervals.
b) The result is close to the speed of sound in air.

10. A person standing between two vertical cliffs and 680m away from the nearest cliff, shouted. He heard the first echo after 4s and the second echo 3s later. Calculate (a) the speed of sound in air and the distance between the two cliffs.

Answer:
1. (a) The first echo comes from the nearest cliff: v = \( \frac{2 \times 680}{4} \) = 340 m/s.
2. (b) The second echo is heard 3 s later, at 7 s: distance to the farther cliff = \( \frac{340 \times 7}{2} \) = 1190 m.
3. Distance between the cliffs = 680 + 1190 = 1870 m.

Teacher's Note:
a) The second echo is heard 4 + 3 = 7 s after the shout.
b) Add the two distances, as the cliffs are on opposite sides.

11. A man standing at 51m from a wall fires a gun. Calculate the time after which an echo is heard. The speed of sound is 340 m/s.

Answer:
1. Total distance = 2 × 51 = 102 m.
2. t = \( \frac{102}{340} \) = 0.3 s.

Teacher's Note:
a) The echo comes after 0.3 s, which is more than 0.1 s, so it is heard distinctly.

12. A man fires a gun towards a hill and hears its echo after 5s. He then moves 340m towards the hill and fires his gun again. This time he hears the echo after 3s. Calculate the speed of the sound.

Answer:
1. First shot: 2d = 5v. Second shot: 2(d - 340) = 3v.
2. Subtracting, 2 × 340 = 2v, so v = 340 m/s.

Teacher's Note:
a) Subtract the two equations to remove d.
b) The distance of the hill is d = 850 m.

13. An engine is approaching a hill at constant speed. When it is at a distance of 0.9km, it blows a whistle, whose echo is heard by the driver after 5s. If the speed of sound is 340m/s, calculate the speed of the engine.

Answer:
1. Let the speed of the engine be u. In 5 s the sound travels 340 × 5 = 1700 m and the engine travels 5u.
2. The sound goes to the hill and returns to the moving engine: 1700 + 5u = 2 × 900 = 1800 m.
3. 5u = 100, so u = 20 m/s.

Teacher's Note:
a) The distance covered by sound plus the distance covered by the engine equals twice the original distance.
b) Convert 0.9 km to 900 m.

14. It takes 2.4 s to record the echo of a sonar. If the speed of sound in water is 1450 m/s, find the depth of the ocean floor.

Answer:
1. d = \( \frac{v t}{2} \) = \( \frac{1450 \times 2.4}{2} \) = 1740 m.

Teacher's Note:
a) The printed time "2.4a" is a typing slip for 2.4 s.
b) Halve the total path of the ultrasound.

15. A ship which is stationary, is at a distance of 2900 m from the seabed. The ship sends an ultrasound signal to the seabed and its echo is heard 4s. Find the speed of sound in water.

Answer:
1. Total distance = 2 × 2900 = 5800 m.
2. v = \( \frac{5800}{4} \) = 1450 m/s.

Teacher's Note:
a) Speed = total distance ÷ time.

16. A ship sends out ultrasound that returns from the seabed and is detected after 3.42s. If the speed of the ultrasound through sea water is 1531 m/s, what is the distance of the seabed from the ship?

Answer:
1. d = \( \frac{1531 \times 3.42}{2} \) = 2618 m (2618.01 m).

Teacher's Note:
a) Halve the total path.

17. A sonar device on a submarine sends out a signal and receives an echo 5s later. Calculate the speed of sound in water if the distance of the object from the submarine is 3625m.

Answer:
1. Total distance = 2 × 3625 = 7250 m.
2. v = \( \frac{7250}{5} \) = 1450 m/s.

Teacher's Note:
a) Double the distance to get the path of the signal.

18. A sonar emits pulses on the surface of water which are detected after reflection from its bottom at a depth of 1531m. If the time interval between the emission and detection of the pulse is 2s, find the speed of sound in water.

Answer:
1. Total distance = 2 × 1531 = 3062 m.
2. v = \( \frac{3062}{2} \) = 1531 m/s.

Teacher's Note:
a) The speed comes out equal to the depth in metres, as the time is 2 s.

19. A sonar device on a submarine sends out a signal and receives an echo 10 s later. Calculate the speed of sound in water if the distance of the object from the submarine is 7650m.

Answer:
1. Total distance = 2 × 7650 = 15300 m.
2. v = \( \frac{15300}{10} \) = 1530 m/s.

Teacher's Note:
a) This is the same method as the previous sonar problems.

20. A man standing at 68m from a wall fires a gun. Calculate the time after which an echo is heard. The speed of sound is 340 m/s.

Answer:
1. Total distance = 2 × 68 = 136 m.
2. t = \( \frac{136}{340} \) = 0.4 s.

Teacher's Note:
a) The time is more than 0.1 s, so the echo is distinct.

