7. WAVES

WAVES This topic is taken from our Book: ISBN : 9789386320070 Product Name : Waves for JEE Main & Advanced (Study Package for Physics) Product Description : Disha's Physics series by North India's popular faculty for IIT-JEE, Er. D. C. Gupta, have achieved a lot of acclaim by the IIT-JEE teachers and students for its quality and indepth coverage. To make it more accessible for the students Disha now re-launches its complete series in 12 books based on chapters/ units/ themes. These books would provide opportunity to students to pick a particular book in a particular topic. Waves for JEE Main & Advanced (Study Package for Physics) is the 7th book of the 12 book set. • The chapters provide detailed theory which is followed by Important Formulae, Strategy to solve problems and Solved Examples.

• Each chapter covers 5 categories of New Pattern practice exercises for JEE - MCQ 1 correct, MCQ more than 1 correct, Assertion & Reason, Passage and Matching based Questions. • The book provides Previous years’ questions of JEE (Main and Advanced). Past years KVPY questions are also incorporated at their appropriate places. • The present format of the book would be useful for the students preparing for Boards and various competitive exams.

Contents 9. Wave - I

Contents 571-632



9.1 Introduction

572



9.2

Pulse and wave

572



9.3

Graphical representation of



9.27

Reflection of sound

607

9.28 Reverberation

607



9.29

Range of hearing

608



9.30

Ultrasound

608



simple harmonic wave

573

9.31 Sonar

609



9.4

Sound

575



9.32

609



9.5

Wave front

575



Review of formulae & important points



9.6

Equation of a travelling wave

576



Exercise 9.1 - Exercise 9.6

612-622



9.7

Plane progressive harmonic wave



Hints & solutions

623-632

Shock waves

611



or sinusoidal wave: y = a sin(wt – kx)

577



9.8

Initial phase

577



9.9

Phase and phase difference

578





9.10

Particle velocity and acceleration

579



sound waves



9.11

Wave equation , y =a sin (kx –wt) 580



Reflection and transmission of



9.12

The speed of a travelling wave

581



9.13

Non-sinusoidal waves

581



9.14

Speed of transverse wave

585



9.15

Sound waves

588



9.16

The speed of longitudinal wave

589



9.17

Speed of sound : newton’s formula

590



9.18

Laplace’s correction

591



10.6 Interference in time : beats

644



9.19

Factors affecting speed of sound in gas 592



10.7

Stationary waves

649



9.20

Energy of a progressive wave

594



10.8

Stationary waves in stretched

654



9.21

Power transmission

595





string fixed at the ends



9.22

Intensity of sound

597



10.9

Stationary longitudinal waves



9.23

Doppler effect

599





9.24

Doppler effect in light

601



9.25

Some important cases of doppler effect 602



9.26

Characteristics of sound

606

10. Wave -II 10.1

10.2

633-698

Reflection and refraction of 634



transverse wave in stretched string

635



10.3

Superposition of waves

637



10.4

Interference

639



10.5

Interference of sound waves:



quinke’s tube

in organ pipes

641

657



Review of formulae & important points

668



Exercise 10.1 - Exercise 10.6

670-684



Hints & solutions

685-698

Chapter10 Wave-II 10.9 STATIONARY LONGITUDINAL WAVES

IN

ORGAN PIPES

When air is blown through the mouth, the sound waves move along the pipe and get reflected at its end (open or close), producing stationary waves. If both the ends of the pipe are open, it is called an open organ pipe. If one end of the pipe closed, it is called closed organ pipe. Close end of the pipe behaves as rigid boundary and open end as the free boundary for the displacement wave, so at the close end, displacement node or pressure antinode forms and at the open end, displacement antinode or pressure node forms.

Open organ pipe Figure shows the various modes of vibrations of organ pipes in the form of displacement and pressure waves.

