written as per the latest textbook prescribed by the

Written as per the latest textbook prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune.

Std. XII Sci.

PERFECT

PHYSICS (Vol. I)

Written as per the new textbook Subtopic-wise segregation for powerful concept building Complete coverage of Textual Exercise Questions, Intext Questions and Numericals Includes relevant board questions of February 2020 Extensive coverage of New Type of Questions ‘Solved Examples’ guide you through every type of problem ‘Apply Your Knowledge’ section for application of concepts ‘Quick Review’ at the end of every chapter facilitates quick revision A compilation of all ‘Important Formulae’ in relevant chapters ‘Competitive Corner’ presents questions from prominent competitive examinations Reading Between the Lines, Enrich Your Knowledge, Gyan Guru, Connections,

NCERT Corner are designed to impart holistic education Topic Test at the end of each chapter for self-assessment Video/pdf links provided via QR codes for boosting conceptual retention QR Code to access the Latest Board Question Papers

Salient Features

© Target Publications Pvt. Ltd. No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical

including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

Balbharati Registration No.: 2018MH0022 P.O. No. 1871TEID: 1947

Printed at: India Printing Works, MumbaiSample

Con

tent

The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve

nearly missed something or want to applaud us for our triumphs, we’d love to hear from you. Please write to us on: [email protected]

Perfect Physics XII, Vol. I is intended for every Maharashtra State Board aspirant of Std. XII, Science. The scope, sequence, and level of the book are designed to match the new textbook of Maharashtra State board. At a crucial juncture of cracking a career defining board examination, we wanted to create a book that not just develops the necessary knowledge, tools and skills required to excel in the examination in students but also enables them to appreciate the beauty of subject and piques their curiosity. We believe the students need meaningful content presented in a way that is easy to read and understand rather than being mired down with facts and information. They do much better when they understand why Physics is relevant to their everyday lives. Comprehension of Physics eventuates naturally when subject is studied systematically with sincere and dedicated efforts. Core of Physics lies in its concepts. To begin with, students should read a concept, contemplate upon its essence and attempt to produce the same in their own words. Students should then attempt theoretical questions based on that concept to gauge the level of understanding achieved. To quote Albert Einstein, “If you can't explain it simply, you don't understand it yourself.” Though Physics is communicated in English, it is expressed in Mathematics. Hence, it is essential to befriend formulae and derivations. These should be learnt and memorized. Once Physical mathematics of concept is ingrained, solved numericals should be studied, starting from simple problems to difficult by escalating level of complexity gradually. Students are required to practise numericals and ascertain their command on problem solving. Calculations at this stage must be done using log table keeping in mind that calculators are not allowed in Board Exams. When it comes to problems in Physics nothing makes students perfect like practice! Frequent revisions of concepts and numericals, help in imbibing the topic learnt and therefore should be allotted definite time. A test on the chapter studied should be taken to check one’s range of preparation. Amongst building concepts, advancing into numbers and equations, it is essential to ponder underlying implications of subject. Students should read from references, visit authentic websites, watch relevant fascinating links and even experiment on their own following proper safety guidelines. As famous hat detective Sherlock Holmes has pointed, people see, they do not observe. By becoming attentive to their surroundings students can easily perceive how Physics has touched entire spectrum of life. The very realization is catalytic enough for students to admire and further dive into this compelling subject. Our Perfect Physics adheres to our vision and achieves several goals: building concepts, developing competence to solve numericals, recapitulation, self-study, self-assessment and student engagement—all while encouraging students toward cognitive thinking. Features of the book presented below will explicate more about the same! We hope the book benefits the learner as we have envisioned. - Publisher Edition: Second

PREFACE

Sample

Con

tent

KEY FEATURES

Reading between the lines provideselaboration or missing fragments ofconcept which is essential for complete understanding of the concept.

NCERT Corner covers information fromNCERT textbook relevant to topic.

Connections enable students to interlinkconcepts covered in different chapters.

QR code provides: i. Access to a video/PDF in order to

boost understanding of a concept oractivity

ii. The Latest Board Question Papers

Connections

QR Codes

Enrich Your Knowledge presentsfascinating information about the conceptcovered.

Enrich Your

Knowledge

Reading between the lines

NCERT Corner

Gyan Guru illustrates real lifeapplications or examples related to theconcept discussed.

GG-Gyan Guru

Apply Your Knowledge includeschallenging questions.

Apply Your

Knowledge

Quick review includes tables/ flow chartto summarize the key points in chapter.

Quick Review

Important Formulae includes all of the keyformulae in the chapter.

Important Formulae

Competitive Corner includes selectivequestions from prominent [NEET (UG),JEE (Main), NEET (ODISHA),MHT CET] competitive exams basedentirely on the syllabus covered in thechapter.

Competitive Corner

Sample

Con

tent

There will be one single paper of 70 Marks in Physics. Duration of the paper will be 3 hours. Section A: (18 Marks)

This section will contain Multiple Choice Questions and Very Short Answer (VSA) type of questions.

There will be 10 MCQs and 8 VSA type of questions, each carrying one mark. Students will have to attempt all these questions. Section B: (16 Marks) This section will contain 12 Short Answer (SA-I) type of questions, each carrying 2 marks. Students will have to attempt any 8 questions. Section C: (24 Marks) This section will contain 12 Short Answer (SA-II) type of questions, each carrying 3 marks. Students will have to attempt any 8 questions. Section D: (12 Marks) This section will contain 5 Long Answer (LA) type of questions, each carrying 4 marks. Students will have to attempt any 3 questions.

Distribution of Marks According to the Type of Questions Type of Questions

MCQ 1 Mark each 10 Marks VSA 1 Mark each 8 Marks SA - I 2 Marks each 16 Marks SA - II 3 Marks each 24 Marks LA 4 Marks each 12 Marks

Percentage wise distribution of marks Theory 63%

Numerical 37%

PAPER PATTERN

Sample

Con

tent

Chapter

No. Chapter Name Marks without option

Marks with option Page No.

1 Rotational Dynamics 5 7 1 2 Mechanical Properties of Fluids 5 7 55 3 Kinetic Theory of Gases and Radiation 5 7 112 4 Thermodynamics 5 7 148 5 Oscillations 4 5 194 6 Superposition of Waves 4 6 239 7 Wave Optics 5 7 286

Scan the given Q.R. Code to access the Latest Board Question Papers.

