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Solutions to

l.B.lradav's 1'101Jlemsin General JJllysics llalu 1

Waves • Optics • Modern Physics

1

'

Second Edition

ABHAY KUMAR SINGH Director Abhay's l.l.T. Physics Teaching Centre Patna-6

CBS

CBS PUBLISHERS & DISTRIBUTORS 4596/1A, 11 DARYAGANJ, NEW DELHI - 11O 002 (INDIA)

ISBN: 81-239-0486-X

First Edition : 1996 Reprint : 1997 Second Edition : 1998 Reprint: 2000 Reprint : 2002 Reprint : 2004 Copyright © Author & Pubfü,her Ail rights rescrved. No part of thi1, book may he rcproduced or transmitted in any form or by any means, electronic or mechanical. including photocopying, recording. or any information torage and rctrieval systcm without permission, in writing, from thc publishcr. Published by S.K. fain for CBS Publi hcrs & Distríbutor . 4596/IA, 11 Darya Ganj, Ncw Delhi - 11 O 002 (India). Printed at : J.S. Offset Printers, Delhi - 110 051

In the memory of Late Shri Arvind Kumar (Ex-Director, The Premier lnstitute, Patna)

The man who taught me how to teach.

FOREWORD

Science, in general, and physics, in particular, have evolved out of man's quest to know beyond unknowns. Matter, radiation and their mutual interactions are basically studied in physics. Essentially, this is an experimental science. By observing appropriate phenomena in nature one arrives at a set of rules which goes to establish sorne basic fundamental concepts. Entire physics rests on them. Mere knowledge of them is however not enough. Ability to apply them to real dayto-day problems is required. Prof. lrodov's book contains one such set of numerical exercises spread over a wide spectrwn of physical disciplines. Sorne of the problems of the book long appeared to be notorious t pose serious challenges to students as well as to their teachers. This book by Prof Singh on the solutions of problerns of lrod.o. v's book, at the outset, seems to remove the sense of awe which at one time prevailed. Traditionally a difficult exercise to solve continues to draw the attention of concemed persons over a sufficiently long time. Once a logical solution for it becomes available, the difficulties associated with its solutions are forgotten very soon. This statement is not only valid for the solutions of simple physical problems but also to various physical phenomena. Nevertheless, Prof Singh's attempt to write a book of this magnitude deserves an all out praise. His ways of solving problems are elegant, straight forward, simple and direct. By writing this book he has definitely contributed to the cause of physics education. A word of advice to its users is hO\wrever necessary. The solution to a particular problem as given in this book is never to be consulted unless an ali out effort in solving it independently has been already made. Only by such judicious uses of this book one would be able to reap better bencfits out of it. As a teacher who has taught physics and who has been in touch with physics curricula at 1.1.T., Delhi for over thirty years, I eamestly feel that this book will certainly be of benefit to younger students in their fonnative years.

Dr. Dilip Kumar Roy Professor of Physics Indian Institute of Technology, Delhi New Delhi-110016.

FOREWORD

A proper understanding of the physical laws and principies that govem nature require solutions of related problems which exemplify tl1e principie in question and leads to a better grasp of the principies involved. It is only through experiments or through solutions of multifarious problem-oriented questions can a student master the intricacies and fall outs of a physical law. According to Ira M. Freeman, professor of physics of the state university of new Jersy at Rutgers and author of "physic--principles and lnsightsn -' 'In certain situations mathematical fommlation actually prometes intuitive understanding....... Sometí mes a mathematical fommlation is not feasible, so that ordinary language must take the place of mathematics in both roles. However, Mathematics is far more rigorous and its concepts more precise titan those of language. Any science that is able to make extensive use of mathematical symbolism and procedures is justly called an exact science". I.E. Irodov's problems in General Physics fulfills such a need. This book originally published in Russia contains about 1900 problems on mechanics, thennodynamics, molecular physics, electrodynamics, waves and oscillations, optics, atomic and nuclear physics. The book has survived the test of class room for many year's as is evident from its number of reprint editions, which have appeared since the first English edition of 1981, including an Indian Edition at affordable price for Indian students. Abhay Kun1ar Singh's present book containing solutions to Dr. I.E. Irodov's Problems in General Physics is a welcome attempt to develop a student's problem solving skills. The book should be very useful for the students studying a general course in physics and also in developing their skills to answer questions nonnally encountered in national level entrance examinations conducted each year by various bodies for admissions to professional colleges in science and teclmology.

B.P. PAL Professor of Physics I.I.T., Delhi

Preface to the Second Edition Perhaps nothing could be more gratifying for an author than seeing his 'brainchild' attain wide acclaim. Fortunately, it happes so with 'Solutions to l. E. Irodov's Problems in General Physics (Volume-II) authored by me. Since inception, it showed signs of excellence amidst its 'peer-group', so much so that it fell victim to Piracy-syndrome. The reported on rush of spurious copies of this vo/ume in the market accelerated the pace of our contemplation for this second edition. Taking advantage of this occassion the book has almost been comptelely vetted to cater to the needs of aspiring students. My heart felt thanks are due to ali those who have directly or indirectly engineered the cause of its existing status in the book-world. Patna June 1997

Abhay Kumar Singh

Preface This is the second volume of my "Solutions to I.E. Irodov's Problems in General Physics." It contains solutions to the Iast three chapters of the problem book ''Problcms in General Physics". As in the first volume, in this second one also only standard methods have been used to solve the proLlems, befitting the standard of the problems sobed. Nothing succeeds like success, they ')ay. From the way my earlier books have been received by physics loving people all over the country, I can only hope that my present attempt too will be appreciated and made use of at a large scale by the physics fratemity. My special thanks are due to my teacher Dr. (Prof.) J. Thakur, Department of Physics, Patna University, who has been my source of energy and inspiration throughout the preparation of this book. I am also thankful to computer operator Mr. S. Shahab Añmad and artist Rajeshwar Prasad of my institute (Abhay's 1.1.T. Physics Teaching Centre, Mahendru, Patna-6) for their pains-taking efforts. I am also than 1-:ful to all my well-wishcrs, friends and family members- for their emotional support.

Abhay Kumar Singh Patna July, 1996

CONTENTS Preface .................................................................................................................................... vii

PARTFOUR OSCILLATIONS ANO WAVES 4.1 4.2 4.3 4.4

Mechanical Oscillations . Electric Oscillations . . . Elastic Waves. Acoustics EJectromagnetic Waves. Radiation

. 1

54 82 103

PART FIVE OPTICS 5.1 5.2 5.3 5.4 5.5 5.6 5.7

Photometry and Geometrical Optics Interference of Light Diffraction of Light ........ . Polarization of Light ....... . Dispersion and Absorption of Light . Optics of Moving Sources .......... . Thermal Radiation. Quantum Nature of Light .

115 149 162 196 . 219

. 229 . 238

PART SIX ATOMIC ANO NUCLEAR PHYSICS 6.1 6.2 6.3 6.4 6.5 6.6 6.7

Scattering of Particles. Rutherford-Bohr Atom ... Wave Properties of Particles. Schrodinger Equation . Properties of Atoms. Spectra Molecules and Crystals Radioactivity . . . . . Nuclear Reactions . . Elementary Particles .

· 259 . 285 . 310 . 337 . 360 . 372 ..... 387

PART FOUR

OSCILLATIONS AND WAVES

4.1 MECHANICAL OSCILLATIONS 4.1

(•) Given, x • a cos ( rot-:)

So, v, •

X • - a ro sin ( ro f -

x· • - a ro2 cos ( ro t - :

and w. •

: )

)

(1)

On-the basis of obtained expressions plots x ( t ) , vx ( t ) and wx ( t) can be drawn as shown in the answersheet, (of the problem book ).

(b) From Eqn (1) vx

•

-

• (ro t a ro sm

:rt) 4

2 2 · 2( ro t - :rt)

=

So, v2x

a ro sm

4

(2)

But from the law x -= a cos ( ro t- :rt/4), so, x2' -= a2 cos2 ( ro t - rr./4) 2

sin

or

or,

2

( co

t - :rtl4 ) • 1 ª2

(3)

Using (3) in (2),

v! • ª2 ro2 ( 1 -

)

v! • ro2 ( ª2 - x2 )

or

(4)

Again from Eqn (4), wx • - a ro2 cos ( ro t- rr./4) • - ro2 x

4.2 (a) From the motion law of the particle 2 x • a sín ( ro t- n:/4) • ; [ 1- cos ( 2 ro

t-;)]

or,

x - ; • - ; cos ( 2 ro t - ; ) • - ; sin 2 ro t • ; sin (2 ro t + n: )

i.e.

x - ; • ; sin ( 2 co t + :rt ).

Now compairing this equation with the general equation of harmonic oscillations : X • A sin ( co0 t + a ) Amplitude, A

= ; and angular frcquency, ro0

Thus the period of one full oscillation, T •

2

•

2 ro.

1t "°"'

roo

l!. ro

(1)

2

(b) Differentiating Eqn (1) w.r.t. time vx = a w cos (2 ro t + re) or v;

= a2 w2 cos2 (2 w t +re)= a2 w2 [ 1- !>in 2 (2 ro t + re)] 2

( x-

From Eqn (1)

2

=

)

(2)

sin 2 ( 2 ro t + r1)

or,

(3) 2

Plot of vx

(

2

2

v, = a ro :x (1 - :) • 4 ro x (a - x)

From Eqns (2) and (3),

x ) is as shown in the answersheet.

4.3 Let the general equation of S.H.M. be x = a cos ( w t + a )

(1)

vx = - a w sin ( e.o t + a )

So, Let us assume that at t

=O

, x

Thus from Eqns (1) and (2) for Therefore

tan u

=-

= x 0 and t = O, x0 and a

v.,

e.o Xo

(2)

vx = vXo. = acosa,

= ,¡ X5

= - a e.o sin a

vXo

and

2

+ ( vXo )

= 35.35 cm

ú)

Under our assumption Eqns (1) and (2) give the sought x and vx t - t = 2 ·40 s , a

=

'V x 2

0

+ ( vx/ro, ) 2

Putting all the given numerical values, we get : x = - 29 cm and vx

4.4 From the Eqn, v; = c.o2 ( a 2 - x2 )

and v Solving these Eqns simultaneously, we get

V( Vi -

= -

(

- rovXxo)

=-

re 4

81 cm/s

(see Eqn. 4 of 4.1)

vi = ol ( a2 - xf )

w=

= tan -1

and u

if

vª ) / ( x'2 -

Ji) ,

=

a=

2

w (

a2 -

x )

V( 1x'2 v

v

Xi ) / ( vi - v

)

4.5 (a) When a particle starts from an extreme position, it is useful to write the motion law as X = a COS C.O t (1) (However x is the displacement from the equlibriVm position) It t1 be the time to cover the distence a/2 then from (1) a 1 re a a = = cos = a cos ro t 1 or cos e.o t 1 = 3 2 2 2 n T Thus t1 = - re = 3 ro = 3 ( 2 n/T) 6

( as

t1 < T14 )

3

x • a cos ro t ,

As

so , xv • - a ro sin m t

v • lvxl • -vx • arosinrot,

Thus

t1 • T/6

for ,

Hence sought mean velocity

J v dt 3a • J •fa ( 2 ,E,/T) sin ro tdt / T/6 • T 6

T/

dt

0·5 m/s

0

(b) In this case, it is easier to write the motion Jaw in the fonn : x • a sin ro t

(2)

If t2 be the time to cover the distance a/2, then from Eqn (2)

a/2

=

a si.nT2 n t2

. 2n t 2

or smT

1 2

=

n

2n

= 6

-t2

Thus

•

T

. sm

n ( as t < .'17.,'/4) 2 6

T 12

or' t2•

Differentiating Eqn (2) w.r.t time, we get

2n v = 1 v,. I

So,

a ro cos ro t • aT cosT t 2n 2n aT cos t, for t s. t2 = T112

•

vx

2n

•

7 Hence the sougbt mean velocity v dt 1

f

• --

fdt

2 1t

r112

• (T/12)

f ªr

2 1t

6a

cos ytdt •

T - lm/s

o

4.6 (a) As x • a sin ro t

vx • a ro cos ro t

so,

!r 8

fa ro cos Tbus < yx > •

J v dt/1dt = o

(2 n/T) tdt

• 2Vn2aro( u.smg T = -;-n)

T

x

3

8

--

(b) In accordance with the problem V

•

Vx

.l., ...

