NCERT Solutions Class 9 Science Chapter 12 Sound
NCERT Solutions Class 9 Science Chapter 12 Sound: National Council of Educational Research and Training Class 9 Science Chapter 12 Solutions – Sound. NCERT Solutions Class 9 Science Chapter 12 PDF Download.
NCERT Solutions Class 9 Science Chapter 12: Overview
Board |
NCERT |
Class |
9 |
Subject |
Science |
Chapter |
12 |
Chapter Name |
Sound |
Topic |
Exercise Solutions |
NCERT Solutions Class 9 Science Chapter 12 – Sound
Part 1
1.) How does the sound produced by a vibrating object in a medium reach your ear?
Answer:
- The sound is the longitudinal wave which is produced by the vibrating object and sound waves required material medium for their propagation. The medium through which they travel it may solid, liquid or gas.
- When a object is vibrating that means the particles of the medium are also vibrating. The particles directly not transmitted to the ear but these vibrations are transferred from one particle to other without actual migration of particles of the medium and finally sound will reach to our ear
- In detail, the particles of the medium which are near to vibrating object are vibrating and hence they get displaced from their mean position so that they exerts the some force on the adjacent particles due to which again the adjacent particles get displaced from their mean position.
- After displacing the adjacent particles the particles which are most near to vibrating object get displaced to their original position again.
- In this way the process of transferring sound or vibrations continues till the vibrations reach to our ear and hence we hear some sound.
Part 2
1.) Explain how sound is produced by your school bell.
Answer:
- In our school, bell produces sound when it is set into vibrations by hammering on it.
- When we hammer on the bell pad it produces vibrations and these vibrations are carried out by the particles of the air medium which are very close to it.
- These particles get displaced from their equilibrium position and they exerts some force on the adjacent particles so that the adjacent particles get displaced again from their mean position.
- And the initial particles displaced reach to their mean position again. In this way the particles of air medium transmits these vibrations produced by the bell and the process of transmitting continues till the sound will reach to our ear.
- And finally we listen the sound of bell.
2.) Why are sound waves called mechanical waves?
Answer:
- Sound waves are called as mechanical waves because they required material medium for their propagation. The material medium may be solid, liquid or gas.
- And sound waves cannot pass through vacuum.
- Sound waves are the longitudinal waves in which particles are vibrating parallel to the direction of propagation.
3.) Suppose you and your friend are on the moon. Will you be able to hear any sound produced by your friend?
Answer:
We all know that, on moon there is no medium and there vacuum is present everywhere.
And we already know that, sound waves requires material medium for their propagation and does not travel through vacuum.
So if I and my friend on moon then also we are not able to hear the sound produced by us.
Part 3
1.) Which wave property determines (a) loudness, (b) pitch?
Answer:
a)
Loudness:
- The loudness of the sound is determined from the amplitude of the sound waves.
- The maximum displacement of the particles of sound wave from its mean position is its amplitude.
- As amplitude is directly proportional to the loudness, more is the amplitude greater is the loudness of sound produced.
b)
Pitch:
- The pitch of the wave is determined from the frequency of the sound wave. Also the pitch of the wave is directly proportional to the frequency of the sound wave.
- As more is the frequency of the sound wave greater is the pitch produced and lower is the frequency of the sound wave lower is the pitch produced.
2.) Guess which sound has a higher pitch: guitar or car horn?
Answer:
- As we know that, pitch of the sound wave depends on the frequency of the sound wave. If the sound wave has higher frequency greater will be the pitch produced.
- The sound produced by guitar has greater frequency than the sound produced by car horn.
- Hence, the sound produced by guitar has high pitch due to higher frequency.
Part 4
1.) What are wavelength, frequency, time period and amplitude of a sound wave?
Answer:
Wavelength of sound wave:
- The distance between two successive compressions or rarefactions is called as the wavelength of the sound wave.
- It is denoted by lambda and has unit as meter.
Frequency of the sound wave:
- The number of oscillations produced per unit time is called as the frequency of the sound wave.
- It is measured in Hz.
Time period of sound wave:
- The time required by the two consecutive compressions or rarefactions to cross a fixed point is called as the time period of sound wave.
- It is denoted by T.
- It is measured in seconds.
Amplitude of sound wave:
- The maximum displacement of the particles of the medium from their mean position is called as the amplitude of the wave.
- It is denoted by a and measured in meter.