[Reference: Maharashtra State Board of Secondary and Higher Secondary Education, Pune - 04]

Note: 1. * mark represents Textual question. 2. # mark represents Intext question. 3. + mark represents Textual examples. 4. symbol represents textual questions that need external reference for an

answer.

CONTENTS

This reference book is transformative work based on textbook Physics; First edition: 2020 published by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. We the publishers are making this reference book which constitutes as fair use of textual contents which are transformed by adding and elaborating, with a view to simplify the same to enable the students to understand, memorize and reproduce the same in examinations. This work is purely inspired upon the course work as prescribed by the Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. Every care has been taken in the publication of this reference book by the Authors while creating the contents. The Authors and the Publishers shall not be responsible for any loss or damages caused to any person on account of errors or omissions which might have crept in or disagreement of any third party on the point of view expressed in the reference book. © reserved with the Publisher for all the contents created by our Authors. No copyright is claimed in the textual contents which are presented as part of fair dealing with a view to provide best supplementary study material for the benefit of students.

Disclaimer Sample

Con

tent

Sample

Con

tent

194

Q.1. Can you recall? (Textbook page no. 109) What do you mean by linear motion and angular

motion? Ans: i. An object when moves in a straight line, then it

is said to have linear motion. ii. An object when revolves around a fixed point or

rotates around itself, then it is said to have angular motion.

Q.2. What is meant by harmonic motion? Ans: An oscillatory motion obeying harmonic

functions is called harmonic motion. Q.3. Explain the term periodic motion. State its

examples. Ans: i. Any motion which repeats itself after a definite

interval of time is called periodic motion. ii. A body performing periodic motion goes on

repeating the same set of movements.

iii. The time taken for one such set of movements is called its period or periodic time.

iv. At the end of each set of movements, the state of the body is the same as that at the beginning.

v. Examples: a. Motion of the moon around the earth and the

motion of other planets around the sun. b. Motion of electrons around the nucleus. Q.4. State true or false. If false rewrite the correct

statement. Uniform circular motion of any object is a

periodic motion. Ans: True. Q.5. Define vibratory motion. Ans: Periodic motion in which a particle repeatedly

moves to and fro along the same path is the oscillatory or vibratory motion.

5.1 Introduction 5.2 Explanation of Periodic Motion 5.3 Linear Simple Harmonic Motion (S.H.M.) 5.4 Differential Equation of S.H.M. 5.5 Acceleration (a), Velocity (v) and

Displacement (x) of S.H.M. 5.6 Amplitude(A), Period(T) and Frequency (n)

of S.H.M. 5.7 Reference Circle Method 5.8 Phase in S.H.M.

5.9 Graphical Representation of S.H.M. 5.10 Composition of two S.H.M.s having same

period and along the same path 5.11 Energy of a Particle Performing S.H.M. 5.12 Simple Pendulum 5.13 Angular S.H.M. and its Differential

Equation 5.14 Damped Oscillations 5.15 Free Oscillations, Forced Oscillations and

Resonance

Contents and Concepts

5 Oscillations

5.2 Explanation of Periodic Motion

For oscillatory motion, the displacement, velocity and acceleration are represented using sine and cosine functions and these functions are part of harmonic functions.

Reading between the lines In uniform circular motion, the object goes on repeating the same set of movements ( along the circular path )at definite interval of time. Hence, the object performs periodic motion.


You have studied in detail about uniform circular motion in Chapter 1- Rotational Dynamics.

Connections

5.1 Introduction

Sample

Con

tent

Page no. 195 to 196 are purposely left blank.

To see complete chapter buy Target Notes or Target E‐Notes

Sample

Con

tent

197

Chapter 5: Oscillations

Q.17. What do you mean by complete oscillation? Ans: A complete oscillation is when the object goes

from one extreme to other and back to the initial position.

Q.18. State the conditions required for simple harmonic motion.

Ans: Conditions for simple harmonic motion: i. Oscillation of the particle is always about a fixed

point. ii. The net force or acceleration is always directed

towards the fixed point. iii. The particle comes back to the fixed point due to

restoring force. Q.19. Explain the following terms: i. Harmonic oscillation ii. Non-harmonic oscillation Ans: i. Harmonic oscillation is that oscillation which

can be expressed in terms of a single harmonic function, such as x = a sint or x = a cost.

ii. Non-harmonic oscillation is that oscillation which cannot be expressed in terms of single harmonic function. It may be a combination of two or more harmonic oscillations such as x = a sinωt + b sin2ωt.

Q.20. Activity. (Textbook page no. 111) Some experiments described below can be

performed in the classroom to demonstrate S.H.M. Try to write their equations.

i. A hydrometer is immersed in a glass jar filled with water. In the equilibrium position it floats vertically in water. If it is slightly depressed and released, it bobs up and down performing linear S.H.M.

ii. A U-tube is filled with a sufficiently long

column of mercury. Initially when both the arms of U tube are exposed to atmosphere, the level of mercury in both the arms is the same. Now, if the level of mercury in one of the arms is depressed slightly and released, the level of mercury in each arm starts moving up and down about the equilibrium position, performing linear S.H.M.

Ans: a. In both the cases, the bob and mercury performs

linear S.H.M.

b. If the level changes by ‘x’ from equilibrium, then restoring force is given by,

f = –kx. [Students are expected to perform the above activity and try to find equations in both the cases on their own.] Q.21. Obtain the differential equation of linear

simple harmonic motion. Ans: i. In a linear S.H.M., the force is directed towards

the mean position and its magnitude is directly proportional to the displacement of the body from mean position.

f –x f = –kx ….(1) where, k is force constant and x is displacement

from the mean position. ii. According to Newton’s second law of motion, f = ma ….(2) From equations (1) and (2), ma = –kx ….(3) iii. The velocity of the particle is given by, v = dx

dt,

Acceleration, a = 2

2

dv d xdt dt

….(4)

iv. Substituting equation (4) in equation (3),

m2

2

d xdt

= –kx

2

2

d xdt

+ km

x = 0

v. Substituting km

= 2, where ω is the angular

frequency,

2

2

d xdt

+ 2x = 0

This is the differential equation of linear S.H.M.

