Chapter 14 383

SOLUTIONS TO PROBLEMS

Section 14.1 The Principle of Superposition

P14.1 y y y x t x t= + = ! + !1 2 3 00 4 00 1 60 4 00 5 0 2 00. cos . . . sin . .a f a f evaluated at the given x values.

(a) x = 1 00. , t = 1 00. y = + + = !3 00 2 40 4 00 3 00 1 65. cos . . sin . . rad rad cma f a f

(b) x = 1 00. , t = 0 500. y = + + + = !3 00 3 20 4 00 4 00 6 02. cos . . sin . . rad rad cma f a f

(c) x = 0 500. , t = 0 y = + + + =3 00 2 00 4 00 2 50 1 15. cos . . sin . . rad rad cma f a f

P14.2

FIG. P14.2

P14.3 (a) y f x vt1 = !a f , so wave 1 travels in the +x direction

y f x vt2 = +a f , so wave 2 travels in the !x direction

(b) To cancel, y y1 2 0+ = : 5

3 4 2

5

3 4 6 22 2x t x t! +

= ++ ! +a f a f

3 4 3 4 6

3 4 3 4 6

2 2x t x t

x t x t

! = + !! = ± + !a f a f

a f

for the positive root, 8 6t = t = 0 750. s

(at t = 0 750. s , the waves cancel everywhere)

(c) for the negative root, 6 6x = x = 1 00. m

(at x = 1 00. m, the waves cancel always)

384 Superposition and Standing Waves

Section 14.2 Interference of Waves

P14.4 Suppose the waves are sinusoidal.

The sum is 4 00 4 00 90 0. sin . sin . cm cma f b g a f b gkx t kx t! + ! + °" "

2 4 00 45 0 45 0. sin . cos . cma f b gkx t! + ° °"

So the amplitude is 8 00 45 0 5 66. cos . . cm cma f ° = .

P14.5 The resultant wave function has the form

y A kx t= FHGIKJ ! +FHG

IKJ2

2 20 cos sin

# " #

(a) A A= FHGIKJ =

!LNMOQP =2

22 5 00

4

29 240 cos . cos .

# $a f m

(b) f = = ="$

$$2

1 200

2600 Hz

P14.6 22

0 0A Acos#FHGIKJ = so

# $2

1

260 0

3

1= FHGIKJ = ° =!cos .

Thus, the phase difference is # $= ° =1202

3

This phase difference results if the time delay is T

f v3

1

3 3= = %

Time delay = =3 000 500

..

m

3 2.00 m s sb g

P14.7 Waves reflecting from the near end travel 28.0 m (14.0 m down and 14.0 m back), while waves

reflecting from the far end travel 66.0 m. The path difference for the two waves is:

&r = ! =66 0 28 0 38 0. . . m m m

Since % = vf

Then & &r r f

v%= = =a f a fa f38 0 246

34327 254

..

m Hz

m s

or, &r = 27 254. %

The phase difference between the two reflected waves is then

# $= = = °0 254 1 0 254 2 91 3. . . cycle radb g a f

Chapter 14 385

P14.8 (a) &x = + ! = ! =9 00 4 00 3 00 13 3 00 0 606. . . . . m

The wavelength is % = = =vf

343

3001 14

m s

Hz m.

Thus, &x%

= =0 606

1 140 530

.

.. of a wave ,

or &# $= =2 0 530 3 33. .a f rad

(b) For destructive interference, we want & &x

fx

v%= =0 500.

where &x is a constant in this set up. fv

x= = =

2

343

2 0 606283

& .a f Hz

*P14.9 (a) #1 20 0 5 00 32 0 2 00 36 0= ! =. . . . . rad cm cm rad s s radb ga f b ga f

#

#1 25 0 5 00 40 0 2 00 45 0

9 00 516 156

= ! =

= = ° = °

. . . . .

.

rad cm cm rad s s rad

radians

b ga f b ga f&

(b) &# = ! ! ! = ! +20 0 32 0 25 0 40 0 5 00 8 00. . . . . .x t x t x t

At t = 2 00. s , the requirement is

&# $= ! + = +5 00 8 00 2 00 2 1. . .x na f a f for any integer n.

For x < 3 20. , ! +5 00 16 0. .x is positive, so we have

! + = +

= !+

5 00 16 0 2 1

3 202 1

5 00

. .

