# Sound : Complete Set of Questions

This set of questions contains all the possible concepts
which could be asked in the examination

Sound

Q.1 Â What is sound?

Sound is a form of energy emitted by a vibrating object producing sensation of hearing.

Q.2 Â Write any three characteristics of sound waves.

Following are the properties of sound waves:

• These are longitudinal waves.
• These are Mechanical waves.
• These are a result of vibrations.

Q.3 Â Energy can neither be created nor be destroyed it can change from one form to another. Does this law apply to sound too?

Yes, sound being a form of energy is created through another form of energy. For example: when we clap our hands we hear sound, this sound is produced due to conversion of muscular energy of our hand to sound energy.

Q.4 Â What are the different types of sound on the basis of frequencies?

On the basis of frequencies, sound can be:

(a) Infrasound â€“ frequencies below than 20 Hz

(b) Audible sound â€“ frequencies between 20 Hz to 20,000 Hz.

(c) Ultrasound - frequencies above 20,000 Hz

Q.5 Â What is a supersonic speed?

When the speed of an object exceeds the speed of the sound, then the speed with which the body travels is supersonic speed. For example, aircrafts, bullet, etc. which travels at a higher speed than the speed of sound.

Q.6 Â When does a body produce sound?

Sound is produced through the to and fro motion of the particles of the medium in which sound travels. These particles vibrate about their mean position forming disturbances in the form of waves.

Q.7 Â Write a short note on propagation of sound.

Or

Write a short note on how does the sound produced by a vibrating object in a medium reaches our ear.

- When sound is produced by the source, for listener to listen the sound passes through certain medium.

- The vibrating object sets the particles of the medium around it into vibration, to and fro motion.

- A particle in contact with the vibrating object moves towards a particular direction. This particle exerts a force on the nearby particles which displaces from its equilibrium position and starts moving towards the same direction in which the original molecule was travelling. After displacing the second particle, the first particle returns to its original position. Similarly, particle second exerts a force on third particle and so on.

- Like this all the particle in between the source and listener vibrates and the sound is conducted to the listener.

Note - the particles of medium do not actually move from the vibrating body, they just vibrates
the particles.

Q.8 Â What is a medium? What are the different mediums in which sound can travel?

Medium is a substance in which sound travels. Without a medium sound cannot propagate. The different medium through which sound travels is gas, liquid and solids.

Q.9Â Â Â Â Â  Give reason:

(a) Astronauts cannot talk to each other in space.

(b) A bomb explosion on moon cannot be heard.

(a) Since sound waves being mechanical in nature, needs a material medium to travel and in space there is no atmosphere present. So, sound could not be heard in space. To communicate in this environment they use radio waves, a type of electromagnetic wave which do not use mediumÂ for its propagation.

(b)Â Since moon does not have atmosphere or a medium in which sound can travel so, if there occurs a bomb explosion on moon, it could not be heard by the persons even present on the moon.

Q.10 Â Which device is used to measure intensity of earthquake?

Seismometers

Q.11 Â What is decibel?

The intensity of a sound is measured in a unit known as decibel which is denoted by dB.

Q.12 Â Sound waves cannot travel through vacuum. Explain.

Sound waves cannot travel through vacuum as these are mechanical waves and needs a material medium to travel.

Q.13 Â (a) What is amplitude?

(b) Name the characteristics of sound.

(a) Amplitude of a wave is the maximum displacement of the particles of the medium from their mean position when a wave passes through the medium. It is denoted by A. the SI unit is metre(m).

(b) The characteristics of sound are:Â Â

(i) Loudness

(ii) Pitch or Shrillness

(iii) Quality or timbre

Q.14 Â On which factors does the speed of sound depend?

Underneath are the factors which have an effect on speed of the sound:

(a) Density of the medium through which sound travels: With the increase in density of medium the speed of sound also increases. i.e. Speed of sound in solids>speed of sound in liquids> speed of sound in air.

(b) Temperature: The speed of sound increases with the increase in the temperature i.e. with every 1o increase in temperature the speed of sound increases by 0.6 m/s.

(c) Humidity of air: As the humidity of air increase the speed of sound also increases.Â

Q.15 Â We hear two distinct sounds when a person strikes a hammer on the railway lines from a distance. Why is it so?

We hear two distinct sounds, when one hammer on the railway line from a distance because due to the striking of hammer the wave produced would travel by two routes one through solid and other through air. Since the speed of sound in air is different from speed of sound in solid, there would be two distinct sounds heard by the listener.

Q.16 Â Give reasons:

(a) Sound travels faster in summer than in winter.

(b) Two friends on the surface of the moon cannot talk to each other.

(c) Presence of an advancing train is felt by sticking our ears to rail lines before its sound approaches to the listener by air.

(a) Sound travels faster in summer season than in winter season because, with the increase in temperature speed of sound also increases and in summers temperature is high with respect to the winters.

(b)Â Two friends on the surface of the moon cannot talk to each other because moon does not have atmosphere or medium for sound to propagate so sound would not produce.

(c)Â Speed of sound depends upon the density of medium in which it travels. Speed of sound will be greater in substances with greater density i.e. speed of sound in solids > speed of sound in liquids > speed of sound in air. Since rails are made of metal (solid) sound would travel faster in ails than in air (gas).Â

Q.17 Â Two persons are holding an iron rod. First person knocked the rod by a hammer. What will be the ratio of times taken by the sound wave in air and in iron to reach the second person?

