7. WAVES

ISBN : 9789386320070

Product Name : Waves for JEE Main & Advanced (Study

Package for Physics)

Product Description : Disha's Physics series by North India's

popu-lar faculty for IIT-JEE, Er. D. C. Gupta, have achieved a lot of

ac-claim by the IIT-JEE teachers and students for its quality and

in-depth 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

Impor-tant Formulae, Strategy to solve problems and Solved Examples.

This topic is taken from our Book:

WAVES

• 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.

9. Wave - I 571-632

9.1 Introduction 572

9.2 Pulse and wave 572

9.3 Graphical representation of

simple harmonic wave 573

9.4 Sound 575

9.5 Wave front 575

9.6 Equation of a travelling wave 576 9.7 Plane progressive harmonic wave

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 9.11 Wave equation , y =a sin (kx –wt) 580 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

9.19 Factors affecting speed of sound in gas 592 9.20 Energy of a progressive wave 594

9.21 Power transmission 595

9.22 Intensity of sound 597

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 9.27 Reflection of sound 607 9.28 Reverberation 607 9.29 Range of hearing 608 9.30 Ultrasound 608 9.31 Sonar 609 9.32 Shock waves 609

Review of formulae & important points 611

Exercise 9.1 - Exercise 9.6 612-622

Hints & solutions 623-632

10. Wave -II 633-698

10.1 Reflection and refraction of

sound waves 634

10.2 Reflection and transmission of

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 641

10.6 Interference in time : beats 644

10.7 Stationary waves 649

10.8 Stationary waves in stretched 654 string fixed at the ends

10.9 Stationary longitudinal waves

in organ pipes 657

Review of formulae & important points 668 Exercise 10.1 - Exercise 10.6 670-684

Hints & solutions 685-698

657

10.9 S

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P

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 = 1 2 l or l₁ = 2L Frequency of vibration f = 1 v l = 1 2 P L g r = f (say) This is the first harmonic.

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(ii) Second mode of vibration :

Here L = 2 2 2 2 l l + = l2 or l₂ = L Frequency of vibration = 2 v l = 2 2 P L g r = 2f

This frequency is called first overtone or second harmonic.

(iii) Third mode of vibration :

Here L = 3 3 3 2 2 2 l l l + + = 3 3 2 l Frequency of vibration f₃ = 3 v l = 3 2 P L g 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

PD = DP_msin(kx- wt)

The reflected sound wave from open end (rigid boundary) is represented by 2

P

D = DP_msin(- - w + pkx t ) = DPmsin(kx+ wt) The resultant stationary wave is given by

DP = D + DP1 P2

= DPmsin(kx- w + Dt) Pmsin(kx+ wt) or DP = 2DP_msin( ) cos( )kx wt

For all values of t, the resultant pressure variation is zero, for which sin kx = 0 or kx = np or 2px l = np x = 2 nl , where n = 0, 1, 2, 3, ...

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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 = ±1 or kx = (2 1) 2 n+ p or 2px l = (2n 1)2 p + x = (2 1) 4 n+ l, where n = 0, 1, 2, 3, ... \ x = ,3 ,5 ,... 4 4 4 l l l

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

0 2 m

P ± DP or P₀±2ABk. Thus pressure at antinodes varies from P₀-2ABk to

0 2

P + ABk.

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

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

B = dP dV V æ_- ö ç ÷ è ø = _{(-dy dx}dP_{/ )} \ strain = dy_dx = dP B -= P B D

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As DP is maximum ( 2± DP_m)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 L_e = L + e, for open organ pipe, the effective length L_e = 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 = 1 4 l or l = 4L₁ Frequency, f₁ = 1 v l = 1 4 P L g r = f (say) This frequency is called first harmonic or fundamental note.

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(ii) Second mode of vibration : In this mode of vibration

L = 2 2 2 4 l l + = 3 2 4 l or l₂ = 4 3 L Frequency f₂ = 2 v l = 3 4 P L g r = 3f (iii) Third mode of vibration : In this mode of vibration

L = 3 3 3 2 2 4 l l l + + = 5 3 4 l or l3 = 4 5 L Frequency , f3 = 3 v l = 5 4 P L g r = 5f 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 = DPmsin(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.).

2

P

D = DP_msin(- - wkx t) = -DPmsin(kx+ wt) The resultant wave is given by

P

D = D + DP₁ P₂

= DP_msin(kx- w - Dt) P_msin(kx+ wt) or _DP = –2DP_mcos( )sin( )kx wt

For all values of t, the resultant pressure variation is zero, for which cos( )kx = 0 or kx = (2 1) 2 n+ p 2 x p l = (2n 1)2 p + x = (2 1) 4 n+ l, where n = 0, 1, 2, 3, ... \ x = ,3 ,5 ,... 4 4 4 l l l

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

cos( )kx = ±1

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These points of maximum pressure variation are called pressure antinodes (see Fig.10.49). The pressure at these points varies from (P – 2D P_m) to (P₀ + 2DP_m).