..

x n

xn

a fa f

$$

, or

The smallest positive value of x occurs for n = 2 and is

x = !+

= ! =3 204 1

5 003 20 0 058 4.

.. .

a f$ $ cm .

P14.10 Suppose the man’s ears are at the same level as the lower speaker. Sound from the upper speaker is

delayed by traveling the extra distance &r L d L= + !2 2 .

He hears a minimum when &r n= ! FHGIKJ2 1

2a f %

with n = 1 2 3, , , �

Then, L d L nvf

2 2 1

2+ ! = !FHG

IKJFHGIKJ

L d nvf

L

L d nvf

nvf

L L

2 2

2 22 2

2

1

2

1

22

1

2

+ = !FHGIKJFHGIKJ +

+ = !FHGIKJFHGIKJ + !FHG

IKJFHGIKJ +

continued on next page

386 Superposition and Standing Waves

d nvf

nvfL2

2 212

212

! !FHGIKJFHGIKJ = !FHG

IKJFHGIKJ (1)

Equation 1 gives the distances from the lower speaker at which the man will hear a minimum. The

path difference "r starts from nearly zero when the man is very far away and increases to d when L = 0. (a) The number of minima he hears is the greatest integer value for which L # 0 . This is the

same as the greatest integer solution to d nvf

# !FHGIKJFHGIKJ

12

, or

number of minima heard greatest integer = = $ FHGIKJ +n dfvmax

12

.

(b) From equation 1, the distances at which minima occur are given by

Ld n

nn nn

vf

vf

=! !

!=

2 12

2 2

122

1 2c h e jc he j

where , , , max� .

P14.11 (a) First we calculate the wavelength: % = = =vf

34421 5

16 0m s Hz

m.

.

Then we note that the path difference equals 9 00 1 0012

. . m m! = %

Therefore, the receiver will record a minimum in sound intensity. (b) We choose the origin at the midpoint between the speakers. If the receiver is located at point

(x, y), then we must solve:

x y x y+ + ! ! + =5 00 5 0012

2 2 2 2. .a f a f %

Then, x y x y+ + = ! + +5 00 5 0012

2 2 2 2. .a f a f %

Square both sides and simplify to get: 20 04

5 002

2 2. .x x y! = ! +% % a f

Upon squaring again, this reduces to: 400 10 016 0

5 002 24

2 2 2 2x x x y! + = ! +..

.% % % %a f

Substituting % = 16 0. m, and reducing, 9 00 16 0 1442 2. .x y! =

or x y2 2

16 0 9 001

. .! =

(When plotted this yields a curve called a hyperbola.)

Chapter 14 387

Section 14.3 Standing Waves

P14.12 y x t A kx t= =1 50 0 400 200 2 0. sin . cos sin cos ma f a f a f "

Therefore, k = =20 400

$%

. rad m % $= =2

0 40015 7

..

rad m m

and " $= 2 f so f = = ="$ $2

200

231 8

rad s

rad Hz.

The speed of waves in the medium is v f fk

= = = = =% %$

$ "2

2200

0 400500

rad s

rad m m s

.

P14.13 The facing speakers produce a standing wave in the space between them, with the spacing between

nodes being

dvfNN

m s

s m= = = =

!%2 2

343

2 8000 214

1e j.

If the speakers vibrate in phase, the point halfway between them is an antinode of pressure at a

distance from either speaker of

1 25

0 625.

.m

2= .

Then there is a node at 0 6250 214

20 518.

..! = m

a node at 0 518 0 214 0 303. . . m m m! =

a node at 0 303 0 214 0 089 1. . . m m m! =

a node at 0 518 0 214 0 732. . . m m m+ =

a node at 0 732 0 214 0 947. . . m m m+ =

and a node at 0 947 0 214 1 16. . . m m m+ = from either speaker.

P14.14 y A kx t= 2 0 sin cos"

''

= !2

2 022

yx

A k kx tsin cos" ''

= !2

2 022

yt

A kx t" "sin cos

Substitution into the wave equation gives ! = FHGIKJ !2

120

2

2 02A k kx t

vA kx tsin cos sin cos" " "e j

This is satisfied, provided that vk

= "

Chapter 14 389

Section 14.4 Standing Waves in Strings

P14.17 L = 30 0. m; µ = ( !9 00 10 3. kg m; T = 20 0. N ; fvL1

2=

where vT=FHGIKJ =

µ

1 2

47 1. m s

so f147 1

60 00 786= =.

