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CBSE Case Study Questions Class 11 Physics PDF Download

Are you a Class 11 Physics student looking to enhance your understanding and prepare effectively for your exams? Look no further! In this comprehensive guide, we present a curated collection of CBSE Case Study Questions Class 11 Physics that will help you grasp the core concepts of Physics while reinforcing your problem-solving skills.

CBSE 11th Standard CBSE Physics question papers, important notes, study materials, Previous Year Questions, Syllabus, and exam patterns. Free 11th Standard CBSE Physics books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 11 Physics Case Study Questions

CBSE Class 11 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 11 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 11 Physics

• Chapter 1: Physical World
• Chapter 2: Units and Measurements
• Chapter 3: Motion in a Straight Line
• Chapter 4: Motion in a Plane
• Chapter 5: Laws of Motion
• Chapter 6: Work, Energy, and Power
• Chapter 7: System of Particles and Rotational Motion
• Chapter 8: Gravitation
• Chapter 9: Mechanical Properties of Solids
• Chapter 10: Mechanical Properties of Fluids
• Chapter 11: Thermal Properties of Matter
• Chapter 12: Thermodynamics
• Chapter 13: Kinetic Theory
• Chapter 14: Oscillations
• Chapter 15: Waves

Class 11 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 11 Physics examinations. Our expert faculty for standard 11 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 11 students understand the concepts and also easy-to-learn solutions.

Class 11 Books for Boards

Why Case Study Questions Matter

Case study questions are an invaluable resource for Class 11 Physics students. Unlike traditional textbook exercises, these questions simulate real-life scenarios, challenging students to apply theoretical knowledge to practical situations. This approach fosters critical thinking and helps students build a deep understanding of the subject matter.

Let’s delve into the different topics covered in this collection of case study questions:

1. Motion and Gravitation

In this section, we explore questions related to motion, velocity, acceleration, and the force of gravity. These questions are designed to test your grasp of the fundamental principles governing motion and gravitation.

2. Work, Energy, and Power

This set of questions delves into the concepts of work, energy, and power. You will encounter scenarios that require you to calculate work done, potential and kinetic energy, and power in various contexts.

3. Mechanical Properties of Solids and Fluids

This section presents case study questions about the mechanical properties of solids and fluids. From stress and strain calculations to understanding the behavior of fluids in different situations, these questions cover a wide range of applications.

4. Thermodynamics

Thermodynamics can be a challenging topic, but fear not! This part of the guide offers case study questions that will clarify the laws of thermodynamics, heat transfer, and thermal expansion, among other concepts.

5. Oscillations and Waves

Get ready to explore questions related to oscillations, simple harmonic motion, and wave characteristics. These questions will deepen your understanding of wave propagation and the behavior of oscillatory systems.

6. Kinetic Theory and Laws of Motion

Kinetic theory and the laws of motion can be complex, but with our case study questions, you’ll find yourself mastering these topics effortlessly.

Discover a wide array of questions dealing with light, lenses, and mirrors. This section will improve your problem-solving skills in optics and enhance your ability to analyze optical phenomena.

8. Electrical Effects of Current

Electricity and circuits are fundamental to physics. The case study questions in this section will challenge you to apply Ohm’s law, Kirchhoff’s laws, and other principles in various electrical circuits.

9. Magnetic Effects of Current

Delve into the fascinating world of magnets and magnetic fields. This set of questions will strengthen your understanding of magnetic effects and their applications.

10. Electromagnetic Induction

The final section covers electromagnetic induction, Faraday’s law, and Lenz’s law. You’ll be presented with scenarios that test your ability to predict induced electromotive forces and analyze electromagnetic phenomena.

In conclusion, mastering Class 11 Physics requires a thorough understanding of fundamental concepts and their practical applications. The case study questions provided in this guide will undoubtedly assist you in achieving a deeper comprehension of the subject.

Remember, practice is key! Regularly attempt these case study questions to strengthen your problem-solving abilities and boost your confidence for the exams. Happy studying, and may you excel in your Physics journey!

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Class 11 Physics Case Study Questions

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Looking for complete and comprehensive case study questions for class 11 Physics? myCBSEguide is just a click away! With extensive study materials, sample papers, case study questions and mock tests, myCBSEguide is your one-stop solution for class 11 Physics exam preparation needs. So, what are you waiting for? Log on to myCBSEguide and get started today!

What is the purpose of physics?

Physics is the study of the fundamental principles governing the natural world. It is a vital part of the scientific enterprise, providing the foundation on which other sciences are built. Physics is essential for understanding how the world works, from the smallest particles to the largest structures in the Universe. In class 11 Physics, students are introduced to the basic concepts of physics and learn about the fundamental principles governing the natural world. Class 11 Physics concepts are essential for understanding the world around us and for further study in physics and other sciences.

What are case study questions in physics?

In physics, case study questions are intended to evaluate a student’s ability to apply theoretical principles to real-life situations. These questions usually ask the student to assess data from a specific experiment or setting in order to discover what physical principles are at play. Problem-solving and critical-thinking skills are developed through case study questions, which are an important aspect of physics education.

CBSE Case Study Questions in Class 11 Physics

CBSE Class 11 Physics question paper pattern includes case study questions. Class 11 Physics case study questions assess a student’s ability to apply physics principles to real-world environments. The questions are usually focused on a situation provided in the Class 11 Physics question paper, and they demand the student to answer the problem using their physics knowledge. Class 11 Physics case study questions are an important aspect of the CBSE physics curriculum. Class 11 Physics case study questions are a useful way to assess a student’s expertise in the subject.

Sample Class 11 Physics Case Study Questions

Expert educators at myCBSEguide have created a collection of Class 11 physics case study questions. The samples of Class 11 physics case study questions are given below. Class 11 physics case study questions are designed to test your understanding of the concepts and principles of physics. They are not meant to be easy, but they should be done if you have a good grasp of the subject. So, take a look at the questions and see how you fare. Good luck!

Class 11 Physics Case Study Question 1

Read the case study given below and answer any four subparts: Potential energy is the energy stored within an object, due to the object’s position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy. Potential energy depends on the force acting on the two objects.

• kinetic energy
• potential energy
• mechanical energy
• none of these
• potential energy decreases
• potential energy increases
• kinetic energy decreases
• kinetic energy increases
• only when spring is stretched
• only when spring is compressed
• both a and b
• 5  ×  10 4  J
• 5  ×  10 5  J

Answer Key:

Class 11 Physics Case Study Question 2

• distance between body
• source of heat
• all of the above
• convection and radiation
• (b) convection
• (d) all of the above
• (a) convection
• (a) increase
• (c) radiation

  Class 11 Physics Case Study Question 3

• internal energy.
• 1 +(T 2 /T 1 )
• (T 1 /T 2 )+1
• (T 1  /T 2 )- 1
• 1 – (T 2  / T 1 )
• increase or decrease depending upon temperature ratio
• first increase and then decrease
• (d) 1- (T 2 / T 1 )
• (b) increase
• (c) constant

Class 11 Physics Case Study Question 4 

• It is far away from the surface of the earth
• Its surface temperature is 10°C
• The r.m.s. velocity of all the gas molecules is more than the escape velocity of the moon’s surface
• The escape velocity of the moon’s surface is more than the r.m.s velocity of all molecules
• T(H 2 ) = T(N 2 )
• T(H 2 ) < T(N 2 )
• T(H 2 ) > T(N 2 )

The given samples of Class 11 Physics case study questions will help Class 11 Physics students to get an idea on how to solve it. These Class 11 Physics case study questions are based on the topics covered in the Class 11 Physics syllabus and are designed to test the student’s conceptual understanding. The questions are of varying difficulty levels and cover a wide range of topics. By solving these Class 11 Physics case study questions, students will be able to develop their problem-solving skills and improve their understanding of the concepts.

Examining Class 11 Physics syllabus

Senior Secondary school education is a transitional step from general education to a discipline-based curriculum concentration. The current curriculum of Class 11 Physics takes into account the rigour and complexity of the disciplinary approach, as well as the learners’ comprehension level. Class 11 Physics syllabus has also been carefully crafted to be similar to international norms.

The following are some of the Class 11 Physics syllabus’s most notable features:

• Emphasis is placed on gaining a fundamental conceptual knowledge of the material.
• Use of SI units, symbols, naming of physical quantities, and formulations in accordance with international standards are emphasised.
• For enhanced learning, provide logical sequencing of subject matter units and suitable placement of concepts with their links.
• Eliminating overlapping concepts/content within the field and between disciplines to reduce the curricular load.
• Process skills, problem-solving ability, and the application of Physics principles are all encouraged.

CBSE Class 11 Physics (Code No. 042)

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Class 11 Physics Case Study Questions PDF Download

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Class 11 Physics Case Study Questions are available here. You can read these Case Study questions by chapter for your final physics exam. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve high marks on your Board exams, practice these questions.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

We are providing Case Study questions for Class 11 Physics based on the Latest syllabus. There is a total of 14 chapters included in the CBSE Class 11 physics exams. Students can practice these questions for concept clarity and score better marks in their exams.

Table of Contents

Class 11th PHYSICS: Chapterwise Case Study Question & Solution

Case study questions play a crucial role in the Class 11 Physics curriculum. They are designed to assess your understanding of various concepts and principles in real-life scenarios. These questions help you apply theoretical knowledge to practical situations, enhancing your problem-solving skills.

Case Study-Based Questions for Class 11 Physics

• Case Study Based Questions on Class 11 Physics Chapter 2 Units and Measurements
• Case Study Based Questions on Class 11 Physics Chapter 3 Motion in a Straight Line
• Case Study Based Questions on Class 11 Physics Chapter 4 Motion in a Plane
• Case Study Based Questions on Class 11 Physics Chapter 5 Laws of Motion
• Case Study Based Questions on Class 11 Physics Chapter 6 Work, Energy, and Power
• Case Study Based Questions on Class 11 Physics Chapter 7 System of Particles and Rotational Motion
• Case Study Based Questions on Class 11 Physics Chapter 8 Gravitation
• Case Study Based Questions on Class 11 Physics Chapter 9 Mechanical Properties of Solids
• Case Study Based Questions on Class 11 Physics Chapter 10 Mechanical Properties of Fluids
• Case Study Based Questions on Class 11 Physics Chapter 11 Thermal Properties of Matter
• Case Study Based Questions on Class 11 Physics Chapter 12 Thermodynamics
• Case Study Based Questions on Class 11 Physics Chapter 13 Kinetic Theory
• Case Study Based Questions on Class 11 Physics Chapter 14 Waves
• Case Study Based Questions on Class 11 Physics Chapter 15 Oscillations

Class 11 Physics MCQ Questions

Before the exams, students in class 11 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be offered in Physics exams for Grade 11. These questions were created by our highly qualified faculty for standard 11 Physics based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 11 in understanding the topics.

Class 11 Books for Boards

Class 11 Physics Syllabus 2024

Unit I: Physical World and Measurement 08 Periods

Chapter–2: Units and Measurements

Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. significant figures. Dimensions of physical quantities, dimensional analysis and its applications.

Unit II: Kinematics 24 Periods

Chapter–3: Motion in a Straight Line

The frame of reference, Motion in a straight line, Elementary concepts of differentiation and integration for describing motion, uniform and non-uniform motion, and instantaneous velocity, uniformly accelerated motion, velocity-time and position-time graphs. Relations for uniformly accelerated motion (graphical treatment).

Chapter–4: Motion in a Plane

Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors. Motion in a plane, cases of uniform velocity and uniform acceleration projectile motion, uniform circular motion.

Unit III: Laws of Motion 14 Periods

Chapter–5: Laws of Motion

Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication.

Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).

Unit IV: Work, Energy and Power 14 Periods

Chapter–6: Work, Energy and Power

Work done by a constant force and a variable force; kinetic energy, workenergy theorem, power. Notion of potential energy, potential energy of a spring, conservative forces: non- conservative forces, motion in a vertical circle; elastic and inelastic collisions in one and two dimensions.

Unit V: Motion of System of Particles and Rigid Body 18   Periods

Chapter–7: System of Particles and Rotational Motion

Centre of mass of a two-particle system, momentum conservation and Centre of mass motion. Centre of mass of a rigid body; centre of mass of a uniform rod. Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications. Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions. Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation).

Unit VI: Gravitation 12 Periods

Chapter–8: Gravitation

Kepler’s laws of planetary motion, universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite.

Unit VII: Properties of Bulk Matter 24 Periods

Chapter–9: Mechanical Properties of Solids

Elasticity, Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear modulus of rigidity (qualitative idea only), Poisson’s ratio; elastic energy.

Chapter–10: Mechanical Properties of Fluids

Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure. Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its simple applications. Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.

Chapter–11: Thermal Properties of Matter

Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv – calorimetry; change of state – latent heat capacity. Heat transfer-conduction, convection and radiation, thermal conductivity, qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law .

Unit VIII: Thermodynamics 12 Periods

Chapter–12: Thermodynamics

Thermal equilibrium and definition of temperature zeroth law of thermodynamics, heat, work and internal energy. First law of thermodynamics, Second law of thermodynamics: gaseous state of matter, change of condition of gaseous state -isothermal, adiabatic, reversible, irreversible, and cyclic processes.

Unit IX:   Behavior of Perfect Gases and Kinetic Theory of Gases 08   Periods

Chapter–13: Kinetic Theory

Equation of state of a perfect gas, work done in compressing a gas. Kinetic theory of gases – assumptions, concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom, law of equi-partition of energy (statement only) and application to specific heat capacities of gases; concept of mean free path, Avogadro’s number.

Unit X: Oscillations and Waves 26 Periods

Chapter–14: Oscillations

Periodic motion – time period, frequency, displacement as a function of time, periodic functions and their application. Simple harmonic motion (S.H.M) and its equations of motion; phase; oscillations of a loaded spring- restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period.

Chapter–15: Waves

Wave motion: Transverse and longitudinal waves, speed of traveling wave, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, Beats.

FAQs about Class 11 Physics Case Studies

What is the best website for a  case   study  of physics  class   11 .

studyrate.in is the best website for Class 11 Physics Case Study Questions for Board Exams. Here you can find various types of Study Materials, Ebooks, Notes, and much more free of cost.

How do you write a case study question for Class 11?

The CBSE will ask two Case Study Questions in the CBSE Class 11th Maths Question Paper. Question numbers 15 and 16 will be case-based questions where 5 MCQs will be asked based on a paragraph.

Are the case study questions based on the latest syllabus?

Yes, the case study questions are curated to align with the latest Class 11 Physics syllabus.

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Class 11 Physics Case Study Question

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NCERT Solutions for Class 11 Physics Chapter 15 Waves

NCERT Solutions for Class 11 Physics Chapter 15 Waves are part of Class 11 Physics NCERT Solutions . Here we have given NCERT Solutions for Class 11 Physics Chapter 15 Waves.

Topics and Subtopics in  NCERT Solutions for Class 11 Physics Chapter 15 Waves :

NCERT Solutions Class 11 Physics Physics Sample Papers

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• NCERT Solutions Class 11 Maths
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Question 15. 7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in a tissue in which the speed of sound is 1.7 km s -1 ? The operating frequency of the scanner is 4.2 MHz. Answer:  Here speed of sound => υ = 1.7 km s-1 = 1700 ms -1 and frequency υ= 4.2 MHz = 4.2 x 10 6 Hz .’. Wavelength, A = υ/V = 1700/(4.2 x 106) =4.1 x 10 -4 m.

Question 15. 18. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3Hz. If the original frequency of A is 324 Hz, what is the frequency of B? Answer:  Let υ 1 and υ 2 be the frequencies of strings A and B respectively. Then, υ 1 = 324 Hz, υ 2 = ? Number of beats, b = 6 υ 2 = υ 1 ± b = 324 ± 6 !.e., υ 2 = 330 Hz or 318 Hz Since the frequency is directly proportional to square root of tension, on decreasing the tension in the string A, its frequency υ 1  will be reduced i.e., number of beats will increase if υ 2 = 330 Hz. This is not so because number of beats become 3. Therefore, it is concluded that the frequency υ 2 = 318 Hz. because on reducing the tension in the string A, its frequency may be reduced to 321 Hz, thereby giving 3 beats with υ 2 = 318 Hz.

Question 15. 19. Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa. (b) bats can ascertain distances, directions, nature and sizes of the obstacles without any “eyes”. (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes. (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium. Answer:   (a) In a sound wave, a decrease in displacement i.e., displacement node causes an increase in the pressure there i.e., a pressure antinode is formed. Also, an increase in displacement is due to the decrease in pressure. (b) Bats emit ultrasonic waves of high frequency from their mouths. These waves after being reflected back from the obstacles on their path are observed by the bats. These waves give them an idea of distance, direction, nature and size of the obstacles. (c) The quality of a violin note is different from the quality of sitar. Therefore, they emit different harmonics which can be observed by human ear and used to differentiate between the two notes. (d) This is due to the fact that gases have only the bulk modulus of elasticity whereas solids have both, the shear modulus as well as the bulk modulus of elasticity. (e) A pulse of sound consists of a combination of waves of different wavelength. In a dispersive medium, these waves travel with different velocities giving rise to the distortion in the wave.

Question 15. 23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium, (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz? Answer:   (a) In a non dispersive medium, the wave propagates with definite speed but its wavelength of frequency is not definite. (b) No, the frequency of the note is not 1/20 or 0.50 Hz. 0.005 Hz is only the frequency ‘ of repetition of the pip of the whistle.

