NCERT Solutions Class 9 Science Chapter 12

NCERT Solutions for Class 9 Science Chapter 12: Sounds

Science is a fascinating subject that demands a higher understanding level and it opens a lot of opportunities for the students who wish to make a career in the field of Engineering and Sciences. Class 9 is a  crucial  stage for the students as it prepares them for the higher  classes and entrance exams.. Therefore, students need to grasp each concept with examples and exercises for an in-depth understanding. 

Class 9 Science Chapter 12: Sounds quenches the students curiosity on sound-related questions. The chapter deals with concepts like sound production with the help of an activity using a tuning fork.   The illustration with the help of the bell jar experiment justifies  the fact that sound needs a medium to travel. In addition, students will learn the properties and characteristics of sound waves, which are also discussed comprehensively. Further, they  will learn the difference between longitudinal waves and transverse waves. 

NCERT Solutions for Class 9 Chapter 12 are available on  Extramarks’ website for  students. It is the best way to enhance one’s conceptual knowledge of the topic. It contains answers to the questions provided in the NCERT textbook so that students can understand easily and  quickly.  . Our NCERT solutions are a reference guide to clarify topic-related doubts. The solutions contain several long, short answers along with MCQs and important questions answered by the subject matter experts. 

Extramarks is  one of the leading online platforms and is most trusted by the lakhs of students. The chapter sound is important  from the examination point of view.. Students can go through our NCERT solutions Class 9 Science Chapter 12 for self-evaluation after studying each chapter. With this help, students will get to know their shortcomings, and need a little more practise  to overcome them before the examination. 

Students can visit our website for the latest information and exam-related news on the NCERT syllabus. Further, students can also refer to  NCERT Solutions Class 10, NCERT Solutions Class 11, and NCERT Solutions Class 12.

Key Topics Covered In NCERT Solutions for Class 9 Science Chapter 12:

NCERT solutions for Class 9 Science Chapter 12 explains new terms and definitions related to sound. Here, students will learn about sound production with the help of activity using a tuning fork.  They will also learn about the characteristics of a sound like pitch, loudness, softness, tone, and noise. It also discusses the values of the speed of sound in different media and reverberations. 

Some of the key topics featured in NCERT Solutions for Class 9 Science Chapter 12 are: 

1. Production of Sound

A ringing bell or thunderclap is one noise, while laughter or rock music is another. All sounds are waves, and therefore, all sounds are the same. Let’s examine how we can apply wave qualities to sound.

A sound is a form of energy transmitted as waves to our ears and for example, when we speak, our vocal cords vibrate. The  string of a guitar vibrates back and forth to produce sound. A tuning fork vibrates and makes a sound. A rattle produces a sound when shaken. Sound waves cannot travel in a vacuum, so they must travel through a medium.

Because sound waves vibrate your ears, you can hear them when they reach your ears. Your brain then relays the vibrations to you via nerves. The brain then converts the messages into sound. Students can refer to NCERT Solutions for Class 9 Science Chapter 12 to understand the production of sound and its fundamentals concepts. 

2. Propagation of Sound

A longitudinal wave is a wave motion where the particles of the medium oscillate around their mean positions in the wave propagation direction.

The most common sound wave type is the longitudinal one.

Let’s take a look at the way sound waves travel. Use a tuning fork to shake. Now, focus on prong A. Figure (a) shows the tuning fork’s normal position and the initial state of air particles. As prong A moves to the right, compressed air particles are generated, as shown in figure (b). Vibrating air layers cause the compression to continue forward.

As prong A returns to its original position, the pressure on its right decreases, creating a rarefaction. This rarefaction is a disturbance that travels forward like compression. As a result, calculating sound propagation can be difficult. Thus, students can solve examples in NCERT Solutions for Class 9 Science Chapter 12. 

3. Reflection of Sound

Sound waves bounce back when they collide with solid or liquid surfaces. Sound waves are subject to the same rules as light waves. Sound waves must reflect on a large surface or obstruction to do so. For example, the rolling of thunder is caused by successive reflections from terrain and clouds. As per the rule of sound reflection, the directions in which sound is incident and reflected would make equal angles with the normal to the reflecting surface. And all the three would lie in the same plane.