.. Hz f f2 12 1 57= = . Hz

f f3 13 2 36= = . Hz f f4 14 3 14= = . Hz

P14.18 L = 120 cm , f = 120 Hz

(a) For four segments, L = 2% or % = =60 0 0 600. . cm m

(b) v f= =% 72 0. m s fvL1

2

72 0

2 1 2030 0= = =.

..a f Hz

P14.19 The tension in the string is T = =4 9 8 39 2 kg m s N2b ge j. .

Its linear density is µ = =(

= (!

!mL

8 101 6 10

33 kg

5 m kg m.

and the wave speed on the string is vT= =

(=!µ

39 2

10156 5

3

..

N

1.6 kg m m s

In its fundamental mode of vibration, we have % = = =2 2 5 10L m ma f

Thus, fv= = =%

156 5

1015 7

..

m s

m Hz

P14.20 (a) Let n be the number of nodes in the standing wave resulting from the 25.0-kg mass. Then

n + 1 is the number of nodes for the standing wave resulting from the 16.0-kg mass. For

standing waves, % = 2Ln

, and the frequency is fv=%

.

Thus, fnL

Tn=2 µ

and also fn

LTn= + +1

21

µ

Thus, n

nT

T

g

gn

n

+ = = =+

1 25 0

16 0

5

41

.

.

kg

kg

b gb g

Therefore, 4 4 5n n+ = , or n = 4

Then, f = =4

2 2 00

25 0 9 80

0 002 00350

.

. .

. m

kg m s

kg m Hz

2

a fb ge j

continued on next page

390 Superposition and Standing Waves

(b) The largest mass will correspond to a standing wave of 1 loop

n = 1a f so 3501

2 2 00

9 80

0 002 00 Hz

m

m s

kg m

2

=.

.

.a fe jm

yielding m = 400 kg

*P14.21 fvL1

2= , where v

T=FHGIKJµ

1 2

(a) If L is doubled, then f L11) ! will be reduced by a factor

1

2.

(b) If µ is doubled, then f11 2) !µ will be reduced by a factor

1

2.

(c) If T is doubled, then f T1 ) will increase by a factor of 2 .

*P14.22 For the whole string vibrating, dNN = =0 642

. m%

; % = 1 28. m. The

speed of a pulse on the string is v fs

= = =% 3301

1 28 422. m m s .

(a) When the string is stopped at the fret, dNN = =2

30 64

2. m

%;

% = 0 853. m

fv= = =%

422

0 853495

m s

m Hz

..

FIG. P14.22(a)

(b) The light touch at a point one third of the way along the

string damps out vibration in the two lowest vibration

states of the string as a whole. The whole string vibrates in

its third resonance possibility: 3 0 64 32

dNN = =. m%

;

% = 0 427. m

fv= = =%

422

0 427990 Hz

m s

m..

FIG. P14.22(b)

P14.23 dNN m= 0 700.

%

%

=

= = =( !

1 40

3081 20 10 0 7003

.

. .

m

m sf vT

e j a f

(a) T = 163 N

(b) f3 660= Hz

FIG. P14.23

392 Superposition and Standing Waves

P14.26 Comparing y x t= 0 002 100. sin cos m rad m rad sa f b gd i b gd i$ $

with y A kx t= 2 sin cos"

we find k = = !2 1$%

$ m , % = 2 00. m, and " $ $= = !2 100 1f s : f = 50 0. Hz

(a) Then the distance between adjacent nodes is dNN m= =%2

1 00.

and on the string are L

dNN

m

m loops= =3 00

1 003

.

.

For the speed we have v f= = =!% 50 2 1001 s m m se j

(b) In the simplest standing wave vibration, d bNN m= =3 00

2.

%, % b = 6 00. m

and fv

ba

b= = =

%100

6 0016 7

m s

m Hz

..

(c) In vT

00=

µ, if the tension increases to T Tc = 9 0 and the string does not stretch, the speed

increases to

vT T

vc = = = = =93 3 3 100 3000 0

0µ µ m s m sb g

Then % cc

a

vf

= = =!300

506 00

1

m s

s m. d c

NN m= =%2

3 00.

and one loop fits onto the string.