+Q.22.A body of mass 0.2 kg performs linear S.H.M. It experiences a restoring force of 0.2 N when its displacement from the mean position is 4 cm. Determine

i. force constant ii. period of S.H.M. and

5.4 Differential Equation of S.H.M.

Hydrometer

Mercuryx x

Solved Examples

[Note: Students can scan the adjacent QR code to get conceptual clarity about oscillations of liquid column in U-tube with the aid of a linked video.]

Sample

Con

tent

198

Std. XII Sci.: Perfect Physics (Vol. I)

iii. acceleration of the body when its displacement from the mean position is 1 cm.

Solution: Given: m = 0.2 kg,F = 0.2 N, 1x = 4 cm = 0.04 m, 2x = 1 cm = 0.01 m To find: i. Force constant (k) ii. Time period of SHM (T) iii. Acceleration (a)

Formulae: i. k = Fx

ii. T = 2

iii. = km

iv. a = 2x

Calculation: From formula (i),

k = 0.20.04

= 5 N/m

From formula (ii),

T = 25 = 0.4 s

From formula (iii),

= 50.2

= 25 = 5 rad/s

From formula (iv), for 2x = 0.01 m, a = 52 0.01 = 0.25 m s2 Ans: i. The force constant is 5 N/m. ii. The time period of SHM is 0.4 s. iii. Acceleration at x = 1 cm is 0.25 m s2. [Note: The solution is modified to get it conceptually correct with respect to given values in the question.] *Q.23. In SI units, the differential equation of an

S.H.M. is 2

2d xdt

= 36x. Find its frequency and

period. Solution:

Given: 2

2

d xdt

= 36 x

2

2

d xdt

+ 36 x = 0

Comparing with differential equation,

2

2

d xdt

+ 2x = 0

2 = 36 = 6 rad/s i. For frequency, = 2n

n = ω2π

= 62π

= 33.142

0.955 Hz

ii. For period, = 2π

T

T = 2πω

= 2π 3.1426 3

1.05 s

Alternate method: Period (T) = 1

n= 1

0.955 1.05 s

Ans: The frequency and period of SHM are 0.955 Hz and 1.05 s respectively.

Q.24. A body of mass 1 kg is made to oscillate on a spring of force constant 16 N/m. Calculate:

i. Angular frequency, ii. Frequency of vibration Solution: Given: m = 1 kg, k = 16 N/m To find: Angular frequency (), Frequency of vibration (n)

Formulae: i. = km

ii. = 2n

Calculation: From formula (i), We have, for S.H.M.

= km

= 161

= 4 rad/s From formula (ii), n =

2

= 4 2 Hz2

n = 23.142

= 0.6365 Hz

Ans: The angular frequency of the body is 4 rad/s and the frequency of vibration is 0.6365 Hz.

*Q.25. Using differential equation of linear S.H.M, obtain the expression for velocity in S.H.M. and acceleration in S.H.M.

Ans: i. Expression for acceleration in linear S.H.M: a. From differential equation,

2

2

d xdt

+ 2x = 0

2

2

d xdt

= 2x ….(1)

b. But, linear acceleration is given by,

2

2

d xdt

= a ….(2)

From equations (1) and (2), a = 2 x ….(3) Equation (3) gives acceleration in linear

S.H.M. ii. Expression for velocity in linear S.H.M: a. From differential equation of linear S.H.M,

2

2

d xdt

= 2x

ddt

dxdt

= 2x

5.5 Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.

Sample

Con

tent

199


dvdt

= 2x …. dx vdt

dvdx

dxdt

= 2x

v dv dx

= 2x …. dx vdt

v dv = 2xdx .…(4) b. Integrating both sides of equation (4), v dv = 2xdx

2 2 2v x = + C

2 2

.…(5)

where, C is the constant of integration. c. At extreme position, x = A and v = 0. Substituting these values in equation (5),

0 = 2 2A 2

+ C

C = 2 2A2

.…(6)

d. Substituting equation (6) in equation (5),

2

2v

= 2 2x2

+

2 2A2

v2 = 2A22x2 v2 = 2(A2 x2) v = ± 2 2A x This is the required expression for velocity in

linear S.H.M. Q.26. Derive expression of displacement of a

particle performing linear S.H.M. Ans: Expression for displacement in linear S.H.M: i. From differential equation of linear S.H.M,

velocity is given by, v = 2 2A x ….(1)

But, in linear motion, v = dxdt

….(2)

From equation (1) and (2),

dxdt

= 2 2A x

2 2

dxA x

= dt ….(3)

ii. Integrating both sides of equation (3),

2 2

dxA x

dt

sin1 xA

= t +

where, is constant of integration which depends upon initial condition (phase angle)

xA

= sin (t + )

x = A sin (t + ) This is required expression for displacement of a

particle performing linear S.H.M. at time t.

Q.27. State an expression for displacement of a particle performing S.H.M. Discuss the formula when particle starts from mean position and extreme position respectively.

Ans: Expression for displacement of a particle performing S.H.M., x = A sin (t + ) ….(1)

i. a. If the particle starts S.H.M. from the mean position, x = 0 at t = 0

= sin–1 xA

= 0 or

Substituting in equation (1), x = A sin(t) This is the expression for displacement at

any instant if the particle starts S.H.M. from the mean position.

b. Positive sign to be chosen if it starts towards positive and negative sign for starting towards negative.

ii. a. If the particle starts S.H.M. from the extreme position, x = A at t = 0

= sin–1 xA

= 2

or 32

Substituting in equation (1),

x = A sin t2

or

x = A sin 3t2

x = A cos (t) This is the expression for displacement at

any instant, if the particle starts S.H.M. from the extreme position.

b. Positive sign for starting from positive extreme position and negative sign for starting from the negative extreme position.

Q.28. State the expressions of displacement, velocity

and acceleration for a body performing linear S.H.M. at time ‘t’.

Ans: i. Displacement, x = A sin (t + )

ii. Velocity, v = dxdt

= A cos (t + )

iii. Acceleration, a = dvdt

= A2 sin (t + )

In the cases (i) and (ii), it is said that, “if the particle starts S.H.M…...” More specifically, it is not the particle that starts its S.H.M., but the observer start counting the time ‘t’ from that instant. The particle is already performing its motion.