Suppose, l = length of the iron rod

Time taken by the sound to travel through the iron rod is given by

${t_1} = {{Distance} \over {Speed}} = {l \over {{V_{Iron}}}}$

Similarly, time taken by the sound to travel through the air is given by

${t_2} = {{Distance} \over {Speed}} = {l \over {{V_{air}}}}$

Therefore, ratio of time taken by the sound wave in Air and Iron

${t_1}:{t_2} = {{{V_{air}}} \over {{V_{Iron}}}}$

Q.18 Â Why is noise different from music?

Q.19 Â Define the terms: (a) Tone (b) Note

(a)Â Tone: A sound produced by single frequency is known as tone.

(b) Note: A sound produced by mixture of frequencies is known as note.

Q.20 Â What changes occurs in the speed of sound when

(a) It travels from iron to air?

(b) Temperature of the air increases?

(a) Speed of sound decreases when sound waves travel from a solid state to gaseous.

(b) With the rise in temperature of air, the speed of sound travelling in it increases.

Q.21 Â With the help of an experiment show that sound needs a material medium for its propagation.

System:

• Suspend an electric bell in an airtight glass bell jar packed with a cork.
• The bell jar is connected to a vacuum pump, from the bottom.

Experiment:

• Press the switch ON, the bell would be heard.
• Now start the vacuum pump.
• With the decrease in the air in the jar through vacuum pump the sound would become paler, with current being flowing constantly.
• When entire air will be removed then no sound will be heard.

Inference:

• Sound needs a material medium to propagate.

Q.22 Â Three friends are made to hear a sound travelling through different media as given below:

Which one will hear the sound first? Andwhy?

Speed of sound in solids > Speed of sound in liquids > Speed of sound in gases

Speed of sound depends upon the elasticity and density of a medium. More the density more is the speed of sound.

Therefore, sound will travel fastest in steel and slowest in oxygen. So, Riya would hear the sound

first.

Q.23 Â Show the propagation of a longitudinal wave showing compression and rarefaction.

Compression are denoted by Â â†’ A, C, E, G

Rarefaction are denoted by â†’ B, D, F

Q.24 Â Differentiate between Loudness and Intensity.

Q.25 Â State the mechanism through which humans can utter sound through their mouth.

The sound produced by our speech is produced through the vibrations of the two vocal cords present in our throat. This vibration is caused by the air approaching from the lungs.

Q.26 Â Name the characteristic of sound involved when a baby distinguishes her motherâ€™s voice with others.

Quality or timber is the characteristic of sound involved here.

Q.27 Give reason for the following:

(a) We hear the sound of a horn of approaching car before the car reaches us.

(b) We hear the sound produced by the humming bees while the sound of vibration of pendulums is not heard.

(c) Â This is so as the speed of car is much less than that of the sound. So, sound of horn travels faster than the car itself and we can hear the sound earlier than actually see the car.

(d) Â This is so as the sound produced by the humming of a bee is in the audible range of our hearing unlike the pendulum, which is so low to be heard by humans.

Q.28 Â How is sound produced when our school bell is struck with a hammer?

Sound is produced when the school bell is struck by the hammer. As this happens the bell starts vibrating. These vibrations then produce disturbances in air, which travels as sound waves to our ear.

Q.29 Â Define the term:

(a) Loudness

(b) Intensity

(c) Pitch

(d) Quality

(a) Loudness: it is the feeling produced in the ear which helps us to differentiate between a loud and a dim sound. The loudness or softness of sound depends upon the amplitude of the wave. The soft sound has small amplitude and louder sound has large amplitude.

(b) Â Intensity: it is the amount of energy passing at every second through a unit area.

(c) Â Pitch: it is the characteristic of sound which helps in differentiating between a harsh sound and a dull sound. It depends upon the frequency of vibration. Low pitch sound have low frequency and high pitch sound have high frequency.

(d) Â Quality: it is the characteristic of sound which differentiate between the two waves of having same pitch and loudness.

Wave

Q.30 Â What is a wave?

A wave is a disturbance or oscillation travelling in a medium with transfer of energy from one point to another without any actual contact of the points.

There are two type of waves:

(a) Longitudinal Waves

(b) Transverse WavesÂ

Q.31 Â What do you mean by oscillation?

OscillationÂ is the repetitive variation of the density about a central value or between two or more different values.Â

Q.32 Â Why are sound waves longitudinal and mechanical waves?

Sound waves are longitudinal waves as the particles of the medium do to and fro motion in the same direction in which wave moves.

Sound waves are mechanical waves as it needs material medium for its propagation.

Q.33 Â Differentiate between longitudinal wave and transverse wave.

Q.34 Â What is wavelength?

The distance between two consecutive compressions and two consecutive rarefaction is known as wavelength. It is denoted by Î».

Q.35 Â Express the units of:

(a) Frequency and

(b) Wavelength.

(a) Frequency â€“ hertz (Hz),

(b) Wavelength â€“ metre (m).

Q.36 Â Define the following terms:

(a) Time period

(b) Frequency of an oscillating body.

(c) State the relation between frequency and time period.

(a) Â Time period (T): The time taken by a vibrating body to complete one oscillation is known as time period. SI unit is second (s).

(b) Â Frequency (Î½ ): The number of oscillations completed by a vibrating body in one second is called its frequency. SI unit of frequency is hertz (Hz).

(c) The no. of waves produced in a single second is known as frequency. This concludes that frequency of a wave with time period T would be 1/T.