EXERCISE QUESTIONS PAGE NO. 174 AND 175

1. What is sound and how is it produced?

Answer: Sound is produced by vibration. When a body vibrates, it forces the neighbouring particles of the medium to vibrate. This creates a disturbance in the medium, which travels in the form of waves. This disturbance, when reaches the ear, produces sound.

Teacher's Note:
a) Sound is a form of energy produced by a vibrating body.
b) It reaches the ear as a series of compressions and rarefactions.

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 creates a region of high pressure in its vicinity. This region of high pressure is known as compressions. 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 backwards, it produces a series of compressions and rarefactions.

[Figure: A vibrating object (such as a tuning fork or a vibrating strip) with air particles drawn as dots; regions where the dots are crowded are labelled compression (C) and regions where they are spread out are labelled rarefaction (R)]

Teacher's Note:
a) Compression is high pressure and rarefaction is low pressure.
b) Label C and R alternately along the direction of the wave in the diagram.

3. Cite an experiment to show that sound needs a material medium for its propagation.

Answer:
1. Take an electric bell and hang this bell inside an empty bell-jar fitted with a vacuum pump.
2. Initially, one can hear the sound of the ringing bell. Now, pump out some air from the bell-jar using the vacuum pump. It will be observed that the sound of the ringing bell decreases.
3. If one keeps on pumping the air out of the bell-jar, then at one point, the glass-jar will be devoid of any air. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating.
4. When there is no air present inside, we can say that a vacuum is produced. Sound cannot travel through vacuum. This shows that sound needs a material medium for its propagation.

[Figure: An electric bell hanging inside a glass bell-jar, connected by a tube to a vacuum pump, with the bell wires leading out to a battery and switch]

Teacher's Note:
a) The bell keeps vibrating, so the sound disappears only because the air is removed.
b) This experiment proves that sound is a mechanical wave.

4. Why is sound wave called a longitudinal wave?

Answer: The vibration of the medium that travels along or parallel to the direction of the wave is called a longitudinal wave. In a sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Hence, a sound wave is called a longitudinal wave.

Teacher's Note:
a) Particle vibration and wave travel are in the same direction.
b) Transverse waves, like water waves, have perpendicular vibrations.

5. Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?

Answer: Quality of sound is that characteristic which helps us identify a particular person. Sound produced by two persons may have the same pitch and loudness, but the quality of the two sounds will be different.

Teacher's Note:
a) Quality is also called timbre.
b) It depends on the mixture of frequencies in the note.

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 × 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.

Teacher's Note:
a) Light is far faster than sound.
b) The delay can be used to estimate the distance of the lightning.

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:
1. For a sound wave, speed = wavelength × frequency, so v = λν and λ = \( \frac{v}{\nu} \).
2. Given that the speed of sound in air = 344 m/s.
3. (i) For ν₁ = 20 Hz, λ₁ = \( \frac{344}{20} \) = 17.2 m.
4. (ii) For ν₂ = 20,000 Hz, λ₂ = \( \frac{344}{20000} \) = 0.0172 m.
5. Hence, for humans, the wavelength range for hearing is 0.0172 m to 17.2 m.

Teacher's Note:
a) Convert 20 kHz to 20,000 Hz first.
b) The lower frequency has the longer wavelength.

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:
1. Let the length of the aluminium rod be d.
2. Speed of sound in aluminium at 25°C, v_Al = 6420 m s⁻¹. Time taken by the sound wave to reach the other end, t_Al = \( \frac{d}{6420} \).
3. Speed of sound in air at 25°C, v_Air = 346 m s⁻¹. Time taken by the sound wave to reach the other end, t_Air = \( \frac{d}{346} \).
4. The ratio of time taken by the sound wave in air and aluminium: \( \frac{t_{Air}}{t_{Al}} = \frac{6420}{346} \) = 18.55 : 1.

Teacher's Note:
a) The distance d cancels in the ratio.
b) Sound is about 18.5 times faster in aluminium than in air.

9. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?

Answer:
1. Frequency is the number of oscillations per second: Frequency = \( \frac{\text{Number of oscillations}}{\text{Total time}} \).
2. Number of oscillations = Frequency × Total time.
3. Given frequency = 100 Hz and total time = 1 min = 60 s.
4. Number of oscillations = 100 × 60 = 6000.
5. Hence, the source vibrates 6000 times in a minute.

Teacher's Note:
a) Convert the minute to seconds before multiplying.
b) 100 Hz means 100 vibrations in every second.

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.

Teacher's Note:
a) State both laws: equal angles and the same plane.
b) The angles are measured from the normal.

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 interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Hence, on a hotter day, the time interval between the original sound and the reflected sound will decrease. Therefore, an echo can be heard only if this time interval is still greater than 0.1 s.

Teacher's Note:
a) Time = 2d ÷ v, so a higher speed gives a shorter time.
b) If the time falls below 0.1 s the echo merges with the original sound.