(i)

First mode of vibration : The lowest frequency note is called fundamental note which is corresponding to the longest wavelength. Thus L = or

Frequency of vibration This is the first harmonic.

l1 f

l1 2

= 2L

=

v 1 = l1 2L

gP = f (say) r

657

658

MECHANICS, HEAT, THERMODYNAMICS & WAVES (ii)

Second mode of vibration : Here

L = l2

or

Frequency of vibration

l2 l2 + = l2 2 2

= L

=

v 2 gP = l2 2L r

= 2f This frequency is called first overtone or second harmonic.

(iii) Third mode of vibration : Here

Frequency of vibration

L =

f3

=

l3 l3 l3 3l3 + + = 2 2 2 2

v 3 gP = l3 2L r

= 3f This frequency is called second overtone or third harmonic. Hence in open organ pipe, the harmonics of frequencies ratio 1 : 2 : 3 : ....... are possible.

Analytical treatment : Consider an open organ pipe of length L lying along the x-axis, with its ends at x = 0 and x = L. The sound wave travelling along the pipe can be represented as DP

=

DPm sin(kx - wt )

The reflected sound wave from open end (rigid boundary) is represented by DP2

=

DPm sin(-kx - wt + p)

=

DPm sin(kx + wt )

The resultant stationary wave is given by DP = = or

DP =

DP1 + DP2 DPm sin(kx - wt ) + DPm sin(kx + wt ) 2DPm sin( kx) cos(wt )

For all values of t, the resultant pressure variation is zero, for which sin kx

= 0

or

kx

= np

or

2p x l

= np

x

=

nl , where n = 0, 1, 2, 3, ........ 2

WAVE - II \

x = 0,

l 3l , l, , .......... 2 2

These points of zero pressure variation are called pressure nodes. On the other hand, the pressure variation is maximum for all values of t, for which sin kx

or

=

(2n + 1)

p 2

=

(2 n + 1)

p 2

x =

(2 n + 1)

l , where n = 0, 1, 2, 3, ................ 4

x =

l 3l 5l , ,........ , 4 4 4

kx = 2p x l

or

\

±1

These points of maximum pressure variation are called pressure antinodes.

Note: 1.

If P0 is the normal pressure in the pipe, then at the positions of pressure nodes, the pressure will be P0 and at the positions of pressure antinodes, it will be P0 ± 2DPm or P0 ± 2 ABk . Thus pressure at antinodes varies from P0 - 2 ABk to P0 + 2 ABk .

2.

The loud sound is heard at pressure antinode or displacement node.

Strain : We know that bulk modulus of medium (air) is B =

\

strain

dP dP = (- dy / dx ) æ dV ö ç÷ V è ø

=

dy dP =dx B

=

-

DP B

659

660

MECHANICS, HEAT, THERMODYNAMICS & WAVES As DP is maximum (±2DPm ) at the antinodes, so strain (positive or negative) at the pressure antinodes and zero at pressure nodes. At antinodes due to the compressions or the rarefactions of the oppositely travelling waves, the strain becomes maximum. At nodes strain becomes zero due to the compression of one wave coming across the rarefaction of the other, as shown in figure.

End corrections : Till now we have assumed that node/ antinode is formed just at the open end. Lord Rayleigh showed that due to inertia the vibrating particles form node/ antinode little above the open end of the pipe. So an end correction is applied which is approximately e = 0.61 r, where r is the radius of the pipe. Thus, for close organ pipe, the effective length

Le = L + e,

for open organ pipe, the effective length

Le = L + 2e.

Close organ pipe Fig. 10.48, shows the various modes of vibration in the form of displacement and pressure waves.

(i)

First mode of vibration : In this mode of vibration L = Frequency,

f1

=

l1 or l1 = 4L 4

v 1 gP = = f (say) l1 4L r

This frequency is called first harmonic or fundamental note.