The frequency of a wave is the reciprocal of its time period.

$f = {1 \over T}$

Where, f = frequency of the wave

T = Time period of the wave

Q.37 Â What is compression and rarefaction pulse.

Compression- That part of longitudinal wave in which particles of the medium are closer to one another than they originally are. This is the region of high pressure.

Rarefaction- that part of a longitudinal wave in which particles of the medium are farther apart than they originally are. This is the region of low pressure.

Q.38 Â Define pulse.

Pulse is a short duration wave generated by a single disturbance in a medium.

Q.39 Â Write any four differences between mechanical and electromagnetic waves.

Q.40 Â State the relation between Time period and Frequency of an oscillating body.

For an oscillating body,

T = time period

Î½ = frequency

No. of oscillations completed in T second = 1

No. of oscillations completed in 1 second =Â ${1 \over T}$

We know no. of oscillations completed in 1 second = frequency (Î½)

Therefore, the relation is shown by,Â $\nu = {1 \over T}$

Q.41 Â A tuning fork oscillates at a frequency of 256 Hz.

(a) What happens to the wavelength of sound generated by tuning fork when the temperature of air increases?

(b) What changes would the wavelength undergo if the temperature rises from 0Â° to 20Â°C?

(a) The wavelength of sound will increase. Since, the speed of sound increases with temperature. As, $v = \nu \lambda$. Therefore, when frequency is constant, wavelength is directly proportional to the speed of sound so when speed of light increases its wavelength also increases.

(b) With every 1o C rise in temperature the speeds of sound increases by 0.6 m/s.

Therefore, sound speed when temperature changes by 20Â°C = 0.6 m/s Ã— 20 = 12 m/s.

Increase in wavelength = increase in the speed/ frequency.

=Â ${{12m/s} \over {256Hz}} = 0.047m$

The increase in wavelength is 0.047m.

Q.42 Â (a) What is meant by â€˜compressionâ€™ and â€˜rarefactionâ€™ of a longitudinal wave?

(b) Give well labeled graphical representation of a longitudinal and transverse wave wave.

(a) When sound is produced by to and fro motion compression and rarefaction are produced.

Compression: it is that part of longitudinal wave in which particles of the medium are closer to one another than they are usually. As a result there occurs temporary reduction in the volume of the medium. These are regions of high density and pressure.

Density â†‘ Â PressureÂ â†‘

Rarefaction: it is that part of longitudinal wave in which particles of the medium are farther from one another than they are usually. As a result there occurs temporary increase in the volume of the medium. These are regions of low density and pressure..

Density â†“Â PressureÂ â†“

Diagram showing compression and rarefaction

(b)

Graphical representation of longitudinal wave.

Graphical representation of transverse wave.

Q.43 Â Diagrammatically show how longitudinal wave travels along a slinky.

- In figure (a) the spring is in its normal position with one end fixed.

- When the loose end of the spring is moved to and fro regularly longitudinal waves are produced.

- There occurs alternatively compressions and rarefactions in the waves produced.

Q.44 Â Which type of waves is produced when:

(a) Wire of a guitar is plucked.

(b) A stone is dropped in a pond.

(a) When wire of a guitar is plucked, a longitudinal wave moves to and fro producing sound and the other wave, transverse wave is produced in the wire of the guitar as the particles vibrate perpendicular to the direction of the motion.

(b) When a stone is dropped in a pond, ripples are produced on the surface of the water. These ripples or water waves are transverse waves. This is as water molecules vibrate up and down perpendicular to the direction of the wave which is a character of transverse wave.

Q.45 Â Establish a relation between, speed, wavelength and frequency of wave.

Here, speed = v

Distance = wavelength =Â Î»

Time period = T

Frequency =Â Î½

Since,

Speed =Â ${{Dis\tan ce} \over {Time}}$

$v = \,{\lambda \over T}$

$v = \lambda \left( {{1 \over T}} \right)$

$v = \,\lambda \nu$ (Since, Î» = 1/T)

The relation between the three is given by, $v = \,\lambda \nu$.

Q.46 Â Graphically represent:

(i) Two sound waves with the same amplitude but different frequencies.Â Â Â Â Â Â

(ii) Two sound waves with the same frequency but different amplitudes.

(iii) Two sound waves with different amplitudes and also different wavelengths.

Q.47 Â Using labeled diagram show how low and high pitch sound are represented.

Q.48 Â What is wave number?

The no. of waves contained in unit length of the medium is known as wave number. It is the reciprocal of wavelength.

Q.49 Â Define sonic boom.

Sonic boom is the sound caused by objects like aircrafts, jet planes, etc. which travels at higher speed than that of sound.

The loud noise produced by high speed aircrafts generates very noisy sound waves known as shock waves in air. These shock waves have a great amount of energy which can even damage buildings and break glass.

A magnificent variation in air pressure created by the shock waves forms a tremendously loud sound Sonic boom.

Reflection of sound

Q.50 Â What is reverberation?

Reverberation is the persistence of sound after the source has stopped emitting sound.

This is due to multiple reflections of sound waves which mostly occur in big auditoriums, cinema halls, concert halls, etc.

Q.51 Â Which type of surfaces reflects the sound waves better?

The substances with hard surfaces reflect the sound waves better.

Q.52 Â How can reverberation be reduced?

Reverberation occurs due to the repeated reflection of the sound waves through the roof or ceiling, floor, walls of the hall or auditorium. To reduce this reflection, the absorption of the sound energy should be enhanced.