SO,

1 < > 1 • 1 < Vx > 1

2V2 aro

Hence, using part (a), 1 < V-> 1 •

3n

(e) We bave got, vx = a ro cos ro t Sov, • 1 v, 1 • a ro cos rot , for t "' T;4

• - a ro cos ro t, for TI 4s

ts

8

l

T

2V2 aro -3 n

4

Hence,

3 T/8

o

T/4

f a oo cos oo t dt +f - a oo cos oo t dt

f dt = f dt

= -

T/4

V

3 T/8

Using ro = 2 rr.lT, and on evaluating the integral we get 2 (4 -Y2) a oo

=------

3rr.

4.7 From tbe motion law, x = a cos oo t,, it is obvious that the time taken to cover the distance equal to tbe amplitude (a), starting from extreme position equals T/4. Now one can write t =

n;

+ 10 ,

(

where r0 O . Toe equilibrium position can in principie be a maximum but then U" ( x0 ) < O and the frequency of oscillations about this equilibrium position will be imaginary. Toe answer given in the book is incorrect both numerically and dimensionally. 4.18 Let us locate and depict the forces acting on the ball at the position when it is at a distance x down from the undeformed position of tbe string.

At this position, the unbalanced downward force on the ball = mg-2Fsin9 By Newton's law,

mx· = m g- 2F sin 8

= mg-2F

• m g - 2F

Thus

..

x=

..,

Sis small ) 4F x • mg- - - x

112

A

1

4F 4F( g-m1x•-m1x-

putting x' = x X=

e ( when

:4Fmr,/)

!!!_8_:l_ T , we get

4T , --x mi

- e-;

Thus T= rt ·y F

.= O·2s

4.19 Let us depict the forces acting on the oscillating ball at an arbitraty angular position 0. (Fig.), relative to equilibrium position where FB is the force of buoyancy. For the baJI from the equation :

-......'

Nz • I z, (wbere we have taken the po,sitiv! sense of Z

Fa

""" •

---

axis in the direction of angular velocity i.e. 8 of the ball - - - -,- - and passes through the point of suspension of the- - - -,- - pendulum O ), we get : - m g l sin e + FB l sin e

- m i1 a·

Using m = : 1t r3 o, FB = : 1t r 3 p and sin 8 • 8 for small 8, in Eqn (1), we get:

(1)

10

8 • -f(1-!)ª Tbus the sougbt time period

T=2

==l====2a

v'f(1-!) .. I

Hence

.T • 2a

1-

::a--11 l

V g ( TI _ 1) • 1·ls

4.20 Obviously for small the ball execute part of S.H.M. Due to tbe perfectly elastic collision the veJocity of balJ simpJy reversed. As the baJI is in S.H.M. (l 8 I < a on the Jeft)its motion law in differential from can be written as

•• g_ 8 • 8 • -

2 ro0 8

(1)

1 If we assume that the hall is released from the extreme position, 8 • of differential equation would be taken in the fonn

8•

fl cos "'º 1

fl cos

•

at t • O, the solution

(2)

1

If t ' be the time taken by the ball to go from the extreme position 8 • f} to the wall i.e. 9 = - a , then Eqn. (2) can be rewritten as

- a •

t'=vf

or

cos-1

Thus the sougbt time T = 2 t ' = 2

=2

vf (;

fl

cos

(-; )- vf

vf(

+ si-n

1

t'

lt - co-s

1

(n-cos-';)

;)

[because si-n 1 x + co-s

; ) ,

1

x - n/2 ]

4.21 net the downward acceleration of tbe elevator car has continued for time t ', then the sought time t

=

w

+ t ' , where obviously

is the time of upward acceleration of the elevator.

w

One should dote that if the point of suspension of a mathematical pendulum moves with an acceleration then the time period of the pendulum becomes

w,

2nv'

1

.....,.

1

g-w

( see 4.30)

In tbis prob]em the time period of the pendulum while it is moving upward with acceleration w becomes

11 2

1t

r-¡-

V

and its time period while the elevator moves downward with the same

magnitude of acceleration becomes 2

t

As the time of upward acceleration equals

w

-w , the total number of oscillations during

this time equals

Y2h/w

21tV l/(g+w)

Y 2 hlw

Thus the indicated time =

·21tv_r;l/-g;-

- hlw -VI (g+w)/ = V-2

g

21tv 1/(g+w> Similarly the indicated time for the time intetval t '

,¡ t '

=

21t

2lt

V //g = t ' ,¡(

g - w )Ig

ll( g- w)

we demand that

V 2 hlw V ( g + w )lg + t' V ( g- w )/g = V 2 h/w + t' t '=

or,

V 2h/w V g +w - Vg vi-vg-w

Hence the sought time

.. r:;-;:

,_-r:¡;-Vg+w-Vg-w V=-!!.. vi-vg-w w

'=V!:.!!..+t w

= _r: V !;:.-.!.;!..: V1+(3-Vl-'3

1-Vl-(3

w

h

'

w ere t-'

A.

=

/

w g

4.22 If the hydromoter were in equlibrium or floating, its weight will be balanced by the buoyancy force acting on it by the fluid. During its small oscillation, let us locate the hydrometer when it is at a vertically downward distance x from its equilibrium position. Obviously the net unbalanced force on the hy rometer is the excess buoyancy force directed upward and equals 1t r2 x p g. Hence for the hydrometer.

••

mx

=-

2

1tr

pgx

.. or,

X

Hence the sought time period T =- 2 a

= -

V

1

n: r p g

= 2· 5 s .

12

4.23 At first Jet us calculate the stiffness K 1 and K 2 of both the parts of tbe spring. If we subject the original spring of stiffness K having the natural length 10 (say), under the deforming forces F -F (say) to elongate the spring by the amount x, then " F = KX (1) Therefore the elongation per unit length of the spring is x / 10 • Now let us subject one of the parts of the spring of natural length T} 10 under the same defonning forces F - F. Then the elongation of the spring will be X

lo TJ lo • TxJ Thus Hence from Eqns (1) and (2) K •

F •

K1

T} K¡

or

Kz •

Similarly

(T}x)

(2)

K¡ • KIT)

(3)

K

1 -T)

Toe position of the block m wben botb the parts of the spring are non-defonned, is its equilibrium position O. Let us displace the block m towards right or in positive x axis by the small distance x. Let us depict the forces acting on the block wben it is at a distance x from its equilibrium position (Fig.). From the second law of motion in projection fonn i.e.

or, Thus

K -+K --)

-

( T) •• X

1-r¡ K

x • mx··

1

= - - mr¡(m) X

Hence the soüght time period T=2n

Y T} ( 1

.!. T) ) m/K

• O· 13 s

4.24 Similar to the Soln of 4.23, the net unbalanced force on the block m when it is at a small horizontal distance x from tbe equilibrium position becomes ( K1 + K2 ) x. From E;" = m wx for the block : Thus Hence the sought time period T •

2n

V

m

K¡ + K2

Alternate : Let us set tbe block m in motion to perfonn small oscillation. Let us locate the block wben it is at a distance x from its equilibrium position. As the spring force is restoring conservative force and deformation of botb the springs are

same, so from the conservation of mechanical energy of oscillation of the spring-block system :

13

m (

!)

2

+

K1

Differentiating with respect to time 1 . ..

2

m2xx

K2

x

1 +

..

or,

2

x2+

2

•

X

-

( K1 + K2 )

Constan!

•

. 2xx

( Kt + K2)

=

O

X

m

Hence the sought time period T • 2 3t

V

m K1 +K2

4.25 During the vertical oscillation Jet us ]ocate the block at a vertical down distance x from its equilíbrium position. At this moment if x1 and are the additional or further elongation of the upper & lower springs relative to the equilibrium position, then the net unbalanced force on the block will be K2 x 2 directed in upward direction. Hence K 2 x2 •

-

..

(1)

mx

We also have x • x1 + x2 (2) As the springs are massless and initially the net force on the spring is also zero so for the spring (3) K1X1 = K2X2 Solving the Eqns (1), ( ) and (3) simultaneous]y, we get K¡ K2 K¡

.. Thus

X

..

x - mx

+ K2 ( K¡ K2IK1 + K2 )

=

X

m

.. I ( K1 Hence the sought time period T = 2 3t V m ---

K2)

K 1 K2

4.26 Toe force F, acting on the weight deflected from the position of equilibrium is 2 T0 sin 0. Since tbe angle 8 is smalt, the net restoring force, F = 2 T0 or,

F

k x , where

X

1

2t

2To k = - 1

So, by using the formula,

wo

=V-l, k

wo

=V -/=z;;rf-;

To

4.27 If the mercury nses m tht. Ieft arm by x it must fall by a slanting length equal to x in the other arm. Total pressure difference in tbe two arms will tben be pgx+pgxcos0 This will give rise to a restoring force

= pgx(l+cos0)

-pgSx(l+cos0) This must equal mass times acceleration which can be obtained from work energy principie.

14

1 m ·x2 2

The K.E. of the mercury in the tube is clearly : So mass times acceleration must be : m Hence m This is S.H.M. with a time period

x·

x· + p g S ( 1 + e.os 8 ) x =- O in

T = 2 :n;

V pgS(l+cos8)

------.

4.28 In the equiJibrium position the C.M. of the rod lies nid way between the two rotating wheels. Let us displace the rod horizontally by sorne small distance and then release it Let us depict the forces acting on the rod when its C.M. is at distance x from its equilibrium position (Fig.). Since there is no net vertical force acting on the rod, Newton's sec.ond law gives :

N1 + N 2 = mg For the translational motion of the rod from the Eqn. : Fx • m wcx

mx·

k N 1 - kN 2 = As the rod experiences no net torque about an axis perpendicular to the plane of the Fig. through the C.M. of tbe rod.

N1 ( I x

) = N2

(1; x

)

Solving Eqns. (1), (2) and (3) simultaneously we get .. 2p x = - k;;:;_s¿x l Hence the sought tim period T •

2rty;;r; - rtyg - l·Ss 2kg

kg

(1) (2)

(3)

15 4.29 (a) Toe only force acting on the ball is the gravitational force F, of magnitude y

j

Jt

p m r,

where y is tbe gravitationa) constant p, tbe density of the Earth and r is the distance of tbe body from tbe centre of the Earth. But, g = y

F =-m

4

x3 p R, so the expression for

g;,-

Fcan be written as,

here R is the radius of the Earth and the cquation of motion in projection

fonn has the fonn, or,

mx· +

g

X

=o

(b) Toe equation, obtained above has tbe fonn of an equation of S.H.M. having the time

T•2"

period,

{'f ,

Hence the body will reach the other end of the shaft in the time, t

=T = x 2

g

= 42

min.

(e) From the conditions of S.H.M., the speed of the body at the centre of the Earth will be maximum, having the magnitude, v

= Rw = RY g/R = VgR =

7·9km/s.

4.30 In the frame of point of suspension the mathematical pendulum of mass m (say) will oscillate. In this frame, the body m will experience the inertial force m ( - w) in addition to the real forces during its oscillations. Therefore in equilibrium position m is deviated by sorne angle say a. In equilibrium position

T0 cos a • m g + m w cos ( Jt - f3 ) and T0 sin a • m w sin ( Jt - f3 ) So, from these two Eqns g-wcos . A wsm.., .. ¡-m-2w - -s2 in_ _2 +_(_m_ g m_w-co_s_ -)2ros a• V and mg-m wcos f3

tan a.,.

w

mw

(1)

U)

7nU)

16

Let us displace the bob m from its equilibrium position by some small angle and then release it Now locate 1he hill atan qular posi1ioo (a+ 8) ftom vertiatl as shoMt in the figure. From the Eqn. : NO& • I z

e·

- m g l sin (a+ 8) - m w cos {,t - fl) l sin (a+ 8) + m w sin {,t - fl) l cos (a+ 8) = m / 2 or,- g (sin a cos 8 + cos a sin 8) - w cos {,t - fl) (sin a cos 8 + cos a sin 8) + w sin (cos a cos 8 - s in a sin 8)

..