There are three applications of the principle of reflection elaborated in NCERT Solutions for Class 9 Science Chapter 12 are:

  • Megaphone: A megaphone refers to a tube that looks like a horn. Consecutive reflections keep sound waves from spreading outside the box, limiting them to the air inside.
  • Hearing Aid : A hearing aid is a device that helps people  who are hard of hearing. The hearing aid’s sound waves are reflected in a smaller area that leads to the ear.
  • Soundboards: Curved surfaces can reflect sound waves. This reflection of sound waves can be used to distribute them around an auditorium evenly. Soundboards are used to reflect sound waves at the source. The speaker’s position is the excellent board’s focal point.

4. Range of hearing 

Vibrating sources emit sound waves that are then carried through the air. The human ear can hear sound waves between 20 Hz to 20 kHz. This is known as the audible range. Ultrasonic waves, also known as ultrasound, are sound waves with frequencies higher than the audible range. Infrasonic sounds are sound waves that have frequencies below the hearing range.

Applications of ultrasound elaborated in NCERT Solutions for Class 9 Chapter 12 

  • It is used for medical diagnosis and treatment as well as surgical procedures.
  • Bats and porpoises use ultrasound to navigate in the dark and find food.
  • It is used to detect a defective foetus.
  • It is used to treat muscular pain.
  • Ultrasonography is a technique that uses ultrasonic waves and creates 3-dimensional images to locate eye tumours.
  • Ultrasound waves are also commonly used to clean spiral tubes and electronic components.
  • To find cracks and flaws in metal blocks, ultrasound is used to inspect them.

5. Sonar

One of the essential uses of sound reflection is oceanographic research. This is done using SONAR. In addition, SONAR technology can detect submerged ships, sunken ships, and icebergs not visible from the surface. Sonar sends ultrasonic waves to the boat in all directions, which are then reflected. To understand the concept of SONAR, students can refer to NCERT Solutions for Class 9 Chapter 12. 

6. Structure of the human ear 

Structure of the Human Ear – The human ear has a sensitive apparatus that allows us to hear. The three main ear sections are the outer, middle, and inner. Further, each component of an ear is responsible in its way for decoding and interpreting sound. For example, Pinna refers to the outer ear. It transports sound from the middle ear to the auditory canal.

The eardrum (also known as the Tympanic Membrane) is a thin membrane located at the terminus of the auditory channel. The eardrum pushes inwardly and outwardly as the rarefaction or compression reaches it. This causes the eardrum to vibrate. These vibrations of the eardrum are amplified by the three bones of the middle ear: the hammer(malleus), the anvil (incus), , and the stirrup(stapes).

The middle ear transmits these vibrations to the inner ear. The inner ear converts pressure fluctuations and vibrations into electrical signals. In addition, the auditory nerve sends these electrical signals to the brain. The brain interprets them as sound.

Students can refer to NCERT Solutions for Class 9 Chapter 12 to further study the functions and structure  of the human ear. 

NCERT Solutions for Class 9 Science Chapter 12: Exercise &  Solutions

NCERT Solutions for Class 9 Chapter 12 Students can go through  Extramarks NCERT solutions to get answers to the questions that they need to practise before exams.  The solutions are prepared by experienced subject matter experts  per the latest CBSE guidelines. It helps students to improve their grades considering that Science is a scoring subject.. Our solutions guide can help students revise the complete syllabus by understanding the key topics and  get excellent results in the examination.. 

Chapter 12 Class 9 Science deals with concepts like sound production, characteristics of sound waves, and concepts of frequency. Students will learn about the difference between longitudinal and transverse waves. In addition, they  will learn about the formula for the period of the sound wave length with its SI unit. It also discusses the characteristics of a sound like pitch, loudness, and softness. 

Students can refer below for specific exercise, examples  and solutions:

  • Class 9 Science Chapter 12: Exercise 12.1 – Examples and Solutions
  • Class 9 Science Chapter 12: Exercise 12.2 – Examples and Solutions
  • Class 9 Science Chapter 12: Exercise 12.3 – Examples and Solutions
  • Class 9 Science Chapter 12: Exercise 12.4 – Examples and Solutions
  • Class 9 Science Chapter 12: Exercise 12.5 – Examples and Solutions

 besides  this, students may  also refer to  solutions for primary and secondary Classes from 1 to 12:

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT Solutions Class 10
  • NCERT Solutions Class 11
  • NCERT Solutions Class 12

NCERT Exemplar for Class 9 Science:

Science is a subject that requires a much deeper understanding to comprehend the concepts. Therefore, solving  questions with varying degrees of difficulty  level  will help students remember the definitions, formulas, concepts  more precisely. In addition, they can evaluate their preparation strategies to master the subject. 