2.) How are the wavelength and frequency of a sound wave related to its speed?
Answer:
The wavelength and frequency of the sound wave are related to speed of sound wave by the following relation.
Speed of the sound wave = frequency of sound wave*wavelength of the sound wave
Thus, v = n*wavelength
3.) Calculate the wavelength of a sound wave whose frequency is220 Hz and speed is 440 m/s ina given medium.
Answer:
Given that,
Frequency of sound wave = 220Hz
Speed of sound wave= 440m/s
We know that,
V= n × wavelength
Thus, wavelength of the sound wave= v/n = 440/220= 2 m
4.) A person is listening to a tone of 500 Hz sitting at a distance of450 m from the source of thesound. What is the time intervalbetween successive compressionsfrom the source?
Answer:
Given that,
Frequency= 500Hz
Wavelength = 450m
We know that,
Time period = 1/f = 1/500= 0.002 seconds
Part 5
1.) Distinguish between loudness and intensity of sound.
Answer:
- The amount of sound energy which is passing each second through unit area is called as intensity of sound wave.
- Sometimes intensity of sound wave is called as loudness but they are not the same.
- Loudness of sound is decided from the response given by our ear to the sound.
- Sometimes two sounds will be produced with same intensity but we can’t hear them with same intensity while we listen one sound more clearly and other with less clarity.
- It means that loudness is depending on our response given by ear.
Part 6
1.) In which of the three media, air, water or iron, does soundtravel the fastest at a particular temperature?
Answer:
As we know that, the speed of sound depends on the nature of the medium. The speed of sound is fastest in solids , slower in liquid and most slowly in gases.
So we can say that , speed of sound is faster in iron than in water or air.
Part 7
1.) An echo is heard in 3 s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?
Answer:
Given that,
Time = 3s
Speed of sound = 342m/s
We know that,
Distance travelled by sound= speed*time = 342*3= 1026m
But this is the total distance travelled by the sound.
Hence, distance of reflecting surface from the source will be 1026/2= 513m
Part 8
1.) Why are the ceilings of concert halls curved?
Answer:
The ceilings of concert halls are made curved because sound produced has to reach the every reach to the every corner of hall after reflection. Hence each person siting in the hall hear sound clearly with no difficulty.
Part 9
1.) What is the audible range of the average human ear?
Answer:
The audible range of the average human ear is from 20Hz to 20000 Hz.
2.) What is the range of frequencies associated with
(a) Infrasound?
(b) Ultrasound?
Answer:
a) The sound with frequency below 20Hz are called as infrasonic sounds.
b) The sounds with frequency greater than 20kHz are called as ultrasound.
Part 10
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:
Given that,
Time= 1.02s
Speed of sound in salt water = 1531m/s
We know that,
Distance= speed*time = 1531*1.02= 1561.62 m
Thus the total distance covered by the sound will be 1561.62m
Hence, the distance of cliff = 1561.62/2= 781.80 m
Exercise Solution:
1.) What is sound and how is it produced?
Answer:
- Sound is the mechanical wave and these are the vibrations produced in a object.
- Sound waves requires material medium for their propagation and hence sound waves cannot travels through vacuum.
- The velocity of sound in solids is greater than liquids and gases.
- When an object is set into vibrations then the particles closest to it get this vibrational energy. And these particles are also set into vibrations. Due to which they exerts some force on the neighbouring particles and returns to their original position.
- After that the neighbouring particles also get displaced and the process of transferring energy in the form of vibrations till continues and finally we hear the sound.
In this way, sound waves are produced due to transfer of vibrations produced from particle to another of the medium.
2.) Describe with the help of adiagram, how compressions andrarefactions are produced in airnear a source of sound.
Answer:
- The following diagram shows the production of sound in air forming compressions and rarefactions.
- When the vibrating object is moving forward, it compresses the air as more pressure is developed at that part and hence creates the compression.
- Compression is the region where there is high pressure created as shown in figure.
- The compression produced starts to move away from the vibrating object and hence vibrating object moves backwards.
- As the vibrating object is moving backwards it creates the region of low pressure which is called as rarefaction as shown in figure.
- As the vibrating object is moving continuously back and forth hence continuously compressions and rarefactions are produced in air medium as shown in figure.
- These successive compressions and rarefactions make the sound wave which propagates through the air medium in the form of alternate compressions and rarefactions as shown in figure.