Sample

Con

tent



Sample

Con

tent

225


pendulum. Other three pendula are having hollow rubber balls as their bobs and will act as the driven pendula. As the pendula A and C are of the same lengths, their natural frequencies are the same. Pendulum B has higher natural frequency as it is shorter and pendulum D is of lower natural frequency than that of A and C.

i. What will happen if pendulum A is set into

oscillations? Explain. ii. Plot the graph between frequency and

square of the lengths, if this activity is repeated for a set of different pendula of different lengths.

Ans: i. a. When pendulum A is set into oscillations,

it oscillates in a plane perpendicular to the string.

b. In the course of time, the other three pendula also start oscillating in parallel planes. This happens due to the transfer of vibrational energy through the string.

c. Oscillations of A are free oscillations and those of B, C and D are forced oscillations of the same frequency as that of A.

d. The natural frequency of pendulum C will be the same as that of A, as it is of the same length as that of A.

e. Among the pendula B, C and D, the pendulum C oscillates with maximum amplitude and the other two with smaller amplitudes.

f. As the energy depends upon the amplitude, pendulum C absorbs maximum energy from the source pendulum A, while the other two absorbs less.

g. It shows that the object C having the same natural frequency as that of the source absorbs maximum energy from the source. In such case, it is said to be in resonance with the source (pendulum A).

h. For unequal natural frequencies on either side (higher or lower), the energy absorbed (hence, the amplitude) is less.

ii. a. If the activity is repeated for a set of pendula of different lengths and squares of their amplitudes are plotted against their natural frequencies, the plot will be similar to that shown in the figure.

b. The peak occurs when the forced frequency

matches with the natural frequency, i.e., at the resonant frequency.

Q.110. State examples where resonance occurs. Ans: i. Most of the traditional musical instruments use

the principle of resonance. ii. Resonance occurs in L.C circuits.

A B

C D (driver

pendulum)

Forced oscillations

You will study more about resonance in AC circuits in Chapter 13 - AC circuits of Part - II.

Connections

f = fr forced frequency (f)

fr = resonant frequency

A2

Resonant frequency

NCERT Corner

The phenomenon of increase in amplitude whenthe frequency ofdriving force is close to thenatural frequency of the oscillator is calledresonance.

Resonance in damped oscillations:

Least damping

Maximum Damping

t 2.0 1.5 1.0 0.5 0

5

4

3

2

1

x(t)

Sample

Con

tent

226


Q.111. A man stands on a weighing machine placed

on a horizontal platform. The machine reads 60 kg. By means of a suitable mechanism, the platform is made to execute harmonic vibrations up and down with a frequency of 2 vibrations per second. What will be the effect on the reading of the weighing machine? The amplitude of vibration of platform is 4 cm.

(g = 10 ms2) Ans: Here, m = 60 kg, n = 2 s1, A = 4 cm = 0.04 m Maximum acceleration, amax = 2 A = (2n)2A = 4 2 n2A

= 4 222

7

(2)2 0.04

= 6.32 ms2 Maximum force on the man = m (g + amax) = 60 (10 + 6.32) = 979.2 N = 97.9 kgf. Minimum force on the man = m (g amax) = 60 (10 6.32) = 220.8 N = 22.08 kgf. Hence, the reading of the weighing machine

varies between 22.08 kgf and 97.9 kgf. Q.112.Two particles performing S.H.M. are shown

below. What will be the displacement of the object in both the cases?

Ans: For figure (a),

As particle starts executing motion from the mean position in anticlockwise direction.

x = r cos t = r cos 2T t

T = 6 s, r = 6 cm

x = 6 cos 26 t

x = 6 cos π3

t

For figure (b), As particle starts executing motion from mean

position in clockwise direction.

x = r sin t = r sin 2T t

T = 4 s, r = 4 cm

x = 4 sin 24 t

x = 4 sin π2

t Q.113. A simple pendulum of length l and having a

bob of mass M is suspended in a car. The car is moving on a circular track of radius R with uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period? (NCERT)

Ans: In this case, the bob of the pendulum experiences two accelerations.

i. Acceleration due to gravity, g, acting vertically downwards.

ii. Centripetal acceleration,

aC =2v

R acting along the horizontal direction.

Thus, Effective acceleration, g = 2 2

Cg a

= 4

22

vgR

new time period,

T = 2gl

= 4

22

2vgR

l

Q.114.Internet my friend. (Textbook page no. 128) i. http://hyperphysics.phyastr.gsu.edu/

hbase/shm.html#c1 ii. http://hyperphysics.phy-astr.gsu.edu/

hbase/pend.html#c1 iii. https://en.wikipedia.org/wiki/

Simple_harmonic_motion iv. https://opentextbc.ca/ v. https://physics.info [Students can use links given above as reference and collect information about oscillations.]

Apply Your Knowledge

6 cmP (t = 0)

y

T = 6 s

x

Figure (a)

Figure (b)

y T = 4 s

x

P (t = 0)

4 cmSample

Con

tent

227


Quick Review

Oscillations

Differential equation

Differential equation for linear S.H.M.

2

2

d xd t

+ 2x = 0

2

2

d x k xdt m

= 0

Linear S.H.M.

In a linear S.H.M.,the force is directedtowards the meanposition and itsmagnitude is directlyproportional to thedisplacement of thebody from meanposition.

Displacement, x = A sint

Velocity, v = 2 2A x Acceleration, a = 2x

Equations of S.H.M. Simple pendulum

T = 2 Lg

When springs are connected in series,

s 1 2 n

1 1 1 1...k k k k

When springs are Connected in parallel, kp = k1 + k2 + …. + kn

Spring constant

Free Oscillations: If an object is allowed tooscillate or vibrate on its own,it does so with its naturalfrequency (or with one of itsnatural frequencies). Forced oscillations: Oscillations in which a bodyoscillates under influence of anexternal periodic force.

Free and forced vibrations

12

E T.E. = constant

P.E. P.E.

K.E. K.E. x = A x = A

x x = 0 Displacement

P.E = 12

kx2 = 12

m2x2

K.E = 12

k (A2 x2)

T.E = 12

kA2 = 12

m2A2

= ConstantGraphical representation of energy (E):

Energy of S.H.M.

Graphical representation of S.H.M.