NCERT Solutions for Class 11 Physics All Chapters

• Chapter 1 Physical World
• Chapter 2 Units and Measurements
• Chapter 3 Motion in a Straight Line
• Chapter 4 Motion in a plane
• Chapter 5 Laws of motion
• Chapter 6 Work Energy and power
• Chapter 7 System of particles and Rotational Motion
• Chapter 8 Gravitation
• Chapter 9 Mechanical Properties Of Solids
• Chapter 10 Mechanical Properties Of Fluids
• Chapter 11 Thermal Properties of matter
• Chapter 12 Thermodynamics
• Chapter 13 Kinetic Theory
• Chapter 14 Oscillations
• Chapter 15 Waves

We hope the NCERT Solutions for Class 11 Physics Chapter 15 Waves help you. If you have any query regarding NCERT Solutions for Class 11 Physics Chapter 15 Waves, drop a comment below and we will get back to you at the earliest.

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Class 11th Physics - Waves Case Study Questions and Answers 2022 - 2023

Class 11th Physics - Waves Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 11 Physics Subject - Waves, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

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Waves case study questions with answer key.

Final Semester - June 2015

A wave motion is a means of transferring energy and momentum from one point to another without any actual transportation of matter between these points. In wave motion, disturbance travels through some medium but medium does not travel along with the disturbance. For propagation of wave, medium must possess two essential property viz. inertia and elasticity. Disturbance produced at one point is communicated to the adjoining particle which also start vibrating simple harmonically about their mean positions. Hence the wave motion travels on and on. Wave motion is categorised as longitudinal and transverse on the basis of mode of vibration of particles of medium. (i) What are elastic waves? (ii) Explain two types of wave motion. (iii) Mention the characteristic of medium in which longitudinal and transverse waves propagate. (iv) What are compression and rarefaction in a longitudinal wave propagation? (v) Why are longitudinal waves called pressure waves? (vi) What type of waves are the sound and the light waves? (vii) An explosion occurs inside a lake. What type of waves are produced inside the water?

Sound travels through a gas in the form of compressions and rarefactions. Newton assumed that the changes in pressure and volume of gas, when sound waves are propagated through it, are isothermal. The amount of heat produced during compression, is lost to the surrounding and similarly the amount of heat lost during rarefaction is gained from the surroundings. So as to keep the isothermal elasticity. Laplace, a french mathematician pointed out that Newton's assumption was wrong. According to Laplace, the changes in pressure and volume of a gas, when sound waves propagated through it, are not isothermal but adiabatic. (i) Write the Newton's formula for velocity of sound in gases and Laplace correction in it. (ii) What is the effect of pressure on velocity of sound in gases? (iii) Find the ratio of velocity of sound in Hydrogen and Oxygen. (iv) Define temperature coefficient of velocity of sound in air. (v) What is effect of humidity on the speed of sound in air? (vi) Explain why propagation of sound in air is an adiabatic process? (vii) If tension of a wire is increased to four times, how is the wave speed changed?

The principle of super position of waves enables us to determine the net waveform when any number of individual waveforms overlap. The net displacement at a given time is the algebraic sum of the displacements due to each wave at that time. When two sets of progressive wave trains of the same type having the same amplitude and same time period travelling with the same speed along the same straight line in opposite directions superimpose a new set of waves are formed. These are called strationary waves or standing waves. The resultant waves do not propagate in any direction, nor there is any transfer of energy in the medium. In stationary waves, there are nodes and anti nodes point where particles are at rest and have largest amplitude respectively. (i) How amplitude of vibration vary in stationary wave? (ii) What is energy of stationary wave? (iii) What is distance between consecutive node, antinode and between node and antinode? (iv) What is phase difference between particles vibrating in a segment of stationary wave and between adjoining segments? (v) Why is a stationary wave so named? (vi) Where will a person hear maximum sound, at node or antinode? (vii) Name the type of stationary wave produced by an organ pipe, open at both ends.

When two sound waves of-nearly same frequency and amplitudes travelling in a medium along the same direction, super-impose on each other, then the intensity of the resultant sound at a particular position rises and falls alternately with time. This phenomenon is known as beat. if intensity of sound is maximum at time t = 0, one beat is said to be formed when intensity becomes maximum again, after becoming minimum once in between. The time interval between two successive beats is called beat period. The number of beats produced per second is called beat frequency. (i) Two sound waves of frequency v 1  and v 2 superimpose to form beats. What is the beat frequency? (ii) What should be the difference in frequency of two sound waves to form beats? Give reason (iii) Write two applications of the phenomenon of beats. (iv) Two sounds of very close fequencies, say 256 Hz and 260 Hz are produced simultaneously. What is the frequency of resultant sound and also write the number of beats heard in one second? (v) A sitar wire and a tabla, when sounded together, produce 5 beats per second. What can be concluded from this? If the tabla membrane is tightened will the beat rate increase or decrease? (vi) A tuning fork of unknown frequency gives 4 beats with a tuning fork of frequency 310 Hz. It gives the same number of beats on filing. Find the unknown frequency.

Whenever there is a relative motion between the source of sound, the observer and the medium, the frequency of sound as received by the observer is different from the frequency of sound emitted by the source. For example to a man standing on a railway platform, when a train blowing its whistle, approaches him, the pitch of the whistle appears to rise and it suddenly appears to drop as the engine moves away from him. Similar effect is observed when the source is at rest and observer moves towards or away from the source. This phenomenon is noticeable only when the relative velocity between the source and the observer is an appreciable fraction of the wave velocity. (i) Name the phenomenon observed in the passage. Define it. (ii) On what factors does the apparent frequency of sound depends? (iii) What physical change occurs when a source of sound moves and the listener is stationary? (iv) What physical change occurs when the source of sound is stationary but the listener moves? (v) A particle travelling with a speed of 0.9 of the speed of sound and is emitting radiations of frquency of 1 KHz and moving towards the observer. What is the apparent frequency of radiation? (vi) Write the condition in doppler effect when apparent frequency of sound increases. (vii) Write two applications of Doppler effect.

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Waves Class 11 Physics Notes And Questions

Please refer to Waves Class 11 Physics notes and questions with solutions below. These revision notes and important examination questions have been prepared based on the latest Physics books for Class 11. You can go through the questions and solutions below which will help you to get better marks in your examinations.

Class 11 Physics Waves Notes and Questions

Angular wave number: It is phase change per unit distance.

Relation between velocity, frequency and wavelength is given as :-  V =uλ Velocity of Transverse wave:- (i)In solid molecules having modulus of rigidity ‘η ’ and density ‘ρ’ is V = √n/p (ii) In string for mass per unit length ’m’ and tension ‘T’ is √T/m

Velocity of longitudinal wave:- (i) in solid V = √Y/ P , Y= young’s modulus (ii) in liquid V = √K/ P , K= bulk modulus (iii) in gases V= √K/P , K= bulk modulus

According to Newton’s formula: When sound travels in gas then changes take place in the medium are isothermal in nature. V = √P/P According to Laplace: When sound travels in gas then changes take place in the medium are adiabatic in nature. V = √PY/P ‘Where Y = Cp /Cv

Factors effecting velocity of sound :-

Effect of humidity :– sound travels faster in moist air (iv) Effect of wind –velocity of sound increasing along the direction of wind. Wave equation:– if wave is travelling along + x-axis

If wave is travelling along –ve x- axis

Phase and phase difference

Phase is the argument of the sine or cosine function representing the wave.

Relation between phase difference (ΔΦ) and time interval (Δt ) is ΔΦ = -2π/T Δt Relation between phase difference ( Δp ) and path difference ( Δx ) is ΔΦ = -2π /λ Δx

Equation of stationary wave:-

Stationary wave formed

(2) For (+ve) sign antinodes are at x=

(3) For (-ve) sign antinodes are at x=

(4) Distance between two successive nodes or antinodes are λ/2 and that between nodes and nearest antinodes is λ/4 (5) Nodes-point of zero displacement- Antinodes- point of maximum displacement

A = Antinodes

Mode of vibration of strings:-

ORGANPIPES 1. In an organ pipe closed at one end only odd harmonics are present

3. Resonance tube: If L 1  and L 2  are the first and second resonance length with a tuning fork of frequency ‘U ’then the speed of sound. v = 4ν(L1 + 0.3D ) Where ,D=internal diameter of resonance tube

Beats formation 1. Beat frequency = No. of beats per second = Difference in frequency of two sources. b = v1 – v2  2. v2 =v1 ± b

3. If the prong of tuning fork is filed, its frequency increases. If the prong of a tuning fork is loaded with a little way, its frequency decreases. These facts can be used to decide about + or – sign in the above equation.

Doppler effect in sound

1. If V, Vo, Vs, and Vm are the velocity of sound , observes, source and medium respectively, then the apparent frequency

3. All the velocity are taken positive with source to observer (S→O) direction and negative in the opposite (S→O) direction

Question and Answer :

Question. A girl is swinging in the sitting position. How will the period of the swing change if she stands up? 4 Answer:  The girl and the swing together constitute a pendulum of time period As the girl stands up her centre of gravity is raised. The distance between the point of suspension and the centre of gravity decreases i.e. length ‘l’ decreases .Hence the time period ‘T’ decreases.

Question. The maximum velocity of a particle, executing S.H.M with amplitude of 7mm is 4.4 m/s. What is the period of oscillation? Answer:

Question. Why the longitudinal wave are also called pressure waves? Answer:  Longitudinal wave travel in a medium as series of alternate compressions and rare fractions i.e. they travel as variations in pressure and hence are called pressure waves.

Question. How does the frequency of a tuning fork change, when the temperature is increased? 7 Answer:  As the temperature is increased, the length of the prong of a tuning fork increased .This increased the wavelength of a stationary waves set up in the tuning fork. As frequency,

Question. An organ pipe emits a fundamental node of a frequency 128Hz. On blowing into it more strongly it produces the first overtone of the frequency 384Hz. What is the type of pipe –Closed or Open? Answer:  The organ pipe must be closed organ pipe, because the frequency the first overtone is three times the fundamental frequency.

Question. Which of the following relationships between the acceleration ‘a’ and the displacement ‘x’ of a particle involve simple harmonic motion? (a) a=0.7x (b) a=-200x 2  (c) a = -10x (d) a=100x 3 Ans: –  (c) reprent SHM.

Question. Can a motion be periodic and not oscillatory? Answer:  Yes, for example, uniform circular motion is periodic but not oscillatory.

Question. Can a motion be periodic and not simple harmonic? If your answer is yes, give an example and if not, explain why? Answer:  Yes, when a ball is doped from a height on a perfectly elastic surface ,the motion is oscillatory but not simple harmonic as restoring force F=mg=constant and not F∝ -x, which is an essential condition for S.H.M.

Question. All harmonic are overtones but all overtones are not harmonic. How? Answer:  The overtones with frequencies which are integral multiple of the fundamental frequency are called harmonics. Hence all harmonic are overtones. But overtones which are non-integrals multiples of the fundamental frequency are not harmonics.

Question. What is the factor on which pitch of a sound depends? Answer:  The pitch of a sound depends on its frequency.

Question. Does the function y = sin 2 ωt represent a periodic or a S.H.M? What is period of motion? Answer:  Displacement y = sin 2 ωt

As the acceleration is not proportional to displacement y, the given function does not represent SHM. It represents a periodic motion of angular frequency 2ω.

Question. All trigonometric functions are periodic, but only sine or cosine functions are used to define SHM. Why? Answer. All trigonometric functions are periodic. The sine and cosine functions can take value between -1 to +1 only. So they can be used to represent a bounded motion like SHM. But the functions such as tangent, cotangent, secant and cosecant can take value between 0 and ∞ (both negative and positive). So these functions cannot be used to represent bounded motion like SHM.

Question. A simple Harmonic Motion is represented by d 2 x/dt 2  +αx = 0. What is its time period? Answer:

Question. The Length of a simple pendulum executing SHM is increased by 2.1%. What is the percentage increase in the time period of the pendulum of increased length? Answer:  

Question. A simple Harmonic motion has an amplitude A and time period T. What is the time taken to travel from x = A to x = A/2. Answer:  Displacement from mean position =

When the motion starts from the positive extreme position, y = A cosωt.

Question. At what points is the energy entirely kinetic and potential in S.H.M? What is the total distance travelled by a body executing S.H.M in a time equal to its time period, if its amplitude is A? Answer:  The energy is entirely kinetic at mean position i.e. at y=0. The energy is entirely potential at extreme positions i.e. y = ± A Total distance travelled in time period T = 2A + 2A = 4A.

Question. A simple pendulum consisting of an inextensible length ‘l’ and mass ‘m’ is oscillating in a stationary lift. The lift then accelerates upwards with a constant acceleration of 4.5 m/s 2 . Write expression for the time period of simple pendulum in two cases. Does the time period increase, decrease or remain the same, when lift is accelerated upwards? Answer:  When the lift is stationary, T = 2π √1/g

Question. An open organ pipe produces a note of frequency 5/2 Hz at 15 0 C, calculate the length of pipe. Velocity of sound at 00C is 335 m/s. Answer:  Velocity of sound at 15 0 C V=V0+0.61xt =335+0.61×15 =344.15 m/s. (Thermal coefficient of velocity of sound wave is .61/ 0 C) Fundamental frequency of an organ pipe

Question. An incident wave is represented by Y(x, t)=20sin(2x-4t).Write the expression for reflected wave (i) From a rigid boundary (ii) From an open boundary. Ans.(i) The wave reflected from a rigid boundary is Y (x, t) = -20sin (2x+4t) (i)The wave reflected from an open boundary is Y (x, t) = 20sin (2x+4t) Explain why (i) in a sound wave a displacement node is a pressure antinode and vice- versa (ii) The shape of pulse gets- distorted during propagation in a dispersive medium. Answer:  (i) At a displacement node the variations of pressure is maximum. Hence displacement node is the a pressure antinode and vice-versa. (ii)When a pulse passes through a dispersive medium the wavelength of wave changes. So, the shape of pulse changes i.e. it gets distorted.

Question. Find the ratio of velocity of sound in hydrogen gas Ƴ 7 /5 to that in helium gas Ƴ 5 /3 at the same temperature. Given that molecular weight of hydrogen and helium are 2 and 4 respectively. Answer:

Question. The equation of a plane progressive wave is, y = 10sin2π(t – 0.005x) where y & x are in cm & t in second. Calculate the amplitude, frequency, wavelength & velocity of the wave. Answer:

Question. Write displacement equation respecting the following condition obtained in SHM. Amplitude = 0.01m Frequency = 600Hz Initial phase = π /6 Answer:

Question. The amplitude of oscillations of two similar pendulums similar in all respect are 2cm & 5cm respectively. Find the ratio of their energies of oscillations. Answer:

Question. What is the condition to be satisfied by a mathematical relation between time and displacement to describe a periodic motion? Answer:  A periodic motion repeats after a definite time interval T. So, y(t) = y (t + T) = Y(t + 2T) etc.

Question. A spring of force constant 1200N/m is mounted horizontal table. A mass of 3Kg is attached to the free end of the spring, pulled sideways to a distance of 2.0cm and released. (i) What is the frequency of oscillation of the mass? (ii) What is the maximum acceleration of the mass? (iii) What is the maximum speed of the mass?

Question. Write any three characteristics of stationary waves. Answer:  (i) in stationary waves, the disturbance does not advance forward. The conditions of crest and trough merely appear and disappear in fixed position to be followed by opposite condition after every half time period. (ii) The distance between two successive nodes or antinodes is equal to half the wavelength. (iii) The amplitude varies gradually from zero at the nodes to the maximum at the antinodes.

Question. Show that the speed of sound in air increased by .61m/s for every 10 C rise of temperature. Answer:

Question. Which of the following function of time represent, (a) simple harmonic (b) periodic but not SHM and (c) non periodic ? Answer:

Question. (a) A light wave is reflected from a mirror. The incident & reflected wave superimpose to form stationary waves. But no nodes & antinodes are seen, why? (b) A standing wave is represented by y=2ASinKxCoswt.If one of the component wave is y1 = Asin(ωt – kx) what is the equation of the second component wave? Answer:  (a) As is known, the distance between two successive nodes or two successive antinodes is λ/2 The wavelength of visible light is of the order of 10 -1m. As such as a small distance cannot be detected by the eye or by a ordinary optical instrument. Therefore, nodes and antinodes are not seen.

According to superposition principle,

Question. Discuss Newton’s formula for velocity of sound in air. What correction was made to it by Laplace and why? Answer:  According to Newton the change in pressure & volume in air is an isothermal process. Therefore he calculated, v = √p/pon substituting the require value he found, the velocity of sound was not in close agreement with the observation value. Then Laplace pointed out the error in Newton’s formula. According to Laplace the change in pressure and volume is an adiabatic process. So he calculated the value of sound as, v = √yr/p on putting require value he found velocity of sound as 332m/s very closed to observed theory.

Question. A train stands at a platform blowing a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle heard by a man running (a)Towards the engine 10 m/s. (b) Away from the engine at 10 m/s? (ii) What is the speed of sound in each case? (iii) What is the wavelength of sound received by the running man in each case? Take speed of sound in still air = 340 m/s. Answer:  (i) (a) When the man runs towards the engine

Question. Explain Doppler’s effect of sound. Derive an expression for the apparent frequency where the source and observer are moving in the same direction with velocity Vs and Vo respectively, with source following the observer. Answer :

Question. For a travelling harmonic wave, y = 2cos(10t – 0.008x + 0.35) where x & y are in cm and t in second. What is the phase difference between oscillatory motions at two points separated by a distance of (i) 4cm (ii) 0.5m (iii) λ/2 (iv) 3λ /4 ? Answer :

Question. (i) A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel? (ii) A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly exited by a 430 Hz source? Will this same source be in resonance with the pipe if both ends are open? (Speed of sound = 340 m/s). Answer : (i) For the fundamental mode

Hence source of frequency 430 Hz will not be in resonance with open organ pipe.