- Thus sound waves travels in the form of alternate compressions and rarefactions through the medium.
3.) Cite an experiment to show thatsound needs a materialmedium for its propagation.
Answer:
- As we know that, sound is the mechanical wave which requires material medium for its propagation.
- The material medium may be solid, liquid or gas.
- The speed of propagation of sound is different in different media.
- To prove sound does not travel through vacuum that means it requires material medium for its propagation we perform an experiment called as bell jar experiment as follows.
- We have taken an electric bell and also the airtight glass bell jar. Now we have suspended the electric bell inside the airtight glass bell jar as shown in figure.
- And the glass bell jar is connected to vacuum pump and if we make switch on then we hear sound of electric bell jar clearly. Now we started the vacuum pump so that there will vacuum is created inside the glass bell jar.
- Due to starring the vacuum pump the air from inside the bell jar is pumping out and in that manner we are listening the sound of decreasing intensity.
- After sometime we observed that there is no sound is hearing although the same current is provided to the electric bell jar
- That means, after sometime there will be no air present inside the glass bell jar that means totally vacuum will be created and due to which we can’t hear the sound of electric bell.
- This experiment proves that, sound wave does not travels through vacuum and it requires material medium for its propagation.
4.) Why is sound wave called alongitudinal wave?
Answer:
Sound waves are called as longitudinal waves because in sound waves the particles of the medium are vibrating parallel to the direction of propagation of the sound wave.
And we already know that, the longitudinal waves are those in which particles of the medium are vibrating parallel to the direction of propagation of wave.
In case you are missed :- Previous Chapter Solution
5.) Which characteristic of thesound helps you to identify yourfriend by his voice while sitting withothers in a dark room?
Answer:
The characteristics properties is sound like pitch of sound and timber of sound helps in identifying the different sounds produced at the same time.
Because, each sound produced has characteristics pitch and timber so that we can identify the sound of our friend in dark room also.
Due to this unique property of pitch and timber of sound produced we can differentiate the sound produced easily.
6.) Flash and thunder are producedsimultaneously. But thunderis heard a few seconds after the flashis seen, why?
Answer:
Although flash and thunder are produced simultaneously in air but we can listen thunder after sometime while we see flash at that time.
Because, the speed of light in air is greater then the speed of sound in air. Due to which the flash reach to our eyes early while due to less speed of sound in air we heard the thunder after sometime or after few seconds.
From this, we may say that sound speed of light in air is greater than speed of sound in air.
7). A person has a hearing rangefrom 20 Hz to 20 kHz. What are the typical wavelengths of sound wavesin air correspondingto these two frequencies? Take the speed ofsound in air as344 m/s.
Answer:
Given that,
Frequency range= 20Hz to 20kHz
Speed of sound = 344m/s
We know that, speed of sound = frequency × wavelength
Hence, wavelength= speed/frequency
When frequency= 20Hz
Wavelength = 344/20= 17.2 m
And when frequency= 20kHz= 20,000Hz
Wavelength = 344/20,000= 0.0172 m
Thus, human ear is audible for the sound waves with wavelength range 0.0172 m to 17.2m
8.) Two children are at opposite ends of an aluminiumrod. Onestrikes the end of the rod with a stone. Findthe ratio of timestaken by the sound wave in air andin aluminium to reach thesecond child.
Answer:
We know that, speed of sound in air = 346m/s
And speed of sound in aluminium rod = 6420m/s
Let us consider the length of rod will be ‘d’
We know that,
Speed = distance/ time
So, time taken by sound wave in air= d/346
And time taken by sound wave in rod = d/ 6420
Now, we have to find the ratio of time taken by sound in air and in aluminium.
Hence, required ratio= (d/346)/(d/6420) = 6420/346= 18.55:1
Thus, the required ration will be 18.55:1.
9.) The frequency of a source of sound is 100 Hz. Howmany times does it vibrate in a minute?
Answer:
We know that, the number of oscillations produced in one second is the frequency of sound wave.
So number of vibrations produced in a minute= 100 × 60= 6000 vibrations.
Thus, in a minute 6000 vibrations will be produced.
10.) Does sound follow the same laws of reflection as light does? Explain.
Answer:
We know the laws of reflection of light which are as follows:
- The incident ray, reflected ray and the normal at the point of incidence all lie in the same plane.
- And, the angle of incidence is always equal to the angle of reflection.
