Sound Class 9 Science Exam Questions

Please refer to Sound Class 9 Science Exam Questions provided below. These questions and answers for Class 9 Science have been designed based on the past trend of questions and important topics in your class 9 Science books. You should go through all Class 9 Science Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

Class 9 Science Exam Questions Sound

Class 9 Science students should read and understand the important questions and answers provided below for Sound which will help them to understand all important and difficult topics.

1 Mark Questions

Question.  What happens to speed of sound when it goes from solid to gaseous state?
Answer. When sound goes from solid to gaseous state it slows down.

Question. How are wavelength, speed and time period related for a sound wave?
Answer. Speed = Wavelength × Time Period

Question. What is the nature of sound waves?
Answer. Sound waves are longitudinal waves which require a material medium to travel.

Question. What is the frequency of wave with time period 0.025 s?
Answer. T = 0.025 s
Therefore, frequency
v = 1/0.025 = 40 Hz

Question. Give one example of transverse and longitudinal wave each.
Answer. Transverse waves – light wave.
Longitudinal waves – sound wave.

Question. On what factor does the quality of the sound depend?
Answer. The quality of sound depend on the frequency and its harmonics.

Question. What are infrasonic and ultrasonic sounds?
Answer. Sound below 20 Hz are called infrasonic and sounds above 20,000 Hz are called ultrasonic.

Question. A baby recognizes her mother by her voice. Name the characteristics of sound involved.
Answer. Quality or timbre

Question. The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?
Answer. Frequency = 100 Hz = 100 vibrations/sec Therefore , in minute no. of vibration =100 × 60 = 6000

Question. How does speed of sound change with temperature of medium?
Answer. The speed of sound increases with temperature.

Question. Name the device which is used to measure intensity of earthquake.
Answer. Seismograph

Question. Why are sound waves called mechanical waves?
Answer. Sound waves are called mechanical waves as they need a material medium to travel.

Question. Find the time period of oscillation if a body vibrates 150 times in a minute.
Answer. Time period

Question. Give two examples of electromagnetic waves.
Answer. X-rays, Gamma rays

Question. Give two examples of mechanical waves.
Answer. Two examples are (i) Sound waves and (ii) Waves produced in a string.

Question. Define amplitude and give its SI unit.
Answer. Amplitude is the maximum displacement of a particle from its mean position, its unit is m.

Question. How is frequency of wave related to its time period?
Answer. v = 1 / T

Question. What is the audible range of frequency in humans?
Answer. Audible range for human beings 20 Hz−20,000 Hz

Question. Distinguish between tone and note of sound.
Answer. Sound of single frequency is called tone whereas sound which is produced by mixture of several frequencies is called note.

Question. What is meant by quality of sound?
Answer. Sound consists of a particular frequency and its multiple frequencies which are called harmonic or overtones. Sound is said to be of good quality if it contain a large number of harmonics.

Question. Name any two animals which produce ultrasound.
Answer. Animals producing ultrasound are : bats and porpoises.

Question. Define reverberation.
Answer. The persistence of sound due to its repeated reflection is called reverberation.

2 Marks Questions

Question. Express the units of:
(a) Frequency and
(b) Wavelength.
Answer. (a) Frequency – hertz (Hz),
(b) Wavelength – metre (m)

Question. What is reverberation?
Answer. 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.

Question. Illustrate the conditions required to hear an echo.
Answer. The minimum distance of the reflecting surface from the source should be 17.2 metres at a temperature of 20 in 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.

Question. Which of the two have shorter wavelength: Infrasonic or ultrasonic waves?
Answer. 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.

Question. Name one application of a sound wave having a frequency of 30 kHz.
Answer. 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.

Question. Why is dog considered most suitable by the police for detective purposes?
Answer. 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.

Question. What is the full form of SONAR?
Answer. Sound Navigation And Ranging’ 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.

Question. Can dolphin detect ultrasonic waves?
Answer. 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.

Question. Write any three characteristics of sound waves.
Answer. Following are the properties of sound waves:
These are longitudinal waves.
These are Mechanical waves.
These are a result of vibrations.

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

Question. What are the different types of sound on the basis of frequencies?
Answer. On the basis of frequencies, sound can be:
(a) Infrasound – frequencies below than 20 Hz
(b) Audible sound – frequencies between 20 Hz to 20,000 Hz.
(c) Ultrasound – frequencies above 20,000 Hz

Question. What is a medium? What are the different mediums in which sound can travel?
Answer. 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.