• 18

8, sin 8 • 8 cos 8 • 1

But for small

- g (sin a+ cosa 8) - w cos (,.: - P) (sin a+ cosa 8) + w sin (cosa - sin a 8)

So,

..

- /8 ( tan a + 8 ) ( w cos

or,

- g ) + w sin

( 1 - tan a 8 ) - --

v_

cosa

Solving Eqns (1) and (2) simultaneously we get cos + w 2> 1 _ _ 2+_w_2 2_w cso p - o t +

a)

moment i.e. at t = O , 0 = 0m than a = O.

If at the initial

Thus tbe above equation takes the fonn 8 = 8m CO'S OOo f =

am cos _r;-y It -

o

3 cos

y_/9-s o-s t

e = 3º cos 3.5 t

Thus

(b) The S.H.M. equation in angular form : 0 = 8m sin ( COo t + a ) If at the initial moment t = O , 0 - O , then a "" O . Then the above equation takes the fonn e= sin roo t Let v0 be the velocity of the Jower end of pendulam at O = O, then from conserved of mechanical energy of oscillaton E mean = Eexrreme or T mean = Uatrem

em

or,

l

Thus 0• •

cos

-1 ( 1

V

-

-1 [

)

2g l =

cos

1 -

2

}2 · 8 x{ O. 9 · 822 xO

=

4 . 5e,

T.trns he sought equation becomes 8 = 0m sin Cüot 4·5° sin 3·5 t :¡¡¡;

(e) Let 00 and v0 be the angular deviation and lin ar veJocity at t = O. As the mecbanical energy of oscillation of the mathematical pendulum is conservation

1 2 m Vo + m g l ( 1 - cos 2 V

or,

2

Thus 0m = c os- 1

cos {

ªº-

=g

ªº ) •

em )

l ( cos 80 - cos 8m )

V2o } 2g l

m g l ( 1 - cos

- 1{

=

cos

0

cos3 -

(

0·22 )2

o

}

2 x 9·8x O· 8 -

5·4

18

Toen from 8 • 5.4º sin (3.5 t + a), we see that sin a =-

3 and cos a < O because 5 .4

the velovity is directed towards the centre. Thus a • ; + 1.0 radians and we get the answer. 4.34 While the body A is at its upper extreme oosition, the spring is obviously elongated by the amount

m;g}

( a-

If we indicate y-axis in. vertically downward direction, Newton's second law of motion in projection form i.e. F, • m w, for body A gives :

m1 g + K( a- m g) • m1oo2 a

or, K( a-

m:g)-

m1( oo2 a - g )

(1)

(Because at any extreme position the magnitude of acceleration of an oscillating body equals o? a and is restoring in nature.) If N be tbe normal force exerted by the floor on the body B, while the body A is at its upper extreme position, from Newton's second law for body B N+ or,

N • m,, g -

+-m:g) •

K( a - m 8 )

• m,, g - mi(

"'2 g 2

00 a

- g ) ( using Eqn. 1 )

HenceN • ( m1 + m2 ) g - m1 ro2 a When the body A is at its lower extreme position, the spring is comprcsed by the distance ( a+ m:g} From Newton's second law in projeciton fonn i.e. F1 -= m w1 for body A at this state: ffl¡ g

-

K( Í, + m:g

) • m¡( -

00

2a) or, K( a + m g ) •

ffl¡ ( g

+ 002 a )

(3)

In this case if N' be the normal force exerted by tbe tloor on tbe body B, From Newton's second law

for body B we get: N ' • K( a + mg

) + m,, g • m1 ( g + ro2 a ) + m2 g ( using Eqn. 3 )

Hence N ' = ( m1 + m2 ) g + m1 ro 2 a From Newton's third law the magnitude of sought forces are N' and N, respectively. 4.35 (a) For the block from Newton's second law in projection form F1 = m w1 N-mg-my' But from

y - a ( 1 - cos ro t )

(1)

19

j," • ol a cos ro t

We get

(2

From Eqns (1) and (2) N•

m g ( 1 + ro;ª cos ro t)

: th

From Newtons's third law the force by wh(ich

g

(3)

dy m cx)crts on the blo k is dircctcd 2

vertically downward and equl - • m g 1 +

cos ro t

(b) When the body m starts, falling behind the plank o¡)o ing contact, N • O, (because the nonnal reaction is the cont.act force). Thus from Eqn. (3)

m g ( 1 + ro;ª cos rot) • O

Hence

ªmio

= g/ ro2

for sorne

t.

• 8 cm.

(e) We observe that the motion takes place about the mean position y = a. At the initial instant y • O. As shown in (b) the normal reaction vanisbes at a height ( g/ ro2 ) above the position of equilibrium and the body Oies off as a free body. The speed of the body at a distance ( g/ ro2 ) from the equilibrium position is ro condition of the problem gives

V a2 - ( g/ ro2) 2 , so tbat the

[ ro V_a_2 -(-g/_ro_2_)_2 J2 _g_

h + 2+a= 2g ro Hence solving tbe resulting quadratic equation and taking the positivc roof,

a•

g - ¡-:¡¡;¡- •20cm. --+V ro2

ro

4.3' (a) Let y ( t) - displacement of the body from the end of tbe unstrecbed position of tbe spring (not the equilibrium position). Then my • -Ky+mg This equation has the solution of the form

y• A+Bcos(oot+a) if

- m ro2 B cos ( ro t + a ) - - K [ A + B cos ( ro t + a ) ] + m g

ro2

Then

=

and

A = !!!..G.. K

m we have

y • O

and

j, • O

at

- roBsin a - O A +Bcosa = O Since B > O and A > O we must have a • 1t

t - O . So

20

y• !!.&.(1-coswt)

and

K

(b) Tension in the spring is T • Ky • mg(l-coscdt) Tmax • 2mg, Tmin • O

so

---

4.37 In accordance with the problem So,

m(

F.• -amr

x· n y" ¡j x· • -

- - a m ( X i1 y ¡j

Thus ax and y· • - ay Hence the solution of the differential equation ax becomes x • a cos ( Wot + b),

x· • -

where ro = a

(1)

So, x = - a Wo sin ( co0 t + a ) From the initial conditions of the problem, vx • O and x • r0 at t ... O

(2)

So from Eqn. (2) a • O, and Eqn takes tbe form x • r0 cos co0 t so, cos Wo t • x/ r0 One of tbe solution of the otber differcntial Eqn ·j, • - a y , becomes y • a' sin ( Wo t + , ) , where co • a From the initial condition, y • O at t • O , so ()' • O and Eqn (4) becomes : y - a' sin ro0

(3)

(4)

t(S)

Differentiating w.r.t. time we get

.

' 0 cos ro0 t y • a 00

(6)

But from tbe initial condition of the problem, j, • v0 at t - O, So, from Eqn (6) v0 = a' ro0 or, a' = v0/ w0 Using it in Eqn (5), we get Vo . . 00oY y• - sm oo t or sm oo t

roo

0

0

Squaring and adding Eqns (3) and (7) we get :

=-

2 2

vo

WoY x sm ro0 t + cos ro0 t = vi +, . 2

(7)

2

2

5

or, i.38 (a) As the elevator car is a translating non-inertial frame, therefore the body m will experience an inertial force m w direct.ed downward in addition to the real forces in the elevator's frame. From the Newton's second law in projection form F1 = m w1 for the body in tbe frame of elevator car:

-K( g +y) + m g + m w mK

m ji

(A)

21 ( Because thc initial elongation in the spring is m g/K )

-K(Y- mKw) ::2(Y-mKw) • -y-mKw)

mj,° • -Ky+mw •

so,

or,

(1)

Eqn. (1) shows lhat the motion of the body m is S.H.M. and its solution becomes

·(-vI;K;;i+a)

y-mw •asm

(2)

Differentiating Eqn (2) w.r.t time

(3) Using the initial condition y ( O ) = O in Eqn (2), we get :

. m w as ma• -- K

and using the other initial condition j,( O ) • O in Eqn (3) a

weget

cosa= O

m

a • -a/2

Thus

and a • -

mw K

Hence using these values in Eqn (2), we get

y.

mKw(

1-oos t)

(b) Proceed up to Eqn.(1). The solution of this differential Eqn be of the form : y-mKw •

y - ...!!.!._ • a sin ( Klm

or,

or,

asin( t+l>)

y-

ª! = a sin (ro ro0

0

t+ m

6)

t + {)) (wher roo =

V=)

From the initial condition that att • O, y (O) =- O , so O = a sin 6 Thus Eqn.(4) takes the from : y -

ª:•

a sin ro0 t

(4)

m

or

6=O

(5)

roo

.

a

Differentiating Eqn. (5) we get : y - 2 roo

= a ro0 cos ro t

(6)

22 But from tbe other initial condition j,( O ) • O at t • O. So, from Eqn.(6)

,-

a COo

- a roo or

3

a - - al

COo

Putting the valuc of a in Eqn. (5), we get the sought y ( t ) . i.e. . ro t ) y -2 at • -3 a. sm ro0 t or y • 3 ª( ro0 t - sm 0 roo roo roo 4.39 There is an important difference between a rubber cord or steel coire and a spring. A spring can be puUed or compressed and in both cases, obey's Hooke's law. But a rubber cord becomes loóse when one tries to compress it and does not then obey Hooke's law. Tbus if we suspend a' body by a rubber cord it stretches by a distance m glK in reaching tbe equilibrium confi uration. If we further strech it by a distance 6 h it will execute hannonic oscillations when released if 11 h s m glK because only in this case wiU the cord remain taut and obey Hooke's law. Tbus 11 '1m_ax = m glK The energy of oscillation in this case is

-1 1 1 -K(l1hum:)2 • -m g 2 2 K

2

4.40 As the pan is of negligible mass, tbere is no loss of kinetic energy evcn though tbe collision is inelastic. Tbe mech nical energy of the body m in the field generated by the joint action of both the gravity force and tbe elastic force is consetved i.e. 11 E • During the motion of tbe body m from the initial to tbe fmal (position of maximum compression of the spring) position A T • O, and Íherefore A U • A Ux, + A Usp • O

o:

. 1

- m g ( h + x ) + - K x2 2 On solving tbe quadratic equation : or

•

O

mg ... J m2g2 2mgh x•--±V 2 + K

As minus sign is not acceptable X -

K

K

1C

+

y

m2g2 2 K

+

2mgh K

1f tbe body m were at rest on the spring, the corresponding position of m will be its equilibrium position and at this position the resultant force on the body m will be zero. Therefore the equilibrium compression Ax (say) due to the body m will be given by K 11 x = m g or /l. x = m g/K Therefore seperation bet\\-een tbe equilibrium position and one of tbe extreme position i.e. the sought amplitude

a - x - !J. x -

y

m2f2 + 2m gh K

K

23 The mechanical energy of oscillation wbich u, conserved equals E = Uatrr: ,because at the extreme position kinetic energy becomes zero. Although the weight of body m is a conseivative force , it is not restoring in this problem, hence Uextl'f!IM is only concerned with the spring force. Therefore 1 2 m2 g2 E= u =- 2 Ka • mgh+ 2K

4.41 Unlike the previous (4.40) problem the kinetic energy of body m decreases dueto the perfectly inelastic collision with the pan.Obviously the body m comes to strike the pan with vclocity v0 = Y 2 g h .If v be the common velocity of the " body m + collision then from the conservation of linear momentum m v0 • (M+m)v

pan " system due to lh

mV2g/i m v0 1) • (M+m) (.A1+m) At the moment the body m strikes the pan, the spring is compressed dueto the weigl-1. o! ·he pan by the amount M g/K . If l be tbe further compression of the spring due to t!J , Jocity acquired by the "pan - body m " system, then from the conservation of 1nedtJnical encrgy of the said system in the field generatad by tbe joint action of both the grav1ty and spring forces

or

V•

(Using 1)

.,., m•--gfl

1

or,

2K f

- m g l -(

=

\" )

m +.t ../ .. /

V

mg±

2

2Kghm

2

m g

O

-r

11

2

+m

l= --------K di

Thus

As minus sign is not acccptable

l

!!!..S.. 1 == K +

K

y

2 2 2K:n=gh m g + (M + m)

If the oscillatiug "pan + body m" system were at rest it corr spond to their equilbr¡ium position i.e. the spring were compressed by ( M + m)g therefore the amplitude of oscillation K

a= 1- mg = K

!!!:.Kv K

1+2/z mg

24 The mechanical energy of oscillation which is only consetved with the restoring forces becomes E = Uextreme =

1Ka

2

(Beca use spring force is the on]y restoring force not the

2

wcight of the body)

1 2 2 E = Tmean = 2 (M + m) a w

Alternately

E =! (M+

thus

m ) a2 (

_.. 4.42 We have F = a ( j,

M:

m

1

)

= -2Ka

2

xJ)

x· i+ j,' Jj - a (y x¡j mx· • ay and mj,' = - ax

or,

m(

So,

From the initial condition,at t = O ,

x=

(1)

O and y = O

So, integrating Eqn,mx· = a y .