WAVE - II (ii)

Second mode of vibration : In this mode of vibration L = l2

or

=

l 2 l2 3l + = 2 2 4 4

4L 3

v 3 gP l 2 = 4 L r = 3f (iii) Third mode of vibration : In this mode of vibration f2

Frequency

=

L =

l3 l3 l3 5l + + = 3 2 2 4 4

or

l3

=

4L 5

Frequency ,

f3

=

v 5 gP = = 5f l3 4L r

This frequency is called second overtone or fifth harmonic. Hence different frequencies produced in a closed organ pipe are in the ratio 1 : 3 : 5 : .......... i.e., only odd harmonics are present in a closed organ pipes.

Analytical treatment : Consider a cylindrical pipe of length L lying along the x-axis with its closed end at x = 0 and open end at x = L. The sound wave sent along the pipe can be represented as DP1 = DPm sin(kx - wt ) The reflected wave from the closed end is represented by (sound wave suffers no phase change due to the reflection from closed end.). DP2

=

DPm sin(-kx - wt )

=

-DPm sin(kx + wt )

=

DP1 + DP2

=

DPm sin(kx - wt ) - DPm sin(kx + wt )

The resultant wave is given by

DP

or DP = –2DPm cos(kx )sin(wt ) For all values of t, the resultant pressure variation is zero, for which cos(kx )

or

= 0 (2n + 1)

p 2

=

(2 n + 1)

p 2

x =

(2n + 1)

l , 4

kx = 2p x l

where n = 0, 1, 2, 3, ..........

l 3l 5l , ,........ , 4 4 4 These points of zero pressure variation are called pressure nodes. On the other hand, the pressure variation is maximum for all values of t, for which

\

x =

or

cos(kx ) = ±1 kx = np

661

662

MECHANICS, HEAT, THERMODYNAMICS & WAVES 2p x l

or

x

= np n

=

l , where n = 0, 1, 2, 3, ............ 2

l 3l , l, ,...... 2 2 These points of maximum pressure variation are called pressure antinodes (see Fig.10.49). The pressure at these points varies from (P – 2D Pm) to (P0 + 2DPm).

\

x

= 0,

Resonance tube : It is used to determine the speed of sound in air with the help of tuning fork of known frequency. It is a close pipe whose length can be changed by changing level of liquid in the tube. When a vibrating tuning fork is brought over its mouth, its air column vibrates longitudinally. If the length of the air column is varies until its natural frequency becomes equal to the frequency of fork, then resonance will occur and loud sound is heard. For the first resonance

L1 + e

l 4

=

… (i)

3l …(ii) 4 Here L1 and L2 are the length of resonance columns and e is the end correction. After solving equations (i) and (ii), we get

and for second resonance

L2 + e

=

l

=

2( L2 - L1 )

L2 – 3L1 2 If f is the frequency of the fork, then speed of sound in air

and

Ex. 24 The first overtone of an organ pipe beats with the first overtone of a close organ pipe with a beat frequency of 2.2 Hz. The fundamental frequency of the closed organ pipe is 110 Hz. Find the lengths of the pipes. Velocity of sound in air = 330 m/s. Sol.

e

=

v

=

f l = 2 f ( L2 - L1 )

Ex. 25 Determine the possible harmonics in the longitudinal vibration of a rod clamped in the middle. Sol. Consider a rod of length L clamped in the middle. It has one node in the middle and two antinodes at its free ends in the fundamental mode.

Suppose Lo and Lc are the lengths of open and close pipes respectively. Frequency of first overtone of open organ pipe,

fo

=

2v v 2 Lo = Lo

Frequency of first overtone of close organ pipe

fc Given \

fo - fc v 3v Lo 4 Lc

=

3v 4 Lc

Fig. 10.52 Here

= 2.2 Hz

or

330 - 3 ´ 110 Lo

l1 or l1 = 2L 4

Frequency of first harmonic = 2.2

f1

v As 4 L = 110 Hz and v = 330 m/s c \

L = 2

v v = l = 2L 1

= 2.2

L o = 0.99m

Ans.