This can be achieved by following ways:

(i) False Ceiling: ceiling could be replaced by false ceiling made of sound absorbing material.

(ii) Floor: carpets could be placed on the floor.

(iii) Walls: the walls could be covered with some sound absorbing materials like, glass wool. In addition furniture is implanted in the room with curtains on the windows and doors.Â

Q.53 Â A certain amount of reverberation is desirable for speeches but the reverberation time should be short. Give reason.

Reverberation is necessary to some limit as this enhances the quality of the speech or music played in the hall.

The time till which the reverberation persists until it becomes indistinct is known as reverberation time.

In an auditorium or a hall, the reverberation time should be so adjusted that the speech could be heard clearly and distinctly. If the reverberation time is increased then the multiple echoes produced interferes with the original sound. As a result, the sound would not be able to hear clearly and distinctly.

Q.54 Â Name one natural phenomenon caused by reflection of sound.

Thunder produced during lightning is a natural phenomenon caused by reflection of sound.

Q.55 Â Name any three practical applications of reflection of sound waves with the help of diagram.Â Â Â Â Â Â Â

The three application of reflection of sound waves are:

Megaphone: The conical shape of megaphone helps in amplifying and directing the sound towards one direction through multiple reflections.

Stethoscope: Itâ€™s an instrument used by the doctor to listen the sound produced by heart and lung in human body.

Sound board: It is a concave board used in auditoriums, cinema halls to spread the sound produced by the speaker evenly in the hall.

Q.56 Â Explain why?

(a) Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen.

(b) An echo is heard faster on a hot day than on a cold day.

(a) Even though thunder and lightning are produced simultaneously, thunder is heard a few seconds after the flash of lightning is seen.

This is so as the velocity of light(3 108Â m/s) is much greater than the velocity of sound (340 m/s). Therefore light travels much faster than the sound.

(b) The speed of sound is greater in hot days as with the increase in temperature the speed of sound also increases. Therefore, an echo is heard faster on a hot day.

Q.57 Â Sound waves follow the same laws of reflection as light waves. State the laws of reflection of sound.

The sound wave and light waves follow the similar laws of reflection.

The Two laws of reflection of sound are:

1. Â First law: The angle of incidence (i) is always equal to the angle of reflection(r).

i = r

2. Second law: The incident sound wave, the reflected sound wave and the normal at the point of incidence, all lie in the same plane.

Q.58 Â How does a sound board work?

In big concert halls, conference halls, auditoriums etc the sound is absorbed by the ceilings, floor, walls, curtains, seats, etc. due to this the speech could not be easily heard by the crowd.

To overcome this sound board is being placed behind the speaker on the stage. This is a concave board which reflects the sound wave produced by the speaker standing on the focus of the soundboard. The reflected waves are directed towards the audience with even distribution in the hall.

The principle behind the sound board is multiple reflection of sound.

Q.59 Â (a) After a snowfall, why does it is quietness all around?

(b) Â Why do empty rooms do sound echoing?

(a) The snow absorbs sound with little reflection of sound. Therefore, the surroundings becomes quite after the snowfall.

(b) Â In an empty room there is very less absorption of the sound as there is no furniture or other material to absorb the sound. Therefore, the reflection dies out more slowly and sound seems to be hollow and echoing.

Q.60 Â What is an echo?

Echo is the recurring sound produced when the sound waves are reflected by an obstacle.

Q.61 Â Illustrate the conditions required to hear an echo. Â  Â  Â  Â Â

• The minimum distance of the reflecting surface from the source should be 17.2 metres at a temperature of 20oin air.
• The minimum time lag between the original sound and the reflected sound should be 0.1 second or 1/10 th of a sound.

Q.62 Â Why do auditoriums have:

(a) Curved roofs?

(b) Curtains, carpets and false ceilings?

(a) The ceilings of an auditorium are curved so that the sound produced reaches all corners of theÂ hall by reflection from the curved surface. This enables the audience sitting in the hall at anyÂ place clear voice of the speaker. Principle behind this is â€˜reflection of sound wavesâ€™.

(b) In big halls, sound is sustained in the hall due to repeated reflections from the walls, ceilings and floor of the hall. This is known as reverberation. A small reverberation is preferred but if it isÂ increased the sound becomes blurred, and could not be easily audible. For removing thisÂ unnecessary reverberation there are certain substances like curtains, carpets and false ceilingsÂ used to absorb sound as a result reverberation is decreased. This is because the soft andÂ porous materials are good absorbers and bad reflectors of sound.

Q.63 Â State reason:

(a) Bats cannot see, then how do they fly in dark and even catch their Prey.

(b) There are Moths of certain families which are able to escape capture from the bat.

(a) Bats produce high frequency squeaks in the direction they fly. The reflected ultrasonic waves are then heard back by them after striking their prey. By this process the position of an obstacle or a prey is judged. So the bat prevents collapsing with the obstacle, on the other hand eats up the prey.

(b) There are certain families of moth which can hear the squeaks produced by the nearby flying bat and could easily escape from being captured.

Q.64 Â When there are some natural calamities to be occurred then some animals escape from the site, before the calamities had happened.Â Â Â Â

There are some animals like dog, elephant, etc. which could hear the low frequency sound waves known as infrasound. These sounds cannot be heard by humans as it low frequency sound less than 20 Hz. As earthquakes produce low- frequency infrasound before the main shock waves begin. Now these sounds can be heard by these animals and they get alert and escape from the site much before.