... . -- ay

we get

Using Eqn (2) in the Eqn m : ..

rny

or

a

x • -y

(2)

m

= - ax , we get ••

ª2

• - m

or y • -

y

a

(

;;;

)2 y

(3)

one of the solution of di_fferential Eqn (3) is y = A sin ( 000 t + a ) , where 000 = a/m . As at t

=

O,

y = O , so the solution takes the fonn y • A sin 000 t

On differentiating w.r.t. time

j, • A ro0 cos ro0 t

From tbe initial condition of the problem, at t =-• O , j, • vo vO = A ro0

So,

y=

Tbus

or

A = v0/ w0

( v0/ oo0 ) sin 000 t

(4)

Tbus from (2) x = v0 sin ro0 t so integrating

Vo

x - B - - cos w0 t Wo

Vo

x = O at t = , O B

On using

Vo

x•-

Hence finally

(5)

= OOo

( 1 - cos ro t)

(6)

0

Wo

Hence from Eqns (4) and (6) we get

[ x - ( vofoo0 wbkh Xo

) ]

2

2

+y

•

( vofoo 0

)2

is tbe equation of a circle of radius ( vofoo0 ) with the centre at tbe point

= vol000 , Yo

• O

4.43 If water has frozen, the system consisting of the light rod and the frozen water in the bollow sphere constitute a compound (pbysical) pendulum to a very good approximation because we can take the whole system to be rigid. For such systems the time period is given by

- Í1

T1 = 2 rt V

g

-

¡ -¡;1 + ;-

V

l

2

k2 =;R

where

is the radius of gyration of the sphere.

Toe situation is different when water is unfrozen. When dissipative forces (viscosity) are neglected, we are dealing with ideal fluids. Such fluids instantaneously respond to (unbalanced) interna} stresses. Suppose the sphere with liquid water actually executes small rigid oscillations. Toen the portion of the fluid above the centre of the sphere will have a greater acceleration than the portion below the centre because the linear acceleration of any element is in this case, equal to angular acceleration of the element multiplied by the distance of thc element from the centre of suspension (Recall that we are considering small oscillations). Toen, as is obvious in a frame moving with the centre of mass, there will appear an unbalanced couple (not negated by any pseudoforces) which will cause the fluid to move rotationally so as to destroy differences in acceleration. Thus for this case of ideal tlui1/1 8

.•

/2

+G

2

8 • - ro2 / 2 8 - G

where ± G is the torque of mutual interactions. We have written tbe restoring forces on each pendulum in thc absencc of the othcr as - ro¡/ 1 8 and -

/ 2 8 respectively. Then

Hence

4.51 Let us locate the rod when it is at small angular position 8 relative to its equilibrium position. If a be the sought distance, then from the conservation 1 . of2 mechanical energy of oscillation m g a ( 1 - cos 8 ) + 1 , constant ( O ) • 2 00 Differentiating w.r.t, time we get : . 1 . .. m g a sin 9 8 + 2 / 00' 2 a a = o m ¡2

But

100,

•U

2

+ ma

and for small O , sin 8 • O, we get

_ a· -( ¡28ª 12+

2

) 8

ª

Hence the time period of oue full osscillation becomes or T 2 = -4-:7t2 ( g

T = 21t

For So,

d a ( ¡ 2a+ a ) = O Tmm , ob.v1ously d 12 2 l l --+ 12a2 1 = O or a = --

2V3

Hi>nce

Trrun = 2n:

Y± gV 3

-¡2-+a) 12a

30 Con 1dcr thc moment of inertia of the triangular plate about AB.

4.52 f

= h

f f./ d m = ff x

2

pdxdy

2 h •f - -·V 2] = f x p, dxh

=

½ ( •- )

...,....

......,

x2 {2to(h -x )d x

l-x r

o

A

h

o

* 6P =

mt

0n using the arca of the triangle /J. A B C •

!_ mh2 if

=

Thus K.E.

and m • p /J. •

2 6

1 82 P.E. = m g ( 1 - oos 8 ) • mgh 3 2 3 Here 8 is the angle that the instantaneous plane of the plate makes witb the equilibrium position which is vertical. (fhe p]ate rotates as a rigid body) h

!.m

E•

Tbus

2 0>2 •

Hence

4.53

m g h / mh 3 6

•

h

T • 2n

So

h2 92+ !.m gh 6 2 3

VA-• n yii.. 2g

g

e2 2

and

lra1u&. -= h/2.

Let us go to the rotating frame, in which the disc is stationary. In tbis frame tbe rod is subjected to coriolis and centrifuga] forces, Fcor and Fe¡, where

Feo, •

f 2 dm (

v' x ro;,

) and

Fe1 •

fd

m

r,

where r is the position of an elemental mass of the rod (Fig.) witb respcct to point O (disc's centre) and

,

V•-

d r' dt

r ·= OP = OA + AP dr d(AP) , (as OA is constant) = --- y dt d t-------As the rod is vibrating transversely, so v' is directed perpendicular to the length of the rod. Hence :. d:,: ( v' x ro) for each elemental mass of the rod is directed along PA. Therefore the net torque of coriolis about A becomes zero. Toe not torque of centrifulgal f(?rce about point So,

-

' .

./'1.

Now,

'i;¡(A.¡ •

J

AP x dmro r =

J

AP x (

7)

d. ro ( OA +AP)

31

f AP (m I ds) •7m a x

=

m20OA

J

=

m ds m2 0 s a.sm 8 ( - k )

1

1

J s ds •

sin 8 ( - k )

So,

'te¡(Z) •

m oo a

sin 8 ( - k )

o

....

'te¡(A) •

k

• 8 m ffio a / sm 2 2

•-

According to the equation of rotationaJ dynamics : t:A (Z)

lA ªz

•

2

l . m 1 •• -meooa2 sm8 • -3- 8 2

or,

2 00 a 0

•• 3 .8 • 2

or,

1

sin 8

2

3 00 a 8·· ■ - -2 --8 21

Thus, for small 8,

This implies that the frequency "'o of oscillation _is ro0

•

vi?i

4.54 The physicaJ system consists with a puJley and the block. Cboosing an intertial frame, Jet us direct tbe x-axis as shown in the figure.

To To

mg '/

r

X

Initially the system IS in equilibrium position. Now from the condition of translation equilibrium for tbe block To• mg Similarly for the rotational equilibrium of the pulley K 1:,./R • ToR or.

(1)

(2)

32

from Eqns. (1) and (2)

(3) Now let us disturb the equilibrium of the system no matter in which way to analyse its motion. At an arbitrary position shown in the figure, from Newton's second law of motion for the block

F"• mw" mg-T• mw• m.i"

(4)

Similarly for the pulley

Nz• 11:Jz TRBut

w=

.. ..

K (4 l + x) R • I 8

(5)

x· = R 8

(6)

fJR

or,

I .. T R - K (6. l + x) R • - x R

from (5) and (6)

(7)

Solving (4) and (7) using the initial condition of the problem

(

/) ..

-KRx• mR•R

x· - -(

or,

Hence the sought time period, T-

x

" 1) x m+R2

2 2 n. 2 _., ../ m + l / R V

ro0

K

Note : we may solve tbis problem by using tbe conse.rvation of meclianical energy also 4. SS At the equilbrium position,

N0

z • O (Net torque about O)

mAgR-mgRsina • O

So,

or mA • msina

(1)

From the-equation of rotational dynamics of a solid body about the stationary axis (say z-axis) of rotation i.e. from Nz • I '3z wben the pulley is rotated by the small angular displacement 8 in clockwise sense relative to the equilibrium position (Fig.), we get : --...,1,. . ,_..._ mA g R - m g R sin ( a +

e)

• [ M: l + m R 2 + mA R 2 ]

e·

Using Eqn. (1)

m g sin a - m g ( sin a cos 8 + cos a sin 8 )

•

{ MR

+ 2m 1+ sina)R} 9·

33 But for small 8, we may write cos 8 • 1 and sin 8 • 8 Thus we have

(

.

. m g sm a - m g sm a + cos a 8

)

=-

{ M R + 2 m ( 1 + sin a ) R }- 8· 2

·· • -2mgcos a 8 --8 M R + 2 m ( 1 + sin a ) R ] (

Hence,

Hence the sougbt angular frequency ro,, •

yM R •

+

2mgcosa

2m R ( 1 +

o

4.56 Let us locate solid cylinder when it is displaced from its

stable equilibrium position by the small angle 8 during its oscillations (Fig.). If ·ev be tbe instantaneous speed of the C.M. (C) of tbe solid cylinder which is in pure rolling, then its angular velocity about its own centre C is ro • vc/r -----(1) Since C moves in a circle of radius (R - r), the speed of C at tbe same moment can be wrinen

.

as Ve=

8(R-r)

(2)

e· ( R r- r )

(3)

Thus from Eqns (1) and (2) (t) -

As the mecbanical energy of oscillation of the solid cylinder is conseived, i.e. E • T +U= constant

mv +

So,

le c..? + m g ( R - r ) ( 1 - cos 8 ) • constant

(Where m is the mass of solid cylinder and le is the moment of inertia of the solid cylinder about an axis passing through its C.M. (C) and perpendicular to tbe plane of Fig. of solid cylinder)

1 or,

2

2 2

mw r +

1 mr 2

-

2

2

2

- ro + m g (R - r) (1 - cos O) - constant

le= m r I .:) 2 3 2 • 2 (R - r ) r ( 8) r 2 4 Differentiating w.r.t. time

4

Thus

e·

=-

Eqn

+ g (R - r) ( 1 - cos 8) • constant, (using Eqn. 3) 3

So,

(using

3

(

. ..

.

( R - r ) 2 8 8 + g sin 8 8

:a::

O

r) 8, (because fer small 8, sin 8 a 8)

... I

2g

roo= V 3 ( R - r )

(1)

and

34

Hence the sought time period

2x • ro0

T• 4.57

1t-V 3(R-r)

2

2g

Let ,c1 and ,c2 be the spring constant of left and right sides springs. As the rolling of tht solid cylinder is purc its lowest point becomes the instantcneous centre of rotation. If 8 lx the small angular displacement of its upper most point relative to its equilibrium position, the defonnation of cach· spring becomes ( 2 R 8 ). Since the mecbanical energy of oscillation of tbe solid cylinder is conserved, E • T + U • 1constant 2 1 ·21 i.e. ,e1 ( 2 R 8 ) 2 + 1C2 ( 2 R 8 ) • constant lp ( 8 ) + 2

2

2

Differentianting w.r.t time 1 · .. 1 2 • 2 lp 2 8 8 +2 ( IC1 + IC2 ) 4 R 2 8 8 .. mR

or,

,

-

(

2

2

- + mR

2) •·

8 + 4R 2 K 8 • O

(Bec.ause lp • Ic + m R ••

2

8

•

mR2

-r

+ mR2)

IC

8 • ---8

Hence

3m

Thus ro0 =

: : and sought time period T-- 2x •21C m -ro0

4.58

o

•X

y[§m -

8 ac

2

,e

In the C.M. frame (which is rigidly attacbed with the centre of mass of tbe two cubes) the cubes oscillates. We know that the kinetic energy of two body system equals

µ v,:. , where

µ is the reduced mass and vra is tbe modulus of velodty of any one body particle relative to otber. From the conservation of mechanical energy of oscillation : 2

2

!2.K x

+ .!_ µ. { !!._ ( lo + x ) } -= constant 2 dt

Here !0 is the natural length of the spring. Differenting the above equation w.r.t time, wc get : 1

2

K

.