Fig. 10.53

WAVE - II In the second mode of vibration L

l2

or

Frequency,

f2

l2 l2 l2 l2 3l 2 + + + = = 4 2 2 4 2 =

2L 3

f1 : f 2 : f3 :.......

f =

v 4L

\

L =

v 400 4 f = 4 ´ 25 = 4m

Sol. If L0 is the length of the pipe at T0, then its length at temperature T is T =

= 1: 3 : 5: ...........

Ex. 26

Three successive frequencies for a string are 75, 125, 175 Hz. (a) State whether the string is fixed at one end or at both ends. (b) What is the fundamental frequency? (c) To which harmonics do these frequencies corresponds? (d) Taking the speed of the transverse wave on the string as 400 m/s, determine the length of the string.

The speed of sound,

(b) (c)

The given harmonics are in the ratio 1 : 3 : 5, so the string is fixed at one end. As the common maximum frequency in the harmonics is 25 Hz, so fundamental frequency = 25 Hz. The given harmonics are the third, fifth and seventh harmonics.

v =

L0[1 + a(T - T0 )] gRT M

We have to find the temperature T0 at which

f (T0 ) gRT0 M 2 L0

\

Sol. (a)

Ans.

Ex. 27 Find the temperature T0 at which the fundamental frequency of an organ pipe is independent of small variation in the temperature in terms of the coefficient of linear expansion (a) of the material of the tube.

v 3v = 3f1 = l = 2 L 2

This is called the third harmonic or first overtone . Similarly for third mode f3 = 5f1. This is the fifth harmonic or second overtone. Hence

(d)

663

T T0

or

=

f (T ) for small (T – T0)

=

gRT M 2 L0 [1 + a (T - T0 )]

= 1 + a(T - T0 )

1/ 2

or

é æ T - T0 ö ù ê1 + ç ÷ú ëê è T0 ø ûú

= 1 + a(T - T0 )

For small (T – T0), we can write

1 æ T - T0 ö 1+ ç ÷ 2 è T0 ø or

T0

= 1 + a(T - T0 ) =

1 2a

Ans.

Kundt’s tube : It is a long glass tube about 5 cm in diameter held horizontally. At one end it carries a disc of cork or board connected with a metal rod which is clamped at its middle. Other end of the tube is closed by a movable piston, so that its length can be adjusted. Lycopodium power is spread on the box of the tube. The free end of the rod rubbed along its length by resin cloth. The rod begins to vibrate longitudinally. These vibrations forced air inside tube through disc. And so stationary longitudinal vibrations are set-up in the tube. At resonance, frequency of vibration of rod becomes equal to frequency of vibration of air column inside tube. For rod : For air : Since \

l rod 2 l air 2 frod vrod l rod

=

Lrod Þ l rod = 2 Lrod Fig. 10.55

=

Lair Þ lair = 2 Lair

= fair =

l L vair v Þ rod = rod = rod l air vair l air Lair

By using Kundt’s tube one can compare the speed of sound in different mediums.

670

MECHANICS, HEAT, THERMODYNAMICS & WAVES

E xercise 10. 1

MCQ Type 1

Level -1 13.

14.

15.

16.

17.

18.

19.

20.

A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a distance 1.21 Å between them. The wavelength of the standing wave is : (a) 1.21 Å (b) 2.42 Å (c) 6.05 Å (d) 3.63 Å Two sinusoidal waves with same wavelengths and amplitudes travel in opposite directions along a string with a speed 10 m/s. If the minimum time interval between two instants when the string is flat is 0.5 s, the wavelength of the wave is (a) 25 m (b) 20 m (c) 15 m (d) 10 m Standing waves are produced in a 10 m long stretched string. If the string vibrates in 5 segments and the wave velocity is 20 m/s, the frequency is (a) 2 Hz (b) 4 Hz (c) 5 Hz (d) 10 Hz A string in musical instrument is 50 cm long and its fundamental frequency is 800 Hz. If a frequency of 1000 Hz is to be produced, then required length of string is (a) 62.5 cm (b) 50 cm (c) 40 cm (d) 37.5 cm A man is watching two trains, one leaving and the other coming in which equal speed of 4 m/s. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air = 320 m/s) will be equal to (a) 6 (b) 3 (c) 0 (d) 12 An open pipe is resonance in its 2nd harmonic with tuning fork of frequency f1. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f1 then again a resonance is obtained with a frequency f2. If in this case the pipe vibrates nth harmonics, then