Q.65 Â Which of the two have shorter wavelength: Infrasonic or ultrasonic waves?

Ultrasonic waves have shorter wavelength than the infrasonic waves. This is so as ultrasonic waves have greater frequencies than infrasonic waves and frequency of a wave is inversely proportional to the wavelength.

Application of Ultrasound

Q.66 Â (a) Write three main properties of ultrasound.

(b) Enlist four medical uses of ultrasound.

(a) Properties of Ultrasound:

• Longitudinal waves of frequencies lie above 20 kHz.
• Humans cannot detect this range of sound but some animals like dog, bat, etc. can.
• It has greater penetrating power so has a wide application in industries and medical purposes.

(b) Medical uses of ultra sound:

• Monitoring fetus inside the mother during pregnancy.
• Medical imaging of internal organs to check their size, structure and physiological functioning in humans.
• Breaking kidney stones into reduced sizes to flush out in the urine.
• Can be used in the neuralgic and rheumatic pains.Â

Q.67 Â Name one application of a sound wave having a frequency of 30 kHz.

Ultrasonic waves are sound waves having a frequency more than 20 kHz (even 30 kHz).

Ultrasonic waves are used in cleaning instruments and electronic devices.

Q.68 Â Explain how ultrasound is used to clean objects used in the industries.

In industries, objects like spiral tubes, electronic component are used which are not easily reachable. Cleaning of these substances is quite hard so ultra sound waves are used to clean them. The objects need to be cleaned are first put in the cleaning solution and then ultrasonic waves are passed through it. Due to high frequency of ultrasonic waves there sets motion in the solution and the dirt particles attached to the objects loosen up and detach and clean the objects.

Q.69 Â Describe one application of ultrasound or ultrasonic waves in industry.

In industries, the large metal blocks used for the construction of bridges, buildings are checked beforeÂ use.

Ultrasound waves are used to detect any fault present in a block of metal by using the principle, â€˜reflection of soundâ€™. This is a technique in which flaws in a block can be detected without being damaging.

Working-The sound waves are made to pass through one side of the metal block. The detectors are placed on the other side which detects the transmitted waves. If there is any flaw or defect like crack or hole, etc. inside the block then ultrasound gets reflected back and does not reach the detector.

Q.70Â  Why is dog considered most suitable by the police for detective purposes?

Certain animals like dogs, bats , dolphins , porpoises ,rats, etc can hear ultrasound waves which cannot be heard by humans. As dogs can hear ultrasonic sound waves of frequencies upto 50,000 Hz they are used by the police for detecting purposes.Â

Q.71 Â Can dolphin detect ultrasonic waves?

There are certain animals like dolphins, bats, porpoises, etc which can even produce ultrasonic waves on addition with hearing the waves. So, dolphin can detect ultrasonic waves.

Q.72 Â Define the terms:

(a) Ultrasound scanner

(b) Ultrasonography

(c) Electrocardiography

(a) Ultrasound scanner is a device used in ultrasonography. Using the reflected ultrasound wave it makes a photo of internal organs of the body. This is used to check the size, structure and physiological functioning of different organs of human body like liver, kidney, uterus, etc. It is also used to monitor the development of the fetus inside the mother.Â

(b) Ultrasonography is the technique used to take pictures of internal organs by the reflection of ultrasonic waves.Â

(c) Echocardiography is used to observe the function of heart through ultrasonic waves.

Q.73 Â What is the full form of SONAR?

â€˜SOund Navigation And Rangingâ€™ is the full form of SONAR. This is a gear used to find the depth of the sea, to locate underwater objects like shoals of fish, icebergs, submarines of the enemies, etc.Â

Q.74 Â What is echo- ranging or sound ranging? State any one application of this technique.

The technique of finding the distance of an object with the concept behind as Echois called echo-ranging or sound ranging. Under this technique, the time taken by the echo to return back is used in finding the positions of substances.

The depth of the sea bed can be determined by this technique.Â

Q.75 Â Why are ultrasonic waves used in SONAR?

- These waves are high frequency and very short wavelength so can easily penetrate in seaÂ water to locate the objects beneath the water while audible range of sound cannot be tooÂ penetrative.

- These waves cannot be misinterpreted by the sounds produced by ships or engine as this rangeÂ of sound could not be heard by the humans.

Q.76 Â Explain the use and working of SONAR

SONAR is a device which uses ultrasonic waves to detect the depth of the sea, underwater substances like shoals of fish, submarines, etc.

Instrument : SONAR consists of:

(i) A transmitter for emitting ultrasonic waves and

(ii) A receiver for receiving reflected ultrasonic waves.

Working:

Transmitter emits ultrasonic waves which travel through water and gets reflected by the object in the path. After getting reflected these waves are detected up by the receiver. The receiver converts the reflected ultrasonic waves into electrical signals and records them.

Let t = time interval between the transmission and reception of the reflected ultrasound waves,

v = speed of sound under sea water.

d = distance of the object that reflected the ultrasound from the SONAR.

Total distance travelled by the ultrasonicwaves = 2 d (from the figure)

As Â Â Â Â Â Â  distance = speed Ã— time,

2 d = v Ã— t

$d = {{vt} \over 2}$

Therefore, the distance of the object from the ship can be calculated from the above formula.

Use:

(i)Â Â Â Â Â Â Â Â  Determine depth of the sea.