(a) Speed of longitudinal wave Speed of longitudinal wave in a medium is given by v = √E/p

where, E is the modulus of elasticity, p is the density of the medium. Speed of longitudinal wave in a solid in the form of rod is given by v = √Y/p where, Y is the Young’s modulus of the solid,

p is the density of the solid. Speed of longitudinal wave in fluid is given by v = √B/p where, B is the bulk modulus, P is the density of the fluid.

(b) Newton’s formula Newton assumed that propagation of sound wave in gas is an isothermal process. Therefore, according to Newton, speed of sound in gas is given by v = √P/p

where P is the pressure of the gas and p is the density of the gas.

According to the Newton’s formula, the speed of sound in air at S.T.P. is 280 m/s. But the experimental value of the speed of soundinairis 3 3 2 m s –1. Newton could not explain this large difference. Newton’s formula was corrected by Laplace.

(c) Laplace’s correction Laplace assumed that propagation of sound wave in gas in an adiabatic process. Therefore, according to Laplace, speed of sound in a gas is given by v = √YP/p

According to Laplace’s correction the speed of sound in air at S.T.P. is 331.3 m/s. This value agrees farily well with the experimental values of the velocity of sound in air at S.T.P.

WAVES TRAVELLING IN OPPOSITE DIRECTIONS

When two waves of same amplitude and frequency travelling in opposite directions y 1  = A sin (kx – ωt) y 2  = A sin (kx + ωt) interfere, then a standing wave is produced which is given by, y = y 1  + y 2 ⇒ y = 2A sin kx cos ωt Hence the particle at location x is oscillating in S.H.M. with angular frequency ω and amplitude 2A sin kx. As the amplitude depends on location (x), particles are oscillating with different amplitude.

Nodes : Amplitude = 0 2A sin kx = 0 x = 0, π/k, 2π/k……. x = 0, λ/2, λ, 3λ/2, 2λ…….. Antinodes : Amplitude is maximum. sin kx = ± 1 x = π/2k, 3π/2k x = λ/4, 3λ/4, 5λ/4 Nodes are completely at rest. Antinodes are oscillating with maximum amplitude (2A). The points between a node and antinode have amplitude between 0 and 2 A. Separation between two consecutive (or antinodes) = λ/2. Separation between a node and the next antinode=λ/4. Nodes and antinodes are alternately placed.

It is clear from the figure that since nodes are, at rest they don’t transfer energy. In a stationary wave, energy is not transferred from one point to the other.

Vibrations in a stretched string

1. Fixed at both ends.

Transverse standing waves with nodes at both ends of the string are formed. So, length of string, l = nλ/2 if there are (n + 1) nodes and n antinodes. Frequency of oscillations is

Fundamental frequency (x = 1) 0v = V 2L It is also called first harmonic. Second harmonic or first overtone

It is also called first harmonic. Second harmonic or first overtone 0v = 2V 2L

The nth multiple of fundamental frequency is known as nth harmonic or (n – 1)th overtone.

2. Fixed at one end Transverse standing waves with node at fixed end and antinode at open end are formed. So, length of string l = (2n – 1) λ/4 if there are n nodes and n antinodes.

Frequency of oscillations

Fundamental frequency, (n = 1) v  0 = V/4L It is also called first harmonic. First overtone or third harmonic.

Only odd harmonics are possible in this case.

Vibrations in an organ pipe

1. Open Organ pipe (both ends open) The open ends of the tube becomes antinodes because the particles at the open end can oscillate freely. If there are (n + 1) antinodes in all, length of tube, l = nλ/2 So, Frequency of oscillations is v = nv/2l

2. Closed organ pipe (One end closed) The open end becomes antinode and closed end become a node. If there are n nodes and n antinodes, L = (2n – 1) λ/4 So frequency of oscillations is

There are only odd harmonics in a tube closed at one end.

Waves having different frequencies

Beats are formed by the superposition of two waves of slightly different frequencies moving in the same direction. The resultant effect heard in this case at any fixed position will consist of alternate loud and weak sounds.

amplitude is 2A cos π (v 1 – v 2 ) t. As the amplitude term contains t, the amplitude varies periodically with time. For Loud Sounds : Net amplitude = ± 2A

Hence the interval between two loud sounds is given as :

⇒ the number of loud sounds per second = v 1  – v 2 ⇒ beat per second = v 1  – v 2 Note that v 1  – v 2  must be small (0 – 16 Hz) so that sound variations can be distinguished.

DOPPLER EFFECT

According to Doppler’s effect, whenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by the listener is different from the actual frequency of sound emitted by the source. Apparent frequency,

Sing Convention. All velocities along the direction S to L are taken as positve and all velocities along the direction L to S are taken as negative. When the motion is along some other direction the component of velocity of source and listener along the line joining the source and listener is considered.

Special Cases :

(a) If the source is moving towards the listener but the listener is at rest, then vs is positive and vL = 0 (figure a). Therefore,

(b) If the source is moving away from the listener, but the listener is at rest, then vs is negative and vL = 0 (figure b). Therefore,

(c) If the source is at rest and listener is moving away from the source, the vs = 0 and vL is positive (figure c). Therefore,

(d) If the source is at rest and listener is moving towards the source, then vs = 0 and vL is negative (figure d). Therefore,

(e) If the source and listener are approaching each other, then vs is positive and vL is negative (figure e). Therefore,

(f) If the source and listener are moving away from each other, then vs is negative and vL is positive, (figure f). Therefore,

(g) If the source and listener are both in motion in the same direction and with same velocity, then vs = vL = v’ (say) (figure g). Therefore,

It means, there is no change in the frequency of sound heard by the listerner. Ap par ent wavelength heard b y the o b ser ver is

CHARACTERISTICS OF SOUND

Loudness of sound is also called level of intensity of sound. In decibel the loudness of a sound of intensity I is 

Pitch  : It is pitch depends on frequency, higher the frequency higher will be the pitch and shriller will be the sound.

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Waves Class 11 Physics Most Important Questions

Many important questions are formed in final examinations from chapter 14 of Physics. In Class 11 Physics Waves Most Important Questions PDF, you will study topics like Transverse and Longitudinal waves, Speed of a Travelling Wave, The Principle of Superposition of Waves, Reflection of Waves, Beats, Doppler Effect.

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NCERT solutions for Physics Class 11 chapter 15 - Waves [Latest edition]

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Solutions for chapter 15: waves.

Below listed, you can find solutions for Chapter 15 of CBSE NCERT for Physics Class 11.

NCERT solutions for Physics Class 11 Chapter 15 Waves Exercises [Pages 387 - 390]

A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s –1 ? (g= 9.8 m s –2 )

A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s –1 .

Use the formula `v = sqrt((gamma P)/rho)` to explain why the speed of sound in air is independent of pressure.

Use the formula `v = sqrt((gamma P)/rho)` to explain why the speed of sound in air increases with temperature.

Use the formula `v = sqrt((gamma P)/rho)` to explain why the speed of sound in air increases with humidity.

You have learnt that a travelling wave in one dimension is represented by a function y= f (x, t)where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

(a) `(x – vt )^2`

(b) `log [(x + vt)/x_0]`

(c) `1/(x + vt)`

A bat emits an ultrasonic sound of frequency 1000 kHz in the air. If the sound meets a water surface, what is the wavelength of the the reflected sound? The speed of sound in air is 340 m s –1  and in water 1486 m s –1 .

A bat emits an ultrasonic sound of frequency 1000 kHz in the air. If the sound meets a water surface, what is the wavelength of the transmitted sound? The speed of sound in air is 340 m s –1  and in water 1486 m s –1 .

A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s –1 ? The operating frequency of the scanner is 4.2 MHz.

A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) 

Where x and y are in cm and t in s. The positive direction of x is from left to right.

(a) Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

(b) What are its amplitude and frequency?

(c) What is the initial phase at the origin?

(d) What is the least distance between two successive crests in the wave?

For the wave described in Exercise 15.8, plot the displacement ( y ) versus ( t ) graphs for  x =  0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

For the travelling harmonic wave

y (x, t) = 2.0 cos 2π (10t – 0.0080x + 0.35)

Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of 4 m.

Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of 0.5 m.

Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of `λ/2`.

Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of `(3λ)/4`.

The transverse displacement of a string (clamped at its both ends) is given by

y(x, t) = 0.06 sin `2/3` x cos (120 πt)

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 × 10 -2 kg.

Answer the following:

Does the function represent a travelling wave or a stationary wave?

Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is `3.0 xx 10^(-2)` kg.

Answer the following :

Determine the tension in the string.

(i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?

Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a traveling wave, (ii) a stationary wave or (iii) none at all:

y = 2 cos (3x) sin (10t)

`"y" = 2sqrt(x - "vt")`

y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t)

y = cos x sin t + cos 2x sin 2t

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10 –2  kg and its linear mass density is 4.0 × 10 –2  kg m –1 . What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?

A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be 2.53 kHz. What is the speed of sound in steel?

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is 340 m s –1 ).

Two sitar strings A and B playing the note ‘ Ga ’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

Explain why (or how) In a sound wave, a displacement node is a pressure antinode and vice versa,

Explain why (or how): Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,

Explain why (or how) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes,

Explain why (or how) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases

Explain why (or how) The shape of a pulse gets distorted during propagation in a dispersive medium.

A train, standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. (i) What is the frequency of the whistle for a platform observer when the train (a) approaches the platform with a speed of 10 m s –1 , (b) recedes from the platform with a speed of 10 m s –1 ? (ii) What is the speed of sound in each case? The speed of sound in still air can be taken as 340 m s –1 .

A train, standing in a station-yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with at a speed of 10 m s –1 . What are the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of 10 m s –1 ? The speed of sound in still air can be taken as 340 m s –1 .

Additional Exercises

A travelling harmonic wave on a string is described by

`y(x,t) = 7.5 sin (0.0050x + 12t + pi/4)`

(a) What are the displacement and velocity of oscillation of a point at  x  = 1 cm, and  t = 1 s? Is this velocity equal to the velocity of wave propagation?

(b) Locate the points of the string which have the same transverse displacements and velocity as the  x =  1 cm point at  t =  2 s, 5 s and 11 s.

A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. (a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? (b) If the pulse rate is 1 after every 20 s, (that is the whistle is blown for a split of second after every 20 s), is the frequency of the note produced by the whistle equal to 1/20 or 0.05 Hz

One end of a long string of linear mass density 8.0 × 10 –3  kg m –1  is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At  t  = 0, the left end (fork end) of the string  x  = 0 has zero transverse displacement ( y  = 0) and is moving along positive  y -direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement  y  as the function of  x  and  t  that describes the wave on the string.

A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h –1 . What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s –1 .

Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse ( S ) and longitudinal ( P ) sound waves. Typically the speed of  S  wave is about 4.0 km s –1 , and that of  P  wave is 8.0 km s –1 . A seismograph records  P  and  S  waves from an earthquake. The first  P  wave arrives 4 min before the first  S  wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?

A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is 40 kHz. During one fast swoop directly toward a flat wall surface, the bat is moving at 0.03 times the speed of sound in air. What frequency does the bat hear reflected off the wall?

NCERT solutions for Physics Class 11 chapter 15 - Waves

Shaalaa.com has the CBSE Mathematics Physics Class 11 CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. NCERT solutions for Mathematics Physics Class 11 CBSE 15 (Waves) include all questions with answers and detailed explanations. This will clear students' doubts about questions and improve their application skills while preparing for board exams.

Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and provide excellent self-help guidance for students.

Concepts covered in Physics Class 11 chapter 15 Waves are Speed of Wave Motion, Reflection of Transverse and Longitudinal Waves, Displacement Relation for a Progressive Wave, The Speed of a Travelling Wave, Principle of Superposition of Waves, Introduction of Reflection of Waves, Standing Waves and Normal Modes, Beats, Doppler Effect, Wave Motion, Speed of Wave Motion, Reflection of Transverse and Longitudinal Waves, Displacement Relation for a Progressive Wave, The Speed of a Travelling Wave, Principle of Superposition of Waves, Introduction of Reflection of Waves, Standing Waves and Normal Modes, Beats, Doppler Effect, Wave Motion.

Using NCERT Physics Class 11 solutions Waves exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE Physics Class 11 students prefer NCERT Textbook Solutions to score more in exams.

Get the free view of Chapter 15, Waves Physics Class 11 additional questions for Mathematics Physics Class 11 CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

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NCERT Solutions for Class 11th: Ch 15 Waves Physics

Ncert solutions for class 11th: ch 15 waves physics science.

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• Physics Important Questions
• Class 11 Physics
• Chapter 14: Waves

Chapter 14 - Waves

Students can find CBSE Important Questions Class 11 Physics Chapter 14 Waves compiled here. These questions are created by experts by referring to the NCERT textbook and syllabus. Solving them will help in preparing for the annual exam. Also, students will get well worse with the questions of this chapter. See the CBSE important questions class 11 physics chapter 14 waves below.

CBSE Important Questions Class 11 Physics Chapter 14 Waves

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RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

Waves Class 11 Important Extra Questions Physics Chapter 15

March 30, 2021 by Prasanna

Here we are providing Class 11 Physics Important Extra Questions and Answers Chapter 15 Waves. Important Questions for Class 11 Physics with Answers are the best resource for students which helps in Class 11 board exams.

Class 11 Physics Chapter 15 Important Extra Questions Waves

Waves important extra questions very short answer type.

Question 1. In a resonance tube, the second resonance does not occur exactly at three times the length at the first resonance. Why? Answer: This is due to the end correction.

Question 2. What is the nature of ultrasonic waves and what is their frequency? Answer: Ultrasonic waves are longitudinal waves in nature and have frequencies greater than 20 k Hz.

Question 3. Is the principle of superposition wave valid in the case of electromagnetic (e.m.) waves? Answer: Yes.

Question 4. A rod is clamped at one end and it is hit by a hammer at the other end (a) at a right angle to its length (b) along the length. What types of waves are produced in each case? Answer: (a) Transverse waves (b) Longitudinal waves.

Question 5. Why do not we hear beats if the frequency of ìwo sounds is widely different? Answer: The beats can’t be heard as separate due to the persistence of hearing if the difference in frequencies is more than 10 Hz.

Question 6. What causes the rolling sound of thunder? Answer: The multiple reflections of the sound of lighting results in the rolling ’ sound of thunder.

Question 7. A tuning fork produces resonance in a closed pipe. But the ‘ same tuning fork is unable to, produce resonance in an open organ pipe of equal length. Why? Answer: It is because the fundamental frequencies of open and closed organ pipes of the same length are different,

Question 8. Thick and big curtains are preferred in a big hall. Why? Answer: To increase the absorption and decrease the reverberation time and hence making the sound uniform,

Question 9. Why female voice is sweeter than that of a man? Answer: This is because the frequency of a lady’s voice is greater than that of a man’s voice.

Question 10. The frequency of the fundamental note of a tube closed at one end is 200 Hz. What will be the frequency of the fundamental note of a similar tube of the same length but open at both ends? Answer: 400 Hz.

Question 11. A wave transmits energy. Can it transmit momentum? Answer: Yes.

Question 12. By how much the wave velocity increases for 1°C rise of temperature? Answer: Wave velocity increases by 0.61 ms for 1°C rise of temperature.

Question 13. Why the sound heard is more in carbon dioxide than in air? Answer: This is because the intensity of sound increases with the increase in the density of the medium.

Question 14. What is the relation between path difference and phase difference? Answer: Phase difference = \(\frac{2π}{λ}\) × path difference.

Question 15. Is it possible to have interference between the waves produced by two violins? Why? Answer: No. This is because the sounds produced will not have a constant, phase difference.

Question 16. The windowpanes of houses sometimes get cracked due to some explosion at a large distance. Which waves are responsible for this? Answer: Shockwaves.

Question 17. Why the velocity of sound is generally greater in solids than in gases? Answer: This is because \(\frac{E}{ρ}\) for solids is much greater than for gases.

Question 18. An observer places his ear at the end of a long steel pipe. He can hear two sounds when a workman hammers the other end of the pipe. Why? Answer: This is because the sound is transmitted both through air and medium.

Question 19. Why a stationary wave is so named? Answer: A stationary wave is so named because there is no net propagation of energy. .

Question 20. How do we identify our friends from his voice while sitting in a dark room? Answer: The quality of sound helps us to identify the sound.

Question 21. Why bells are made of metal and not of wood? Answer: This is because the wood has high damping.

Question 22. What is the nature of light waves? Answer: Electromagnetic waves.

Question 23. When a vibrating tuning fork is moved speedily towards a wall, beats are heard. Why? Answer: This is due to the difference in the frequency of the incident wave and the apparent frequency of the reflected wave.

Question 24. The weight suspended from a sonometer wire is increased by a factor of 4. Will the frequency of the wire be increased exactly by a factor of 2? Justify your answer. Answer: No. There will be a slight increase in the length of the wire. So the frequency shall become slightly less than double.

Question 25. Are sound waves in air longitudinal or transverse? Answer: Longitudinal.

Question 26. Can you notice the Doppler effect if both the listener and the source of sound are moving with the same velocity in the same direction? Why? Answer: No. This is because there is no relative motion between the source of sound and the listener.

Question 27. If oil of density higher than that of water is used in place of water in a resonance tube, how does the frequency change? Answer: The frequency is governed by the air column and does not depend, upon the nature of the liquid. So frequency would not change.

Question 28. Which property of the medium enables the transverse waves to pass through it? Answer: Modulus of rigidity.

Question 29. In a thunderstorm why flash of light is seen earlier than the sound of thunder? Answer: The speed of light is much higher than the speed of sound waves.

Question 30. What is the main difference between the brazing of a honeybee and the roaring of a lion? Answer: Brazing of honeybee is of high pitch and lion’s roar has high intensity.

Question 31. If you set your watch by the sound of a distant siren, will it go fast or slow? Answer: It will go slow due to the low value of the speed of sound in the air.

Question 32. What is a dispersive medium? Answer: A medium in which the wave velocity depends on the frequency of the wave is called a dispersive medium.