Question. A person produced a sound with a siren near a cliff and heard echoes after six seconds. Find the distance of the siren from the cliff if velocity of sound waves produced is 330 m/s?
Answer. Time of echo = 6 sec
Velocity V= 330 m/s
Distance between source and cliff = v X t / 2 = 330 x 6 / 2 = 900m

Question. (a) Define frequency of a sound wave and give its SI unit.
(b) Why are the roof and walls of an auditorium/ hall generally covered with sound absorbent materials?
Answer. (a) Frequency is the number of vibrations produced in 1 second. Its unit is Hertz (Hz)
(b) To reduce reverberation of sound.

Question. What is ultrasound? What is its frequency? Give its uses.
Answer. Waves of frequency above 20,000 Hz is called ultrasound. It is used to clean parts, detect flaws, in echocardiography, ultrasonography, etc.

Question. Why do we see streaks of lightning much before we hear the sound of thunder?
Answer. Thunder is heard after the flash of light is seen because sound and light travel at different speed.
Light travels at the speed of about
3 x 108 m / s whereas sound travels at about 300 m/s.

Question. An echo is returned in 6 seconds. What is the distance of reflecting surface from source? Given that speed of sound is 342 m/s.
Answer. Echo takes 6 sec to return.
Speed of sound = 342 m/s

Question. (a) Sound is produced when your school bell is struck with a hammer. Why?
(b) Which characteristic of sound helps to identify your friend by his voice while sitting with others in a dark room?
Answer. (a) When the bell is struck with a hammer, the bell starts vibrating . These vibrations set the air near it to vibrate which spread as longitudinal waves in all direction. These vibrations are sound waves.
(b) The quality or timbre of sound helps us to identify our friends.

Question. Calculate the frequency of sound, where source of sound produces 500 compressions and 500 rarefactions in air in 25 seconds.
Answer. Since, a compression and rarefaction combines to form a wave therefore no. waves = 500.

Question. Give one difference between transverse and longitudinal wave. Give one example for each.
Answer. In transverse waves particles move perpendicular to the direction of propagation eg. light waves. In longitudinal waves the particles vibrate along the direction of propagation, eg. sound wave.

Question. Sound of explosions taking place on other planets is not heard by a person on the earth. Give reason.
Answer. Sound of explosions taking place on other planets is not heard on earth because sound cannot travel in vacuum.

Question. (a) Which wave property determines (i) Loudness, (ii) Pitch?
(b) How are wavelength and frequency related to speed of sound waves?
Answer. (a) 1) Loudness – Amplitude (2) pitch – Frequency
(b) speed = wavelength × frequency

Question. Three persons, A, B and C are made to hear a sound travelling through different media as given below:

Who will hear the sound first? and why?
Answer. A will hear the sound first because sound travels fastest in solids.

Question. What is SONAR? Write two uses of SONAR technique?
Answer. The acronym SONAR stands for sound, navigation and ranging. It is a technique used to determine depth of sea end locate underwater objects like hills, valleys, sunken ships etc.

Question. Differentiate between mechanical waves and electromagnetic waves.
Answer. Mechanical waves are waves which need a material medium to travel . They may be transverse or longitudinal. Electromagnetic waves do not require a material medium .They can travel in vacuum. They are always transverse waves.

Question. Give four properties of sound waves.
Answer. Properties of sound –
(i) It is a mechanical wave.
(ii) It travels fastest in solids.
(iii) The speed of sound depends on temperature
(iv) It travels as longitudinal wave.

Question. We receive heat and light from the Sun, but don’t hear the sound of explosions occurring on it. Why?
Answer. Sound cannot travel in vacuum. So explosions in sun cannot be heard on earth. Light travels to earth since it can travel in vacuum.

Question. How does speed of sound change with (a) temperature of medium (b) physical state of medium?
Answer. (a) With increase in temperature, speed of sound increases.
(b) Denser is the medium , greater is the speed of sound . Hence it is maximum in solid and minimum in gases.

Question. How does sonic boom occur?
Answer. When any object travels at a speed greater than sound, shock waves are produced which travel at the speed of sound . The sound produced is called sonic boom , which sounds as an explosion.

Question. The string of a guitar is plucked. What type waves are produced in a) guitar (b) air?
Answer. In the guitar transverse waves are produced, whereas in air longitudinal waves are produced.