12. Give two practical applications of reflection of sound waves.

Answer:
1. (i) Reflection of sound is used to measure the distance and speed of underwater objects. This method is known as SONAR.
2. (ii) 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.

Teacher's Note:
a) Megaphones and curved ceilings of concert halls are other applications.
b) Name two applications and explain each in one line.

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:
1. Height of the tower, s = 500 m, velocity of sound, v = 340 m s⁻¹, g = 10 m s⁻² and the initial velocity of the stone u = 0.
2. Time taken by the stone to fall to the base, t₁: s = ut₁ + \( \frac{1}{2} \) g t₁², so 500 = 0 + \( \frac{1}{2} \) × 10 × t₁².
3. t₁² = 100, so t₁ = 10 s.
4. Time taken by the sound to reach the top from the base, t₂ = \( \frac{500}{340} \) = 1.47 s.
5. The splash is heard at the top after t = t₁ + t₂ = 10 + 1.47 = 11.47 s.

Teacher's Note:
a) Add the time of fall and the time taken by the sound to travel up.
b) The two times are found separately because the two motions are different.
c) Use u = 0 as the stone is dropped.

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:
1. Speed of sound, v = 339 m s⁻¹ and wavelength, λ = 1.5 cm = 0.015 m.
2. Speed = wavelength × frequency, so ν = \( \frac{v}{\lambda} \) = \( \frac{339}{0.015} \) = 22600 Hz.
3. 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.

Teacher's Note:
a) Convert 1.5 cm to metres before dividing.
b) The wave is an ultrasonic wave.

15. What is reverberation? How can it be reduced?

Answer: Persistence of sound (after the source stops producing sound) due to repeated reflection is known as reverberation. As the source produces sound, it starts travelling in all directions. Once it reaches the wall of a room, it is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Due to this, sound can be heard even after the source has ceased to produce sound.
To reduce reverberations, sound must be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibreboard, rough plastic, heavy curtains, and cushioned seats can be used to reduce reverberation.

Teacher's Note:
a) Excess reverberation makes speech unclear in a hall.
b) Soft, rough materials absorb sound instead of reflecting it.

16. What is loudness of sound? What factors does it depend on?

Answer: A loud sound has high energy. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Teacher's Note:
a) Loudness depends mainly on amplitude and also on the sensitivity of the ear.
b) Doubling the amplitude makes the loudness four times.

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 its prey.

Teacher's Note:
a) The bat finds the distance from the time between the squeak and its echo.
b) This is the same principle as SONAR.

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.

Teacher's Note:
a) It cleans parts in hard-to-reach places such as spiral tubes.
b) Dust, grease and dirt drop out of the object.

19. Explain the working and application of a sonar.

Answer:
1. 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.
2. A beam of ultrasonic sound is produced and transmitted by the transducer (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.
3. 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'.

Teacher's Note:
a) Remember the formula 2d = vt.
b) The transmitter sends the pulse and the detector receives the echo.
c) Applications: depth of the sea, submarines, icebergs and sunken ships.

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:
1. Time taken to hear the echo, t = 5 s and the distance of the object from the submarine, d = 3625 m.
2. Total distance travelled by the sonar waves during the transmission and reception in water = 2d.
3. Velocity of sound in water, v = \( \frac{2d}{t} \) = \( \frac{2 \times 3625}{5} \) = 1450 m s⁻¹.

Teacher's Note:
a) Use 2d because the signal goes and returns.
b) The speed of sound in water is about 1450 m/s.

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.

[Figure: A metal block with an ultrasound source on one side and a detector on the opposite side; a flaw inside the block reflects the waves]

Teacher's Note:
a) Ordinary X-rays and visible light are not suited for thick metal blocks.
b) A cracked region reflects the ultrasound, so the detector shows a gap.

22. Explain how the human ear works.

Answer: Different sounds produced in our surroundings are collected by pinna that sends these sounds to the ear drum via the ear canal. The ear drum starts vibrating back and forth rapidly when the sound waves fall on it. The vibrating eardrum sets the small bone hammer into vibration. The vibrations are passed from the hammer to the second bone anvil, and finally to the third bone stirrup. The vibrating stirrup strikes on the membrane of the oval window and passes its vibration to the liquid in the cochlea. This produces electrical impulses in nerve cells. The auditory nerve carries these electrical impulses to the brain. These electrical impulses are interpreted by the brain as sound and we get a sensation of hearing.

[Figure: Parts of the human ear: pinna, ear canal, ear drum, hammer, anvil, stirrup, oval window, cochlea and auditory nerve]

Teacher's Note:
a) Remember the order: pinna, ear canal, eardrum, hammer, anvil, stirrup, cochlea, auditory nerve.
b) The three small bones amplify the vibrations.

Please click the link below to download CBSE Class 9 Science Sound Notes.