Question 33. Will sound be louder at node or antinode in a stationary wave? Answer: It will be louder at node due to large pressure variation there as ΔP = – strain × elasticity.

Question 34. Even if a powerful thermonuclear explosion takes place on a planet, the sound is not heard on Earth. Why? Answer: The interplanetary space is devoid of continuous material. In the absence of a material, the medium sound is not heard on the earth.

Question 35. In older days messages were conveyed to distant villages by beating of big drums. Why? Answer: It has a larger area and the intensity of sound (I) is directly proportional to the area of an oscillator (A) i.e. I ∝ A.

Question 36. Which is the most basic property of the wave? Answer: Frequency is the basic property of the wave.

Question 37. You can make waves in a pond by throwing a stone in it. What is the source of energy of the wave? Answer: The kinetic energy of the stone is the source of wave energy.

Question 38. Why echoes are not heard in a small room? Answer: Because the minimum distance between the obstacle reflecting sound waves and the source of sound is less than 17 m in a small room, so echoes are not heard.

Question 39. Name two properties that are common to all types of mechanical waves? Answer:

• They require a material medium for their propagation.
• The medium itself does not move with the wave.

Question 40. In sound, beats are heard when two independent sources are sounded together. Is it possible in the case of sources of light? Answer: No. This is because the phase difference between two independent sources of light is random.

Question 41. Mention a condition when Doppler’s effect in sound is not applicable. Answer: When the velocity of the source or listener exceeds the velocity of sound.

Question 42. What will be the velocity of sound in a perfectly rigid rod and why? Answer: Infinite because the value of Young’s modulus of elasticity is infinite for a perfectly rigid rod.

Question 43. Sound is simultaneously produced at one end of two strings of the same length, one of rubber and the other of steel. In which string will the sound reach the other end earlier and why? Answer: In the case of steel-string as \(\frac{Y}{d}\) ratio is larger for steel than rubber.

Question 44. Two persons cannot talk on the moon just as they do on the earth. Why? Answer: Due to the absence of air on the moon.

Question 46. Why the bells of colleges and temples are of large size? Answer: The larger the area of the source of sound more is the energy transmitted into the medium. Consequently, the intensity of sound is more and loud sound is heard.

Question 47. State the limitations of Doppler’s effect. Answer: Doppler’s effect is applicable only when there is the relative velocity between the source and the listener and is less than the velocity of sound. The effect is not applicable if the relative velocity is greater than the velocity of sound i.e. if the source or observer moves with supersonic velocity.

Question 48. Animals and human beings have two ears. What help do they render in listening sound from a distant source? Answer: The two ears receive sound in different phases, thus they help in locating the direction of the source of the sound. Turning of head and receiving sound in phase turns the direction.

Question 49. An observer at a sea-coast observes waves reaching the coast. What type of waves does he observe? Why? Answer: Elliptical waves while the waves on the surface of the water are transverse, the waves just below the surface of the water are longitudinal. So the resultant waves are elliptical.

Question 50. Why the reverberation time is larger for an empty hall than a crowded hall? Answer: It is due to the fact that the energy absorption in an empty hall is very small as compared to that of the crowded one.

Waves Important Extra Questions Short Answer Type

Question 1. Here are the equations of three waves: (a) y (x, t) = 2 sin (4x – 2t) (b) y (x, t) = sin (3x – 4t) (c) y (x, t) = 2 sin (3x – 3t). Rank the waves according to their (A) wave speed and (B) maximum transverse speed, greatest first. Answer: (A) b, c, a. Standard wave equation is y (x, t) = A sin (ωt – kx) ∴ (a) v a = \(\frac{\omega}{k}=\frac{2}{4}=\frac{1}{2}\) unit = 0.5 unit.

(b) v b  = \(\frac{\omega}{k}=\frac{4}{3}\)unit = 1.33 unit.

(C) V c   = \(\frac{\omega}{k}=\frac{3}{3}\) = 1 unit. clearly v b > v c > v a , so order is b, c, a.

(b) Transverse speed (vt) = \(\frac{\mathrm{d} y}{\mathrm{dt}}\) ∴ |(v t ) a | = 2 × 2 = 4 |(v t ) b | = 4 |(v t ) c | = 2 × 3 = 6

∴ clearly (c), (a) and (b) tie.

Question 3. When are the tones called harmonics? Answer: The tones are called harmonics if the frequencies of the fundamental tone and other overtones produced by a source of sound are in the harmonic series.

Question 4. What will be the effect on the frequency of the sonometer wire if the load stretching the sonometer wire is immersed in water? Answer: Due to the upthrust due to buoyancy experienced by the load, the effective weight will decrease, so tension and hence frequency will decrease as v ∝ \(\sqrt{T}\).

Question 5. An organ pipe is in resonance with a tuning fork. If the pressure of air in the pipe is increased by a factor of 139, then how should the length be changed for resonance? Answer: We know that the velocity of sound is independent of pressure, so there is no change in frequency and hence there is no need to change the length of the pipe.

Question 6. Sound waves travel through longer distances during the night than during the day. Why? Answer: Earth’s atmosphere is warmer as compared to the surface of the earth at night. The temperature increases with altitude and thus the velocity of sound increases. It is a case of reflection from denser to rarer medium.

The sound waves get totally internally reflected.

Question 7. Water is being continuously poured into a vessel. Can you estimate the height of the water level reached in the vessel simply by listening to the sound produced? Answer: Yes, the frequency of the sound produced by an air column is inversely proportional to the length of the air column. As the level of water in the vessel rises, the length of the air column in the vessel decreases, so the frequency of sound increases, and hence shrillness of sound increases.

From the shrillness of sound, we can have a rough estimate of the level of water in the vessel.

Question 8. A sonometer wire resonates with a tuning fork. If the length of the wire between the bridges is made twice even then it can resonate with the same fork. Why? Answer: When the length of the wire is doubled, the fundamental frequency is halved and the wire vibrates in two segments so the sonometer wire will still resonate with the given tuning fork.

Question 9. Doppler’s effect in sound is asymmetric. Explain. Answer: Sound waves require a material medium for their propagation.

The apparent frequency is different whether the source moves towards the stationary observer or an observer moves towards the stationary source. Thus the Doppler’s effect is said to be asymmetric. No such asymmetry occurs in light because apparent frequency remains the same in either the case whether the source or the listener moves.

Hence Doppler’s effect is said to be symmetric in light.

Question 10. What is redshift? Answer: It is due to Doppler’s effect in the case of light waves. It is known that all stars are moving away from each other. So apparent frequency of light from a star as received by an observer on earth is less than the actual frequency. Since wavelength is inversely proportional to the frequency, the apparent wavelength of light from stars is more than the actual wavelength.

In other words, due to the Doppler effect, the wavelength of light shifts towards a longer end i.e. towards red color and so it is called redshift.

Question 11. A sitar wife and a tabla when sounded together give 4 beats/ sec. What do we conclude from this? As the tabla membrane is tightened, the beat rate increases or decreases, explain. Answer: When sitar and tabla are sounded together, they give 4 seats/ sec. From this, we conclude that the frequencies of the two sounds differ by 4. If the frequency of tabla is greater than that of a sitar, then on tightening the tabla membrane, the frequency of tabla will further increase and hence the difference in frequencies will increase.

Thus beat rate will increase. If the frequency of tabla is less than that of sitar, then on tightening the tabla membrane, the frequency of tabla will increase and the difference in frequencies will decrease. So beat rate will decrease.

Question 12. Explain why frequency is the most fundamental property of a wave. Answer: When a wave passes from one medium to another, its velocity and wavelength change but the frequency remains the same. Hence frequency is said to be the most fundamental property of a wave.

Question 13. Sound is produced by vibratory motion, explain why then a vibrating pendulum does not produce sound? Answer: The sound which we can hear has a frequency from 20 Hz to 20,000 Hz.

The frequency of the vibrating pendulum does not lie within the audible range and hence it does not produce sound.

Question 14. Explain why stringed instruments are provided with hollow boxes. Answer: The hollow boxes are set into forced vibrations along with the strings. The loudness is higher if the area of the vibrating body is more. So hollow boxes attached to increase the loudness of sound.

Question 15. When we start filling an empty bucket with water, the pitch of sound produced goes on changing. Why? Answer: An empty bucket behaves as a closed organ pipe. The frequency of fundamental note produced by it is given by v = \(\frac{v}{4l}\).

As the bucket starts filling, the length (l) of the resonating air column decreases, and hence frequency increases. Since the pitch of a sound depends upon the frequency. So it changes with the change in frequency.

Question 16. Two loud-speakers have been installed in an open space to listen to a speech. When both are operational, a listener sitting at a .particular- place receives a very faint sound. Why? What will happen if one loud-speaker is kept off? Answer: When the distance between two loud-speakers from the position of listener is an odd multiple of \(\frac{λ}{2}\), then due to destructive interference between sound waves from two loud-speakers, a feeble sound is heard by the listener.

When one loud-speaker is kept off, no interference will take place and the listener will hear the full sound of the operating loud-speaker.

Question 17. Why is the sound produced in the air not heard by a person deep inside the water? Answer: The velocity of sound in water is much lesser than the velocity of sound in air. So the sound waves are mostly reflected from the surface of the water. Only little refraction of sound from air to water takes place. Moreover, the refracted sound waves die off after traveling a small distance in the water. Hence no sound waves reach deep inside the water.

Question 18. The reverberation time is larger for an empty hall than a crowded hall. Explain why? Answer: The sound prolongs for a longer time in an empty hall as it does not get absorbed. In the case of a crowded hall, the sound is absorbed by the audience. Hence reverberation time is less in the crowded hall.

Question 19. If a balloon is filled with C0 2 gas, then how will it behave for sound as a lens? On filling it with hydrogen gas, what will happen? Answer: The speed of sound in CO 2 is less than in air (∴ v ∝ \(\frac{1}{\sqrt{\rho}}\)). So the balloon filled with CO 2 gas behaves as a concave lens.

When it is filled with hydrogen, the speed of sound in H2 gas becomes greater than in air. So it will behave as a convex lens.

Question 20. Why on reflection from the rigid surface, the wave does not change type, and only phase change occurs, while on reflection from a smooth surface, type changes while phase does not change? Answer: When a compression strikes the rigid surface, the surface does not move but pushes the compression back as such. On the other hand, the direction of vibration of the particles of the medium is reversed. Thus the type of wave does not change while phase change occurs.

When a compression strikes the smooth surface (i.e. yielding surface), the compression continues to move forward with the surface. But due to refraction, a part of the wave moves backward. Thus type changes while phase does not change.

Question 21. Explain why transverse elastic waves can’t propagate through a fluid? Answer: When a transverse elastic wave travels through a solid, a shearing strain develops which is supported by the elastic solid because of the development of a restoring force. Thus elastic waves can propagate through solids.

But a liquid or a gas (i.e. fluid) can’t support a shearing strain. So there will be no restoring forces when there are transverse displacements and so transverse vibrations are not possible.

Question 23. Distinguish between transverse waves and longitudinal waves. Answer: Longitudinal waves:

• Particles of the medium vibrate along the direction of propagation of the wave.
• They travel in the form of alternate compressions and rare¬factions.
• They can be formed in any medium i.e. solid, liquid, or gas.
• When these waves propagate, there are pressure changes in ‘ the medium.
• They can’t be polarised.

Transverse waves:

• Particles of the medium vibrate, to the direction of propagation of the wave.
• They travel in the form of alternate crests and troughs.
• These can be formed in solids and on the surfaces of liquids only.
• There are no pressure changes due to the propagation of these waves in the medium.
• They can be polarised.

Question 24. Distinguish between progressive waves and stationary waves. Answer: Progressive waves:

• The disturbance travels onward. It is1 handed over from one particle to the next.
• Energy is transported in the medium along with the propagation of waves.
• Each particle of the medium executes S.H.M. with the same amplitude.
• No particle of the medium is permanently at rest.
• Changes in pressure and density are the same at all points of the medium.

Stationary waves:

• The disturbance is confined to a particular region and there is no onward motion.
• No energy is transported in the medium.
• All the particles of the medium except at nodes execute S.H.M. with different amplitude.
• The particles of the medium at nodes are at rest.
• The changes of pressure and density are maximum at nodes and minimum at antinodes.

Question 25. Distinguish between musical sound and noise. Answer: Musical sound:

• It produces a pleasant effect on the ear.
• It has a high frequency.
• There are no sudden changes in the amplitude of the musical sound waves.
• It is a desirable sound.
• It produces an unpleasant effect on the ear.
• It has a low frequency.
• There are sudden changes in the amplitude of noise waves.
• It is an undesirable sound.

Question 26. What are the characteristics of wave motion? Answer:

• Wave motion is a form of disturbance that travels in a medium due to repeated periodic motion of the particles of the medium.
• The wave velocity is different from the particle velocity.
• The vibrating particles of the medium possess both K.E. and P.E.
• The particle velocity is different at different positions of its vibrations whereas wave velocity is constant throughout a given medium.
• Waves can undergo reflection, refraction, diffraction, dispersion, and interference.

Putting v o = 332 ms -1 at T0 i.e. 0°C, we get α = \(\frac{332}{546}\) = 0.61 ms -1 °C -1

Question 28. An electric bell is put in an evacuated room (a) near the center (b) close to the glass window, in which case the sound is heard (i) inside the room, (ii) out of the room. Answer:

• Sound is not heard in cases (a) and (b) inside the room as the medium is not there for the propagation of sound.
• In case (a) sound cannot be heard outside for the reason given in (i) above.

In case (b) since the bell is very close to the window, the glass pane picks up its vibrations which are conveyed to the eardrum through the air outside the room. So, the sound can be heard in condition (b).

Question 29. One of the primitive musical instrument is flute, yet produces good musical sound, how? Answer: The flute is an open organ pipe instrument having some holes which determine the wavelength and hence the frequency of sound. produced. By closing one or more holes, the length of the vibrating air column is changed and thus different harmonics are produced. The harmonic rich sound is a good musical sound.

Question 30. Write basic conditions for the formation of stationary waves. Answer: The basic conditions for the formation of stationary waves are listed below:

• The direct and reflected waves must be traveling along the same line.
• For stationary wave formation, the superposing waves should either be longitudinal or transverse. A longitudinal and a transverse wave cannot superpose.
• For the formation of stationary waves, there should not be any relative motion between the medium and oppositely traveling waves.
• The amplitude and period of the superposing waves should be the same.

Question 31. What is the difference between interference and stationary waves? In which phenomenon, out of the two, energy is not propagated? Why there is no energy at interference minimum? Answer: The superposition of two waves close to each other traveling in the same direction produces interference. The energy gets redistributed. It is minimum or zero at points of destructive interference and maximum at points of constructive interference. It is to be noted that the interference minima may not be points of zero energy unless •, the frequencies and amplitudes of the superposing waves are exactly equal.

The superposition of two similar waves (waves having the same amplitude and period) traveling in opposite direction produce stationary waves The nodes have no vibration of particles but antinodes have a maximum amplitude of vibration.

Question 32. Write the applications of beats. Answer: Beats are used to:

• Determine an unknown frequency by listening to the best frequency Δv. Then unknown frequency V’ = v ± Δv where v is known and it is close to the unknown frequency. The exact value of v’ is found by loading and filling the tuning fork of unknown frequency from which + or – sign is chosen.
• Tune musical instruments by sounding them together and reducing beats number to zero.
• Make a sound rich in musical effect by the deliberate introduction of beats between different musical instruments.
• To produce very low-frequency pulses which otherwise cannot be produced. The beat frequency is the low-frequency sound.
• Receive radio program by the superheterodyne method.
• Detect harmful gases in mines.

Waves Important Extra Questions Long Answer Type

Let vs = velocity of a source moving towards O at rest and let S reaches to S’ in one second. Thus the sound waves will be crowded in length (v – vs). So if λ’ be the new wavelength, Then λ’ = \(\frac{v-v_{s}}{v}\)

if υ’ be the apparent frequency, then υ’ = \(\frac{v}{\lambda^{\prime}}=\frac{v}{v-v_{s}}\)υ

∴ υ’ > v i.e. when S moves towards O, the apparent frequency of sound waves is greater than the actual frequency.

Let v o = velocity of an observer moving towards S at rest. As the observer moves towards S at rest, so the velocity of sound waves w.r.t. the observer is v + v o .

If υ’ = apparent frequency, then υ’ = \(\frac{\mathbf{v}+\mathbf{v}_{0}}{\lambda}=\frac{\mathbf{v}+\mathbf{v}_{0}}{\mathbf{v}}\)υ clearly v’ > υ.

When O moves towards S, then υ” = \(\left(\frac{v+v_{0}}{v}\right) v^{\prime}=\left(\frac{v+v_{0}}{v-v_{s}}\right) v\)

Question 2. Give the analytical treatment of beats. Answer: Consider two simple harmonic progressive waves traveling simultaneously in the same direction and in the same medium. Let

• A be the amplitude of each wave.
• There is no initial phase difference between them.
• Let v 1 and v 2 be their frequencies.

If y 1 and y 2 be displacements of the two waves, then y 1 = A sin 2π v 1 t and y 2 = A sin 2π v 2 t

∴ Amplitude becomes maximum at times given by t = \(0, \frac{1}{v_{1}-v_{2}}, \frac{2}{v_{1}-v_{2}}, \frac{3}{v_{1}-v_{2}}\)…..

∴ Time interval between two consecutive maxima is = \(\frac{1}{v_{1}-v_{2}}\)

∴ Beat period = \(\frac{1}{v_{1}-v_{2}}\) ∴ Beat frequency = v 1 – v 2 ∴ no. of beasts formed per sec. = v 1 – v 2 .