Phase difference between:

Velocity and displacement: 2

Velocity and acceleration: 2

Displacement and acceleration:

Particles starts from mean positionParticles starts from extreme position

Disp

lace

men

t

O

A

A /2

3/2 2

t

Acc

eler

atio

n

2A

2A

O /2 3/2 2t

Vel

ocity

A

A /2

3/22

tO

Angular S.H.M.Angular acceleration isdirectly proportional tothe angular displacementand directed opposite tothe angular displacement. α

Period of vibrationsof the magnet,

T = 2

T = 2 IB

Magnet performingangular S.H.M

Differential equation

m2

2

d x dxb kxd t d t

= 0

Angular frequency

= 2k b

m 2m

Period of oscillation

T = 2

= 2

2

k bm 2m

Damped S.H.M.

Sample

Con

tent

228


1. Restoring force: f = kx = m2x 2. Differential equation of a linear S.H.M:

2

2

d xdt

+ 2x = 0 3. Displacement in S.H.M: i. General equation: x = A sin (t + ) ii. x = A sin t (from mean position) iii. x = A cos t (from extreme position) 4. Velocity in linear S.H.M: v = ± 2 2A x 5. Acceleration in linear S.H.M: a = 2x

6. Period in S.H.M: T = 2

= 2 mk

7. Frequency in S.H.M: n = 1T

=2

= 1 k2 m

8. Energy in S.H.M: i. Potential energy:

P.E = 12

kx2 = 12

m2x2

ii. Kinetic energy:

K.E = 12

k (A2 x2) = 12

m2 (A2 x2)

iii. Total energy:

T.E = 12

kA2 = 12

m2A2 = 2m2n2A2 9. Composition of S.H.M’s: i. Resultant equation of two S.H.M’s: x = R sin (t + ) ii. Resultant amplitude: R = 2 2

1 2 1 2 1 2A A 2A A cos

iii. Phase: = tan1 1 1 2 2

1 1 2 2

A sin A sinA cos A cos

10. Oscillating spring: i. Force, F = mg = kx

ii. Period, T = 2 mk

iii. When connected in series,

s 1 2 3

1 1 1 1 ....k k k k

iv. When connected in parallel, kp = k1 + k2 + k3 + ….

11. Simple pendulum: Period, T = 2 Lg

12. Magnet vibrating in uniform magnetic field:

i. Time period, T = 2 IB

ii. Angular acceleration, = BI

13. Damped force: Fd = bv (In magnitude) 14. Equation of damped S.H.M:

m2

2

d xdt

+ b dxdt

+ kx = 0 15. Amplitude of damped oscillation: Ad = Aebt/2m

16. Angular frequency of damped oscillation:

= 2k b

m 2m

17. Time period of damped oscillation: T =

2

2

k bm 2m

5.2 Explanation of Periodic Motion 1. Give reason: Every oscillatory motion is

periodic but every periodic motion need not be oscillatory.

Ans: Refer Q.8 5.3 Linear Simple Harmonic Motion (S.H.M.) 2. What is the concept of restoring force in spring

mass oscillator? Ans: Refer Q.13 5.4 Differential Equation of S.H.M. 3. Define linear S.H.M. Obtain differential

equation of linear S.H.M. [Feb 20] Ans: Refer Q.15 and Q.21 4. The differential equation for linear S.H.M of a

particle of mass 2 g is 2

2

dtxd + 16x = 0. Find the

force constant. Ans: 32 dyne/cm 5. The differential equation for linear S.H.M is

given by 2

2

d xdt

= –4x. If the amplitude is 0.5 m

and initial phase is 6 radian, obtain the

displacement equation of S.H.M and find the velocity at x = 0.3 m for the particle in S.H.M.

Ans: x = 0.5 sin m6

t2

, 0.8 m/s

Important Formulae

Exercise

Sample

Con

tent

229


5.5 Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.

6. Using differential equation of linear S.H.M,

derive an expression for (i) acceleration in S.H.M., (ii) velocity in S.H.M.

Ans: Refer Q.25 7. Obtain an expression of displacement of a

particle performing linear S.H.M. Ans: Refer Q.26 8. State an expression for the maximum and

minimum values of displacement, velocity and acceleration of a particle performing S.H.M.

Ans: Refer Q.29 9. A needle of a sewing machine moves in a path of

amplitude 2 cm and has a frequency of 10 Hz in S.H.M. Find its acceleration 1

40 second after it has

crossed the mean position. Ans: 78.88 m/s2

10. The maximum velocity of a particle performing

S.H.M.is 6.28 cm/s. If the length of its path is 8 cm, calculate its period.

Ans: 4 s 11. A particle performs S.H.M of amplitude 10 cm,

its maximum velocity during oscillations is 100 cm/s. What is its displacement, when the velocity is 60 cm/s2?

Ans: 8 cm 5.6 Amplitude (A), Period (T) and Frequency (n)

of S.H.M. 12. Define the following terms: i. Amplitude of S.H.M. ii. Period of S.H.M. iii. Frequency of S.H.M. Ans: Refer Q.43(Only definitions) 13. Write short note on series combination of springs. Ans: Refer Q.44 14. Write short note on parallel combination of

springs. Ans: Refer Q.45 15. A S.H.M. is given by the equation

x = 8 sin (4t) + 6 cos (4t) m. Find its i. amplitude ii. initial phase

iii. period iv. frequency. Ans: i. 8 2 m ii. 0 radian iii. 0.5 sec iv. 2 Hz 5.7 Reference Circle Method 16. Prove that a linear S.H.M. is the projection of a

U.C.M. along any of its diameter. Ans: Refer Q.49

5.8 Phase in S.H.M. 17. What does the following phase angle indicates? i. = 0 ii. = 180 Ans: Refer Q.53 (only i and ii) 5.9 Graphical Representation of S.H.M. 18. Give graphical representation of displacement,

velocity and acceleration against phase angle, for a particle performing linear S.H.M. from (a) the mean position (b) the positive extreme position.

Ans: Refer Q.56 5.10 Composition of two S.H.M.s having same

period and along the same path 19. Derive expression for resultant displacement of

two S.H.M.s having same period and parallel to each other. Find the expression for resultant amplitude and resultant phase. Also, obtain the resultant amplitude when phase difference is

i. 0 rad ii. π2

iii. Ans: Refer Q.57 5.11 Energy of a Particle Performing S.H.M. 20. Obtain the expression for total energy of a

particle performing S.H.M and show that the total energy is conserved.