2xx+

1

2

. .. [ d(lo+x) ·] µ 2 x x • O becomes dt • x

Tbusx· • - .!.x ( where µ • mim 2 ) µ m1 + m2

Hence tbe natural frequency of oscillation : co 0

:s

-V1K -;; µ

where µ •

m1m2

.

ffl¡+

3S

4.59 Suppose the balls 1 & 2 are displaced by xi, from their initial position. Then tbe energy . 1 2 ·2 1s: E • 1 1 2 ' 2 + "'2 x + k (x1 - x:z) • m 1v 1 m r ¡ 1 2 2 2 Also total momentum .is : m 1

x1

•m1v1

+m2

+ ffl1 X2 , X• X¡- X2 m1 + "'2 "'2 x1 - X+ ------x, x2 - X - - m1-x m + mi m1 + m 2 1 1 1 2 + 1 m1 m2 x·2 E • (m1 + m2) ;,:z A + kx 2 m¡ + "'2 2 2 ffl1X1

X•

Define Then

.

x-

Hence

1 2

So

ffl¡ V¡

m¡

mz

1k

'2

1

2

+ m2x + 2 x • (a) From the above equation m

2 mi

1

We see ro •

2

1

Vi-

2

2

2

m

1

+ mi •

x 24 • 6 s -1 , when µ • 2

•

µ (b) The ene,gy of oscillation is 1 m1 "'2 2 2 "'2 + "'2 V 1

1 2 •

2J

1

m1 V1

m1m2

+mi

m

1

•

v2

1

+ m2

2

kg.

3

-4

2 X

2

m1 m2

(0.12) • 48

• 4.8 mJ

10

X

We have x • a sin (ro t + a)

x • O at t • O so a • O

Initially

x • a sin rot. Also x • v1 at t • O. V¡ 12

Then So

and bence a • ;;- •

ro a • v1

6

•

2 cm.

4.60 Suppose the disc 1 rotates by anglc 81 and the disc 2 by anglc 82 in the opposite sense. Then total torsion of tbe rod • 81 + 82 and torsional P.E. •

K ( 91 +

82 ) 2 {1)

(2)

The K.E. of tbe system (neglecting the moment of inertia of the rod) is

So total energy of the rod 1

"2

1

·2

1

2

E - 2 11 81 +-2 12 82 +2 K ( 81 + 82 ) We can put the total angular momentum of tbe rod equal to zero since thc frequency associated with the rigid rot.ation of tbe wbole system must be zero (and is known).

38

.

.

Thus

or

.

.

82 81+ 82 1// -1/1 + 1/ /

81

1/fi

2

1

2

So

and The angular oscillation, frequency corresponding to tbis is

_r;-

/

co2 • 4.61

K

/1/2 J, •

11 +

Kll' and

T • 2 1t V !.. , where K

2

In tbe first mode tbe c.arbon atom retn=-ins fixed and the oxygen atoms move in equal opposite steps. Then total energy is (1)

e

1

o

1

·2

2

2 m0 x- + 2Kx2 2 where x is tbe displacement of one of the O atom (say left one). Thus coi • K/mo.

e

o

(2)


't subject to x ( O ) =- x ( O ) • O where F is constant. The solution of this equation will be sought in the fonn F x • k + A cos ( ro0 t + a ) , O :s t :S 't

x = B cos ( ro0 ( t - -c ) +

), t>-c

A and o. will be determined from the boundary condition at t = O. F O• -+Acosa k O= a = O and A =

Thus

--k

and

- m0 A sin a F

x=k

( 1 - cos ro

t ) O s t < 't.

F

0

B and

tJ will F

be detennined by the continuity of x and

F

x at t

=

-r•

Thus

k ( 1 - cos roo -r ) = B cos f3 and

'fo

k sin ro0 -r = - o cos Wo t -

o

C V0 A•B•--and 2

Thus

C V0 10 --+- - o 2

Wo

C V0 q 1 = - - ( 1 + cos 000 t )

so

2

C V0 q 2= - - ( 1 -

COS Wo

2

t)

4.100 Toe flux in the coil is cl,(r) ..

{g,o , O I

f ldt Toe equation of the current is

(1)

d2/

Thi1, mean that o

-L d/ = _o_ dt C LC- 2 1=0 dt

r

with

2

COo -

1

Le

I ( sin w0 =

1 0

r + a)

58 Putting in (1)

lo

-C [ cos ( ro0 t + a ) - cos a ] Olo 1 = ± 1 0 cos 000 t . From Faraday's Jaw

- L 1 0 ro0 cos ( ro0 t + a ) = -

This implies cos a • O

E•

ointegrating from t = -

to -

E

d1

,,-,ro

A

.

(P+ iro)e-fh+i0>t • gm 1,,,

1

•

2 + O} sin ro t + w cos ro t

f3

P

- fh

p2 +

me

= l ,,.e

2

2 O}

_ p, sin ( ro t + 6 ) .J

1.. ro , tanu=A.

V p2 + 002

t'

( An arbitrary constant of integration has been put equal to z ro.)

Thus V

= Qe • 1,.

V(O) • 1,,, C

4.104 / • 1m

e -

P' sin ( ro t + l>)

e -

e

sin6 • I,. C

"

ro 0}2 + p2

P' sin ro t A • t'

.!_

R = (L1+ L2 )2. also

1n

o•

!e !... + Ld

1

+ R, I l= +

For the critica) case R - 2 Thus LC

-v7i

i:i + 2 V LC q +

q-O

eª'

Look for a solution with q a

a=-VL1 C. An independent solqtion is t eª ' . Thus q = (A+ Bt) At

e-,/VLC,

t • O q • C V0 thus A • C V0 t•Oq•l•O

Also at

o

= B -A-

vlC

B z

vov"fL

85 Thus finally

I-

Tr - 'c L "º V 7' e

dn dt

.::..i-

1 (cTT "º+ - V Le •

_

-,/ .¡-j;"c

v.-'c) - /VLC Vf t e

o

e-,/"LC

t L

The current has becn defined to increase the chargc..Hcnce the miaus sign. The current is maximum when

di dt•-Y yo e-t1v!c (

1-dt c) -o

This gives t • ./LC and thc magnitudc of thc maximum current is

Vo _ JC

1 /mu J •

-¡ V t •

Toe equation of the circuit is ( I is the current)

di

L dt + RI • V,..cos rot From thc thcory of diffcrcntial cquations I • lp+lc where /p is a particular integral and le is the complemcntary function (Solution of the differential equation with the RHS = O ). Now

1e• 1co e

-tR/L

and for /p wc writc lp • I,,. cos ( ro t - cp ) Substituting wc gct

Thus Now in an inductivc circuit J • O at t • O becausc a current cannot change suddenly. Thus

and so

leo • -

V,..

--;:::==== cos VR2+

ro2 L2

(í)

66

Here tbe equation is (Q is charge. on the capacitor)

CQ + R dt • V,., cosrot A so]ution subject to Q • O at t • O is of the fonn (as in the previous problem)

Q

= Q,,. [ cos ( ro t -

tp ) -

cos ip-e

t/Re

]

Substituting back

C Q,,. cos ( ro t -

. ( ro t - -cp) -cp) - ro R Q,.. sm

- v.

coscot

• V,,.{ cos cp cos ( ro t - cp ) - sin ipsin ( ro t - ip ) } Q,.. =- C V,,. cos cp

so

ro R Q,..

Tbis leads to

• V 111 sin cp

CV Q,,. -

'V

_

, tan cp •

111

ro R C

l+(ooRC)2 Hence 1-

dQ = dt

l

[- sin ( ro t - cp ) + c s2 cp e - ,1Re

V,,.

V (role)'

smcp

R2+

Toe solution given in the book satisfies / • O at t • O. Then Q • O at t • O but this will not satisfy the equation at t • O. Thus J - O, ( Equation will be satisfied with J • O only if Q • O at t = O) With our J, Toe current lags behind the voltage by the phase angle -1

q, - tan Now L = J!o n 2 :TC a 2 l, R•

But Toen

-r¡o¡L-

l = Jengtb of the solenoid

p·2:TCan·l , 2b = diameter of the wirc :TCb 2

1 b • 2n 2 2 µ n ! :TCa · 2 :TCv 1 q> = ta'f\-l 0-------x :TC-p ·2nanl 4 n2 2bn • 1

= tan-

:.

J1o 1 ---

n2 a V

4pn

·

67 HercV • V,.. cos ro t

I • 1,.. cos ( ro t + cp) where

Now

Thus the current is abead of tbe vo)tage by

cp - ta-n

1 ro! e

-

tan

-1 V(

)2 -1

v.

- 60·

Rlm

, Herc

or

V•

.

f Idt

IR.,- ·_o

e. .' V = - ro V0 sin ro t

1

R I +e I =

-

1

Ignoring transients, a so)ution has the form I = J 0 sin ( ro t - a)

l ro R 10 cos ( ro t - a ) +

1 - ---

R

,.......

o------------------

sin ( ro t - a ) • - ro V0 sin ro t

• - ro V 0 { sin ( ro t - a ) cos a + cos ( ro t - a ) sin a }

R /0

so

-

V0 sin a

a

=

•

lo

- - V 0 cos a roe -

re + tan-

1

( ro

R C)

Vo

lo = ---:;;::;::====

,/ R2+( wlcf I • 1 0 sin ( ro t-

,

Then It satisfies

f

Q • o I dt •

tan-

Q0 +

l

ro R C - n;) = - / 0 sin ( ro t - tan-

lo -

ro cos ( ro t - tan-

Vo ( 1 + cos ro t) = R

1

+

1

ro R C )

l

ro R C)

68

Y0 ( 1 + cos co t ) • - R10 sin ( co t-

if

ta-n

lo

Qo

+ C+

1

co R C ) 1

coCC06(cot- tan- coRC)

Qo • CVo

Thus

!.J,_ • Yo/ fu(

and

coC R 10 =

co R C )2

V0 ro R C

checks

- 1;:::==== + ( coR e) V

2

,

V •

Hence

Q

Vo • V0 + -;::::====cos ( cot- a)

...¡ 1 + < co R e )2

e Vo (b) - = Tt

v

V0 2

+

1

or

T}2 -

e co R e)

Re

or

-

1 •

V T}2 -

co2 (R

C )2

1 / co - 22 ms.

ct1rrent VR 0 Curr ent (a)

(b) coL - _l_ roe

tan cp •

(b) as

R 2

1

ro

I

v; iu>t oe . = iw Toen le = 1

cuv e

Ic imt

0

iro e l

= L,R

V

v. eimt

_o R+iroL

I • Le+ IL,R-

_ V. R - i ro L + i ro C ( R 2 + oi L 2 ) iut o 2 2 2 e R +w L

Toen taking the real part

Yo"R 2 + { ro C ( R 2 + ro 2 L 2 ) - w L } 2

where

I • ---------------------------------------------------- cos ( ro t - denotes time average. Now

/i • locos w t

• Real part of 10 eiwt

d/2 d/1 Toe current in the coil 2 satisfies R 12 + L2 dt = - L12 t

d

.l2

or

=R

iwt -iwL12 .L ( in the complex case ) lo e +lW

2

taking the real part /2

=-

w L12 2 2 R + w L2 2

( m L2

)

mLu = - -;:::==== /0 cos ( m t + cp )

...jR2 + 002 L2

't

Where tan cp = -L R O)

cos m t - R sin oo t

. i.aking ti.me average, we get 2

. The repulsive nature of the force is also consistent with Lenz's law, assuming, of comse, that L12 decreases with x.

.