The NCERT exemplar is designed especially for competitive exams; it comprises  MCQs, new concepts, and skill-developing questions. It will help students generalise the concepts and expose them to multiple concept-based questions. The Exemplar prepares  students  for problems of higher aptitude level and focuses more on applying concepts learned in class. Students can refer to NCERT exemplar textbooks and start their preparation. NCERT books won’t be enough to get an outstanding result, you need to supplement your preparation with NCERT solutions followed by exemplar to maximise your potential and ace the exam with flying colours. 

Some of the questions from NCERT Solutions for Class 9 Chapter 12 are taken and provided in the  exemplar. It has detailed explanations and answers. The Exemplar covers basics on all topics for students besides advanced level information.  

Key Features of NCERT Solutions for Class 9 Science Chapter 12:

Class 9 Science NCERT Solutions Chapter 12 includes all essential topics mentioned earlier in this article such as the production of sound, characteristics of sound waves, and longitudinal waves. In addition, the students will learn more about the sound wave’s frequency, amplitude, and speed. Sound also includes concepts like the reflection of sound, echo, reverberations, and uses of multiple reviews of sounds, ultrasound, and sonar. 

These are the key concepts from NCERT Solutions for Class 9 Science Chapter 12: 

  • Students can refer to solutions and learn the main topics and sub topics  with an in-depth understanding and conceptual clarity.  
  •  To solve the problems using different methods are provided.
  • The problems are solved using their expertise,  skills, and thematic understanding of the chapter.
  • Solutions can be used as a reference for further study, review, and practice examples and in text and end text exercises to assess their understanding  of the chapter.They can cross check their answers from solutions and gauge their preparation level.

Q.1 What is sound and how is it produced?

Ans-

Vibration in the particles causes sound. When a body vibrates, it forces the nearby particles of the medium to vibrate. Hence, a disturbance is created in the medium, which travels in the form of waves. This disturbance is called sound.

Q.2 Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.

Ans-

When a body vibrates, it creates a region of high pressure and low pressure in its surrounding. These regions of high pressure and low pressure are called compressions and rarefactions. As the body continues to move forward and backward, it produces a series of compressions and rarefactions.

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

Ans-

For this, take an electric bell and hang it inside an empty bell-jar fitted with a vacuum pump. Now, start removing the air from the bell-jar with the help of a vacuum pump. It can be observed that the sound of the ringing bell decreases. On further pumping off air, vacuum will be created inside the jar. At this moment, no sound can be heard from the ringing bell although one can see that the prong of the bell is still vibrating. This experiment shows that sound cannot travel through vacuum. Hence, sound requires a material medium for its propagation.

Q.4 Why is sound wave called a longitudinal wave?

Ans-

The vibration of the medium that moves along the direction of the wave is known as longitudinal wave. In the case of sound wave, the particles of the medium vibrate in the direction parallel to the direction of the propagation of disturbance. Therefore, a sound wave is called a longitudinal wave.

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

Ans-

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

Q.6 Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Ans-

The speed of sound is very less as compared to the speed of light. Hence, sound of thunder takes more time to reach the Earth as compared to light. Therefore, a flash is seen before than the thunder is heard.

Q.7 A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s–1.

Ans-

Given,  v 1  = 20 Hz,  v 2  = 20 kHz = 20000 Hz           Speed of sound in air, v = 344 ms -1 On using the relation,       v = v× λ or,   λ =  v v Let, the corresponding wavelengths for frequencies 20  Hz and 20 kHz are λ 1  and  λ 2 . ∴       λ 1  =  v v 1  =  344 20  = 17.2 m          λ 2  =  v v 2 =  344 20000  = 0.0172 m MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8srps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@57CE@

Q.8 Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.