Question 3. What conditions are necessary for the good acoustical properties of the building? How are they met? Answer: An acoustically good building is one in which the sound is heard clearly in every nook and corner, some conditions must be fulfilled.

These are: (a) the building should have proper reverberation time, the reverberation time is given by Sabine’s formula which for a hall is T = \(\frac{0.166 \mathrm{~V}}{\sum \alpha \mathrm{A}}\) The reverberation time is adjusted by:

• changing the volume (V): This can be changed little due to the size of the hall already fixed.
• changing effective absorbing area: This can be done artificially by putting heavy curtains, paintings, providing open windows, wall coverings, etc.
• providing sufficiently energetic sound: The sound should be sufficiently loud and intelligible at every point.
• eliminating echo: Except for the desired one, echoes must be eliminated.
• properly focussing the sound: The sound has to be properly focussed to avoid the source of silence and also unreliable focussing.
• avoiding unique reinforcement: No single overtone should uniquely be reinforced then the total quality of the note will be affected.
• avoiding extra noise: Extra noises including resonance within the building has to be avoided.
• eliminating smooth curved surfaces: Smooth surfaces reflect sound sharply which may cause several problems.

To achieve these goals following steps need to be taken: (a) Properly design the building to optimize its acoustic condition. (b) Decorate selectively the building with paintings, floral designs, perforations. (c) Use false perforated or cardboard ceilings, and perforated structures at the curved walls. (d) Use carpets, upholster seats with holes in the bottom. (e) Use carpets or mats etc. on floors. (f) Heavy curtains, wall hangings, etc. should be used. (g) Properly place the mike and loud-speakers in the hall. (h) Avoid sharp corners in the hall and make the stage back parabolic with mike at its focus.

Numerical Problems:

Question 1. How far does the sound travel in the air when a tuning fork of frequency 280 Hz makes 15 vibrations? Given the velocity of sound is 336 ms -1 . Answer: Velocity of sound, v = 336 ms -1 frequency, v = 280 Hz

∴ Time of one complete vibration, T = \(\frac{1}{v}=\frac{1}{280}\) s. Time to complete 15 vibrations, t = 15 T = \(\frac{1}{280}\) × 15

If x be the distance covered in time t, then x = v × t = 336 × \(\frac{15}{280}\) = 18 m x = 18 m.

Question 2. Audible frequencies have a range of 20 Hz to 20 × 10 3 Hz. Express this range in terms of (i) period T, (ii) wavelength λ, (iii) angular frequency CD. Given the velocity of sound = 340 ms -1 . Answer: V = 340 ms -1 Using v 1 = 20 Hz v 2 = 20 × 10 3 Hz (i) T = \(\frac{1}{v}\), we get T 1 = \(\frac{1}{20}\) = 0.005 s and T 2 = \(\frac{1}{20}\) × 10 3 = 0.00005 s = 5 × 10 -5 s.

(iii) Angular frequency, ω = 2πv ∴ ω 1 = 2πv 1 = 2π × 20= 40π rads -1 ω 2 = 2πv 2 = 2π × 20 × 10 3 = 40π × 103 rad s -1 ∴ angular frequency range is 40π rad s -1 to 40π × 10 3 rad s -1

Question 3. A copper wire is held at the two ends by rigid supports. At 30°C, the wire is just taut with negligible tension. Find the speed of transverse waves in the wire at 10°C. α = 1.7 × 10 -5 °C -1 , λ = 1.4 × 10 11 Nm -1 and ρ = 9 × 10 3 kg m -3 Answer: When the temperature changes from 30°C to 10°C, then the change in length of the wire is Δl = l ∝ Δt = l × 1.7 × 10 -5 × 20 = 3.4 × l × 10 -4 m

Here, ρ = 9 × 10 3 kg m -3 Y = 1.4 × 10 11 Nm -2 Δt = 30 – 10 = 20°C α = 1.7 × 10 -5 °C -1

Question 4. The speed of sound in hydrogen is 1270 ms -1 . What will be the speed in a mixture of oxygen and hydrogen mixed in a volume ratio of 1:4? Answer: Let V 1 and V 2 be the volumes of O 2 and H 2 respectively in the mixture. ρ 1 and ρ 2 be their respective densities.

If m1 and m2 be the mass of oxygen and hydrogen respectively, then m 1 = V 1 ρ 1 and m 2 = V 2 ρ 2

Question 5. A tuning fork A of unknown frequency is sounded with a tuning fork B of frequency 288. Four beats are heard in one second. Tuning fork A is then loaded with a little wax and again sounded with fork B. Again 4 beats are heard in one second. What is the frequency of A? Answer: Since tuning fork A of known frequency say v gives 4 beats/ sec with B of frequency 288, so frequency of tuning fork A is given by ∴ v = 288 ± 4 = 284 or 292. On loading A with wax, it again produces 4 beats/s with B of frequency 288.

If the frequency of A is 292, it means on loading A with wax, its frequency falls to 284 which can produce 4 beats/s with a B frequency of 288. So the frequency of A maybe 292.

If the frequency of A is 284, on loading it with wax, its frequency falls below 284 so that when it is sounded with B of frequency 288, it forms more than 4 beats/s which is not consistent with the given condition.

So the frequency of A can’t be 284. Hence the frequency of A = 292.

Question 6. The first overtone of an open organ pipe beats with the first overtone of a closed organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of a closed organ pipe is 110 Hz. Find the lengths of the pipes. The velocity of sound in air = 330 ms -1 . Answer: Fundamental frequency of closed organ pipe is v c = \(\frac{\mathrm{v}}{4 \mathrm{~L}_{\mathrm{c}}}\) L c = length of closed pipe v or L c = \(\frac{\mathrm{v}}{4 \mathrm{~v}_{\mathrm{c}}}\)

Here, ν c = 110 Hz v = 330 ms -1

∴ L c = \(\frac{330}{4 \times 110}\) = 0.75m = 75 cm. 0.4 × 110 Now frequency of first overtone of open pipe is v 1 = \(\frac{\mathrm{v}}{\mathrm{L}_{0}}\) where L o = length of open pipe Frequency of first overtone of closed pipe is

Question 8. A set of 24 tuning forks is so arranged that each gives 4 beats per second with the previous one and the last sounds the octave of the first. Find the frequency of the first and last. Answer: As the last is the octave (double) of the first, so the frequencies are ¡n the increasing order. Let frequency of the first = x frequencyofthe2nd = x + 4 frequency olthe3rd = x + 2(4) frequency of the 24th = x + 23 (4) But this is the octave of the first ∴ frequency of the 24th = 2(x) (given) ∴ x + 23(4) = 2x or x = 92 ∴ frequency of last tuning fork 2 × 92 = 184 Hz.

Question 9. A drop of water 2 mm in diameter falling from a height of 50 cm in a bucket generates a sound that can be heard from a 5 m distance. Take all gravitational energy difference equal to sound energy, the transformation being spread in time over 0.2s, deduce the average intensity. Take g = 10 ms 2 . Answer: Here, ρ of water = 103 kg m -3 D = 2mm

∴ radius = r = \(\frac{2}{2}\) = 1 mm = 10 -3 m h = 50cm = 0.50 m g = 10 ms -2 d = 5m ∴ a = area in which sound is heard =4πd 2 I = intensity of sound =?

Question 10. A train moves towards a stationary observer with a velocity of \(\frac{1}{30}\)th of the velocity of sound. The whistle of the engine regularly blows after one second. What ¡s the interval between successive sounds of the whistle as heard by the observer? Answer: Actual time interval between successive blows of whistle = 1 sec.

∴ Actual frequency of whistle, v = 1 Hz Let v = velocity of sound ∴ velocity of source, vs = \(\frac{v}{30}\) ∴ apparent frequency of the whistle is ν’ = \(\frac{v}{v-v_{s}}\) × ν = \(\frac{v}{v-\frac{v}{30}}\) × 1 or ν’ = \(\frac{30}{29}\) Hz

∴ Apparent time interval between two successive sounds of whistle is t = \(\frac{1}{v^{\prime}}=\frac{29}{30}\) s.

Question 11. The splash is heard 4.2,3 s after a stone is dropped into a well which is 78.4 m deep. Find the velocity of sound in the air. Answer: Here, Depth of well, h = 78.4 m v = velocity of sound = ? t = total time after which splash is heard = 4.23 s. t 1 = time taken by stone to hit water surface in the well. t 2  = time the sound takes in reaching top of the well. ∴ t 1 +t 2 = 4.23 s u = 0, g = 9.8 ms -2 , h = 78.4m, t = t 1

Question 12. What is the intensity level in dB of sound whose intensity is 10 -6 watt m -2 . Take zero levels of intensity = 10 -2 watt m -2 . Answer: Here, I = 10-6 w m -2 I 0 = 10 -12 w m -2

L =? L = log10 \(\left(\frac{I}{I_{0}}\right)\) = log10 \(\left(\frac{10^{-6}}{10^{-12}}\right)\) = log 10 10 6 = 6 log 10 10 = 6 × 1 = 6B As 1B = 10 dB ∴ L = 6 × 10 = 60 dB

Question 14. A police-man on duty detects a drop of 15% in the pitch of the horn of a motor car as it crosses him. If the velocity of sound is 330 ms -1 , calculate the speed of a car. Answer: v s = speed of car = ? v = velocity of sound = 330 ms -1 v = frequency of sound emitted by horn If v’ be the apparent frequency’of sound when car approaches the police man, then ν’ = \(\frac{\mathrm{v}}{\mathrm{v}-\mathrm{v}_{\mathrm{s}}}\) × ν …(i)

Question 15. A wire of length 80 cm has a frequency of 250 Hz. If the length of the wire is increased to 100 cm and tension is reduced to \(\frac{1}{4}\)th of its original value, then calculate the new frequency. Answer: l 1 = initial length = 80 cm ν 1 = 250 Hz

Let T 1 = T be the tension l 2 = 100 cm T 2 = \(\frac{1}{4}\)T 1 = \(\frac{T}{4}\). ν = ?

Question 16. A tuning fork A of unknown frequency is sounded with a tuning fork B of frequency 512 Hz. 8 beats are heard in 1 second. Tuning fork A is then loaded with a little wax and sounded together with B. Again 8 beats are heard in 1 second. Find the frequency of Answer: Let ν = frequency of tuning fork A No. of beats formed/sec. =8 ∴ ν = 512 ± 8 = 520, 504

On loading A with wax, it again produces 8 beats/second with B of frequency 512. Now if ν A = 520, then it means its frequency decreases on loading. If v. falls to 504 then it will produce 8 beats/sec. with B. ∴ ν A = 520

If ν A is 504, then on loading its frequency falls below 504, so it forms more than 8 beats/sec. when sounded together with B which is not consistent with the given condition.

So the frequency of A cannot be 504. ∴ ν A = 520 Hz.

Question 17. Two engines pass each other in opposite directions with a velocity of 60 km ph. One of them is emitting a note of frequency 540 Hz. Calculate the frequencies heard in the other engine before and after they have passed each other. The velocity of sound = 316.67 ms -1 . Answer: Here, v s = v o = 60 kmph = 60 × \(\frac{5}{18}\) = 16.67 ms -1 ν = 540 Hz ν’ = ? v = velocity of sound = 316.67 ms -1

Case A: Before passing each other: v = 316.67 ms -1 v s = v o = 16.67 ms -1 ∴ ν’ = ?

Question 18. The audible frequency range of a human ear is 20 Hz to 20 kHz. Convert this into the corresponding wavelength range. Take the speed of sound in the air to be 340 ms -1 . Answer: Here, ν 1 = 20 Hz ν 2 = 20000 Hz v = 340 ms -1 λ 1 = ? λ 2 = ?

∴ λ 1 = \(\frac{v}{v_{1}}=\frac{340}{20}\) = 17 m and λ 2 = \(\frac{v}{v_{2}}=\frac{340}{20000}\) = 0.017 m

∴ Corresponding wavelength range is 0.017 to 17 m.

Question 19. For aluminum, the bulk modulus and the modulus of rigidity are 7.5 × 10 10 Nm -2 and 2.1 × 10 10 Nm -2 . Find the velocity of longitudinal and transverse waves in the medium. The density of aluminum is 2.7 × 10 3 kg m -3 . Answer: Here, for aluminium, ρ = 2.7 × 10 3 kg m -3 . η = 2.1 × 10 10 Nm -2 k = 7.5 × 10 10 Nm -2 v l = velocity of longitudinal waves = ? v t = velocity of transverse waves = ?

Question 20. The velocity of sound at 27°C is 350 ms -1 . If the density of air at N.T.P. is 1.293 kg m -3 , calculate the ratio of the specific heats of air. Answer: Here, T = 27 + 273 = 300k v 27 =. velocity of sound at 27°C = 350 ms -1 ρ = 1.293 kg m -3 T o = 273 k v o = velocity at N.T.P. = ? P = 76 cm of Hg = 0.76 × 13600 × 9.8 Nm -2

Question 21. A tuning fork B produces 6 beats per second with another tuning fork of frequency 288 Hz. The tuning fork of unknown frequency is filed and the number of beats now produced is 4 per second. Calculate the frequency of the tuning fork. Answer: Known frequency of the fork =288 Hz No. of beats/second = 6 ∴ Unknown frequency = 288 ± 6 = 294 or 282.

When the prongs of tuning fork B are filed, its frequency increases. If the frequency of B is originally 294 Hz, on filing it will increase. So the difference between the frequencies will increase and hence no. of beats will also increase.

If the frequency of B is originally 282 Hz, then on filing it, its frequency will increase, so the difference between the frequencies will decrease, and hence the no. of beats will decrease.

In this case, when the prongs of B are filed, the no. of beats decreases from 6 to 4. Thus the frequency of the given tuning fork is 282 Hz.

Question 22. Find the change in the frequency observed by a listener when a source approaching him with a velocity of 54 km h -1 starts going away from him with the same velocity. Take velocity of sound = 330 ms -1 and v = 300 Hz. Answer: Here, v s = 54 km h -1 = 54 × \(\frac{5}{18}\) = 15 ms -1 v = 330 ms -1 ν = 300 Hz Case I: When the source is approaching, then ν’ 1  = \(\frac{\mathbf{v}}{\mathbf{v}-\mathbf{v}_{\mathbf{s}}}\)ν = \(\frac{330}{330-15}\) × 300 = 314 Hz

Case II: When the source reduces then ν’ 2 = \(\frac{\mathbf{v}}{\mathbf{v}+\mathbf{v}_{\mathbf{s}}}\)ν = \(\frac{330}{330+15}\) × 300 = \(\frac{330}{345}\) × 300 = 287 Hz

∴ Change in frequency, ν’ 1 – ν’ 2 = 314 – 287 = 27 Hz.

Question 23. A progressive wave is given by y = 12 sin (5t – 4x) On this wave, how far away are the two points having a phase difference of \(\frac{π}{2}\)? Answer: Here, ΔΦ = phase difference = \(\frac{π}{2}\)

Question 25. Two sound waves y 1 = A 1 sin 1000 π(t – \(\frac{x}{220}\) ) and y 2 = A 2 sin 1010 π(t – \(\frac{x}{220}\) ) are superposed. What is the frequency with which the amplitude varies? Answer: Rate of variation of amplitude is equal to the beat frequency. Here, 2πν 1 = 1000 π or ν 1 = 500 and 2πν 2 = 1010 π or ν 2 = 505

∴ beat frequency = ν 2 – ν 1 = 505 – 500 = 5.

Value-Based Type:

Question 1. A group of students went to a place on an excursion. While boating on seawater, the students identified a submerged Torpedo-shaped structure. The boys debated among themselves on what they saw. A student by the name of Sharath considering it as a threat informed the police. The police took necessary steps to protect the country from the enemy submarine. Sharath was rewarded. (a) What can you say about the qualities exhibited by Sharath? Answer: Navigator is a responsible citizen, he is duty-minded, having a presence of mind.

(b) A SONAR system fixed in a submarine operates at a frequency of 40 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km/hr. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to bel450 m/s. Answer: Here, frequency of SONAR (Source) = 40 k Hz = 4 × 10 3 Hz Speed of the sound wave, V = 1450 m/s Speed of the observer, V o = 360 kM/h = 100 m/s

∴ Apparent frequency received by an enemy submarine ; V’ = \(\left(\frac{V+V_{0}}{V}\right) \cdot V=\left(\frac{1450+100}{1450}\right)\) × 40 × 10 3 =4.276 × 10 4 Hz.

This frequency is reflected by the enemy submarine (Source) and observed by SONAR (Now observer). In this case, Apparent frequency is given by V” = \(\left(\frac{V}{V-V_{s}}\right) \times V^{\prime}=\left(\frac{1450}{1450-100}\right)\) × 4.276 × 10 4 =45.9k Hz

Question 2. Jagat and Ram are working in the same company. Jagat has noticed that Ram is suffering from Cancer. Ram is not aware of this. When Jagat asks him to go for a checkup, Ram refused. He gets convinced how even when he realizes normal, it is very important to get the checkup done once a year. (a) What according to you, are the values displayed by Jagat in helping Ram. Answer: He has concern for hi^ friend, also he has the knowledge of medical facilities available

Question 3. ‘Preeti a student of class XI was reading the newspaper, The Headlines in the Newspaper were about the earthquake that had taken place in Assam on the previous day. She was very depressed seeing the loss of life and property. She approached her physics teacher and got the information about how an earthquake occurs. (a) What can you say about the inquisitiveness of Preeti? Answer: She has concern for society and is sympathetic towards others.