(ii)Â Â Â Â Â Â Â  Trace underwater hills, icebergs, submarines and sunken ships.

(iii) Â Â Â Â Â  Helps in communication between ships.

Figure showing the working of SONAR

Human ear

Q.77 Â What is persistence of hearing?

The time period till which the sensation of sound persists in our brain i.e. 0.1 second is known as persistence of hearing.

Q.78 Â What is the range of frequency which is audible to a average human?

The sound frequency ranging from 20 Hz to 20,000 Hz is audible to human. But, the children below 5 years of age can hear frequencies upto 25,000 Hz.

Q.79 Â Why does sound wave also known as Pressure wave?

As sound wave consists of an alternating pattern of high pressure (compressions) and low pressure (rarefactions) regions travelling through the medium, it is known as a pressure wave.

Q.80 Â How does the ear drum vibrates?

Sound wave is a longitudinal wave i.e. the particles of the medium vibrate to and fro parallel to the direction of wave. Through this there generates a low pressure and high pressure areas alternately known as rarefaction and compression respectively. When this sound wave of different pressure reaches the ear drum, the low pressure region or the rarefaction pulls the ear drum outward whereas the high pressure region or compression pushes the ear drum inwards. This impinges by alternate regions on the drum sets the ear drum in vibratory motion.

Q.81 Â Write a short note on Human ear.

Definition: Ear in humans is the sense organs which helps them to hear sound. It converts pressure variations in air of audible frequencies into electric signals which travel to brain through the auditory nerve.

Structure of ear:

(a) Outer part â€“

• Pinna collects sound waves from surroundings.
• Ear canal acts as long passage.
• Ear drum (tympanum) is present at the end of ear canal which is a circular and elastic membrane.

(b) Middle part â€“It consist of three bones, hammer, anvil and stirrup,

(c) Inner Part â€“It has a coiled tube, a cochlea which is filled with liquid containing nerve cells that are sensitive to sound.

Working:

• Pinna collects sound waves from the surroundings and made to fall on the ear drum, after which the eardrum starts vibrating back and forth quickly due to compression and rarefaction.
• These vibrations are then amplified many times by three bones in the middle ear.
• These vibrations are then passed through the liquid in cochlea which begins to vibrate and the pressure variations are turned into electric signals.
• These electric signals are carried by auditory nerve to brain. The brain interprets them as sound and humans get the sensation of hearing.

Q.82 Â In the old age people lost their hearing ability. Which device can be useful for them to support them hearing.

People which have problem in hearing, or which hear hard are given hearing aid for solving the problem. This is an electronic, battery operated device. It consists of a microphone and a speaker. The microphone converts the sound into electric signals which are then amplified by the amplifier. Now, these amplified signals are directed to the speaker where they are again converted to sound. This sound can now be heard by the person distinctly and clearly.

Numericals :Â

Q.1 Â Calculate the time taken by a sound wave to travel 1.5 km, when its frequency and wavelength are 2000 Hz and 0.35 m respectively.

Given,

Distance, s = 1.5 km = 1500 m

Time =?

Speed of sound, v =?

$Time = \,{{Dis\tan ce} \over {Speed}}$

For speed,

Frequency,Â $\nu \, = \,2000Hz$

Wavelength, Â Î» = 35 cm

$v = \,\nu \lambda$

= 2000 Â Ã— 0.35 = 700 m/s.

Therefore,Â $Time = \,{{Dis\tan ce} \over {Speed}}$

= ${{1500} \over {700}}$ = 2.15 second

Q.2Â  In a concert a pianist played some music on piano. Compare the Two Note A and B played when the speed of sound is 340 m/s and their wavelengths are 1.5 m and 1.33 m respectively.

Given,

Speed of sound, v= 340 m/s

Wavelength of Note A, ${\lambda _A}$Â = 1.5 m

Wavelength of Note B, ${\lambda _B}$Â = 1.33 m

Frequency of Note A, $\nu {\,_A}$ =?

Frequency of Note B,Â $\nu {\,_B}$ =?

$Frequency\,\,of\,\,the\,\,sound\,\,wave,\nu = {v \over \lambda }$

${\nu _A} = {v \over {{\lambda _A}}} = {{340} \over {1.5}}\, = \,226.66Hz$

${\nu _B} = {v \over {{\lambda _B}}} = {{340} \over {1.33}}\, = \,255.63Hz$

The frequency of Note B is more than frequency of Note A.

Q.3 Ocean waves have speed 15 m/s and time period 0.01. Find out the wavelength and the distance between the adjacent crest and the trough.

WavelengthÂ $(\lambda ) = Time\,\,period(T) \times Speed\,\,of\,\,sound(v) = 15 \times 0.01 = 0.15m$

Distance between crest and trough is half the wavelength = $\lambda /2 = 0.075m$.

Q.4 Â Calculate the frequency at which the boat rocks when waves of wavelength 100 m travelling at a speed of 20 m/s strikes it.

Given,

Wavelength, $\lambda$Â = 100 m,

Speed of sound, v = 20 m/s,

Frequency, $\nu$Â =?