Ans: Refer Q.62 21. State the expression of kinetic energy and

potential energy at i. mean position ii. extreme position Ans: Refer Q.63(i and ii) 22. Graphically represent the variation of K.E., P.E.,

and T.E. with displacement. Ans: Refer Q.66 23. A particle of mass 0.2 kg performs S.H.M of

amplitude 0.1 m and period 3.14 second. Find its kinetic energy and potential energy when it is at a distance of 0.03 m from mean position.

Ans: 3.64 10–3 J, 3.6 104 J 24. The total energy of a particle of mass 200 g

performing S.H.M is 10–3 joule. Find the maximum velocity and the period if amplitude is 4 cm.

Ans: 0.1 m/s, 0.8 s 5.12 Simple Pendulum 25. Derive an expression for period of simple

pendulum. Ans: Refer Q.77(i-viii) 26. Mention the laws of simple pendulum. Ans: Refer Q.81

Sample

Con

tent

230


27. Difference between conical pendulum and simple pendulum.

Ans: Refer Q.84 28. The bob of a simple pendulum of length 1 metre

performs linear S.H.M of amplitude 2 cm. If mass of the bob is 100 gram and its total energy is 3 105 J, find the periodic time of S.H.M.

Ans: 1.57 s 29. A simple pendulum of length 1 m has a mass of

10 g and oscillates freely with amplitude 2 cm. Find its potential energy at extreme point. [g = 980 cm /s2]

Ans: 196 erg 30. A simple pendulum of length 1 m performs

S.H.M with amplitude 2 cm. Find the angular displacement of the bob from the mean position when the potential energy and kinetic energy of the bob becomes equal.

Ans: 1.414 102 rad 31. When the length of a simple pendulum is

increased by 22 cm, the period changes by 20 %. Find the original length of simple pendulum. [g = 9.8 m/s2]

Ans: 0.5 m 32. Determine the length of a second’s pendulum at

a place, where the acceleration due to gravity is 9.77 m/s2.

Ans: 0.9806 m 33. The period of a simple pendulum increases by

10% when its length is increased by 21 cm. Find the original length and period of the pendulum. [g = 9.8 m/s2]

Ans: i. 1 m ii. 2.006 s 34. The period of simple pendulum is found to

increase by 50% when the length of the pendulum is increased by 0.6 m. Calculate the initial length and initial period of oscillation at a place where g = 9.8 m/s2.

Ans: i. 0.48 m ii. 1.318 s 5.13 Angular S.H.M. and its Differential Equation 35. Derive an expression for the period of a magnet

vibrating in a uniform magnetic field and performing S.H.M.

Ans: Refer Q.96 36. A circular disc of mass 10 kg is suspended by a

wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire.

Ans: 2.0 Nm rad-1

37. A magnetic needle has magnetic moment 6.7 × 10–2 Am2 and moment of inertia I = 7.5 × 10–6 kg m2. It performs 10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field?

Ans: 0.01 T 5.14 Damped Oscillations 38. Obtain the differential equation of damped

harmonic oscillations. Ans: Refer Q.104 39. Draw displacement-time graph of damped

harmonic oscillations. What can be concluded from it?

Ans: Refer Q.106 5.15 Free Oscillations, Forced Oscillations and

Resonance 40. With the help of an example, explain the concept

of free oscillation and forced oscillation. Ans: Refer Q.107 1. The displacement of a particle performing linear

harmonic motion in one time period is (A) A (B) 2A (C) zero (D) 4A 2. If the maximum velocity and acceleration of a

particle executing S.H.M are equal in magnitude, the time period will be

(A) 1.57 s (B) 3.14 s (C) 6.28 s (D) 12.56 s 3. Which of the following quantities are always

negative in a simple harmonic motion?

(A) F

a

(B) v

r

(C) F

r

(D) None of these 4. The total work done by a restoring force in

simple harmonic motion of amplitude A and angular speed , in one oscillation is _______.

(A) 12

mA22 (B) zero

(C) mA22 (D) 12

mA 5. Which one of the following is not a

characteristics of S.H.M? (A) Its acceleration is maximum in the

extreme position. (B) It is the projection of a uniform circular

motion on a diameter. (C) Its velocity is maximum at the mean

position. (D) Its velocity time graph is a straight line.

Multiple Choice Questions

Sample

Con

tent



Sample

Con

tent

233


37. Period of a simple pendulum will be doubled if (A) its length is increased to four times. (B) the acceleration due to gravity is doubled. (C) the mass of the bob is doubled. (D) the length of the pendulum and mass of

the bob are doubled. 38. For small amplitudes, the force constant of a

simple pendulum (A) is independent of acceleration due to

gravity. (B) is inversely proportional to the

acceleration due to gravity. (C) independent of the mass of the bob and

the length of the pendulum. (D) directly proportional to the mass of the

bob. 39. If a simple pendulum oscillates with an

amplitude 50 mm and time period of 2 s, then its maximum velocity is

(A) 0.10 m/s (B) 0.15 m/s (C) 2.8 m/s (D) 1.16 m/s 40. In a damped harmonic oscillator, periodic

oscillations have _______ amplitude. (A) gradually increasing (B) suddenly increasing (C) suddenly decreasing (D) gradually decreasing 41. Which of following figure(s) represent(s)

damped simple harmonic motions? (A) Fig. 1 alone (B) Fig. 2 alone (C) Fig. 4 alone (D) Fig. 3 and 4 42. Presence of damping in an oscillator (A) reduces the amplitude and frequency. (B) reduces the amplitude and time period. (C) increases the time period and amplitude. (D) increases frequency and decreases

amplitude.

*43. A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is

(A) 3 A

2 (B)

2 A3

(C) A2

(D) 1 A2

*44. A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is

(A) 36 J (B) 9 J (C) 27 J (D) 18 J

*45. The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6th of that on the earth’s surface]

(A) 1/6 m (B) 6 m

(C) 1/36 m (D) 1 m6

*46. Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio

(A) 1 : 4 (B) 1 : 2 (C) 2 : 1 (D) 4 : 1

*47. The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?

(A) The acceleration is maximum at time T. (B) The force is maximum at time 3T/4. (C) The velocity is zero at time T/2. (D) The kinetic energy is equal to total energy

at time T/4.