(b) Earthquake generates sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about 4 km/s, and that of the P wave is 8 km/s. A seismograph records P and S waves from an earthquake. The first P wave arrives 4 min before the first S wave. Assuming the waves travel in a straight line, how far away does the earthquake occur? Answer: LetV 1 and V 2 be the velocities of S waves and P waves and t 1 , t 2 be the time taken by these waves to travel to the position of seismograph. If l is the distance of occurrence of earthquake from the seismograph, . ∴ l = V 1 t 1 = V 2 t 2 ⇒ 4t 1 = 8 t 2 or t 1 = 2 t 2 [∴ V 1 = 4 kmh -1 ,V 2 = 8 kmh -1 ]

Also, t 1 – t 2 = 4 min = 240 S. 2t 2 – t 2 = 240 [Using (i)] or t 2 = 240 S and t 1 = 2 × 240 = 480 S. l = v 1 t 1 = 4 × 480 = 1920 km

Question 4. Rajesh was waiting for the train on the platform with his parents. They were going to Mumbai and the train arrived half an hour late. He felt that when the train was coining towards the station, the intensity of the sound of the whistle was gradually increasing and on the platform the sound was maximum and when the train passes away the intensity of the whistle was decreasing. Rajesh got confused and asked his father the reason behind it. His father could not give the correct answer. So, he decided to ask his physics teacher. (i) What are the values displayed here by Rajesh? Answer: Values displayed by Rajesh are: Awareness, curiosity, creativity, and intelligence,

(ii) Write the answer given by your teacher. Answer: The teacher explained that it is due to “Doppler’s effect”. Whenever there is a relative motion between the source and observer O. There is a change in observed frequency. So, the apparent frequencies of sound heard by the listener are different from the actual frequency of the sound of the source, which is given by. V = V o \(\left(\frac{V+V_{0}}{V+V_{s}}\right)\)

Here, V is the speed of sound through the j medium, V o is the velocity of the observer relative to the medium, and Vs is the source velocity relative to the medium.

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NCERT Solutions for Class 11 Physics Chapter 14 - Waves

Wave is a quivering disruption that passes through a medium because of repetition of regular motion of particles of any medium. Ch 14 Physics Class 11 helps create a solid base for future courses like engineering and medical science.

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NCERT solutions for Class 11 Physics Chapter 14 introduce students to many new concepts like effects of Doppler, types of waves and their interrelationships. Students in class 11 have to gain in-depth knowledge about different topics on this subject to score better in their exams, as well as rank higher in competitive tests.

Vedantu’s PDF on Physics Class 11 Waves NCERT solutions will help students get familiar with the increasing importance of waves in various mediums.

Mechanical Properties Of Solids Chapter at a Glance - Class 11 NCERT Solutions

A wave is a disturbance that propagates in space, transport energy and momentum from one point to another without the transport of matter.

Mechanical transverse waves are produced in such type of medium which have shearing property, so they are known as shear wave or S-wave 

A crest is a portion of the medium, which is raised temporarily above the normal position of rest of particles of the medium, when a transverse wave passes.  

A trough is a portion of the medium, which is depressed temporarily below the normal position of rest of particles of the medium, when a transverse wave passes.

Longitudinal Wave Motion:   Longitudinal waves have oscillatory motion of the medium particles that produce regions of compression (high pressure) and rarefaction (low pressure) which propagate in space with time (see figure).

The regions of high particle density are called compressions and regions of low particle density are called rarefactions.

Wavelength $\lambda $ (length of one wave): Distance traveled by the wave during the time interval in which any one particle of the medium completes one cycle about its mean position. We may also define wavelength as the distance between any two nearest particle of the medium, vibrating in the same phase

Phase: Phase is a quantity which contains all information related to any vibrating particle in a wave. For equation $y= A\;sin\left ( \omega t-kx \right ); \left ( \omega t-kx \right )=phase$

Wave number $(v\bar{})$ : it is defined as $(v\bar{})=\frac{1}{\gamma \lambda }=\frac{k}{2\pi }=$ number of waves in unit length of the wave pattern.

Differential equation of Harmonic Progressive Waves:

Wave velocity: The velocity with which the disturbance, or planes of equal (wave front), travel through the medium is called wave (or phase) velocity

Transverse wave: A transverse wave is a moving wave whose oscillations are perpendicular to the direction of the wave

The speed of a wave on a string is given by 

$v=\sqrt{\frac{T}{\mu }}$

where T is tension in the string (in Newtons) and $\mu$  is mass per unit length of the string (kg/m). 

When a traveling wave s established on a string, energy is transmitted along the direction of propagation of the wave, in form of potential energy and kinetic energy

Intensity of Sound Waves: The amount of energy carried per unit time by a wave is called its power and power per unit area held perpendicular to the direction of energy flow is called intensity.

Loudness: Audible intensity range for humans: The ability of human to perceive intensity at different frequencies are different. The perception of intensity is maximum at 1000 Hz and perception of intensity decreases as the frequency decreases or increases from 1000Hz.

Decibel Scale: The logarithmic scale which is used for comparing two sound intensity is called decibel scale. The intensity level $\beta$

described in terms of decibels is defined as $\beta = 10 \;log\left ( \frac{I}{I_{0}} \right )(dB)$

Superposition of Waves : The phenomenon of intermixing of two or more waves to produce a new wave is called Superposition of waves. Therefore, according to the superposition principle. 

The resultant displacement of a particle at any point of the medium, at any instant of time is the vector sum of the displacement caused to the particle by the individual waves.

Coherence: Two sources are said to be coherent if the phase difference between them does not change with time. In this case their resultant intensity at any point in space remains constant with time. Two independent sources of sound are generally incoherent in nature, i.e. phase difference between them changes with time and hence the resultant intensity due to them at any point in space changes with time.

Standing Waves: Standing waves can be transverse or longitudinal, e.g., in strings (under tension) if reflected wave exists, the waves are transverse-stationary, while in organ pipes waves are longitudinal-stationary.

Beats: When two sound waves of same amplitude and different frequency superimpose, then intensity at any point in space varies periodically with time. This effect is called beats. Beat phenomenon can be used for determining an unknown frequency by sounding it together with a source of known frequency.

Doppler’s Effect: The apparent change in frequency or pitch due to relative motion of source an observer along the line of sight is called the Doppler Effect.

Assumptions: (i) The velocity of the source, the observer and the medium are along the line joining the positions of the source and the observer. 

(ii) The velocity of the source and the observer is less than velocity of sound.

Access NCERT Solutions for Class 11 Physics Chapter 14 – Waves

1. A string of mass \[\mathbf{2}.\mathbf{50}\text{ }\mathbf{kg}\] is under a tension of \[\mathbf{200}\text{ }\mathbf{N}\]. The length of the stretched string is \[\mathbf{20}.\mathbf{0m}\]. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?

Ans: It is provided that,

Mass of the string, \[M=2.50\text{ }kg\]

Tension in the string, \[T=200\text{ }N\]

String length, \[\text{l}=20.0\text{ }m\]

Mass per unit length, \[\mu =\frac{M}{l}=\frac{2.50}{20}=0.125kg{{m}^{-1}}\]

The transverse wave’s velocity in the string is given by:

\[v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{200}{0.125}}=\sqrt{1600}=40m{{s}^{-1}}\]

Clearly, the time taken by disturbance to reach the other end is, \[t=\frac{l}{v}=\frac{20}{40}=0.50s\]

2. A stone dropped from the top of a tower of height \[\mathbf{300}\text{ }\mathbf{m}\]high splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\]? (\[\mathbf{g}=\mathbf{9}.\mathbf{8}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{2}}}\])

Tower height, \[s=300\text{ }m\]

Stone’s initial velocity, \[u=0\]

Acceleration, \[a=\mathbf{g}=\mathbf{9}.\mathbf{8}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{2}}}\]

Sound speed in air \[=340\text{ }m/s\]

The time that stone takes to strike the water in the pond can be estimated using the motion’s second equation, as:

\[s=u{{t}_{1}}+\frac{1}{2}gt_{1}^{2}\]

\[\Rightarrow 300=0+\frac{1}{2}\times 9.8\times t_{1}^{2}\]

\[{{t}_{1}}=\sqrt{\frac{300\times 2}{9.8}}=7.82s\]

Time taken by the sound to reach the tower top, \[{{t}_{2}}=\frac{300}{340}=0.88s\]

Therefore, the time after which the sound of splash is heard, \[t={{t}_{1}}+{{t}_{2}}\]

\[\Rightarrow t=7.82+0.88=8.7s\]

The time after which the sound of splash is heard is $8.7s$.

3. A steel wire has a length of \[\mathbf{12}.\mathbf{0}\text{ }\mathbf{m}\] and a mass of \[\mathbf{2}.\mathbf{10}\text{ }\mathbf{kg}\]. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at \[20{}^{0}C=343m{{s}^{-1}}\].

Steel wire’s length, \[l=12\text{ }m\]

Steel wire’s mass, \[m=2.10\text{ }kg\]

Velocity of the transverse wave, \[v=343\text{ }m{{s}^{-1}}\]

Mass per unit length, \[\mu =\frac{m}{l}=\frac{2.10}{12}=0.175kg{{m}^{-1}}\]

For tension T, transverse wave’s velocity can be calculated using the relation:

\[v=\sqrt{\frac{T}{\mu }}\]

\[\Rightarrow T={{v}^{2}}\mu \]

\[\Rightarrow T={{(343)}^{2}}\times 0.175=20588.575\approx 2.06\times {{10}^{4}}N\]

Tension in the wire is \[2.06\times {{10}^{4}}N\].

4. Use the formula \[v=\sqrt{\frac{\gamma P}{\rho }}\] to explain why the speed of sound in air

(a) is independent of pressure,

Ans: We have,

\[v=\sqrt{\frac{\gamma P}{\rho }}\]                   …...(i)

Where, 

Density, \[\rho =\frac{Mass}{Volume}=\frac{M}{V}\]

M = Molecular weight of gas

V = Volume of gas

Hence, equation (i) becomes:

\[v=\sqrt{\frac{\gamma PV}{M}}\]              ……(ii)

Now ideal gas equation for \[n=1\] is:

For constant T, \[PV=Constant\]

Both \[M\] and \[\gamma \] are constants, \[v=Constant\]

Hence, the speed of sound is independent of the change in the pressure of the gas at a constant temperature.

(b) increases with temperature,

\[v=\sqrt{\frac{\gamma P}{\rho }}\]               ……(i)

\[P=\frac{RT}{V}\]               ……(ii)

Substituting (ii) in (i), we get:

\[v=\sqrt{\frac{\gamma RT}{V\rho }}=\sqrt{\frac{\gamma RT}{M}}\]  ……(iii)

Mass, \[M=\rho V\]is a constant

\[\gamma \]and \[R\]are also constants.

We get from equation (iii),

\[v\propto \sqrt{T}\].

Hence, the sound speed in a gas is directly proportional to the square root of the gaseous medium’s temperature, i.e., the sound speed increases with rise in the gaseous medium’s temperature and vice versa.

(c) increases with humidity.

Ans: Let \[{{v}_{m}}\] and \[{{v}_{d}}\] are the sound speed in moist air and dry air respectively and \[{{\rho }_{m}}\] and \[{{\rho }_{d}}\] are the densities of moist air and dry air respectively.

\[v=\sqrt{\frac{\gamma P}{\rho }}\]

The speed of sound in moist air is:

\[{{v}_{m}}=\sqrt{\frac{\gamma P}{{{\rho }_{m}}}}\]           ……(i)

The speed of sound in dry air is:

\[{{v}_{d}}=\sqrt{\frac{\gamma P}{{{\rho }_{d}}}}\]           ……(ii)

On dividing equations (i) and (ii), we get:

\[\frac{{{v}_{m}}}{{{v}_{d}}}=\sqrt{\frac{\gamma P}{{{\rho }_{m}}}\times \frac{{{\rho }_{d}}}{\gamma P}}=\sqrt{\frac{{{\rho }_{d}}}{{{\rho }_{m}}}}\]       ……(iii)

However, the presence of water vapour decreases the density of air, i.e.,

\[{{\rho }_{d}}<{{\rho }_{m}}\] 

\[\Rightarrow {{v}_{m}}>{{v}_{d}}\]

Hence, the speed of sound in moist air is higher than it is in dry air. Thus, in a gaseous medium, the sound speed increases with humidity.

5. You have learnt that a travelling wave in one dimension is represented by a function \[y=f(x,t)\] where x and t must appear in the combination \[x-vt\] or \[x+vt\], i.e. \[y=f(x\pm \nu t)\]. Is the converse true? Examine if the following functions for y can possibly represent a travelling wave:

(a) \[{{(x-vt)}^{2}}\]

Ans:  No. 

For \[x=0\] and \[t=0\], the function \[{{(x+vt)}^{2}}\] becomes \[0\].

Hence, for \[x=0\] and \[t=0\], the function represents a point.

(b)  \[log\left[ \frac{x+vt}{{{x}_{0}}} \right]\]

Ans: Yes. 

For \[x=0\] and \[t=0\], the function \[\log \left( \frac{x+vt}{{{x}_{0}}} \right)=\log 0=\infty \]

Since the function does not converge to a finite value for \[x=0\] and \[t=0\], it does not represent a travelling wave.

(c) \[\frac{1}{(x+vt)}\]

Ans: No. 

For \[x=0\] and \[t=0\], the function

 \[\frac{1}{x+vt}=\frac{1}{0}=\infty \]

The converse is not true. The requirement for a wave function of a travelling wave is that for all x and t values, wave function should have a finite value. Therefore, none can represent a travelling wave.

6. A bat emits ultrasonic sound of frequency \[\mathbf{1000}\text{ }\mathbf{kHz}\] in air. If the sound meets a water surface, what is the wavelength of 

(a) the reflected sound, 

Frequency of the ultrasonic sound, \[\nu =1000kHz={{10}^{6}}Hz\]

Speed of sound in air, \[{{v}_{a}}=340m{{s}^{-1}}\]

The wavelength \[({{\lambda }_{r}})\]of the reflected sound is given by:

\[{{\lambda }_{r}}=\frac{{{v}_{a}}}{\nu }=\frac{340}{{{10}^{6}}}=3.4\times {{10}^{-4}}m\]

(b) the transmitted sound? Speed of sound in air is \[340m{{s}^{-1}}\] and in water \[1486m{{s}^{-1}}\].

Ultrasonic sound’s frequency, \[\nu =1000kHz={{10}^{6}}Hz\]

Sound speed in water, \[{{v}_{w}}=1486m/s\]

The wavelength is given as: \[{{\lambda }_{t}}=\frac{1486}{{{10}^{6}}}=1.49\times {{10}^{-3}}m\]

7. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is \[\mathbf{1}.\mathbf{7}\text{ }\mathbf{km}{{\mathbf{s}}^{-\mathbf{1}}}\]? The operating frequency of the scanner is \[\mathbf{4}.\mathbf{2}\text{ }\mathbf{MHz}.\]

Ans: It is provided that, 

Sound speed in the tissue, \[v=1.7Km{{s}^{-1}}=1.7\times {{10}^{3}}m{{s}^{-1}}\]

Scanner’s operating frequency, \[\nu =4.2MHz=4.2\times {{10}^{6}}Hz\]

The wavelength of sound wave in the tissue is given by:

\[\lambda =\frac{v}{\nu }=\frac{1.7\times {{10}^{3}}}{4.2\times {{10}^{6}}}=4.1\times {{10}^{-4}}m\]

The wavelength of sound in the tissue is \[4.1\times {{10}^{-4}}m\].

8. A transverse harmonic wave on a string is described by \[y(x,t)=3.0\sin \left( 36t+0.018x+\frac{\pi }{4} \right)\]

Where x and y are in cm and t in s. The positive direction of x is from left to right.

(a) Is this a travelling wave or a stationary wave?

If it is travelling, what are the speed and direction of its propagation?

Ans: The given equation is the equation of a travelling wave, moving from right to left because it is an equation of the type

\[y(x,t)=A\sin (\omega t+kx+\phi )\]

Here, \[A=3.0cm\], \[\omega =36ra{{d}^{-1}},k=0.018cm\]and \[\phi =\frac{\pi }{4}\]

\[\therefore \]Speed of wave propagation is given by,

 \[v=\frac{\omega }{k}=\frac{36rad{{s}^{-1}}}{0.018c{{m}^{-1}}}=\frac{36rad{{s}^{-1}}}{0.018\times {{10}^{2}}{{m}^{-1}}}=20m{{s}^{-1}}\]

The speed of wave propagation is \[20m{{s}^{-1}}\].

(b) What are its amplitude and frequency?

Ans: Amplitude of wave, \[A=3.0cm=0.03m\]

Frequency of wave, \[\nu =\frac{\omega }{\,2\pi }=\frac{36}{2\pi }=5.7Hz\]

(c) What is the initial phase at the origin?

Ans: Initial phase at origin, \[\phi =\frac{\pi }{4}\]rad

(d) What is the least distance between two successive crests in the wave?

Ans: Least distance between two successive crests in the wave,\[\Rightarrow \lambda =\frac{2\pi }{k}=\frac{2\pi }{0.018}=349cm=3.49m\]

9. For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for \[\mathbf{x}=\mathbf{0},\text{ }\mathbf{2}\text{ }\mathbf{and}\text{ }\mathbf{4}\text{ }\mathbf{cm}\]. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase?

Ans: All the waves have different phases.

The given transverse harmonic wave is:

\[y(x,t)=3.0\sin \left( 36t+0.018x+\frac{\pi }{4} \right)\]         ……(i)

For \[x=0\], the equation becomes:

\[y(0,t)=3.0\sin \left( 36t+\frac{\pi }{4} \right)\]

Also, \[\omega =\frac{2\pi }{T}=36rad/s\]

\[\therefore T=\frac{2\pi }{36}s\]

Now, plotting y vs t graphs using the different values of t, as listed in the given table.

For \[x=0,\text{ }x=2,\text{ }and\text{ }x=4\], the phases of the three waves will get altered. This is because amplitude and frequency are same for any change in x. The \[y-t\] plots of the three waves are shown in the given figure.