$\nu = {v \over \lambda } = {{20m/s} \over {100m}} = 0.2\,\,Hz$

Q.5 Â In a cancer treatment hospital an ultrasound scanner is used to locate tumors in a tissue. It operates at frequency of 4.2 MHz. Calculate the wavelength of sound in a tissue? (Speed of sound in the tissue is 1.7 km/s)

Speed of sound, v = 1.7 km/s,

Frequency of wave, $\nu$ = 4.2 MHz = 4.2 Ã— 106 Hz

Wavelength,Â $\lambda = {v \over \nu } = {{1.7km/s} \over {4.2 \times 1{0^6}/s}} = {{1700m/s} \over {4.2 \times 1{0^6}/s}} = 4 \times 1{0^{ - 4}}m$

Q.6 Calculate the frequency of sound, where source of sound produces 500 compressions and 500 rarefactions in air in 25 seconds.

Since, a compression and rarefaction combines to form a wave therefore no. waves = 500.

Frequency =Â ${{No.\,\,of\,\,waves} \over {Time}} = {{500} \over {25}} = 20Hz.$

Q.7 Â What will be the wavelength and velocity of the wave produced when a stone is thrown in a pond. After the fall, stone produces 12 full ripples in 1 second in water. Also, the distance between a crest and a trough is 10 cm.

Given,

Frequency = ${{No.\,of\,waves} \over {Time}}$ = ${{12} \over 1}$ = 12

Distance between the crest and next trough, Î» /2 = 10 cm

Therefore, wavelength, Î»Â = 20 cm = 0.20 m

Since, velocity = frequency Ã—Â wavelength

Therefore, velocity = 12 Â Ã— 0.20 = 2.40 m/s

Q.8 Â A kid watching Dusshera celebration from a distance watches the statue of Ravana burn into flames and hears the explosion after 2 sec. If the speed of sound in air was 335 m/s what would be the distance of statue from the kid?

Given,

Speed of sound, v = 335 m/s

Time, t = 2 sec

Distance, d = v Ã— t

= 335 Ã— 2 = 670 m.

Q.9 Â Find the angle of incidence when 110o is the angle between incident wave and reflected wave.

Given,

$\angle i + \angle r = 110^\circ$Â

Since, Â $\angle i$ (angle of incidence) = $\angle r$Â (angle of reflection)

$\Rightarrow 2\angle i = 110^\circ$

$\Rightarrow \angle i = {{110} \over 2} = 55^\circ$

The angle of incidence = 55o

Q.10 Â A boy standing on a tower of height 500 m drops a stone into a pond of water at the base of the tower. Calculate:

(a) Time taken by the stone to reach the pond.

(b) Time taken by the splash to be heard by the boy.

[Given, g = 10 ms-2 and speed of sound = 340 ms-1].

(a) Time taken by the stone to reach the pond.

According to the equation of motion-

$s = ut + {1 \over 2}g{t^2}$

Where, s = height of the tower = 500m

u = Initial velocity of stone = 0

t = time taken by the stone to reach the base of the tower =?

g = acceleration due to gravity = 10 m/s2

By putting the values in equation,

$500 = 0 \times t + {1 \over 2} \times 10{t^2}$

500 = 5 t2

t = $\sqrt {100}$Â = 10 seconds.

(b) Time taken by the splash to be heard by the boy = Time taken by the stone to reach the pond + Â  time taken by the splash to reach at the top of the tower.

Time taken by the splash to reach to the top of the tower:

${t^!}\, = \,{{Dis\tan ce\,\,travelled} \over {Speed}}$

${{500} \over {340}} = \,1.47\sec onds$

Therefore,

Time taken by the splash to be heard by the boy = 10 + 1.47 seconds = 11.47 seconds

Q.11 Â Calculate the distance between a source of sound and the wall when echo returns from the surface of the wall in 1.5s. (Speed of sound= 334 m/s)

Given,

Time taken to receive echo= 1.5 s

Speed of sound= 334 m/s

Since, echo travels twice the distance between the source and the object.

Therefore,

$Distance = {{Speed\,\,of\,\,sound\,\, \times \,\,Time\,\,taken\,\,to\,\,receive\,\,echo} \over 2}$

$Distance = {{334 \times 1.5} \over 2} = 250.5\,\,m$

Q.12 Â Ranvijay is standing at a distance of 51m from a wall. He fires a gun. Find out the time after which an echo is heard. (Speed of sound in air = 340 m/s)

Given,

Distance, d = 51 m,

Speed of sound in air, v = 340 m/s.

Since,

Speed = ${{Distance} \over {Time}}$

Therefore, Time =Â ${{Distance} \over {Speed}}$

For echo,Â ${{Distance \times 2} \over {Time}}$

$t = {{2d} \over v} = {{2 \times 51} \over {340}} = 0.3\,\,s$

The time after which echo is heard by Ranvijay is 0.3 second.

Q.13 Â A girl standing in the middle of a big square field explodes a cracker. She heard an echo 0.4 later when there is a tall building on one side of the field. What is the size of the field? (Given, Speed of sound in air is 330 m/s).

Given,

Speed of the sound, v = 330 m/s

Time taken in hearing the echo, t = 0.4 second

Distance travelled by the sound, d = v Ã— t

= 330 Ã— 0.4 = 132 m.

The distance travelled by the sound is twice of the original distance as in echo produced sound need to travel two times.

Therefore, distance = 132/2 = 66 m.

Side of the square field = twice the distance between the girl and building

= 132 m

Size of square field = side Ã— side

= 132 Ã— 132 = 17424 m2

The size of square field is 17424 m2.

Q.14 Â A Ship emits ultrasound waves from SONAR fitted down. This wave returns from an underwater iceberg in 1.02 seconds. Calculate the distance of iceberg from the ship when the speed of sound in sea water is 1531 m/s.