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 = 4 l

… (i) and for second resonance L₂+e = 3

4 l

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

l = 2(L₂-L₁)

and e = 2– 3 1

2

L L

If f is the frequency of the fork, then speed of sound in air

v = fl = 2 (f L2-L1)

Ex. 24

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.

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

o f = ₂2 o v L = _o v L

Frequency of first overtone of close organ pipe

c f = ₄3 c v L Given f_o-f_c = 2.2 Hz \ ₄3 o c v v L - L = 2.2 As 4vL = 110 Hz and v = 330 m/s_c \ 330 3 110 o L - ´ = 2.2 or L_o = 0.99m Ans.

Ex. 25

vibration of a rod clamped in the middle.

Sol.

Fig. 10.52

Here L = 2 1

4 l

or l = 2L₁ Frequency of first harmonic

1 f = 1 v l = 2 v L Fig. 10.53

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In the second mode of vibration

L = l₄2+l₂2+l₂2+l₄2 = 3l₂2 or l2 = 2 3 L Frequency, f₂ = 2 v l = ₂3v_L = 3f1

This is called the third harmonic or first overtone . Similarly for third mode

f₃ = 5f₁.

This is the fifth harmonic or second overtone. Hence f1:f2:f3:... = 1: 3 : 5: ...

Ex. 26

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.

Sol.

(a) The given harmonics are in the ratio 1 : 3 : 5, so the string is fixed at one end.

(b) As the common maximum frequency in the harmonics is 25 Hz, so fundamental frequency = 25 Hz.

(c) The given harmonics are the third, fifth and seventh harmonics.

(d) f = ₄v_L

\ _{L = 4}v_{f = 4 25}400_´ = 4m Ans.

Ex. 27

the material of the tube.

Sol.

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

T = L0[1+ a -(T T0)]

The speed of sound, v = RT

M

g We have to find the temperature T₀ at which

0 ( ) f T = f T for small (T – T( ) ₀) \ 0 0 2 RT M L g = 0 0 2 [1 ( )] RT M L T T g + a -or 0 T T = 1+ a -(T T0) or 1/ 2 0 0 1 T T T é ₊æ - öù ê ç ÷ú ê è øú ë û = 1+ a -(T T0)

For small (T – T₀), we can write

0 0 1 1 2 T T T æ - ö + ç ÷ è ø = 1+ a -(T T0) or T₀ = _2a1 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 : rod

2 l

= LrodÞ lrod = 2Lrod

For air : air

2 l

= L_airÞ l = _air 2L_air

Since f_rod = f_air

\ rod

rod v

l = air_air rod_air

v v v Þ l = rod_air l l = rod_air L L

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

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13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

(a) n = 3, f₂ = 3₄ f₁ (b) n = 3, f₂ = 5₄ f₁

(c) n = 5, f₂ = 5₄ f₁ (d) n = 5, f₂ = 3₄ f₁

19. 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

20. Two pulses in a stretched string whose centres are initially 8 cm

Level -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

(a) Zero (b) Purely kinetic (c) Purely potential

(d) Partly kinetic and partly potential

21. 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 :

(a) f t t0 (b) f t t0 (c) f t t0 (d) f t t0

22. 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

O y B A C D wt 2p 3p/2 p/2 p

E xercise 10. 1

MCQ Type 1

Answer Key

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(a) The wave C is ahead by a phase angle of ₂p and the wave B lags behind by a phase angle of ₂p

(b) The wave C is behind by a phase angle of ₂p and the wave B lags ahead by a phase angle of ₂p

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

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

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13. (a) The wavelength l l = = 1.21 Å 14. (d) T/2 = 0.5 s, \ T = 1 s T f n l = _{= n = 10 × 1 = 10 m.} 15. (c) 5 2 l = 10, \ l = 4 m. Now f = n_{l =} 20₄ = 5 Hz. 16. (c) As f = 1 / 2l F m , \ 1 2 f f = 2₁ l l or l2 = 1 1 2 800 ₅₀ 1000 f f l = ´ = 40 cm. 17. (a) f₁ = s f n n - n = 320 240 320 4 æ ö ç _- ÷ è ø= 243 Hz. and f₂ = s f n n + n = 320 240 320 4 æ ö ç ₊ ÷ è ø= 237 Hz. \ Beats frequencyf = f_{b 1}~f₂ = 6 Hz. 18. (c) f₁ = 2æç₂n_lö÷ è ø 2 f = næç_è4n_lö÷_ø

\ f2 = ₄nf ; (where n is odd number.)1

As f > f2 1, \ n = 5.

19. (c) A person can hear sound of frequency f ³20000Hz. \ n ×1500 = 20000

or n = éê_ë20000₁₅₀₀ ùú_û; where n is an odd number = 13.33

\ n = 13.

It is 13th _{harmonic or 6 overtones.}

20. (b) 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. 21. (a) f₁ = s f v æ n ö ç ÷ ç_{n -} ÷ è øand f2 = s f v æ n ö ç ÷ ç_{n +} ÷ è ø; so the frequency of whistle suddenly changes from f₁ to f₂.

22. (b) For wave B, y = A and so j = p/ 2. For wave C, y = – A and so j = - p ./ 2