82

4.3 ELASTIC WAVES. ACOUSTICS 4.1S0 Since the temperature varies linearly we can write the temperature as a function of x, which is, the distance from the point A towards B. T2-T1 i.e., T • T + x, [O< x < l] 1 1

hence,

dT = (

72

(1)

; Ti ) dx

In order to travel an elemental distance of dx which is at a distance of x from A it will takc a time dx _,-; (2) dt av T From Eqns (1) and (2), expressing dx in terms of dT, we get

d t-

l ( arr

ldT ) Tz-T1

Which on integration gives

or, 21 Hence the sought time t • ------

Y'i; +V'i;)

(

4.151 Equation of plane wave is given by

; ( r , t ) = a cos ( ro t -

k- r) ,

where

'Te• V

i

called the wave vector

I\

and n is the unit vector normal to the wave surfacc in the direction of the propagation of wave.

P (x y.,i ) 1

;, "n"'

z

83 ; ( x , y , z ) • a cos ( ro t - k% x - ky y - kz z )

or,

; ( x1 , y1 , z1 -, t )

Thus

s(

and

=

.

• a cos ( ro t - k x cos a - k y cos f:l - k z cos y ) a cos ( ro t - k x1 cos a - ki>71 cos '3 - k z1 cosy)

x2 , y2 , z2 , t ) = a cos ( ro t - k x2 cos a - k y2 cos

- k z-i cos y )

Hence the sought wave phase ifference (J)2 - q>¡ •

or A q> •

k [ ( Xi -'X2) cosa.+ ( Y1 -

1 q>i - «J>d •

- ;r-o ¡ [(

X¡ -

Y2)

cos '3 + ( Z1 - Z2) COS

y]

k 1 [ ( x 1 - x 2 ) cos a + ( y1 - y2 ) cos fJ + ( z1 - z2 ) cos y ] 1

X2)

cos Cl + ( Y1 -

Y2 )

cos '3 + ( Z1 -q

) cos y ] 1

4.152 The phase of tbc oscillation can be written as

et,= rot-k-r When the wave moves along the x-axis 4> • ro t - kx x (On putting /c,

•

kz • O).

Since the velocity associated with this wave is v1 We have Similarly Thus 4.1!3 The wave equation propagating in the direction of +ve x axis in medium K is give as ; - a cos ( ro t- kx)

So,

s • a cos k ( v t - x ) ,

where k • ro and, v 'is tbe wave vclocity V

In the refrence frame K' , the wave velocity will be ( v - V) propagating in the direction of +ve x axis and x will be x'. Tbus the sought wave equation. ; • acosk[(v-V)t-x']

or,

;

• a cos [ ( ro - :

V) -t

h'

] • a ros [ ro t ( 1 -:

)- kx' ]

1

This follows on actually putting

;=f(t+ax)

él; - 1 ax2

in tbe wavc equation

a2; v2 rJ t 2

(Wc have writtcn tbe one dimensional form of the wave equation.) Tben

.!.. f" ( t + ax v2

) • a 2 f" ( t + ax ) .

84 so the wave equation is satisfied if a - ::t-

1

v'

That is the physical meaning of the constant a . Toe given wave equation ; • 60cos(1800t- 5·3x) is of the type ; = a cos ( ro t-

kx),

where a = 60 x 10- 6 m

ro • 1800 per sec and k - 5·3 per metre 2n k•so A•- 2,.; ;_, k

As

(J.)

(J.)

k • -

and also

so v • - = 340 mis k

V'

(a) Sought ratio =

ªA = ª2nk

= 5·1 x 10-s

(b) Since ; • a cos ( ro t - k x )

!s. • - a ro sin ( ro t - k x )

éJ t So velocity oscillation amplitude u

( t

or vm • a ro • 0·11 mis

),,.

(1)

and the sought ratio of velocity oscillation amplitude to the wave propagation velocity

=

=

Vm

V

0·11=

3·2xlo-4

340

(e) Rclative defonnation = : ; • a k sin ( oo t - kx) So, relative defonnation amplitude

-(:;)._•ale • ( 60 x 10-

6

4

x 5·3 ) m • 3·2 x 10- m

From Eqns (1) and (2)

( il) dX

Thus ( ) éJX

-

• ak • ª

"'

00 V

. ! (E.S) V

éJt

"'

!V ( éJt ) , where v • 340 mis is the wave vclocity.

"'

"'

(a) Toe given equation is, ; =

acos(oot-kx)

(2)

85 So at ; • a coskx Now,

• - a ro sin ( ro t - k x )

dt

- a ro sin kx, at t • O.

and

; • + a k sin ( ro t - k x )

Also,

t = O,

and at ;

- - a k sin k x .

Hence ali the graphs are similar having different amplitudes, as shown in thc answersheet of the problem book.

(b) At the points, where -. O, the velocity direction is positive, i.e., along + ve x - axis in the case of longitudinal and + ve y- axis in the case of transvel'l'ie wavcs, where is positive and vice versa. For sought plots see the answer-sheet of the problem book. In the given wave equation the particle's displacement amplitude = a e- 1" Let two points x1 and x2 , between which the displacement amplitude differ by 11 • 1 % So, or or

In ( 1 - T] )

or, So path diffcrence =

X2-X1 •

y X1

-

•

-

Y Xi

Jn ( 1 -11)

-

y

- ln ( 1 -1)) y

and phase differencfJ • = -

2 Ax x path difference 2

1t

)..

!l) •

In( 1-

y

2

TJ = A·y 1t

0·3 rad

Let S be the source whos, position vector relative to the reference point O is Since intensities are inversely proportional to the square of distanccs,

88

-

lntensity at P (11 ) Intensity at Q ( / 2 ) where d1

PS and di

•

- Jf • QS.

But intensity is proportional to the square of amplitude.

ai

So, 2

ª2

Thus

•

or a 1 d1

.;i

•

aidi •

k(say)

ªi d1

•

k

ª1

k and 4i, • ª2

/\

Let n be thc unit vector along PQ dirccted from P to Q. _,. I\ k" Then PS • d1 n - -n ª1

le " •din=-n

-+

and

I\

SQ

ª2

From the 1riangle law of vector addition. -+

OP

-+

+ PS

-+

....

• OS

k,.

....

or r1 + - n = r ª1

....

I\

....

a1r1+k n • a1 r

or

(1)

(2)

Similarly Adding (1) and (2),

........

ª1 '1 + "2 '2 r • -----

....

Hence

(a) We know that the equation of a spheric.al wave in a homogeneous absorbing medium of wave damping coefficient y is : a I 0 e -yr • ---cos(rot-kr)

r

Thus particle's displacement amplitude equals

a,0 e-yr r According to the conditions of the problem, a'o at

r =

r0 , a0 :a:

e-yro

ro

(1)

,-yr

and when

r • r,

-

- ªº

r

(2)

87

Thus from Eqns (1) and (2) el

or,

= rJ-

T

y ( r - To ) = In ( l] r0 ) - In r

or,

y= ,

(b)

To

( r- r:0 )

aOe

In 3 + In 5 - In 10

In l] + In To - In r

= --------------------------- 0·08m-

1

5

r- r0

-yr

As;= -cos(rot-kr) r

a'oe-yr

at • -

So,

ro sin ( ro t - k r )

T

-yr a, Oe (O

(

rlut at point A, So,

(*L =

a' O e -yr

r

- -ao ll

ªo ro

a0 2 x

fl

fl

-- -

-

50 X 103

(a) Equation of the resultant wave,

1; = l;1 +

r

).a

= 2 a cos k (

6 X

y; X)

• a' cos { ro t - k ( x + Y) } , where 2 Now, the equation of wave pattem is,

2

X

22

7

X

1•45

X

3

10 = 15m/s

cos { ro t - k ( xz+ y) }•

a' • 2 a cos k ' ( Y ;x

)

x + y • k, (a Const .)

For sought plots see the answer-sbeet of the problem book. For antinodes, i.e. maximum intensity cos k(y-x) • 2

:t

1 • cosnx

2nx

±(x-y)=-k-=n>.. or,

y•x:tn>..,n•0,1,2,

Hence, the particles of the medium at tbe points, lying on the solid straight lines ( y • x ± n >.. ) , oscillate with maximum amplitude. For nodes, i.e. minimum intensity, cosk(y-x)•

O

2

or,

k(y-x) ±

2

=

(2 l)x n+ 2

88

y • X ± ( 2. n + 1 ) A./2 ,

or,

and hence the particles at the points, lying on dotted lincs do not oscillatc. (b) When the waves are longitudinal, For sought plots see the answer-shect of the problem book. - 1 Sl k ( y-x ) • cos - cos- 1 -;2 a a or,

.

s1-;;•cos{

k(y- x)+cos

-1 -s¡2;}

ºk( • -S;2cos k( y - x ) - sm

y - x )' sm ( cos -1S --2;¡\)

s2 cos k ( y - x ) - s.1n k ( y - X )

• -

a2

a from (1), if

(1)

sin k ( y - x ) • O sin ( n ,w; )

;1 • ;2 (- 1t thus, the particles of the medium at the points lying on the straight lines, y • x ±n will oscillate along those lines (even n), or at right angles to them (odd n). Also from (1), if

cosk(y-x) •O• ;{

2

• 1-

a

,t

cos ( 2 n + 1 )

;22/ a2,

2

•

a cuele .

Thus the particles, at the points, where y • x ± ( n ± 1/ 4 ) A , will oscillate along circles. In general, all other particles will move along cllipses. The displacement of oscillations is given by • a cos ( ro t - k x ) Without loss of generality, we confine ourselves to x • O. Then the displacement maxima occur at co t = n :1t• Concentrate at ro t = O • Now the eneigy density is given by w =- p a 2.co2 s in2 co t at x • O 2 T/6 time later (where T • n: is the time period) than t • O. ro 3 2 2.2:1t 2 2 w • p a ro sin - • - p a ro • w0

3 Thus

.•

1

4

2 2 2wo p a ro • - - . 2 3

2

)..

89

The power output of the source much be 4 ,r, P 1 0

Q Watt.

•

The required flux of accoustic power is then : Q - C 4,r,

s Where '2 is the soJid angle subtended by the disc enclosed by the ring at S. This solid angle is '2 • 2 ,r, (1 - cos a) So flux 4> = lo lo # 'btf

:r)

(1-

Substitution gives 4> • 2 Il x 30 [1 -

l ,.

l

µ W • 1.99 µ w.

y:-:¡ 1

4

Eqn. (1) is a well known result stich is derived as follows; Let SO be the polar axis. Toen the required solid angle is the area of that part of the surface a sphere of much radius whose colatitude iss a. a

Thus

C•

f

2 it sin 8 d 8 • 2 it (1 - cos a).

o From the result of 4.162 power flowing out througb anyone of the opening

PV (l

h/2 =

)

2 -

= ; ( l-

R 2+ ( h/2 )2

V 4R:+ h

2)

As total power output equals P, so the power reaching the lateral surface must be.

• P - 2p· ( 1 2

h ) V 4 R 2 + h2

Ph

-

- O· 07W

V 4 R 2 + h2

We are given • acoskxrot so Thus

!asx .• - a k sin k x cos ro t 0=

( ; ), •

( éJ X),•

ut

a cos k x , ( ; )•,

• a k sin k x , ( : O

S • - a ro cos k x sin oo t

and

v 2 = - a cos k x

!) • t•

T/2

a k sin k x

90 (a) Toe graphs of ( l;) and ( ;) (b)

are as shown in Fig. (35) of the book (p.332).

We can calculatc the density as foJlows : Takc a parallelopipcd of cross section unity and lcngth dx with its cdges at x and x+dx. After thc oscillation the edgc at x goes to x +!; ( x ) and the edge at x + d x goes to x+dx+;(x+dx) = x +d x

+; ( x ) + il dx . Thus the volume of the element (originally dx) becomes éJx

(1+}:)dx Pao

p•

and hencc the density bccomes

f: •

1+ éJx

f On substituting we get for the density p ( x ) the curves shown in Fig.(35). referred to above. (e) The velocity v (x) at time t = T/4 is

• -arocoskx

E s_ )

(

éJt

t • T/4

On plotting we get the figure (36). =- a cos k x cos ro t

Given ;

(a) The potential energy density (per unit volume) is the energy of longitudinal strain. This is 2

wp •

(

stress x strain ) •

E ( :;

2

E

ro2

-

k2

Thus

wP

•

is the Jongillldinal strain)

( :;

E a2 ¡¿. s in k x cos2 ro t

wp •

But

) ,

=-

p

2

or E k

2

= p ro

1 p a2 e.o2 s.m2 k x cos2 ro t 2

(b) The kinetic energy density is 2

•

p(

) •

2

2

2

pa ro c os kxs in

2

ro r.