Question. Find the distance between a surface and the source of sound, if speed of sound is 334 m/s and echo returns from the surface in 1.5 s.
Answer. Speed of sound = 334 m/s
Time of Echo = 1.5 sec

Question. Define wavelength and relate it with frequency or sound and its velocity.
Answer. Wavelength is the distance travelled between consecutive compressions or rarefactions.
Velocity = distance / time
If the distance travelled = 1 wave length = λ
and time to travel = Time period = T
V = λ / T = λ x v
where v = 1 / T

Question. Why can’t we hear the sound of squeaks produced by bats?
Answer. Since the sound produced by bats is of frequency greater than 20,000 Hz. Which is not audible to the human ear, we cannot hear the squeaks.

Question. State the mechanism through which humans can utter sound through their mouth.
Answer. 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.

Question. Why are sound waves longitudinal and mechanical waves?
Answer. Sound waves are longitudinal waves as the particles of the medium do to and promotion in the same direction in which wave moves. Sound waves are mechanical waves as it needs material medium for its propagation.

Question. How is sound produced when our school bell is struck with a hammer?
Answer. 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.

Question. Define the terms: (a) Tone (b) Note
Answer. (a) Tone: A sound produced by single frequency is known as tone.
(b) Note: A sound produced by mixture of frequencies is known as note.

Question. What is the range of frequency which is audible to a average human?
Answer. 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.

Question. What changes occurs in the speed of sound when
(a) It travels from iron to air?
(b) Temperature of the air increases?
Answer. (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.

Question. Why does sound wave also known as Pressure wave?
Answer. 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.

3 Marks Questions

Question. How does a sound board work?
Answer. 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.

Question. (a) After a snowfall, why does it is quietness all around?
(b) Why do empty rooms do sound echoing?
Answer. (a) The snow absorbs sound with little reflection of sound. Therefore, the surroundings becomes quite after the snowfall.
(b) 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.

Question. (a) Write three main properties of ultrasound.
(b) Enlist four medical uses of ultrasound.
Answer. (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.

Question. Explain how ultrasound is used to clean objects used in the industries.
Answer. 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.

Question. Define the terms:
(a) Ultrasound scanner
(b) Ultrasonography
(c) Electrocardiography
Answer. (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.
(b) Ultrasonography is the technique used to take pictures of internal organs by the reflection of ultrasonic waves.
(c) Echocardiography is used to observe the function of heart through ultrasonic waves.

Question. 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.
Answer. Given,
Distance, s = 1.5 km = 1500 m
Time =?
Speed of sound, v =?
Time = Distance / speed
For speed
Frequency, v = 2000Hz
Wavelength, λ = 35 cm
u = vλ
= 2000 × 0.35 = 700 m/s.
Therefore, Time = Distance / speed
= 1500/700 = 2.15 second

Question. 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)
Answer. 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,

Question. 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?
Answer. (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.

Question. Using labeled diagram show how low and high pitch sound are represented.
Answer. 

Question. Define sonic boom.
Answer. 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

Question. How can reverberation be reduced?
Answer. 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.

Question. A certain amount of reverberation is desirable for speeches but the reverberation time should be short. Give reason.
Answer. 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

Question. Explain why?
Answer. (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 10 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.

Question. On which factors does the speed of sound depend?
Answer. 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 1 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.

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

Question. 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?
Answer. Suppose, l = length of the iron rod
Time taken by the sound to travel through the iron rod is given by Similarly, time taken by the sound to travel through the air is given by Therefore, ratio of time taken by the sound wave in Air and Iron

Question. Why is noise different from music?
Answer. 

Question. Three friends are made to hear a sound travelling through different media as given below

Which one will hear the sound first? And why?
Answer. 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.

Question. Show the propagation of a longitudinal wave showing compression and rarefaction.
Answer. Compression are denoted by → A, C, E, G
Rarefaction are denoted by → B, D, F

Question. Differentiate between Loudness and Intensity.
Answer. 

S.no	Loudness	Intensity
1	it is subjective in nature	it is objective in nature
2	it is not completely physical quantity	it is physical quantityty
3	it depends upon the sensitivity of ear	it does not not depend upon the sensitivity of ear
4	unit of measurement: Decibel (dB)	unit of measurement: waits per square metre (W/m2)

Question. What is a wave?
Answer. 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 Waves.

Question. Differentiate between longitudinal wave and transverse wave.
Answer. 