Thus, the similar laws are also followed by the sound waves which are given below:
- Incident sound wave, reflected sound wave and the normal at the point of incident sound wave all lies in the same plane.
- And, the angle of incidence is always equal to the angle of reflection which are made by sound waves with 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 productionremains the same. Do youhear echo sound on a hotter day?
Answer:
- As we know that, the velocity of sound depends on the temperature. More is the temperature greater is the velocity of sound.
- In the hotter day, the temperature will be more due to which the velocity of sound will be more. Since velocity of sound is directly proportional to the temperature of the medium through which it is traveling.
- And hence the time interval between the sound incident and reflected will be less.
- We know that, human ear can hear the echo with minimum time interval of 1/10th of a second.
- But, on hotter day, as the velocity of sound is greater hence the time interval between the sound incident and reflected will be greater or less than 0.1s.
- If it is greater than 0.1 s then we can hear the echo but if the time interval is less than 0.1s then we can’t hear the echo.
- But, on hotter day velocity of sound is more so time interval must be less than 0.1s hence we can’t hear echo sound.
- So all these consequences are due to the temperature dependence of velocity of sound.
12.) Give two practical applications of reflection ofsound waves.
Answer:
- In SONAR method the process of reflection of sound is used to measure the distance of objects which under water. By using which we can assume the distance of the aquatic animals from the surface of water and we can detect them easily.
- The working of stethoscope is also depending on the reflection of sound waves. When we check heartbeats using stethoscope then due sound produced which we heard.
13.) A stone is dropped from the top of a tower 500 mhigh into apond of water at the base of the tower. When is thesplashheard at the top? Given, g = 10 m/s² andspeed of sound =340 m/s.
Answer:
Given that,
Height of tower = 500m
Acceleration due to gravity g= 10m/S2
Speed of sound = 340m/s
Initial velocity of stone u= 0
Let, t’ be the time required to stone to reach the base of tower from the top of the tower.
We have,
S= ut + 1/2at2
500= 0 + ½ × 10× t²
500= 5× t²
Hence, t²= 500/5= 100
Taking square root on both side
t= 10s
This is the time taken by stone to reach the base of the tower from top of the tower.
When stone reach the base of tower it creates the sound in air whose speed is 340m/s.
Let t” be the time taken by sound to reach the top of tower from the base of the tower.
Then, speed = distance/time
Time t”= 500/340= 1.47s
Thus, the total time taken by stone to reach the base of tower and then to reach the top of tower will be = 10 + 1.47= 11.47 seconds.
14.) A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
Answer:
Given that,
Speed of sound = 339m/s
Wavelength = 1.5cm = 0.015m
We know that,
Speed = frequency*wavelength
Frequency= speed/ wavelength = 339/0.015= 22600Hz
And we know that, human ear is audible for sound with frequency 20 Hz to 20 kHz.
So here the sound produced is audible.
15.) What is reverberation? How can it be reduced?
Answer:
- In big halls or auditorium halls the sound produced will reflect continuously and the after many more reflections it reach to the receiver whose intensity is very small.
- Since as there are many reflection then intensity of sound obviously goes on decreasing. Due to which the sound reached after last reflection will be of small intensity which we can’t hear.
- This repeated reflections of sound which results in persistence of sound is called as reverberations.
- In big halls and auditorium the excessive reverberations produced are highly undesirable. To reduce this reverberations the roof of such halls is sound absorbing materials like compressed fibreboard, rough plaster or draperies.
- If the reverberations are not reduced then every person sitting in the hall can’t listen the sound properly due reverberations.
16.) What is loudness of sound? What factors does it depend on?
Answer:
- The amount of sound energy passing each second through a unit area is called as the intensity of sound or sometimes loudness.
- Loudness is decided from the response given by our ear to the intensity of sound produced.
- The loudness of the sound mainly depends on the amplitude of the sound wave. And it is directly proportional to the square of the amplitude of the sound wave.
- It means, greater is the amplitude of the sound wave produced greater is the loudness.
- We are listening a silent song on TV and something rally is going out from our house with huge sound then we say that the loudness of sound in rally is more than the song on TV. Due to which we can’t hear song properly.
- Thus, when the amplitude of the sound is more it’s loudness is also more.
17.) Explain how bats use ultrasound to catch a prey.