1. (C) 2. (C) 3. (D) 4. (B) 5. (D) 6. (A) 7. (B) 8. (A) 9. (B) 10. (D) 11. (B) 12. (D) 13. (C) 14. (C) 15. (B) 16. (D) 17. (C) 18. (A) 19. (B) 20. (A) 21. (D) 22. (B) 23. (B) 24. (A)

T/4 T/23T/4

T0

x

t

Answers to Multiple Choice Questions

Disp

lace

men

t

Time

(2)

Time

Disp

lace

men

t

(1)

Disp

lace

men

t

(4)

Time

Disp

lace

men

t

Time

(3) Sample

Con

tent

234


25. (C) 26. (B) 27. (C) 28. (B) 29. (C) 30. (D) 31. (C) 32. (A) 33. (A) 34. (A) 35. (D) 36. (B) 37. (A) 38. (D) 39. (B) 40. (D) 41. (A) 42. (A) 43. (A) 44. (D) 45. (A) 46. (B) 47. (B)

7. K.E. = 1

4 T.E.

2 2 21 1 1k A x kA2 4 2

A2 x2 = 2A

4

x2 = A22A

4

x2 = 23A

4

x = 3A2

9. a = 2ω x When x = 0, a = 0 11. Resultant amplitude R = 2 2

1 2 1 2A A 2A A cos

= 2 210 5 2 50 cos( / 6) = 14.56 m Resultant phase

= tan1 1 1 2 2

1 1 2 2

A sin A sinA cos A cos

Here 1 = 0, 2 = /6

= tan1 10(sin0) 5(sin / 6)10(cos0) 5(cos / 6)

= tan1

2.5

10 2.5 3

954 28. As can be seen from the graph, one of the

S.H.M.s starts from the extreme position (X = A) corresponding to initial phase of

2 and

the second S.H.M. starts from a point midway

between O and A, Ax2

corresponding to

initial phase of 4 .

Thus, 1 = 2 and 2 =

4

Phase difference, = 12 = 2

4 =

4 rad

43. v = 2 2A x

Given, v = maxv A2 2

2 2A A x2

2 2A A x2

2

2 2A A x4

x2 = A2 – 2A

4 =

23A4

x = 3 A

2

44. K.E. = 21 kA2 = 2 21 m A

2

= 12 1 (100)2 (6 10–2)2

= 18 J 45. Since, L = 2

g

m

e

LL

= m

e

gg

Lm = 16

Le …. em

gg6

= 16 1 = 1

6m

46. In series, ks = k

2

In parallel, kp = 2k Frequency,

n = 1 k2 m

n k

s

p

nn

= s

p

kk

= k 12 2k =

12

1. A particle executes simple harmonic motion and is

located at x = a, b and c at times t0, 2t0 and 3t0 respectively. The frequency of the oscillation is

[JEE (Main) 2018]

(A) 0

12 t

cos–1 a c2b

(B) 0

12 t

cos–1 a b2c

Competitive Corner

Hints to Multiple Choice Questions

Sample

Con

tent

235


(C) 0

12 t

cos–1 2a 3cb

(D) 0

12 t

cos–1 a 2b3c

Hint: We have a = A sin t0;b = A sin 2t0 and c = A sin 3t0 Now a + c = A sin t0 + A sin 3t0 a + c = 2 A sin 2t0 cos t0 a + c = 2 b cos t0

t0 = cos–1 a c2b

= 0

12 t

cos–1 a c2b

2. A mass is suspended from a vertical spring

which is executing S.H.M. of frequency 5 Hz. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is (acceleration due to gravity g = 10 m/s2)

[MHT CET 2018] (A) 2 m/s (B) m/s

(C) 1 m/s2

(D) 1 m/s

Hint: n = 5 Hz, T = 1s5

T = 2 mk

The restoring force is equal to the weight of the spring.

kx = mg

m xk g

T = 2 xg

T = 2 Ag

….( At highest position, x = A)

1 A25 g

21 A425 g

A = 2 2 2

g 10 1100 100 10

vmax = A = 2 5 2

1 110

m/s 3. The displacement of a particle executing simple

harmonic motion is given by y = A0 + Asin t + Bcos t. Then the amplitude of its oscillation is given by:

[NEET (UG) 2019]

(A) 2 20A (A B) (B) A + B

(C) 2 20A A B (D) 2 2A B

Hint:

y = A0 + A sin t + B sin t y A0 = A sin t + B cos t Resultant amplitude, R = 2 2 oA B 2AB cos 90

R = 2 2A B 4. Average velocity of a particle executing SHM in

one complete vibration is: [NEET (UG) 2019]

(A) 2A

2

(B) zero

(C) A2 (D) Aω

Hint: In one complete vibration, displacement is zero. Average velocity, vavg = 0 5. The radius of circle, the period of revolution,

initial position and sense of revolution are indicated in the figure.

y-projection of the radius vector of rotating

particle P is: [NEET (UG) 2019]

(A) y(t) = 3cos 3 t2

, where y in m

(B) y(t) = 3cos t2

,where y in m

(C) y(t) = –3cos 2πt, where y in m

(D) y(t) = 4sin t2

,where y in m

Hint: The projection of the particle on Y-axis represents simple harmonic motion.

t = 0, displacement y is maximum, so equation will be cosine function. y = a cos t = 2

T = 2

4 =

2 rad/s

A

B 2 2A B

A0

t

X 3 m

T = 4 sP(t = 0)

Y

Sample

Con

tent

236


From figure, a = 3 m

y = 3cos t2

6. A mass falls from a height ‘h’ and its time of

fall ‘t’ is recorded in terms of time period T of a simple pendulum. On the surface of earth it is found that t = 2T. The entire set up is taken on the surface of another planet whose mass is half of that of earth and radius the same. Same experiment is repeated and corresponding times noted as t and T. Then we can say

[NEET (Odisha) 2019] (A) t = 2T (B) t = 2 T (C) t> 2T (D) t< 2T Hint: On the surface of the earth,

time taken for falling, t = 2hg

Time period of simple pendulum,

T = 2gl

On the surface of other planet, Time taken for falling,

t = 2hg

= 2hg / 2

= 2t ….( g = g2

)

Time period of simple pendulum,

T = 2gl

= 2g / 2

l

= 2 T ….( g = g2

)

tT = 2 t

2 T= 2 ….(Given: t = 2T t

T= 2)

t = 2T 7. A body performing a simple harmonic motion

has potential energy ‘P1’ at displacement ‘x1’. Its potential energy is ‘P2’ at displacement ‘x2’. The potential energy ‘P’ at displacement (x1 + x2) is [MHT CET 2019]