10. For the travelling harmonic wave \[y(x,t)=2.0\cos 2\pi \left( 10t-0.0080x+0.35 \right)\]

Where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of \[\mathbf{4}\text{ }\mathbf{m},\text{ }\mathbf{0}.\mathbf{5}\text{ }\mathbf{m},\] \[\frac{\lambda }{2}\], \[\frac{3\lambda }{4}\].

(a) \[\mathbf{4}\text{ }\mathbf{m}\]

Ans: Equation for a travelling harmonic wave is given by:

\[y(x,t)=2.0\cos 2\pi \left( 10t-0.0080x+0.35 \right)\]

\[\Rightarrow y(x,t)=2.0\cos (20\pi t-0.016\pi x+0.70\pi )\]

Where, Propagation constant, \[k=0.0160\pi \]

Amplitude, \[a=2cm\]

Angular frequency, \[\omega =20\pi rad/s\]

Phase difference is given by:

\[\phi =kx=\frac{2\pi }{\lambda }\]

for \[x=4m=400\text{ }cm\]

\[\Rightarrow \phi =0.016\pi \times 400=6.4\pi \] rad

(b) \[\mathbf{0}.\mathbf{5}\text{ }\mathbf{m}\]

Ans: Phase difference is given by:

\[\phi =kr=\frac{2\pi }{\lambda }\]

For \[x=0.5\text{ }m=50\text{ }cm\]

\[\Rightarrow \phi =0.016\pi \times 50=0.8\pi \] rad

(c) \[\frac{\lambda }{2}\]

For \[x=\frac{\lambda }{2}\]

\[\Rightarrow \phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{2}=\pi \] rad

(d) \[\frac{3\lambda }{4}\]

For \[x=\frac{3\lambda }{4}\]

\[\Rightarrow \phi =\frac{2\pi }{\lambda }\times \frac{3\lambda }{4}=1.5\pi rad\]

11. The transverse displacement of a string (clamped at its both ends) is given by \[y(x,t)=0.06\sin \frac{2\pi x}{3}\cos (120\pi t)\] Where x and y are in m and t in s. The length of the string is \[\mathbf{1}.\mathbf{5}\text{ }\mathbf{m}\] and its mass is \[\mathbf{3}.\mathbf{0}\text{ }\times \text{ }\mathbf{1}{{\mathbf{0}}^{-\mathbf{2}}}\mathbf{kg}\].

Answer the following:

(a) Does the function represent a travelling wave or a stationary wave?

Ans: The general equation of stationary wave is given by:

\[y(x,t)=Asin(kx)cos(\omega t)\]

This given equation is similar to the equation of stationary wave:

\[y(x,t)=0.06\sin \frac{2\pi x}{3}\cos (120\pi t)\]

Hence, the given function is a stationary wave. 

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

Ans: The transverse displacement of the string is given as:

\[y(x,t)=0.06\sin \left( \frac{2\pi }{3}x \right)\cos (120\pi t)\] 

From above equation, $k=\frac{2\pi }{3}$

\[\therefore \] Wavelength, $\lambda =\frac{2\pi }{k}=\frac{2\pi }{\frac{2\pi }{3}}=3m$

It is given that:

\[120\pi =2\pi \nu \]

Frequency, \[\nu =60\]Hz

Wave speed, \[v=\nu \lambda =60\times 3=180m{{s}^{-1}}\]

(c) Determine the tension in the string.

Ans: The transverse wave’s velocity travelling in a string is given by the relation:

\[v=\sqrt{\frac{T}{\mu }}\]    ……(i)

Where, Velocity of the transverse wave, \[v=180m{{s}^{-1}}\]

Mass of the string, \[m=3.0\times {{10}^{-2}}kg\]

Length of the string, \[l\text{ }=\text{ }1.5\text{ }m\]

Mass per unit length of the string, \[\mu =\frac{m}{l}=\frac{3.0}{1.5}\times {{10}^{-2}}=2\times {{10}^{-2}}kg{{m}^{-1}}\]

Tension in the string \[=\text{ }T\]

From equation (i), tension can be obtained as:

\[T={{v}^{2}}\mu \]

\[\Rightarrow T={{(180)}^{2}}\times 2\times {{10}^{-2}}\]

\[\Rightarrow T=648N\]

The tension in string is $648N$.

12.(i) For the wave on a string described in Question 11, do all the points on the string oscillate with the same 

(a) frequency, 

Ans: Yes, all the points on the string vibrate with the same frequency, except at the nodes which are having zero frequency.

Ans: Yes, all the points in any oscillating loop have the same phase, except at the nodes.

(c) amplitude? Explain your answers. 

Ans: No, all the points in any oscillating loop have different vibration amplitudes.

(ii) What is the amplitude of a point \[\mathbf{0}.\mathbf{375}\text{ }\mathbf{m}\] away from one end?

Ans: The given equation is:

\[y(x,t)=0.06\sin \left( \frac{2\pi }{3}x \right)\cos \left( 120\pi t \right)\]

For x = 0.375 m and t = 0

Amplitude\[=0.06\sin \left( \frac{2\pi }{3}x \right)\cos 0\]

\[\Rightarrow a=0.06\sin \left( \frac{2\pi }{3}\times 0.375 \right)\times 1\]

\[\Rightarrow a=0.06\sin \left( 0.25\pi  \right)=0.06\sin \left( \frac{\pi }{4} \right)\]

\[\Rightarrow a=0.06\times \frac{1}{\sqrt{2}}=0.042m\]

The value of amplitude is \[0.042m\].

13. Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent, (i) a travelling wave, (ii) a stationary wave or (iii) none at all:

(a) \[y=2\cos (3x)sin(10t)\]

Ans: This equation demonstrates a stationary wave because the harmonic terms \[kx\] and \[wt\] seem separately in the equation.

(b) \[y=2\sqrt{x-vt}\]

Ans: This equation is not having any harmonic term. Therefore, it is not representing either a stationary wave or travelling wave.

(c) \[y=3\sin (5x-0.5t)+4\cos (5x-0.5t)\] 

Ans: This equation demonstrates a travelling wave as it is having harmonic terms \[kx\] and \[wt\] are in \[kx~wt\] combination.

(d) \[y=\cos x\sin t+\cos 2x\sin 2t\]

Ans: This equation demonstrates a stationary wave because it is having harmonic terms \[kx\] and \[wt\] separately in the equation. It is the superposition of two stationary waves.

14. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of \[\mathbf{45}\text{ }\mathbf{Hz}\]. The mass of the wire is \[\mathbf{3}.\mathbf{5}\text{ }\times \text{ }\mathbf{1}{{\mathbf{0}}^{-\mathbf{2}}}\mathbf{kg}\] and its linear mass density is \[\mathbf{4}.\mathbf{0}\text{ }\mathbf{x}\text{ }\mathbf{1}{{\mathbf{0}}^{-\mathbf{2}}}\mathbf{kg}\text{ }{{\mathbf{m}}^{-\mathbf{1}}}\]. What is 

(a) the speed of a transverse wave on the string, and 

Ans: Provided that,

Mass of the wire, \[m=3.5\times {{10}^{-2}}kg\]

Linear mass density, \[\mu =\frac{m}{l}=4.0\times {{10}^{-2}}kg{{m}^{-1}}\]

Frequency of vibration, \[\nu =45Hz\]

 Length of the wire, \[l=\frac{m}{\mu }=\frac{3.5\times {{10}^{-2}}}{4.0\times {{10}^{-2}}}=0.875m\]

The wavelength of the stationary wave \[(\lambda )\] is given by: 

\[\lambda =\frac{2l}{n}\] where, \[n=\]number of nodes

 For fundamental node, \[n=1\]: 

\[\lambda =2l\]

\[\lambda =2\times 0.875=1.75m\]

The speed of the transverse wave in the string is given as: 

\[v=\nu \lambda =45\times 1.75=78.75m{{s}^{-1}}\]

The speed of transverse wave is \[78.75m{{s}^{-1}}\].

(b) the tension in the string?

Ans: The tension produced in the string is given by the relation: 

\[\Rightarrow T={{(78.75)}^{2}}\times 4.0\times {{10}^{-2}}=248.06N\]

The tension in the string is $248.06N$.

15. A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency \[\mathbf{340}\text{ }\mathbf{Hz}\]) when the tube length is \[\mathbf{25}.\mathbf{5}\text{ }\mathbf{cm}\text{ }\mathbf{or}\text{ }\mathbf{79}.\mathbf{3}\text{ }\mathbf{cm}\]. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.

Frequency of the turning fork, \[\nu =340\text{ }Hz\]

Since the given pipe is attached with a movable piston at one end, it will behave as a pipe with one end closed and the other end open, as shown in the given figure.

Such a system gives odd harmonics. The relation of fundamental note in a closed pipe is given by:

\[{{l}_{1}}=\frac{\lambda }{4}\]

Where, length of the pipe \[{{l}_{1}}=25.5cm=0.255m\]

\[\Rightarrow \lambda =4{{l}_{1}}=4\times 0.255=1.02m\]

The relation of sound speed is given by:

\[v=\nu \lambda =340\times 1.02=346.8m{{s}^{-1}}\]

16. A steel rod \[\mathbf{100}\text{ }\mathbf{cm}\] long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be \[\mathbf{2}.\mathbf{53}\text{ }\mathbf{kHz}.\] What is the speed of sound in steel? 

Length of the steel rod, \[l\text{ }=\text{ }100\text{ }cm\text{ }=\text{ }1\text{ }m\]

Fundamental frequency of vibration, \[\nu =2.53kHz=2.53\times {{10}^{3}}Hz\]

An antinode (A) is formed at its centre, and nodes (N) are formed at its two ends when the rod is plucked at its middle, as shown in the given figure.

The distance between two successive nodes is \[\frac{\lambda }{2}\]

\[\therefore l=\frac{\lambda }{2}\]

\[\lambda =2l=2\times 1=2m\]

The sound speed in steel is given by:

\[v=\nu \lambda \]

\[\Rightarrow v=2.53\times {{10}^{3}}\times 2\]

\[\Rightarrow v=5.06\times {{10}^{3}}m{{s}^{-1}}\]

\[\Rightarrow v=5.06km{{s}^{-1}}\]

The speed of sound in steel is \[5.06km{{s}^{-1}}\].

17. A pipe \[\mathbf{20}\text{ }\mathbf{cm}\] long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a \[\mathbf{430}\text{ }\mathbf{Hz}\] source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is  \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\]).

Length of the pipe, \[l=20\text{ }cm=0.2\text{ }m\]

Source frequency \[=\text{ }{{n}^{th}}\] normal mode of frequency, \[{{\nu }_{n}}=\text{ }430\text{ }Hz\]

Speed of sound, \[v=340\text{ }m/s\]

In a closed pipe, the n th normal mode of frequency is given by the relation:\[{{\nu }_{n}}=(2n-1)\frac{v}{4l}\]

Where, n is an integer \[=\text{ }0,1,2,3\ldots ..\]

\[\Rightarrow 430=\frac{(2n-1)340}{4\times 0.2}\]

\[\Rightarrow 2n-1=\frac{430\times 4\times 0.2}{340}\]

\[\Rightarrow 2n-1=1.01\]

\[\Rightarrow n\approx 1\]

Clearly, the vibration frequency’s first mode is resonantly excited by the given source. In a pipe open at both ends, the \[{{n}^{th}}\] mode of vibration frequency is given by:

\[{{\nu }_{n}}=\frac{nv}{2l}\]

\[\Rightarrow n=\frac{2l{{v}_{n}}}{v}\]

\[\Rightarrow n=\frac{2\times 0.2\times 430}{340}=0.5\]

Since the number of the vibration mode has to be an integer, the given source does not generate a resonant vibration in an open pipe.

18. Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency \[\mathbf{6}\text{ }\mathbf{Hz}\]. The tension in the string A is slightly reduced and the beat frequency is found to reduce to \[\mathbf{3}\text{ }\mathbf{Hz}\]. If the original frequency of A is \[\mathbf{324}\text{ }\mathbf{Hz}\], what is the frequency of B?

Frequency of string A, \[{{f}_{A}}=324Hz\]

 Frequency of string B \[={{f}_{B}}\]

Beat’s frequency, \[n=6\text{ }Hz\]

Beat’s frequency is given as:

\[n=\left| {{f}_{A}}\pm {{f}_{B}} \right|\]

 \[\Rightarrow 6=324\pm {{f}_{B}}\]

\[\Rightarrow {{f}_{n}}=\text{ }330\text{ }Hz\text{ }or\text{ }318\text{ }Hz\]

Frequency decreases with a reduction in the tension in a string because frequency is directly proportional to the tension’s square root. It is given as: 

\[\nu \propto \sqrt{T}\]

 Hence, the beat frequency cannot be \[330\text{ }Hz\]

\[\therefore {{f}_{B}}=318Hz\]

19. Explain why (or how): 

(a) In a sound wave, a displacement node is a pressure antinode and vice versa, 

Ans: A node is a point where the vibration amplitude is the minimum and pressure is the greatest. An antinode is a point where the maximum vibration amplitude and pressure are the lowest. Therefore, a displacement node is a pressure antinode and vice versa.

(b) Bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”,

Ans: Bats emit powerful high-frequency ultrasonic sound waves. These waves get reflected toward them by obstructions. A bat takes a reflected wave and measures the distance, size, direction, and nature of an obstacle with the aid of its brain senses.

(c) A violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, 

Ans: The overtones generated by a sitar and a violin, and the powers of these overtones, are different. Hence, one can differentiate between the notes produced by a sitar and a violin even if they have a similar vibration frequency.

(d) Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and 

Ans: Solids have a shear modulus. They can provide shearing stress. Since fluids do not hold any definite shape, they produce shearing stress. The transverse wave propagation is such that it provides shearing stress in a medium. The propagation of such a wave is probable only in solids and not in gases. Both solids and fluids have their respective bulk moduli. They can provide compressive stress. Hence, longitudinal waves can pass through solids and fluids.

(e) the shape of a pulse gets distorted during propagation in a dispersive medium. 

Ans: A pulse is a combination of waves having various wavelengths. These waves move in a dispersive medium with varying velocities, depending on the medium's nature. This results in the shape distortion of a wave pulse.

20. A train, standing at the outer signal of a railway station blows a whistle of frequency \[\mathbf{400}\text{ }\mathbf{Hz}\] in still air. 

(i) What is the frequency of the whistle for a platform observer when the train 

(a) approaches the platform with a speed of \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\], 

Frequency of the whistle, \[\nu =400\text{ }Hz\]

Speed of the train, \[{{v}_{T}}=10\text{ }m/s\]

The whistle’s apparent frequency \[\left( \nu  \right)\] as the train approaches the platform is given by: 

\[\nu '=\left( \frac{v}{v-{{v}_{T}}} \right)\nu \]

\[\Rightarrow \nu '=\left( \frac{340}{340-10} \right)\times 400=412.12Hz\]

(b) recedes from the platform with a speed of \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\]?

Ans: The apparent frequency (\[\nu ''\]) of the whistle as the train recedes from the platform is given by the relation: 

\[\nu ''=\left( \frac{v}{v+{{v}_{T}}} \right)\nu \]

\[\Rightarrow \nu ''=\left( \frac{340}{340+10} \right)\times 400=388.57Hz\]

The apparent frequency of the whistle is $388.57Hz$.

(ii) What is the speed of sound in each case? The speed of sound in still air can be taken as \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\].

Ans: The apparent change in sound frequency is caused by the relative motions of the source and the observer. These relative motions generate no effect on the sound speed. Therefore, the sound speed in the air in both cases remains the same, i.e., \[340\text{ }m/s.\]

21. A train, standing in a station-yard, blows a whistle of frequency \[\mathbf{400}\text{ }\mathbf{Hz}\] in still air. The wind starts blowing in the direction from the yard to the station with at a speed of \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\]. What is the frequency, wavelength, and speed of sound for an observer standing on the station’s platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of \[\mathbf{10}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\]? The speed of sound in still air can be taken as \[\mathbf{340}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}.\]

Ans: For the stationary observer:

Frequency of the sound generated by the whistle, \[\nu =400\text{ }Hz\]

Speed of sound \[=\text{ }340\text{ }m/s\]

Velocity of the wind, \[v=10\text{ }m/s\]

As there is no relative motion between the observer and the source, the sound’s frequency heard by the observer will be  similar to that produced by the source, i.e., \[400\text{ }Hz\]. 

The wind is blowing towards the observer. Hence, the sound’s effective speed increases by \[10\] units, i.e., \[{{v}_{e}}=340+10=350m/s\]

The wavelength \[(\lambda )\] of the sound heard by the observer is given by: 

\[\lambda =\frac{{{v}_{e}}}{\nu }=\frac{350}{400}=0.875m\]

For the running observer: 

Velocity of the observer, \[{{v}_{o}}=10\text{ }m/s\]

The observer is moving towards the source. As a result, the relative motions of the source and the observer, there is a frequency change (\[\nu '\])

 This is given by the relation:

 \[\nu '=\left( \frac{v+{{v}_{o}}}{v} \right)\nu \]

\[\Rightarrow \nu '=\left( \frac{340+10}{340} \right)\times 400=411.76Hz\]

Since the air is still so, the effective speed of sound \[=340+0=340\text{ }m/s\]

The source is at rest. Hence, the sound’s wavelength will not change, i.e., $\lambda $remains \[0.875\text{ }m.\]

Hence, the two given situations are not exactly identical.

Additional Exercise:

22. A travelling harmonic wave on a string is described by

 \[y(x,t)=7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right)\]

(a) What are the displacement and velocity of oscillation of a point at \[\mathbf{x}\text{ }=\text{ }\mathbf{1}\text{ }\mathbf{cm},\text{ }\mathbf{and}\text{ }\mathbf{t}\text{ }=\text{ }\mathbf{1}\text{ }\mathbf{s}\]? Is this velocity equal to the velocity of wave propagation? 