.Question. Define the following terms:
(a) Time period
(b) Frequency of an oscillating body.
(c) State the relation between frequency and time period.
Answer. (a) 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) Frequency (ν ): 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.
Where, f = frequency of the wave
T = Time period of the wave

Question. What is compression and rarefaction pulse.
Answer. 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.

Question. Write any four differences between mechanical and electromagnetic waves.
Answer. 

Question. State the relation between Time period and Frequency of an oscillating body.
Answer. For an oscillating body,
T = time period
ν = frequency
No. of oscillations completed in T second = 1
No. of oscillations completed in 1 second = 1/T
We know no. of oscillations completed in 1 second = frequency (ν)
Therefore, the relation is shown by, V = 1/T

Question. 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 0° to 20°C?
Answer. (a) The wavelength of sound will increase. Since, the speed of sound increases with temperature. As. 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 1 C rise in temperature the speeds of sound increases by 0.6 m/s.
Therefore, sound speed when temperature changes by 20°C = 0.6 m/s × 20 = 12 m/s.
Increase in wavelength = increase in the speed/ frequency.

The increase in wavelength is 0.047m.

Question. Diagrammatically show how longitudinal wave travels along a slinky.
Answer. – 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.

Question. Which type of waves is produced when:
(a) Wire of a guitar is plucked.
(b) A stone is dropped in a pond.
Answer. (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.

Question. Establish a relation between, speed, wavelength and frequency of wave.
Answer. Here, speed = v
Distance = wavelength = λ
Time period = T
Frequency = ν
Since,
Speed = distance/time

Question. Calculate the frequency at which the boat rocks when waves of wavelength 100 m travelling at a speed of 20 m/s strikes it.
Answer. Given,
Wavelength, λ = 100m
Speed of sound, v = 20 m/s,
Frequency, v = ?
v = v / λ = 20m/s / 100m = 0.2 Hz

Question. 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?
Answer. Given,
Speed of sound, v = 335 m/s
Time, t = 2 sec
Distance, d = v × t
= 335 × 2 = 670 m.

Question. Find the angle of incidence when 110^o is the angle between incident wave and reflected wave.
Answer. Given,
∠ i + ∠r = 110^o
Since, ∠i (angle of incidence) = ∠r (angle of reflection)
⇒ 2∠i = 110^o
⇒ ∠i = 110 / 2 = 55^o
The angle of incidence = 55^o

Question. A hospital uses an ultrasonic scanner to locate tumors in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 km/s. The operating frequency of the scanner is 4.2 MHz (MHz = 106 Hz).
Answer. Speed of sound 1.7 km/s = 1.7 x 103 m/s
Frequency = 4.2 MHz = 4.2 x 106 Hz

Question. How are ultrasonic waves different from ordinary sound waves? State two applications of ultrasound.
Answer. Ultrasonic waves have frequency above 20,000 Hz. They are not audible to human beings . They are able to travel in well defined paths even in the presence of obstacles as their wavelength is very small.
Application –
(i) Used to clean parts of equipment located in hard to reach places.
(ii) Used to detect cracks and flaws in metal blocks

Question. A sonar device on a submarine sends out a signal and receives an echo 5 seconds later. Calculate the speed of the sound in water if the distance of the object from the submarine is 3625 m.
Answer. Speed =Total distance travelled / Total time
Here object distance = 3625 m
Therefore total distance travelled by sound = 3625 × 2 m
Time = 5 s
Speed = 3625 x 2 / 5 = 1450 m / s

Question. What is echo ranging? State any one application of this technique.
Answer. Echo ranging is the technique used to determine the distance and location of object by using ultrasound. The time interval between transmission and reception of sound is noted and distance can be calculated by the formula d = v X t / 2 where v is the velocity of sound and t is time for echo to be heard. It is used in SONAR to locate depth of the sea and other underwater objects.

Question. Distinguish between echo and reverberation.
Answer. When sound reflected from object returns to the listener and can be heard clearly, it is called an echo. It can be heard only if the reflected sound is heard after 0.1 s. In time less than 0.1 s, the listener is unable to identify the sound and its reflection. Sound created in a big hall is reflected continuously from its wall. This repeated reflection causes persistence of sound for some time. This is called reverberation.

Question. Give significance of ultrasound in metal industry.
Answer. Ultrasound is used in metal industry to detect flaws in metals . The cracks and holes inside metal blocks reduce its strength. Ultrasound is passed through the metal and if there is a flaw , it is detected by the reflection of these sound waves.
Hence, ultrasound has a significant role to play in the metal industry as it ensures the strength of the material by detection of flaws.