(A) P1 + P2 (B) 2 2

1 2P + P

(C) P1 + P2 + 2 1 2P +P

(D) 1 2P P

Hint: P1 = 21

1 kx2

21x = 12P

k

P2 = 22

1 kx2

22x = 22P

k

P = 21 2

1 k x x2

= 2 21 2 1 2

1 k x x 2x x2

= 1 2 1 21 2P 2P 2P 2Pk 22 k k k k

P = 1 2 1 21 2k P P 2 P P2 k

P = P1 + P2 + 1 22 PP 8. A small mass ‘m’ is suspended at the end of a

wire having (negligible mass) length ‘L’ and cross-sectional are ‘A’. The frequency of oscillation for the S.H.M. along the vertical line is (Y = Young’s modulus of the wire)

[MHT CET 2020]

(A) 121 YA

2π mL

(B) YA2πmL

(C) 12YA2π

mL

(D) 2πYAmL

Hint: Frequency of oscillation for S.H.M.

n = 1 g2π l

But Y = mgLAl

l = mgLYA

n = g YA12π mgL

= 1 YA2π mL

= 121 YA

2π mL

9. The displacement of the particle executing

linear S.H.M. is x = 0.25 sin (11t + 0.5) m. The

period of S.H.M. is ( = 227

)

[MHT CET 2020] (A) 3

7s (B) 2

7s

(C) 47

s (D) 17

s

Hint: x = 0.25 sin(11t + 0.5) Comparing with x = A sin (t + ) = 11

Time period, T = 2πω

= 2π11

= 2 2211 7

= 47

Sample

Con

tent

237


10. The displacement time graph of particle executing S.H.M. is given in figure: (sketch is schematic and not to scale) [JEE (Main) 2020]

Which of the following statements is/are true for

this motion?

i. The force is zero at t = 3T4

ii. The acceleration is maximum at t = T

iii. The speed is maximum at t = T4

iv. The P.E. is equal to K.E. of the oscillation

at t = T2

(A) i, ii and iv (B) i and iv (C) i, ii and iii (D) ii, iii, and iv

Hint: For option (i), when t = 3T4

, x = 0

F = ma = m2x ....( |a| = 2x) At x = 0, F = 0 For option (ii), when t = T, x = A |a| = 2A i.e., acceleration is maximum

For option (iii), when t = T4

= T, x = 0

|v| = 2 2ω A x |v| = A ….(at x = 0) i.e., speed is maximum. For option (iv),

If K.E = P.E, then x = A2

x = A cos t

A2

= A cos t

t = T8

Hence, option (iv) is incorrect. 11. The phase difference between displacement and

acceleration of a particle in a simple harmonic motion is [NEET (UG) P-I 2020]

(A) 32 rad (B)

2 rad

(C) zero (D) rad 12. Identify the function which represents a periodic

motion. [NEET (UG) P-II 2020] (A) e–ωt (B) eωt (C) loge(ωt) (D) sinωt + cosωt Hint: For periodic function f(t) = f(t + T) where T is time period of function Here, sin(ωt + 2π) + cos(ωt + 2π) = sinωt + cosωt

Q.1. Select and write the correct answer: [04] i. A seconds pendulum is suspended in an elevator moving with constant speed in downward

direction. The periodic time (T) of that pendulum is _______. (A) less than two seconds (B) equal to two seconds (C) greater than two seconds (D) very much greater than two seconds ii. A simple pendulum performs simple harmonic motion about x = 0 with an amplitude A and time

period T. The speed of the pendulum at x = A/2 will be

(A) A 3 /T (B) A/T (C) 3

2 A/T (D) 32 A/T

iii. The acceleration of a particle in S.H.M. is _______. (A) maximum = a2, when it passes through mean position (B) maximum = a2, when it attains the extreme position (C) minimum = a, when it passes through its mean position (D) minimum = a, when it attains the extreme position iv. A mass m attached to a light spring oscillates with a period of 2 s. If the mass is increased by 2 kg, the

period increases by 1 s, then value of m is (A) 1 kg (B) 1.6 kg (C) 2 kg (D) 2.4 kg

Time: 1 Hour 30 Min Total Marks: 25TOPIC TEST

SECTION A

O time (s)

2T4

T4

3T4

T 5T4di

spla

cem

ent

Sample

Con

tent

238


Q.2. Answer the following: [03] i. When can a oscillation said to be a complete oscillation? ii. Define ideal simple pendulum. iii. Describe the state of oscillation if the phase angle is 1125°. Q.3. A simple pendulum of length 1.25 m has a mass of 50 g and oscillates freely with amplitude 5 cm.

Find its P.E. at extreme positions. (g = 980 cm/s2) Q.4. Obtain an expression of displacement of a particle performing linear S.H.M. Q.5. Derive the differential equation of linear simple harmonic motion. Q.6. What is the value of kinetic energy and potential energy at i. mean position? ii. extreme position? What can be concluded about total energy from it. Q.7. Compare the kinetic energy and potential energy of a particle performing S.H.M. when particle is at

distance of x = ± A2

. Q.8. A particle performs S.H.M of period 12 second and amplitude 8 cm. If initially the particle is at

positive extremity, how much time will it take to cover a distance of 6 cm from the extreme position?

Q.9. Draw graphs of displacement, velocity and acceleration against phase angle, for a particle

performing linear S.H.M. from the mean position Q.10. The equation of motion of a body performing linear S.H.M is x = 3 sin ππt +

3

cm. Determine its

i. displacement ii. velocity iii. acceleration Q.11. Obtain the expression for the period of a magnet vibrating in a uniform magnetic field and

performing S.H.M.

Q.12. Using differential equation of linear S.H.M, obtain the expression for velocity in S.H.M. and

acceleration in S.H.M. Q.13. i. State the laws of simple pendulum. ii. A particle executes S.H.M. with a period 8 s. Find the time in which half the total energy is

potential.

SECTION B

Attempt any Four: [08]

Attempt any Two: [06]

Attempt any One: [04]

SECTION C

SECTION D

Download the answers of the Topic Test by scanning the given Q.R. Code.

Sample

Con

tent

https://www.targetpublications.org/std-12-physics-book-vol-1-hsc-science-maharashtra-board-perfect-series.html

written as per the latest textbook prescribed by the

Documents