Ans: The given harmonic wave is:

\[y(x,t)=7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right)\]

For \[x\text{ }=\text{ }1\text{ }cm\] and \[t\text{ }=\text{ }1\text{ }s,\]

\[y(1,1)=7.5\sin \left( 0.0050+12+\frac{\pi }{4} \right)\]

\[\Rightarrow y(1,1)=7.5\sin \left( 12.0050+\frac{\pi }{4} \right)\]

\[\Rightarrow y(1,1)=7.5\sin \theta \]

\[Where,\theta =12.0050+\frac{\pi }{4}=12.0050+\frac{3.14}{4}=12.78rad\]

\[\Rightarrow \theta =\frac{180}{3.14}\times 12.78={{732.81}^{o}}\]

\[\therefore y(1,1)=7.5\sin ({{732.81}^{o}})\]

\[\Rightarrow y(1,1)=7.5\sin (90\times 80+{{12.81}^{o}})=7.5\sin {{12.81}^{o}}\]

\[\Rightarrow y(1,1)=7.5\times 0.2217\]

\[\Rightarrow y(1,1)=1.6629\approx 1.663cm\]

The velocity of the oscillation at a given point and time is given as:

\[v=\frac{d}{dt}y(x,t)=\frac{d}{dt}\left[ 7.5\sin \left( 0.0050x+12t+\frac{\pi }{4} \right) \right]\]

\[\Rightarrow v=7.5\times 12\cos \left( 0.0050x+12t+\frac{\pi }{4} \right)\]

At \[x\text{ }=\text{ }1\text{ }cm\] and \[t\text{ }=\text{ }1s\], 

\[v=y(1,1)=90\cos \left( 12.005+\frac{\pi }{4} \right)\]

\[\Rightarrow v=90\cos ({{732.81}^{o}})=90\cos \left( 90\times 8+{{12.81}^{o}} \right)\]

\[\Rightarrow v=90\cos ({{12.81}^{o}})\]

\[\Rightarrow v=90\times 0.975=87.75cm/s\]

Now, the equation of a propagating wave is given by:

\[y(x,t)=asin(kx+\omega t+\phi )\]

Where, \[k=\frac{2\pi }{\lambda }\]

\[\lambda =\frac{2\pi }{k}\] and \[\omega =2\pi \nu \]

\[\Rightarrow \nu =\frac{\omega }{2\pi }\]

\[Speed,v=\nu \lambda =\frac{\omega }{k}\]

Where, \[\omega =12rad/s\]

\[k=0.0050{{m}^{-1}}\]

\[\Rightarrow v=\frac{12}{0.0050}=2400cm/s\]

Hence, the velocity of the wave oscillation at \[x=1\text{ }cm\] and \[t=1\text{ }s\] is not equal to the wave propagation’s velocity. 

(b) Locate the points of the string which have the same transverse displacements and velocity as the \[\mathbf{x}=\mathbf{1}\text{ }\mathbf{cm}\] point at \[\mathbf{t}=\mathbf{2}\text{ }\mathbf{s},\text{ }\mathbf{5}\text{ }\mathbf{s}\text{ }\mathbf{and}\text{ }\mathbf{11}\text{ }\mathbf{s}.\]

Ans: The relation of propagation constant with wavelength is given by:

\[k=\frac{2\pi }{\lambda }\]

\[\Rightarrow \lambda =\frac{2\pi }{k}=\frac{2\times 3.14}{0.0050}=1256cm=12.56m\]

Therefore, all the points at distances \[n\lambda (n=\pm 1,\pm 2,...and\text{ }so\text{ }on)\], i.e., \[\pm 12.56m,\pm 25.12m,..\ldots \] and so on for \[x=1\text{ }cm\], will have the same displacement as the \[x=1\text{ }cm\] points at \[t=2\text{ }s,\text{ }5\text{ }s,\text{ }and\text{ }11\text{ }s\].

23. A narrow sound pulse (for example, a short pip by a whistle) is sent across a medium. 

(a) Does the pulse have a definite (i) frequency, (ii) wavelength, (iii) speed of propagation? 

Ans: (a) The sound pulse does not have a constant wavelength or frequency. However, the sound speed vibration remains the same, equal to the sound speed in that medium.

(b) If the pulse rate is \[\mathbf{1}\] after every \[\mathbf{20}\text{ }\mathbf{s}\], (that is the whistle is blown for a split of second after every \[\mathbf{20}\text{ }\mathbf{s}\]), is the frequency of the note produced by the whistle equal to \[\frac{1}{20}\] or \[\mathbf{0}.\mathbf{05}\text{ }\mathbf{Hz}\]?

Ans: The short pip produced after every \[20\text{ }s\] does not implies that the frequency of the whistle is \[\frac{1}{20}\] or \[0.05\text{ }Hz\]. It means that \[0.05\text{ }Hz\] is the frequency of the repetition of the pip of the whistle.

24. One end of a long string of linear mass density  \[\mathbf{8}.\mathbf{0}\text{ }\mathbf{x}\text{ }\mathbf{1}{{\mathbf{0}}^{-\mathbf{3}}}\mathbf{kg}\text{ }{{\mathbf{m}}^{-\mathbf{1}}}\]is connected to an electrically driven tuning fork of frequency \[\mathbf{256}\text{ }\mathbf{Hz}\]. The other end passes over a pulley and is tied to a pan containing a mass of \[\mathbf{90}\text{ }\mathbf{kg}\]. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At \[\mathbf{t}=\mathbf{0}\], the left end (fork end) of the string \[\mathbf{x}=\mathbf{0}\] has zero transverse displacement (\[\mathbf{y}=\mathbf{0}\]) and is moving along positive y-direction. The amplitude of the wave is \[\mathbf{5}.\mathbf{0}\text{ }\mathbf{cm}\]. Write down the transverse displacement y as function of x and t that describes the wave on the string.

Ans: The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation:

\[y(x,t)=a\sin (\omega t-kx)\] ……(i)

Linear mass density, \[\mu =8.0\times {{10}^{-3}}kg{{m}^{-1}}\]

Frequency of the tuning fork, \[\nu =256\text{ }Hz\]

Amplitude of the wave, \[a=5.0\text{ }cm=0.05\text{ }m\]……(ii)

Mass of the pan, \[\text{m}=90\text{ }kg\]

Tension in the string, \[T=mg=90x9.8=882\text{ }N\]

The transverse wave’s velocity is given by:

\[\Rightarrow v=\sqrt{\frac{882}{8.0\times {{10}^{-3}}}}=332m/s\]

Angular frequency, \[\omega =2\pi \nu \]

\[\Rightarrow w=2\times 3.14\times 256\]

\[\Rightarrow w=1608.5=16\times {{10}^{3}}rad/s\]……(iii)

Wavelength, \[\lambda =\frac{v}{\nu }=\frac{332}{256}m\]

\[\text{Propagation constant},k=\frac{2\pi }{\lambda }\]

\[\Rightarrow k=\frac{2\times 3.14}{{}^{332}/{}_{256}}=4.84{{m}^{-1}}\]……(iv)

Substituting the values from equations (ii), (iii), and (iv) in equation (i), we get,

\[y(x,t)=0.05\sin (1.6\times {{10}^{3}}t-4.84x)m\]

This describes the wave of the string.

25. A SONAR system fixed in a submarine operates at a frequency \[\mathbf{40}\text{ }\mathbf{kHz}\]. An enemy submarine moves towards the SONAR with a speed of \[\mathbf{360}\text{ }\mathbf{km}{{\mathbf{h}}^{-\mathbf{1}}}\]. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be \[\mathbf{1450}\text{ }\mathbf{m}{{\mathbf{s}}^{-\mathbf{1}}}\].

SONAR system’s operating frequency, \[\nu =40\text{ }kHz\]

Speed of the enemy submarine, \[{{v}_{e}}=360km/h=100\text{ }m/s\]

Sound speed in water, \[v=1450\text{ }m/s\]

The source is at rest and the observer is moving towards it. Hence, the apparent frequency (\[\nu '\]) received and reflected by the submarine is given by:

\[\nu '=\left( \frac{v+{{v}_{e}}}{v} \right)\nu \]

\[\Rightarrow v'=\left( \frac{1450+100}{1450} \right)\times 40=42.76kHz\]

The frequency (\[\nu ''\]) received by the enemy submarine is given by:

\[\nu ''=\left( \frac{v}{v-{{v}_{s}}} \right)\nu '\]

Where, \[{{v}_{s}}=\text{ }100\text{ }m/s\]

\[\Rightarrow \nu ''=\left( \frac{1450}{1450-100} \right)42.72=45.93kHz\]

26. Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can experience both transverse (S) and longitudinal (P) sound waves. Typically, the speed of S wave is about \[\mathbf{4}.\mathbf{0}\text{ }\mathbf{km}{{\mathbf{s}}^{-\mathbf{1}}}\], and that of P wave is \[\mathbf{8}.\mathbf{0}\text{ }\mathbf{km}{{\mathbf{s}}^{-\mathbf{1}}}\]. A seismograph records P and S waves from an earthquake. The first P wave arrives \[\mathbf{4}\] min before the first S wave. Assuming the waves travel in straight line, at what distance does the earthquake occur?

Ans: Let \[{{v}_{s}}\] and \[{{v}_{p}}\] be the velocities of S and P waves respectively. 

Let L be the distance between the seismograph and the epicentre. 

We have, \[L={{v}_{s}}{{t}_{s}}\]    ……(i)

\[L={{v}_{p}}{{t}_{p}}\]    ……(ii)

Where, \[{{t}_{s}}\] and \[{{t}_{p}}\] are the respective times taken by the S and P waves to reach the seismograph from the epicentre. 

It is given that, \[{{v}_{p}}=8km/s\]

\[{{v}_{s}}=4km/s\]

From equations (i) and (ii), we have:

\[{{v}_{s}}{{t}_{s}}={{v}_{p}}{{t}_{p}}\]

\[4{{t}_{s}}=8{{t}_{p}}\]

\[{{t}_{s}}=2{{t}_{p}}\]   ……(iii)

It is also given that:

\[{{t}_{s}}-{{t}_{p}}=4\min =240s\]

\[\Rightarrow 2{{t}_{p}}-{{t}_{p}}=240\]

\[\Rightarrow {{t}_{p}}=240\]

And \[{{t}_{s}}=2\times 240=480s\]

From equation (ii), we get: 

\[L=8\times 240=1920\text{ }km\]

Clearly, the earthquake occurs at a distance of \[1920\text{ }km\] from the seismograph.

27. A bat is flitting about in a cave, navigating via ultrasonic beeps. Assume that the sound emission frequency of the bat is \[\mathbf{40}\text{ }\mathbf{kHz}\]. During one fast swoop directly toward a flat wall surface, the bat is moving at \[\mathbf{0}.\mathbf{03}\] times the speed of sound in air. What frequency does the bat hear reflected off the wall?

Ans: Bat’s ultrasonic beep frequency, \[\nu =40\text{ }kHz\]

Velocity of the bat, \[{{v}_{b}}=0.03v\]

 Where, v is the velocity of sound in air

 The apparent frequency of the sound striking the wall is given by:

\[\nu '=\left( \frac{v}{v-{{v}_{b}}} \right)\nu \]

\[\Rightarrow \nu '=\left( \frac{v}{v-0.03v} \right)\times 40\]

\[\Rightarrow \nu '=\frac{40}{0.97}kHz\]

This frequency is reflected by the stationary wall (v s =0) toward the bat. 

The frequency (\[\nu ''\]) of the received sound is given by the relation:

\[\nu ''=\left( \frac{v+{{v}_{b}}}{v} \right)\nu '\]

\[\Rightarrow \nu ''=\left( \frac{v+0.03v}{v} \right)\times \frac{40}{0.97}\]

\[\Rightarrow \nu ''=\frac{1.03\times 40}{0.97}=42.47kHz\]

The frequency of the received sound is \[42.47kHz\].

NCERT Solutions for Class 11 Physics Chapter 14 Free PDF Download

The PDF of NCERT Class 11 Physics Chapter 14 solutions have all the essential topics, sub-topics with comprehensive explanations which assist students of class 11 to grasp the concepts better.

Waves Class 11 NCERT solutions PDF provides students with many practise exercises as well as a detailed explanation of the questions and solutions provided in the chapter.

Class 11 Physics Ch 14 NCERT solutions lay the foundation for class 12 CBSE boards, which determines the basis of their admission in higher education institutes. The focus on concepts and theoretical aspects of this chapter is essential to build a proper foundation for the Class 12 CBSE board exam. It has been designed with the latest guidelines and syllabus of CBSE and NCERT.

Physics is a subject that has many numerical, formulas and graphical representations. Students often require easy references before the final exam. The PDF contains many forms of practise questions like short-questions, long questions, MCQ’s, diagrammatic and illustrative representations.

Students can quickly recap and revise different formulas and numerical valuations before class 11 final exams without any rush or stress.

NCERT Solutions for Class 11 Physics Chapter 14

The Waves Chapter of Class 11 PDF covers many topics such as:

Nature of waves in different mediums.

Graphical representation of waves and its properties.

Comparison between waves.

Speed of sound.

Harmonic motion.

Resonance among waves and calculation.

Displace and velocity of oscillations.

Waves are connected with rotating electrons, protons, neutrons and other fundamental particles. Waves are also associated with molecules and atoms. The chapter describes many features of a wave, and it’s characteristics like frequency, amplitude, wavelength, phase, resonance and wave displacement.

The waves are in various forms like sounds, lights, radio signals and microwaves. This chapter of Waves Class 11 NCERT solutions has many detailed sections as well, which are also crucial for engineering and college admission entrance tests. The chapter will further teach students about the idea of wave motion.

Vedantu’s experienced team of teachers has designed the syllabus according to the latest guidelines of NCERT and CBSE.

Benefits of NCERT Solutions for Class 11 Physics Chapter 14 PDF

There are many benefits to using PDF that can be downloaded from Vedantu’s website.

The benefits are:

The PDF will help students approach their revision for the final class 11 exam in a more efficient way.

The PDF helps students so that they are in an advantageous position for their competitive exams. Most of the students appear in IIT-JEE, AIEEE and medical entrance exams. The material can help them prepare additionally.

PDF of NCERT solutions for Class 11 Physics Chapter 14 is available for free from Vedantu’s website.

NCERT solutions For Class 11 Physics Chapter Waves is designed by an expert team of teachers, who have many years of experience in teaching science and engineering students around the country.

Students can easily download the PDF from anywhere and everywhere. It is time-efficient and custom-made for easy grasping.

Define Overtone

The term ‘overtone’ is applied to standing waves and is an essential part of this chapter. An overtone is a frequency that is more than the primary frequency of sound. Overtone is put in an application to higher-frequency standing waves.

It is a type of frequency that is usually provided by instruments. Overtone is also called second harmonic. One of the main advantages is that overtone can adapt any valuation of the fundamental frequency.

FAQs on NCERT Solutions for Class 11 Physics Chapter 14 - Waves

1. What is the Fundamental Property of Waves?

Every wave has several different types of fundamental properties. The basic properties of waves can be described as reflection, refraction, diffraction and interference. The waves also consist of wavelength, frequency, amplitude and speed. Its amplitude, length and frequency generally explain a wave.

2. What are Some Characteristics of a Wave?

Some of the characteristics of a wave includes frequency, amplitude, wavelength, phase, resonance and wave displacement.

3. What are the Topics Covered Under Chapter 14, Waves?

The topics covered under this chapter includes:

4. What is a transverse wave motion?

A transverse wave motion is where the particles of the medium are seen to vibrate at right angles towards the direction that the wave propagates. Some examples of these are waves on a string, electromagnetic waves and surface water waves.  It is to be noted that in the case of the electromagnetic wave, the disturbance caused is not the result of the vibration that the particles produce but is the result of the oscillation that occurs in the electric and magnetic fields taking place at right angles to the direction in which the wave travels. 

5. What are the factors that influence the velocity of the sound?

Some factors affecting the velocity of the sound are as follows:

 The velocity of the gas is always inversely proportional to the square root of the density of the respective gas.

The velocity of the sound is always directly proportional to the square of its absolute temperature in a gas

It is usually independent of the change in the pressure of the gas but on the condition that the temperature remains constant. 

In the case of moist air, the velocity of the sound is greater as compared to the dry air.

6. What is the Doppler effect?

The Doppler effect propagates the philosophy that in the case of the relative motion between the source of the sound and the listener, the frequency that the source emits is different from the frequency of the sound that the listener encounters. 

For more such insight into the chapter, it is important that the students refer to the NCERT solutions and practice the exercises that it provides. It will help the student to dwell deeper into the topics and get a better look at the subject matter. 

7. Is Class 11 Physics Chapter 14 hard?

No, Class 11 Physics Chapter 14 is not hard. But to be able to do well in the exam, good preparation is required. To avail this, it is important that the student read all the chapters thoroughly and mark all the important areas so that it gets easier to refer to them in the time of need. They should also cultivate the habit of making their own notes so that they are able to retain all the concepts and ideas that the chapter presents. Lastly, it goes without saying that the student should get their hands on the NCERT solutions and practice all the given exercises. With the practice of these simple steps, they will definitely be able to score and secure good grades in the exam. 

8. From where you can download the NCERT Solutions for Class 11 Physics Chapter 14?

The students can download the NCERT Solutions easily from the website of Vedantu for free. They can also be downloaded from the Vedantu app. These NCERT solutions will prove to be the best source of guidance for the students as they present the exercises with the aim of preparing the student for the exam. With the regular practice of the contents of the NCERT solutions, the student will develop the idea of how to answer any question in the examination. 

NCERT Solutions for Class 11 Physics

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