Question. Give the function of (a) ear drum (b) cochlear fluid in human ear.
Answer. (a)The eardrum is a membrane which vibrates when sound reaches it . It helps in amplifying sound by means of 3 small bones attached to it.
(b) The cochlear fluid is set into vibrations. There are tiny hair like structures in it which convert these
vibrations into electrical impulses that are carried to the brain by the auditory nerve.

Question. Aditi clapped her hands near a cliff and heard the echo after 4 seconds. What is the distance of the cliff from her if the speed of sound is taken as 346 m/s .
Answer. Time of echo = 4 s
Speed of sound = 346 m/s

Question. When a sound is reflected from a distant object, an echo is produced and heard in a winter night. Now, if the distance between the reflecting surface and the source of sound production remains the same, will you hear the echo of the same sound on a summer day? Explain.
Answer. Time taken = Total distance velocity. On a hotter day the velocity of sound is more. So reflection will reach in lesser time. If the time taken by reflection is less than 0.1 sec. Echo will not be heard.

Question. What is meant by intensity of sound? How is it different from loudness?
Answer. Intensity is the amount of sound energy passing per second per unit area. It is proportional to square of amplitude. Loudness and intensity both depend upon the amplitude of sound. But loudness is the physiological response of our ears to a particular frequency. Our ears are more sensitive to some frequencies as compared to others.

Question. A sound wave has frequency of 3 kHz and a wavelength 45 cm. How long will it take to travel 1.8 km?
Answer. Frequency ν = 3 KHz = 3000 Hz
Wavelength λ = 45 cm = 0.45 m
Distance d = 1.8 km = 1800 m
V = νλ = 3000 × 0.45 m/s
Time for wave = d / v = 1800 / 3000 x 0.45 = 1.33 s

Question. Describe two uses of multiple reflections of sound.
Answer. Two uses of multiple reflection of sound are –
(i) Musical instruments like trumpets and Shehnais are designed in such a way that sound moves with the tube of the instrument by multiple reflection and comes out from the conical opening at its end. The sound waves are thus directed in the direction of audience in front.
(ii) In stethoscopes the sound of the patients heart beat reaches the doctors ear by multiple reflection of sound within the stethoscope tube.

Question. Distinguish between transverse waves and longitudinal waves.
Answer. Transverse wave
1. The particles moves or vibrate about their mean position in a direction perpendicular to the direction of propagation
2. It travels in the form of crests and troughs.
3. Light waves are transverse waves and may travel in vacuum.
Longitudinal waves
1. The particles vibrate about their mean position in a direction parallel to the direction of propagation
2. It travels in the form of compression and rarefaction.
3. Sound waves are longitudinal waves and can be produced only in a material medium.

Question. Differentiate between low and high pitch sound using neat and labelled diagram.
Answer. 

Question. Draw a neat labelled structure of human ear, depicting the auditory parts only.
Answer. 

Question. Define amplitude of a wave. Hew does it help to explain loudness of sound produced?
Answer. In a wave, the particles vibrate about their mean position. The maximum displacement of the particle about its mean position is called amplitude. Larger is the amplitude , greater is the energy carried by the wave. Loudness depends on amplitude, A loud sound will have greater amplitude as compared to a soft sound.

Question. An observer stands between two distant cliffs and claps his hands. He receives echo after 2 s and 2.5 s respectively. If speed of sound is 330 m/s, find the distance between the cliffs ?
Answer. Time of echo = 2 s and 2.5 s
speed of sound = 330 m/s

Distance between cliffs = 300 + 412.5 = 712.5 m

5 Mark Questions

Question. Jiya’s sister is music lover, she used to listen songs the whole daylong.
Answer. 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 Jiya’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) of 8 h, i.e. 1h to damage the hearing power.
(b) Jiya’s sister is ignorant about the incapacity to hear. She didn’t take things seriously, which could lead her to a permanent loss of hearing.

Question. 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)
Answer. Given,
Distance, d = 51 m,
Speed of sound in air, v = 340 m/s.
Since,

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

Question. 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).
Answer. 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 × t
= 330 × 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 × side
= 132 × 132 = 17424 m²
The size of square field is 17424 m².