Section 14.5 Standing Waves in Air Columns

P14.27 (a) For the fundamental mode in a closed pipe, % = 4L , as

in the diagram.

But v f= % , therefore Lvf

=4

So, L = =!

343

4 2400 357

1

m s

s m

e j.

(b) For an open pipe, % = 2L , as in the diagram.

So, Lvf

= = =!2

343

2 2400 715

1

m s

s m

e j.

%/4

L

NA

%/2

AA N

FIG. P14.27

Chapter 14 393

P14.28 dAA m= 0 320. ; % = 0 640. m

(a) fv= =%

531 Hz

(b) % = 0 085 0. m; dAA mm= 42 5.

*P14.29 Assuming an air temperature of T = ° =37 310C K , the speed of sound inside the pipe is

v = + * ° ° =331 0 6 37 353 m s m s C C m s. a f .

In the fundamental resonant mode, the wavelength of sound waves in a pipe closed at one end is

% = 4L . Thus, for the whooping crane

% = = (4 5 0 2 0 101. . ft fta f and fv= =

(FHG

IKJ =%

353

2 0 10

3 28157 9

1

m s

ft

ft

1 m Hz

b g.

.. .

P14.30 The wavelength is % = = =vf

3431 31

m s

261.6 s m.

so the length of the open pipe vibrating in its simplest (A-N-A) mode is

dA to A m= =1

20 656% .

A closed pipe has (N-A) for its simplest resonance,

(N-A-N-A) for the second,

and (N-A-N-A-N-A) for the third.

Here, the pipe length is 55

4

5

41 31 1 64dN to A m m= = =%. .a f

P14.31 For a closed box, the resonant frequencies will have nodes at both sides, so the permitted

wavelengths will be L n= 1

2% , n = 1 2 3, , , �b g .

i.e., Ln nv

f= =%

2 2 and f

nvL

=2

.

Therefore, with L = 0 860 m. and + =L 2 10. m, the resonant frequencies are

f nn = 206 Hza f for L = 0 860. m for each n from 1 to 9

and + =f nn 84 5. Hza f for + =L 2 10. m for each n from 2 to 23.

394 Superposition and Standing Waves

P14.32 The air in the auditory canal, about 3 cm long, can vibrate with a node at the closed end and

antinode at the open end,

with dN to A cm= =34

%

so % = 0 12. m

and fv= = ,%

343

0 123

m s

m kHz

.

A small-amplitude external excitation at this frequency can, over time, feed energy

into a larger-amplitude resonance vibration of the air in the canal, making it audible.

P14.33 For both open and closed pipes, resonant frequencies are equally spaced as numbers. The set of

resonant frequencies then must be 650 Hz, 550 Hz, 450 Hz, 350 Hz, 250 Hz, 150 Hz, 50 Hz. These are

odd-integer multipliers of the fundamental frequency of 50 0. Hz . Then the pipe length is

dvfNA = = = =%

4 4

340

4 501

m s

s.70 mb g .

P14.34 The wavelength of sound is % = vf

The distance between water levels at resonance is dvf

=2

- = =Rt r dr v

f$ $2

2

2

and tr vRf

=$ 2

2.

P14.35 %2

= =dLnAA or L

n= %2

for n = 1 2 3, , , �

Since % = vf

L nvf

=FHGIKJ2

for n = 1 2 3, , , �

With v = 343 m s and f = 680 Hz,

L n n=FHG

IKJ =

343

2 6800 252

m s

Hz ma f a f. for n = 1 2 3, , , �

Possible lengths for resonance are: L n= 0 252 0 252. . m, 0.504 m, 0.757 m, , m� a f .

P14.36 The length corresponding to the fundamental satisfies fvL

=4

: Lvf1

4

343

4 5120 167= = =a f . m .

Since L > 20 0. cm, the next two modes will be observed, corresponding to fv

L= 3

4 2

and fv

L= 5

4 3

.

or Lvf2

3

40 502= = . m and L

vf3

5

40 837= = . m .

Chapter 14 395

P14.37 For resonance in a narrow tube open at one end,

f nvL

n= =4

1 3 5, , , �b g .

(a) Assuming n = 1 and n = 3 ,

3844 0 228

= v.a f and 384

3

4 0 683= v

.a f .

In either case, v = 350 m s .

(b) For the next resonance n = 5 , and Lvf

= = =!