Answer:
- Bats are only those who has capacity to listen ultrasound. They produces ultrasonic waves which are incident on the prey and after reflection of this ultrasonic waves from prey will be received by the bats. And from which they find the prey and their distance from them.
- In this way, by using ultrasonic waves and reflection of sound waves bats catch the prey.
Bats are having ability of producing ultrasound and also they can hear the ultrasound which helps them to find the prey nearby them. Since they are not having sensation of vision.
18.) How is ultrasound used for cleanings
Answer:
- We know that, ultrasonic waves are having frequency of sound.
- While cleaning the objects they are put into the cleansing solutions and through that solution ultrasonic waves are passed and due to high frequency of ultrasonic waves it helps in removing the dirt from the objects easily.
- Thus, the high frequency ultrasonic waves are used to remove the dirt from the objects which is located at difficult pars we can’t remove it.
- In this way ultrasonic waves can also be used in cleaning.
19.) Explain the working and application of a sonar.
Answer:
Working of SONAR:
- The full form of SONAR is Sound Navigation And Ranging. It is the device which uses ultrasonic waves to measure the distance, direction and also the speed of underwater objects.
- It consist of transmitter and a detector both are installed in a boat or a ship. The transmitter is helps in producing and transmitting the ultrasonic waves.
- After production of ultrasonic waves, this waves are travels through water and strikes the object which is in water.
- After striking the ultrasonic waves from the object under water get reflected back and it is detected by the detector.
- The detector is used to convert ultrasonic waves into the electrical signals which are again appropriately interpreted.
- In this way, the distance of the object from which ultrasonic waves are reflected in water can be determined from the speed of ultrasonic waves in water and the time interval between transmission and reception of ultrasonic waves.
- If t is the time interval and v is the speed of ultrasonic waves in water then total distance covered will be 2d.
- So we can write, 2d= v × t
Applications of SONAR:
- This method is also called as echo ranging. And the sonar technique is mainly used to determine the depth of the sea and to locate the underwater hills, valleys, submarines, icebergs, sunken ships etc.
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.
Ans:
Given that,
Time = 5s
Distance = 3625 m
We know that, the sound wave produced travels the distance from point of production and then to the point of reception.
So total distance covered will be 2d.
Hence, speed of sound = 2d/ 5= 2 × 3625/5= 1450 m/s
Thus, the speed of sound in water will be 1450 m/s.
21.) Explain how defects in a metal block can be detected using ultrasound.
Ans:
- Ultrasound can be used to detect cracks and flaws in metal blocks also. The cracks or holes which are not visible and located inside the metal block decreases the strength of structure of metal.
- For constructing buildings we required metal blocks having greater strength so ultrasound can be used to detect this defects.
- In this process, ultrasonic waves are allowed to pass through the metal blocks and some detectors are placed which receives the transmitted ultrasonic waves from the metal block.
- If there is a small defect is present metal block then also the reflected wave shows the defect and it can be detected easily. And after this detection the metal blocks are used in construction of buildings to increase the strength.
- In this way, ultrasound is used in detecting defects in metal blocks in order to increase their strength.
22.) Explain how the human ear works.
Ans:
The following figure shows the human ear.
Human ear is most sensitive organ in the body. It responds to sound waves in audible range.
Working of human ear:
- Human ear consist of outer ear, inner ear and middle ear. The outer ear consist of pinna. The pinna is connected to 2 to 3 centimetre long canal.
- Canal is connected to ear drum. The middle ear is made up from bones named hammer, anvil and stirrup. Hammer is connected to ear drum and stirrup is connected to inner ear.
- The anvil is between hammer and stirrup. The middle ear has large empty part and contains air. The middle ear connect to through by a tube known as Eustachian tube.
- The structure of inner ear is like a coil named cochlea. It is made up from liquid. The construction of ear is as,
- The sound waves enter in ear by pinna. It reaches at ear drum through canal. As we know that, sound waves consist of compressions and rarefaction.
- In compressions density of molecules is high but in rarefaction the density of molecules is low. When compressions strike on ear drum then it pushes ear drum inward. Similarly ear drum pushes outward because of rarefaction.
- The ear drum get continuous vibrations due to sound waves falling on it. The vibrations are passed to cochlea through hammer bone, anvil and stirrup bone. The cochlea converts vibrations into electrical signals. This electrical signals reach to brain via nerves cells. In this way human eye works.
In case you are missed :- Next Chapter Solution