Question. Prove that sound waves are mechanical waves with the help of an experiment.
Answer. Take an electric bell and an airtight glass bell jar. The electric bell is suspended inside the airtight bell jar. The bell jar is connected to a vacuum pump, as shown in Fig. If you press the switch you will be able to hear the bell. Now start the vacuum pump. When the air in the jar is pumped out gradually, the sound becomes fainter, although the same current is passing through the bell. After some time when less air is left inside the bell jar you will hear a very feeble sound. When all the air is removed, no sound can be heard. This shows that sound require a material medium to travel.

Question. (a) What is meant by ‘Compression’ and ` Rarefaction’ of a longitudinal wave?
(b) Give well labelled graphical representation of a longitudinal wave.
Answer. (a) Compression : When air vibrates, it is pushed away by the vibrating object . In that region air particles come very close to each other. This is called compression.
Rarefaction : When the air particles are compressed in a region they have a tendency to push the particles is the forward and backward direction simultaneously. Particles move away from that region. Now there are less air particles then normal . This is called rarefaction.

Question. (a) Give three medical uses of ultrasound.
(b) A ship which is stationary is at a distance of 2800 m from the sea-bed. The Ship sends an ultrasound signal to the sea-bed and its echo is heard after 4 s. Find the speed of sound in water.
Answer. (a) Ultrasound can be used in:
(i) Monitoring foetus during pregnancy.
(ii) Medical imaging of internal organs to check their size, structure and physiological functioning in humans.
(iii) Breaking kidney stones into smaller sizes to flush out in the urine.
(b) Given,
Distance, d = 2800 m
Time, t = 40 sec
Velocity, v =?
Total distance travelled by sound = speed of sound × time
So, 2d = v × t
⇒ v=2d / t
⇒ v = 2 × 2800 / 4
= 1400 m/sec

Question. A body is vibrating 6000 times in one minute. If the velocity of sound in air is 360 m/s, find:
(a) Frequency of vibration in hertz,
(b) Wavelength of the wave produced.
Answer. (a) Frequency of vibration in hertz
Given,
Number of vibration in one minute = 6000
Number of vibrations in one sec = 6000 vib / 60s = 100Hz
Therefore, Frequency, ν = 100 Hz
(b) Wavelength of the wave produced
Given,
Velocity of speed in air, v = 360 ms
Frequency, ν= 100 Hz
Wavelength, λ = ?

The wavelength of the wave is 3.6 m

Question. Explain how human ear works in the transmission of sound wave to the brain.
Answer. The human ear consists of three parts – the outer ear, middle ear and inner ear.
Outer ear : This is also called ‘pinna’. It collects the sound from the surrounding and directs it towards auditory canal.
Middle ear : The sound reaches the end of the auditory canal where there is a thin membrane called eardrum or tympanic membrane. The sound waves set this membrane to vibrate. These vibrations are amplified by three small bones- hammer, anvil and stirrup.
Inner ear : These vibration reach the cochlea in the inner ear and are converted into electrical signals which are sent to the brain by the auditory nerve, and the brain interprets them as sound.

Question. 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.
Answer. (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) Two friends on the surface of the moon cannot talk to each other because moon deoes not have atmosphere or medium for sound to propagate so sound would not produce.
(c) 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).

6 Mark Questions

Question. Name any three practical applications of reflection of sound waves with the help of diagram.
Answer. 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: It’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.

Question. (a) What is meant by ‘compression’ and ‘rarefaction’ of a longitudinal wave?
(b) Give well labeled graphical representation of a longitudinal and transverse wave wave.
Answer. (a) When sound is produced by to and formation 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 ↑ Pressure ↑
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 ↓ Pressure ↓

(b)

Question. Explain the use and working of SONAR
Answer. 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 ultrasonic waves = 2 d (from the figure)
As distance = speed × time,
2 d = v × t
d = vt/2
Therefore, the distance of the object from the ship can be calculated from the above formula.
Use:
(i) Determine depth of the sea.
(ii) Trace underwater hills, icebergs, submarines and sunken ships.
(iii) Helps in communication between ships.

Question. 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.
Answer. [Given, g = 10 ms 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 / 2)gt²
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 x t + 1/2 x 10t²
500 = 5t²
t = √100 = 10 second.
(b) Time taken by the splash to be heard by the boy = Time taken by the stone to reach the pond + 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^！ = Distance travelled / Speed
500 / 340 = 1.47 seconds
Therefore,
Time taken by the splash to be heard by the boy = 10 + 1.47 seconds = 11.47 seconds

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