91

0n plotting we get Fig. 37 given in the book (p. 332). For example at t • O 1 2 2 • 2 p a ro si n k x w • wp + wk • 2

"íc so wc do get the figure.

and the displacement nodes are at x • :t

2

4.16' Let us denote the displacement of the elements of the string by - a-sin kx cos ro t

since the string is 120 an long we must have k-120 • n ff If x1 is the distancc at which the displacement ,amplitude f1rst equals 3·5 mm tben a sin k x1 Toen

•

3·5 • ásin ( k x1 + 15 k )

k X1 + 15 k - 3t - k X1

k x1-

or

1t -15

k

2 One can convince ourself that the string has the form shown below

· 1·5 7-5 15

15

15

It shows that

k x 120 = 4 3t,

so

15

k • -1ct m 30

-1

Thus we are dealing witb tbe tbird overtone Also

14.

'7 We bave n • ;

k x1

=

•

a • 3·5 V2 mm • 4·949 mm .

so

y[;• ; ,/'fJ

Where M • total mass of tbe wire. W!ien tbe wire 1 1 is stretched, total mass of thc wire remains constant. For the flfSt wire the new lengtb • l + Tti l and for the record wire, the length is l + Tt21. Also T1 • a ( Tti /) where a is a constant and T2 = a ( TJ2 l ). Substituting in the above formula. 1 .../ ( a Tt1 l ) ( l + Tt1 l) V¡=

2 ( l + 111 l) =

•

2 ( l + TJ2 l) V2

V¡ V2 =

v1

y

=

V

M

y y

1 + fll 1 + T)2

1 + TJ1 ) = Tl2 ( 1 + T)2 ) Tl2(

M

.. / ( a Tt2 l ) ( l + TJ2 l )

1 V2

V

fl2 •

T)¡

1 + 1')2 1 + T)¡

O· 04 ( 1 + O· 02 ) = 1·4 O· 02 ( 1+ O· 04 )

92

Let initial length and tension be l and T respectively.

So,

v1

12,- --vJT

•

In accordance with the problem, thc ncw length 35 l' • l - l x • O· 65 l

.

Tx70

and ncw tcns10n, T' • T+ Thus the new frcquency

v

100

100 • 1· 7 T

1--rr 2 lV '

2•

V2

1

'yill

2 x O· 65 l

.¡-:¡;;j

P1

1•3

-------2 O· 65 O- 65

Hence

V¡

Obviously in this case the velocityof sound propagation V • 2 V ( 12 - 11) wherc 12 and 11 are consecutivc lengths at which resonance occur

(a)

In our problem, ( /i

- 11 )

So

v • 2 v l • 2 x 2000 x 8· 5 an/s • O· 34 km/s.

l

•

Wben the tube is closed at one end v= =

V

41 ( 2 n + 1 ) 340 x O·

4

,

where

n • O , 1 , 2 , ...

85 ( 2 n + 1 ) • 100 ( 2 n + 1)

n • O , 1 , 2 , 3 , 4 , 5 , 6 , ... , we get

Thus for n1 •

100 1 Hz , ni • 300 1 Hz , n3

•

500 1 Hz , n4

•

700 1 Hz ,

ns • 900 1 Hz , n6 • 1100 1 Hz, n1 • 1300 1 Hz

Since v should be < v0 • 12501 Hz, we need not go beyond Hes• Thus 6 natural oscillations are possible.

(b) Organ pipe opened from both ends vibrates with ali harm.onics of the fundamental frequency. Now, the fundamental mode frequency is given as V= V/A.

or,

V

•V/2 [

Here, also, end correction has been neglected. So, the frequencies of higher modes of vibrations are given by v • n (v/2 l)

(1)

93 or, It may be checked by putting the values of n in the equation (1) that below 1285 Hz, there are a total of six possible natural oscillation frequencies of air column in the open pipe. Since the copper rod is clamped at mid point, it becomes a mode and the two free ends will be anitinodes. Thus the fundamental mode formed in the rod is as shown in the Fig. (a). .>,

4.171 (a)

4.171 (b)

l= A

In this case,

2 V

So,

0

=:i =.!_\fiyf p 21

21

e

where E = Young's modules and p is the density of the copper Similarly the second mode or tbe first overtone in the rod is as shown above in Fig. (b). l = 3A

Here

2 21

2n+l V ==

21

vr

3v = _l_

V¡=

Hence

21

p

IEwhere n = O, 1, 2 ...

V

Putting the given values of E and p in the general equation

v = 3·8 ( 2 n + 1 ) k Hz Hence v0 = 3 · 8 k.Hz, v1 = ( 3·8 x 3 ) k Hz, v2 = ( 3·8 ) x 5 = 19 k Hz, v3 = ( 3·8 x 7 ) = 26·6 k Hz, v4 = ( 3·8 x 9 ) = 34·2 k Hz, v5 = ( 3·8 x 11) = 41·8 k Hz, v6 = ( 3·8) x 13 k Hz= 49·4 k Hz and V7 = ( 3·8) x 14 k Hz> 50 k Hz. Hence the sought number of frequencies between 20 to 50 k Hz equals 4. Let two waves 1 = a cos ( ro t - k x ) and 2 = a cos ( ro t + k x ), supcrpose and as a result, we have a standing wave (the resultant wave ) in the string of the form = 2 a cos kx cos ro t . According to the problem 2 a = ªm·

94

Hence the standing wave excited in the string is

• a,,. cos kx cos ro t

(1)

• - ro a,,. cos kx sin ro t

or,

(2)

So the kinetic energy confined in the string element of length dx, is given by :

dr-!(7m)(*) dT• 1

or,

(m dx) 2 1

m a2 ro2

dT=

"' 21

2

• 2 ro t a,2,. ro2 cos2 kx sm

2

s in2 rot cos2 ; ..,'Jt x dx

Hence tbe kinetic energy confined in tbe string corresponding to the fundamental tone 2

2

T•

I

2

J

).12

cos2 2"A.tr, x dx

dT • m a "' 00 s in rot 21 o

Because, for the fundamental tone, lengtb of the string l • ; 1 2 . 2 T • 4 m a,2,.ro sm oot

lntegrating we get,

! m a!

Hence the sought maximum kinetic energy equals, Tmax =

ro2 ,

because for Tmax, sin2 oot = 1

(ii) Mean kinetic energy averaged over one oscillation period 2 ,t/(1)

J s in

2

=

f Tdt-= !.ma2 ol J dt 4

ootdt

o

m

f

dt

o

or, We have a standing wave given by the equation - a sin k x cos oot • k . ;¡ 'j::. So, = - a ro sm x stn ro t

ª'

and

• akcoskx cosoot

(1)

(2)

9S

The kinetic energy confined in an element of length dx of the rod 2

2

(pSdx)( ; ) •

dT •

2

2

p S a2 m sin rot sin k x dx

So total kinetic energy confined into rod ).12

T•

f dT • !

pS

f t

a2 m2 sin2 mt

sin 2

2

xdx

o T _ n:S a2 ro2 p sin2 rot

or,

(3)

4k

The potential energy in the above rod element

Jau - - J lj

dU -

o

a2; a,

wbere F • ( pSdx)-:-:-1

F1 d;,

F¡ = - ( p S dx ) ro2 ;

or,

¡

dU • 002 p S dx

so,

J; el; o

or,

dU•

p 00 SS 2

2

p ro2 S a2 cos 2 rot s in2 k x dx

dx•

2

2

J

Thus the total potential energy stored in the rod U =

dU

)./2

or,

2

2

2

U • pm ; S a cos mt

J t sin

2 2

x dx

o U =

So,

npSa

22 2 00 cos ro t

4k

To find the potential energy stored in the rod element we may adoptan easier way. We know that the potcntial energy density confined in a rod under elastic force equals : 1 . 1 1 2 Un = ( stress x stram ) = a E = Y E 2 2 2 1

2 2

2

1 p ro 2

·2PVE•27E

!. ( ax )

=

2 k2

2

=

1

p a 2 o>2 cos2 rot cos2 k x

2

96

Hence the total potential energy stored in the rod

U

vi

=f

f

Uv d V =

pa

2

ci cos2 co t cos2 k x S d x

o 2

=

n: p S a 2 w cos2 w t 4k

(4)

Hence the sought mechanical energy confined in the rod between the two adjacent nodes -

1t

e oo2 ª2 s

E= T+ U -

4k

Receiver R1 registers the beating, due to the sound waves reaching directly to it from source and the other due to the reflection from the wall. Frequency of sound reaching directly from S to R1

=v0-

vs-R

when S moves towards R1

V-U

l

and

V-

vs' _ R

1

= v0 _v_

when S moves towards tbe wall

v+u

Now frequency reaching to R 1 after reflection from wall vw-R

= v0 1

and vw ' _R = v l

V -

, when

-

v+u

-- V 0

S moves towards R1

when S moves towards the wall

,

V-U

Thus the sought beat frequency /l. V = ( Vs-Rl-

= v

Vw-R¡)

v

-

-

0

v

0

v-u

v -

v+u

or . ( v' w-Ri- Vs-R¡) 2 Vo v u 2 u Vo = 1 Hz

= ---2 v

-

u

2

• --

v

Let the velocity of tuning fork.is u. Thus frequency reaching to tbe observer due to the tuning fork that approaches the observer

v, = v0 -- V [ v = velocity of sound ] v-u Frequency reaching the observer due to the tunning fork that recedes from tbe observer V

11

=

V

--

V

+u 1 vv(-

OV

So, Beat frequency v - v" = v =

0

or, So,

V=

1 - - )

-_ v-u

v+u

2 v0 v u v2-u2 VU

2

+ ( 2 V Vo ) U

-

V

2

V= Ü

97

- 2 V Vo ±

Hence

u-

...¡ 4 V V 2 + 4 v2 V 2 2v

Hence the sought value of u, on simplifying and noting that u > O

Obviously the maximum_ frequcncy will be heard when the source is moving with maximum velocity towards the receiver and mínimum frequency will be heard when the source recedes with maximum velocity. As the source swing hannonically its maximum velocity equals a e.o. Hence V

max

-

Vo ---

V

v-aro

So the frequency band width Av • or,

and

V ....:n -

Vo ---

....

v+aro 2

vmin • v 0

Vmax -

V

v( 2 a ) v -a ro2

( Av a 2 ) ro2 + ( 2 v0 v a ) ro - Av v 2 • O - 2 Vo v a ± 4 v v 2 a 2 + A v 2 a2 v 2

'V

So,

ro=------2-a------Av

On simplifying (and taking + sign as ro -. O if Av -. O)

(V

2

1 +(

ro - v vo Ava

Av ) v0

_

1)

lt should be noted that the frequency emitted by the source at time t could not be received at the same moment by thc receiver, becousc till that time the source will cover the distancc

!

w t 2 and the sound wave will take the further time

!

w t 2/v to reach the receiver. Therefore

the frequency noted by the receiver at time t should be emitted by the source at the time t1< t . Thercforc t¡ + (

W

tt/

V ) •

t

(1)

and the frequency noted by the receiver V

V•

Vo

·

Solving Eqns (1) and (2), we get V•

V+ W t1

Vo

/

V

• 1 · 35 kHz.

1+ 2 t V

(2)

9B (a) When tbe observer receives the sound, the source is closest to him. It means, that frequency is emitted by the source sometimes before (Fig.) Figure shows that the source approaches the stationary observer with velocity vs cos 0. Hence the frequency noted by the observer V

=

Vo (

V - V COS

s'-

:t.-- ..

e

8)

•

vo( v - T:I

1 - ;:,.,. e

ros 8) •

Vl 2 + x 2

x But

-sv

..