Given,

Time taken by the ultrasonic wave to return back to ship after reflection from the iceberg, t = 1.02 seconds

Speed of sound in sea water = 1531 m/s

Total distance travelled by the ultrasonic waves = 2 Ã— Distance between the ship and the iceberg, d

Distance = Speed Ã— Time

2 d = 1531 Ã— 1.02

âˆ´ Â d =${{1531{\rm{ }} \times {\rm{ }}1.02} \over 2}$ = 780.81m

Therefore, the distance between the ship and iceberg is 780.81 m.

Value Based Questions: -

Q.1 Â Rahul went to Russia for holidays there he watched an opera performance in Novosibirsk Opera and Ballet Theatre, in Novosibirsk. He admired the architecture and furnishing of the auditorium. There were curved ceiling, swags, curtains and cushions positioned all around. The floor was also carpeted. After the performance was over all were taken to watch the auditorium. During that he noticed a sound board behind the stage. After watching all this he doubted whether all these things where just for increasing the beauty of the hall or also had a scientific reason.

On the basis of above passage answer the following questions:

(a) What could be the reason for placing of curtains, cushions and carpets in the opera house?

(b) Discuss the use of curved ceiling and sound board?

(c) Which values are possessed by Rahul?

(a) To avoid reverberation (multiple reflection of sound) in the auditoriums, materials which can absorb sound energy is used.

(b) Sound boards and curved ceilings are present in the auditorium to generate maximum reflections to facilitate the sound in every corner of the auditorium.

(c) Rahul has a keen interest in art but also has scientific approach towards the happenings around.

Q.2 Â Shikha went to a Himachal Pradesh to meet her friend Rita. She was astonished to watch high mountain ranges, waterfalls, and the terrific greenery all around during her journey. After reaching their Rita took her to a peak and asked her to call upon her name at the top of her voice. By doing this Shikha could hear the echoes of her own voice which filled her with joy. After returning to Ritaâ€™s home, she tried to hear her echo in the same way in the room but failed. She searched about this on internet and got answers to her questions and went to bed.

(a) What is Echo?

(b) Why Shikha could not heard the echo at Ritaâ€™s room?

(c) Which qualities do Shikha have?

(a) Echo is a phenomenon which produces recurrences of sound from a source by reflections through the obstacle.

(b) For hearing an echo, there should be some requirements to be fulfilled like the distance between the source of sound and obstacle should be minimum 17.2m. Also most of the sound is being absorbed by the furnishing in the room.

(c) She is curious to learn about the natural phenomenon for which she makes use of latest technology.Â

Q.3 Â Vartika was conceiving for the first time. Her mother -in -law did not wanted the first child of her to be a girl. Vartikaâ€™s mother -in -law took her to a gynecologist for ultrasonography to determine whether the child is a boy or a girl. Doctor denied telling the gender of the child.

(i) Define ultrasonography? Why is ultrasonography used in pregnancy?

(ii) Which principle is involved behind its working?

(iii) Why did Doctor denied the in telling the gender of the child?

(i) Ultrasonography is a technique in which ultra sound waves are used for imaging of internal organs of body, like heart, uterus, etc. to detect any dis functioning in the organ. It is used in the diagnosis of congenital defects and abnormalities which could be present in the child.

(ii) Reflection of sound.

(iii) To reveal the gender of a child before birth is against the law. Under this law the parent of the child and the doctor are charged with a fine and also could be put in jail.

Q.4 Â On D.D. channel Akhil was watching a documentary based on usage of submarine in Indian navy. The navy officials were using a device to track down their enemies submarines present in the water at a distance of few kilometers from their ship. They also used this device for underwater communication and to avoid obstacles in the way.

(i) Which device are they using?

(ii) What is the principle used behind this device?

(iii)What are the other uses of this device?

(i) SONAR-Sound Navigation and Ranging.

(ii) Principle used reflection of ultrasonic sound waves.

(iii) SONAR can be used to trace the shoal of fish, a sunken submarine or ship, icebergs present in the sea or oceans.

Q.5 Â Jiyaâ€™s sister is music lover, she used to listen songs the whole daylong.Â  She likes to listen music at high intensity levels. Family members used to warn her for not being aware about her health as she sometimes has pain in her ears. But she ignores listening to her family. Continuous hearing of loud music can damage her hearing power. Damage to hearing ability depends on two factors: the sound intensity level (dB) and the exposure time. In general, at 90 dB, it takes 8 hours or less for the damage to receptor nerves to occur. Moreover, it is discovered that if the sound level is increased by 5 dB, the safe exposure limit is dropped to half.

From reference to the above passage, answer the following questions:

(a) What time would be safe to expose the ear to a sound of 95 dB and 105 dB respectively?

(b) What nature of Jiyaâ€™s sister is being reflected in the passage?

(a) As it is found that by increasing the sound intensity by 5 dB from 90 dB then the time taken to damage the hearing power would decrease to half. Therefore the time for safe exposure of sound of intensity 95 dB would be half of 8 h which is less than 4h. Similarly, with sound Intensity level of 105 dB, which is 15 dB i.e., 3 times 5 dB above 90 dB, therefore will have (1/2)3 of 8 h, i.e. 1h to damage the hearing power.Â

(b) Jiyaâ€™s sister is ignorant about the incapacity to hear. She didnâ€™t take things seriously, which could lead her to a permanent loss of hearing.

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