3 f 4 1

(a)

n = 3, f2 =

(c)

5 n = 5, f2 = f 4 1

apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 s, the total energy of the pulses will be

8 cm

21.

n = 3, f2 =

(d)

3 n = 5, f2 = f 4 1

f

f (a)

(b)

t0

t

(c)

t

t0

t

The figure shows four progressive waves A, B, C and D with their phases expressed with respect to the wave A. It can be concluded from the figure that y

A

B

An organ pipe is closed at one end has fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is (a) 14 (b) 13 (c) 6 (d) 9 Two pulses in a stretched string whose centres are initially 8 cm

t

(d)

t0 22.

t0

f

f

5 f 4 1

(b)

(a) Zero (b) Purely kinetic (c) Purely potential (d) Partly kinetic and partly potential A man is standing on a railway platform listening to the whistle of an engine that passes the man at constant speed without stopping. If the engine passes the man at time t0. How does the frequency f of the whistle as heard by the man changes with time :

O

C

D

p

p/2

3p/2

wt

2p

Answer Key

13

(a )

15

(c )

17

(a )

19

(c )

21

(a )

Sol. from page 685

14

(d )

16

(c )

18

(c )

20

(b )

22

(b )

WAVE - II (a)

The wave C is ahead by a phase angle of lags behind by a phase angle of

(b)

p 2

The wave C is behind by a phase angle of lags ahead by a phase angle of

p and the wave B 2

p 2

(c) (d)

p and the wave B 2

671

The wave C is ahead by a phase angle of p and the wave B lags behind by a phase angle of p The wave C is behind by a phase angle of p and the wave B lags ahead by a phase angle of p

WAVE - II

685

S olutions Exercise10.1Level -1 13. 14.

(a)

The wavelength

(d)

l = l = 1.21 Å T/2 = 0.5 s, \ T = 1 s

l=

15.

(c)

5

17.

(c)

(a)

l = 10, \ l = 4 m. 2

f1 =

As

f

=

\

f1 f2

=

or

l2

=

f1 = f

\

æ n ö 2ç ÷ è 2l ø

f2

=

æ n ö nç ÷ è 4l ø

f2

=

n f1 ; (where n is odd number.) 4

As f 2 > f1, \ n = 5.

n 20 = = 5 Hz. l 4

19.

(c)

1 F /m , 2l

800 f1 l1 = ´ 50 = 40 cm. 1000 f2

20.

(b)

21.

(a)

22.

(b)

n æ 320 ö = 240 ç ÷ = 237 Hz. n + ns è 320 + 4 ø

Beats frequency fb = f1~f2 = 6 Hz.

A person can hear sound of frequency f ³ 20000 Hz. \ n ×1500 = 20000

é 20000 ù ê 1500 ú ; where n is an odd number ë û = 13.33 n = 13. \ It is 13th harmonic or 6 overtones. After 2 s, the each wave travels a distance = 2 × 2 = 4 m. The wave shape is shown in figure. Thus energy is purely kinetic. or

l2 l1

æ 320 ö n = 240 ç 320 - 4 ÷ = 243 Hz. è ø n - ns

and f2 = f

\

(c)

n = nT = 10 × 1 = 10 m. f

Now f = 16.

18.

n =

æ n ö æ n ö ÷÷ ; so the frequency of f1 = f çç ÷÷ and f2 = f çç n v è n + vs ø s ø è whistle suddenly changes from f1 to f2. For wave B, y = A and so j = p / 2 . For wave C, y = – A and so j = - p / 2 .