Ans-

Consider the length of the aluminium rod = l. Speed of sound wave in aluminium at 25 °C, v 1 =6420 ms -1 Hence, time taken by the sound wave to reach the other end, t 1 = d v 1 = d 6420 … (i) Speed of sound in air at 25 °C, v 2 = 346 ms -1 Hence, the time taken by sound wave to reach the other end, t 2 = d v 2 = d 346 … (ii) From (i) and (ii), t 2 t 1 = d 346 d 6420 = 6420 346 =18.55 . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeaabaWaaqaafaaakq aabeqaaiaaboeacaqGVbGaaeOBaiaabohacaqGPbGaaeizaiaabwga caqGYbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGSbGaaeyzai aab6gacaqGNbGaaeiDaiaabIgacaqGGaGaae4BaiaabAgacaqGGaGa aeiDaiaabIgacaqGLbGaaeiiaiaabggacaqGSbGaaeyDaiaab2gaca qGPbGaaeOBaiaabMgacaqG1bGaaeyBaiaabccacaqGYbGaae4Baiaa bsgacaqGGaGaaeypaiaabccacaqGSbGaaeOlaaqaaiaabofacaqGWb GaaeyzaiaabwgacaqGKbGaaeiiaiaab+gacaqGMbGaaeiiaiaaboha caqGVbGaaeyDaiaab6gacaqGKbGaaeiiaiaabEhacaqGHbGaaeODai aabwgacaqGGaGaaeyAaiaab6gacaqGGaGaaeyyaiaabYgacaqG1bGa aeyBaiaabMgacaqGUbGaaeyAaiaabwhacaqGTbGaaeiiaiaabggaca qG0bGaaeiiaiaabkdacaqG1aGaaeiiaiaabclacaqGdbGaaeilaaqa aiaadAhadaWgaaWcbaGaaGymaaqabaGccqGH9aqpcaaI2aGaaGinai aaikdacaaIWaGaaeiiaiaab2gacaqGZbWaaWbaaSqabeaacaqGTaGa aeymaaaaaOqaaiaadIeacaWGLbGaamOBaiaadogacaWGLbGaaiilai aabccacaqG0bGaaeyAaiaab2gacaqGLbGaaeiiaiaabshacaqGHbGa ae4AaiaabwgacaqGUbGaaeiiaiaabkgacaqG5bGaaeiiaiaabshaca qGObGaaeyzaiaabccacaqGZbGaae4BaiaabwhacaqGUbGaaeizaiaa bccacaqG3bGaaeyyaiaabAhacaqGLbGaaeiiaiaabshacaqGVbGaae iiaiaabkhacaqGLbGaaeyyaiaabogacaqGObaabaGaaeiDaiaabIga caqGLbGaaeiiaiaab+gacaqG0bGaaeiAaiaabwgacaqGYbGaaeiiai aabwgacaqGUbGaaeizaiaabYcacaqGGaGaaeiDamaaBaaaleaacaaI Xaaabeaakiabg2da9maalaaabaGaamizaaqaaiaadAhadaWgaaWcba GaaGymaaqabaaaaOGaeyypa0ZaaSaaaeaacaWGKbaabaGaaGOnaiaa isdacaaIYaGaaGimaaaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiai aabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGa aeiiaiaabccacaqGUaGaaeOlaiaab6cacaqGGaGaaeikaiaabMgaca qGPaaabaGaae4uaiaabchacaqGLbGaaeyzaiaabsgacaqGGaGaae4B aiaabAgacaqGGaGaae4Caiaab+gacaqG1bGaaeOBaiaabsgacaqGGa GaaeyAaiaab6gacaqGGaGaaeyyaiaabMgacaqGYbGaaeiiaiaabgga caqG0bGaaeiiaiaabkdacaqG1aGaaeiiaiaabclacaqGdbGaaeilaa qaaiaadAhadaWgaaWcbaGaaGOmaaqabaGccqGH9aqpcaqGGaGaae4m aiaabsdacaqG2aGaaeiiaiaab2gacaqGZbWaaWbaaSqabeaacaqGTa GaaeymaaaaaOqaaiaadIeacaWGLbGaamOBaiaadogacaWGLbGaaiil aiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiDaiaabMgacaqGTb GaaeyzaiaabccacaqG0bGaaeyyaiaabUgacaqGLbGaaeOBaiaabcca caqGIbGaaeyEaiaabccacaqGZbGaae4BaiaabwhacaqGUbGaaeizai aabccacaqG3bGaaeyyaiaabAhacaqGLbGaaeiiaiaabshacaqGVbGa aeiiaiaabkhacaqGLbGaaeyyaiaabogacaqGObGaaeiiaiaabshaca qGObGaaeyzaaqaaiaab+gacaqG0bGaaeiAaiaabwgacaqGYbGaaeii aiaabwgacaqGUbGaaeizaiaabYcacaqGGaGaaeiDamaaBaaaleaaca aIYaaabeaakiabg2da9maalaaabaGaamizaaqaaiaadAhadaWgaaWc baGaaGOmaaqabaaaaOGaeyypa0ZaaSaaaeaacaWGKbaabaGaae4mai aabsdacaqG2aaaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeii aiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGa GaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabcca caqGGaGaaeiiaiaab6cacaqGUaGaaeOlaiaabccacaqGOaGaaeyAai aabMgacaqGPaaabaGaaeOraiaabkhacaqGVbGaaeyBaiaabccacaqG OaGaaeyAaiaabMcacaqGGaGaaeyyaiaab6gacaqGKbGaaeiiaiaabI cacaqGPbGaaeyAaiaabMcacaqGSaaabaGaaeiiaiaabccacaqGGaGa aeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccaca qGGaGaaeiiaiaabccacaqGGaWaaSaaaeaacaqG0bWaaSbaaSqaaiaa bkdaaeqaaaGcbaGaaeiDamaaBaaaleaacaqGXaaabeaaaaGccqGH9a qpdaWcaaqaamaalaaabaGaamizaaqaaiaaiodacaaI0aGaaGOnaaaa aeaadaWcaaqaaiaadsgaaeaacaaI2aGaaGinaiaaikdacaaIWaaaaa aacqGH9aqpdaWcaaqaaiaaiAdacaaI0aGaaGOmaiaaicdaaeaacaaI ZaGaaGinaiaaiAdaaaGaeyypa0JaaGymaiaaiIdacaGGUaGaaGynai aaiwdacaqGGaGaaeOlaaaaaa@8128@