'

'I (1)

v

x

,

or,

So,

,¡l

2+x 2

'

II

'\J

v,

= -v =

t o

TJ

(2)

cos 0 = r¡

Hence from Eqns. (1) and (2) the sought frequency V

=

Vo

• 5 kHz

l -TJ2 (b) When tbe source is right in front of O, the sound emitted by it will not be Doppler shifted because 0 = 90º. This sound will be received at O at time t = .!_ after the source has V

passed it. Toe source will by then have moved abead by a distance vs t = l r¡. Toe distance between the source and the observer at this time will be l Vl + r¡2 = 0.32 km. Frequency of sound when it reaches the wall , V

v+u -v-V

wall will retlect the sound with same frequency v'. Thus frequency noticed by a stationary observer after retlection from wall

v" = v' _v - , since waJI behaves as a source of frequency v'. v-u Thus, or,

So,

,, v+u v v+u v =v--·---=v-v v-u v-u 11 A v-u '),.,,"= A u V-= -or A v+u v+u A" 2U V - U

1--=l---=-A v+u v+u

Hence tbe sought percentage11 change in wavelength A- A 2u x 100 = -x 100 % = 0.2% decrease. = r,,1 v+u

99

Frequency of sound reaching the wall.

V Vo(V U)

(1)

s

Now for the observer the wall becomes a source of frequency v receding from it with velocity u Thus, the frequency reaching the observer v'-

V

(-V ) -

Vo

v +u

(

v)+u

[Using (1)]

Hence the beat frequency registered by the receiver (obsetver) , 2 UVo /iv • v 0 - v • -• 0.6 Hz. v+u Intensity of a spberical sound wave emitted from a point source in a homogeneous absorbing medium of wave damping coefficient y is given by

I - !. p a2-e

2

2

Y' o>2

v

So, Intensity of sound at a distance r1 from the source 11 1/2 p a 2 e- 2Y "i oi2 v =2= '1

'1

and intensity of sound at a distance r2 from the source 2 1/2 p a 2-e 2Y' 2 oiv = J2 / r 2 = ------...-----

'2

1 11

But according to the problem

h

-;f • -;J

Tl

n ,2

--,¡ • 1

So,

l .

or,

(a)Loudncss icvel in bells = log

fo. (1

0

is tbe theshold of audibility.)

So, loudness level in decibells, L • 10 log

I 10

Tbus loudness level at x

Similarly Thus

=

x1

= Lx

lx..

= 10 log

10

1

lx.¡_ Lx.¡_ - 10 log lo

r

L JCi - L %1 • 10 log

lx.¡_ %1

100

l/2pa2002ve-2yx2 ----------- = L + lOlog e1/2 p a2 ú)2 V e -2 y Xi Xi

lOlog

or,

2y (x z- x >

1

LXz = Lxi - 20 y ( x2 - xi) log e L' = L - 20 y x log e [ since ( x2 - x1 ) = x ]

Hence

= 20 d B - 20 x O· 23 x 50 x O· 4343 d B • 60dB-10dB • 50dB (b) Toe point at which the sound is not heard any more, the loudness level should be zero. Thus L 60 300m O = L - 20 y x log e or x = 20 y log e= 20 x O· 23 x O· 4343= (a)

As there is no damping, so 2

1/2 p a 002 v/ r log ----:--2-:2 -- = - 20 log r 0 1/2 p a 00 V

J

Lr0 = 10 log To = 10 -'

Similarly Lr =

1

- 20 log r

So, or, Lr = Lr 0 + 20 log

(7r)o

= 30 + 20 x log 2lO0

= 36 d

B

(b) Let r be the sought distance at wbich the sound is not heard. So,

L = L + 20 log ro = O or, Lr = 20 log r ro r r0 o

So,

r = 3/2 20

log10

Thus Thus for r > O· 63 km

or 10

( 312)

or

30 = 20 log

20

= r/20

r = 200V10 = O· 63 Km. no sound will be héard.

We treat the fork as a point source. In the absence of damping the oscillation has tbe form Const. -cos ( ro t - k r ) r Because of the damping of the fork the amplitude of oscillation decreases exponentially with the retarded time (i.e. the time at which the wave started from tbe source.). Thus we write for the wave amplitude.

e- (r-;) - (,+ - ) - (,+ - ) =

This means that

e

-=

Const. r

e

p

p+clp

----

x+dr

101

Thus

e-

--) ( 't rB-'A

ln-

V

or

= ,.._

rB

rA

rn - 'A

=

0.12. s-1

"t+--v

(a) Let us consider the motion of an element of'the medium of thickness dx and unit area of cross-section. Let 1; -= displacement of the particles of the medium at location x. Then by the equation of motion pdx;° = -dp where dp is the pressure increment over the length dx Recalling the wave equation •• ;

¡/t

2 =V

ax 2

we can write the foregoing equation as

;l2 t

p v2 dx 2

= -dp

ax

Integrating this equation, we get A p = surplus pressure =

- p v2!x5.

+ Const.

O

In the absence of a deformation (a wave), the surplus pressure is Ap = O. So 'Const' = O and 2

Ap=p v.

a 'j:

ax

(b) We have found earlier that w = wk + wP = total energy density w1 = ;

p (

r

wP= ;

f!

E (:

= ; p

v2 ( :

!f

lt is ea&y ,,, see that the space-time average of both densities is the same and the space time average of total en rgy density is then

Toe intensity of the wave is 2

< ( > Ap > pv

J = V =

Using

< ( A p ) > = 12 ( A p ),,. 2

J

2

we get

( A p );.,

2p v

102

The intensity of the sound wave is ( A p );. = 2pv

I =

( Ap );. 2pv)..

Using v • v A, p is the density of air. Thus the mean energy flow reaching the ball is

R re 1t

2/

=

R re

2(

-.,.._.- .,../

Ap );_ 2pv)..

R 2 being the effective arca (arca of cross section) of the hall.

Substitution gives 10.9 mW.

p 4x r

We have

.

.

mtens1ty-

2-

or

.v

( Ap );_ 2p v A ) ( p,,.

3

1·293 kg/ m x 340 mis x 0·80w

/pvP

V

1

-V

1·293 x 340 x ·8 ( kg kg m

2 re x l ·S x l ·S m2

; 3 m s-

2xx1·5x1·5

• 4·9877 ( kg m - 1 -s 2

)= 5 Pa .

),,= . 5 X 10-S

( Ap p

Ap = - p v 2

(b) We have

ax

( A p ),,. = p v2 k ;,,. = p v 2 x v ;,,. ( Ap )m 5

m•a•---2 3t p V

.

2 3t X 1•293 X 340 X 6()()

V

•3µm

6

S,,,• )..

3 X lQ 340/600

•

18()f} X l0 - 6 • S 34

X

l0 - 6

Express L in beis. (i.e. L = 5 beis). Toen the intensity at the relevant point (at a distance r from the source) is : J0•1 cf Had there been no damping the intensity would have been : e 21 ' 10•1 = 3.3 µ W/ m

•

2

4.198 For the Poynting vector we can derive as in (196)

=l 2

¡-;:;:; E ,,2.

-V

al ong the d"rre.ction of propag.ation.

Hence in time t (wbicb is mucb longer tban tbe time period T reaching tbe ball is

--

r;:;:;E,,,2 x t = 5 kJ.

1-

2

of tbc wave), tbe energy

x V 4.199 Here E = Em cos kx cos wn:R t 2

-

From div E Also

= O we get Emx = O

- --

aB

ar -+.

so Where

-

so

- VXE

Em is in tbe y - z plane.

=-

V cos k X

X

-

E,,, cos 00 t

= F"x E,. sin kx cos oo t

F"xE,,,

B - ---

co

E,,,

- -

e

_,.

sin kx sin co t • B,,, sin kx sin co t

-

_.,

and B,,. l. E,,, in tbe y - z plane.

--

At

t - O, B • O, E - E,,, cos kx

At

t - T / 4 E • O , B • B,,.sin k x

106

- - --

E • E,,. cos k x ro t - = --rxE,,. sin k x sin ro t (exactly as in 199) H J.lo (1) - E,,. x ( l?'x E,,. ) 1 S • ExH - ----- sin2kx sin2rot Jlo O) 4

- -

!

Sx •

Thus

E0 c E

-

1

--

sin2kxsin2rot < s" > = o

Inside the condenser the peak electrical energy W,: =

C V,; 2

!. V. 2

=-

2

Eo nR d

"'

(d = separation between the plates, reR 2

area of each plate.).

•

V • V,,. sin ro t, V,,. is the maximum voltage

Changing electric field causes a displacement current

. = -¡¡¡ éJD =

E ,,. 00 COS 00 t

Eo

}d,s

Eo ro V,,. d cos ro t

=

This gives rise to a magnetic field B ( r) (ata radial distancc r from the centre of the plate) 2

B ( r ) · 2 1t r - Jlo re r jdis • J.lo rc r B •

1 2

fd

r

2 J.lo =

=

8

E0

1 16 """'E

d

cos oo t

r

Eo J.loOOd V,,,cos oot

Energy associated with this field is 2 3 B 1 2 =

2 Eo ro V,,.

00

2

IR

µ0 2

2 re

d

2

2

r r dr x d x V,,. cos ro t o

ro2 R 4 2 2 • -V. cos ro t O ru d m 2

Thus the maximum magnetic energy W = "'

Hence

1 W - ,,, = We

8

R2

E2

oJ!o (

16

R)2 y2

ro

d

"' 2

Eo l.lo (

co R )2

=1

8

The apptoximation ate va\id only if ro R .

2

n be along the z axis. Then Sin • E1" H1, -E1, H1"

and Using the boundary condition E 1 , that

E2 ,

•

,

H1 ,

•

H2t at the boundary ( t • x or y) we see

P · a IP::::-\.i when I 2

e-r_• .• I

p_.• I

_. e I ¡ •r• •·-em

m• ' '

' '

e¡

if

-

e

• -

m

m;

d2

But

-

dt

2

I m¡ r;•

p -

' '

• fixed

o.

l 4 3t Eo

1p

Thus

m

'

O for a closed system

Hence

1.212 p =

m-r-•--

2

2

2

2

1 - ( e 00 a ) c os e.o t 1 e2ro 4 2 < P > • -- - ( e 002 a )2 x - • a • 5.1 x 10- 15 W. 3 3 4 1t Eo 3 e 2 12 n e0 e 1

2

1.213 Here

e e 2q l p • - x force • . 2 2 m mR 4 1tEo

.:..¡.

p.

Thus

( e

1 ( 4 3t Eo )

3

q) _L 2

mR 2

3 e 3•

.214 Most of the radiation occurs when the moving particle is closest to the stationary particle. In that region, we can write and apply the previous problem's formula Tbus

AW

1

2

... (4nEo)3 J?

00( .!J!!.:2.. )2 m

(the integral can be taken between ± 00 with little error.)

dt (b2+v2t2)2

112 4.217 P is a fixed point at a distance / from the equilibrium position of the particle. Because l > a , to first order in

the distance between P and the instantaneous position of the particle

is still l. For the first case y = O so t = T/ 4 Toe corresponding retarded time is t ' •

j," ( t ' ) • -

Now

2

00 a

! !-! ) • -

cos ro (

2

00 a

1 sin :

at t - O so at the retarded time t' • -

For the second case y • a

001 e

••

2 y(t,)•-r a

Thus

ro l co s-

c Toe radiation fluxes in the two cases are proportional to ( ·j, ( t ' ) )2 so b . . 2 rol -SS1 • tan - • 3 .06 on su sti tutl on.

e

2

Note : Toe radiation received at P at time t depends on tbe acceleration of tbe charge at the retarded time. 4.218 Along the circle x • R sin w t , y • R cos e.o t where w = ;

. If t is e

1

parameter in x ( t ) , y ( t ) and

t' is the observer time t'hen t , = t + _l-_x ( t_._) e

o

where we have neglected tbe effect of tbe y--cordinate which is of second order. Toe observed cordinate are -V- ,. . x'(t') = x(t), y'(t') = y(t) Then

... ft. dt'

-

dt =

dt' dtd't

sin w t wR

1--cos cot =

and

2

-----cox

1-

e

e

2

v (vc-R R

=--1-

cR This is the observed acceleration.

-vx/R

1-

!!!_!!_(-V x/R) dt' dt 1 _!_l cR

1

p

l.)

=-----

cR