5

4

5 350

4 3841 14

1

m s

s m

b ge j

. .

22.8 cm

68.3 cm

f = 384 Hz

warm

air

FIG. P14.37

*P14.38 (a) For the fundamental mode of an open tube,

Lvf

= = = =!

%2 2

343

2 8800 195

1

m s

s m

e j. .

(b) v = + *° ! ° =331 0 6 5 328 m s m s C C m s. a f

We ignore the thermal expansion of the metal.

fv v

L= = = =

% 2

328

2 0 195842

m s

m Hz

.a f

The flute is flat by a semitone.

Section 14.6 Beats: Interference in Time

P14.39 f v T) ) fnew Hz= =110540

600104 4.

&f = 5 64. beats s

P14.40 (a) The string could be tuned to either 521 Hz or 525 Hz from this evidence.

(b) Tightening the string raises the wave speed and frequency. If the frequency were originally

521 Hz, the beats would slow down.

Instead, the frequency must have started at 525 Hz to become 526 Hz .

continued on next page

396 Superposition and Standing Waves

(c) From fv T

L LT= = =

%µ

µ2

1

2

ff

TT

2

1

2

1

= and Tff

T T T22

1

2

1

2

1 1

5230 989=

FHGIKJ = FHG

IKJ = Hz

526 Hz. .

The fractional change that should be made in the tension is then

fractional change = ! = ! = =T TT

1 2

1

1 0 989 0 011 4 1 14%. . . lower.

The tension should be reduced by 1.14% .

P14.41 For an echo + =+!

f fv vv v

s

s

b gb g the beat frequency is f f fb = + ! .

Solving for fb .

gives f fv

v vbs

s=

!2b gb g when approaching wall.

(a) fb =!

=2562 1 33

343 1 331 99a f a fa f

.

.. Hz beat frequency

(b) When he is moving away from the wall, vs changes sign. Solving for vs gives

vf vf fs

b

b=

!=

!=

2

5 343

2 256 53 38

a fa fa fa f . m s .

Section 14.7 Nonsinusoidal Wave Patterns

P14.42 We evaluate

s = + + +

+ + +100 157 2 62 9 3 105 4

51 9 5 29 5 6 25 3 7

sin sin . sin sin

. sin . sin . sin

. . . .. . .

where s represents particle displacement in nanometers

and . represents the phase of the wave in radians. As .

advances by 2$ , time advances by (1/523) s. Here is the

result:

FIG. P14.42

400 Superposition and Standing Waves

P14.51 Call L the depth of the well and v the speed of sound.

Then for some integer n L n nvf

n= ! = ! =

!!

2 14

2 14

2 1 343

4 51 5

1

11

a f a f a fb ge j

% m s

s.

and for the next resonance L n nvf

n= + ! = + =

+!

2 1 14

2 14

2 1 343

4 60 0

2

21

a f a f a fb ge j

% m s

s.

Thus, 2 1 343

4 51 5

2 1 343

4 60 01 1

n n!=

+! !

a fb ge j

a fb ge j

m s

s

m s

s. .

and we require an integer solution to 2 1

60 0

2 1

51 5

n n+ = !. .

The equation gives n = =111 5

176 56

.. , so the best fitting integer is n = 7 .

Then L =!

=!

2 7 1 343

4 51 521 6

1

a f b ge j

m s

s m

..

and L =+

=!

2 7 1 343

4 60 021 4

1

a f b ge j

m s

s m

..

suggest the best value for the depth of the well is 21 5. m .

P14.52 v =(

=!48 0 2 00

4 80 10141

3

. .

.

a fa f m s

dNN m= 1 00. ; % = 2 00. m; fv= =%

70 7. Hz

% aavf

= = =343

70 74 85

m s

Hz m

..

P14.53 (a) Since the first node is at the weld, the wavelength in the thin wire is 2L or 80.0 cm. The

frequency and tension are the same in both sections, so

fL

T= =(

=!1

2

1

2 0 400

4 60

2 00 1059 9

3µ .

.

..a f Hz .

(b) As the thick wire is twice the diameter, the linear density is 4 times that of the thin wire.

+ =µ 8 00. g m

so + =+

Lf

T1

2 µ + =

LNM

OQP (

=!L1

2 59 9

4 60

8 00 1020 0

3a fa f.

.

.. cm half the length of the

thin wire.