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

Ans-

Given, Frequency of sound = 100 Hz          Total time = 1 min = 60 s On using the relation, Frequency = Number of oscillations Total time Here, Number of oscillations = Frequency × Total time                                     = 100 Hz × 60 s = 6000  Thus, the source vibrate 6000 times in a minute. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0le9yqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6D92@

Q.10 Does sound follow the same law of reflection as light does? Explain

Ans-

Yes, the incident sound wave and the reflected sound wave make the same angle with the normal to the surface at the point of incidence. Moreover, the incident sound wave, the reflected sound wave, and the normal to the point of incidence all lie in the same plane.

Q.11 When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?

Ans-

An echo can be heard when the time interval between the original sound and the reflected sound is at least 0.1 s. The speed of sound in a medium increases with an increase in temperature. Therefore, echo cannot be heard on a summer day as the time interval between the original sound and the reflected sound decreases.

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

Ans-

(i) Sound board: It is used to send the sound towards audience in a big hall or auditorium. This works on the basis of laws of reflection of sound waves.

(ii) Stethoscope: It is also based on reflection of sound. In a stethoscope, the sound of a patient’s heartbeat reaches to the doctor’s ear by multiple reflection of sound.

Q.13 A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s–2 and speed of sound = 340 m s–1.

Ans-

Given, h = 500 m, g = 10 ms -2 , Speed of sound, v = 340 ms -1 The intial velocity (u) of the stone when it is dropped is zero, i.e., u = 0 Let, it takes (t) time to reach on the surface of water On using the relation, s = ut + 1 2 gt 2 ∴ 500 = 0+ 0 .5 ×10×t 2 or, 100 = t 2 or, t = 10 s Now, let time taken by the sound to reach at the top the tower be t’ then, h =v×t’ or, 500 = 340×t’ or, t’ =1.47 s ∴ Total time = t+t’ = 10+1.47 = 11.47 s. Hence, after 11.47 s later, the splash will be heard at the top of the tower. 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Q.14 A sound wave travels at a speed of 339 m s–1. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Ans-

Given, Speed of sound, v = 339 ms -1 Wave length, λ = 1.5 cm = 0.015 m On using the relation, v = λ × v ∴ v = v λ = 339 0.015 = 22600 Hz As the frequency range of audible sound for humans lies between 20 Hz to 20000 Hz. Since, the frequency of the given sound is more than 20000 Hz, it is not audible. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@3490@

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

Ans-

The reverberation is the persistence of sound due to repeated reflection. When the source produces sound, it starts moving in all the directions. It is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Thus, sound can be heard even after the source has stopped producing the sound.

To minimise reverberation, sound should be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibre board, heavy curtains, and cushioned seats can be used to reduce reverberation.

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

Ans-

Loudness depends on the amplitude of the sound. The louder a sound, the more energy it has. Loudness is proportional to the square of the amplitude of vibrations.

Q.17 Explain how bats use ultrasound to catch a prey.

Ans-

Bats generate high-pitched ultrasonic shrill. These high-pitched shrill are reflected by objects such as preys and comes back to the bat’s ear. Thus, a bat can easily determine the distance of a prey to catch it.

Q.18 How is ultrasound used for cleaning?

Ans-

To clean an object when it is put in a cleaning solution, the ultrasonic sound waves are passed through the solution. The high frequency of these ultrasound waves detaches the dirt from the objects.

Q.19 Explain the working and application of a sonar.

Ans-

It is a device that uses ultrasound propagation to navigate, communicate or detect underwater objects such as submarine, a sunken ship, an iceberg, etc. A SONAR apparatus consists of two parts:

  1. A transmitter (for emitting ultrasonic waves).
  2. A receiver (for detecting the reflected ultrasonic waves).

Both these parts are installed in a ship or a boat. The transmitter sends ultrasonic waves towards the ocean floor.

These waves when reflected back by an object or the ocean floor in form of ‘echo’ are detected by a detector.

Depth of an object can be calculated using time interval between generation of wave and reception of its echo and the speed of sound in water. This method is known as sonar.

Q.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, Time taken to hear to echo, t = 5 s Distance of the object from the submarine, s = 3625 m ∴ Total distance travelled = 2 s ∴ Velocity, v = 2 s t = 2 × 3625 m 5 = 1450 ms −1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXafv3ySLgzGmvETj2BSbqeeuuDJXwAKbsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@D30A@

Q.21 Explain how defects in a metal block can be detected using ultrasound.

Ans-

The ultrasounds cannot pass through a defective metal block but they are reflected back. This fact is used to detect defects in metal blocks. Ultrasound is passed through one end of a metal block and detectors are placed on the other end. Thus, defects in metal blocks can be detected by using ultrasound.

Q.22 Explain how the human ear works.

Ans-

Pinna collects different sounds produced in our surroundings and sends these sounds to the ear drum through ear canal. At the end of ear canal, there is a thin membrane called the eardrum or tympanic membrane. When the compression of sound waves strikes the eardrum, it is pushed inward. When the rarefaction of sound waves strikes the ear drum, it moves outward. Thus, eardrum vibrates. Middle ear consists of three bones – hammer, anvil and stirrup. These bones amplify the vibrations from the eardrum and transmit these vibrations to the inner ear. Inner ear consists of cochlea that converts vibrations or pressure variations into electrical signals. These electrical signals sent to brain via auditory nerve. Brain interprets them as sound.

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FAQs (Frequently Asked Questions)

1. Which are the essential topics covered in NCERT Solutions for Class 9 Chapter 12?

The chapter sound includes essential sub- topics such as reflection of sound, echo, reverberations, uses of multiple sound reviews, ultrasound, and sonar. The main topics are listed below: :

  • Production of Sound
  • Propagation of Sound
  • Reflection of Sound
  • Range of Hearing
  • Applications of Ultrasound
  • Sonar
  • Structure of Human Ear

2. What is sound, and how is it produced?

The vibration of objects produces sound, which is a form of energy. Vibration is the rapid to-and-fro movement of an object. According to the theory, vibration is caused by particles causing a disturbance in the medium. For example, a rubber band can vibrate and produce sound by being stretched. The vibrating body can travel through the medium, but the particles cannot move forward. This is how sound is created.

3. What is the application of SONAR?

SONAR stands for sound, navigation, and range. SONAR is an acoustic device that measures underwater objects’ speed, direction, and depth. Submarines, for example, use SONAR to communicate or detect   different objects like submarines, sunken ships etc.  underwater. It produces an ultrasonic sound and is designed to travel below the seawater. The transducer transmits this sound. The detector records the echo produced when an object reflects it. The sensor converts the detected signal into electrical signals that can be used for measurement.