21.1. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:

The graph at t = 1 s differs from the graph at t = 0 s in that the left wave has moved to the right by 1 m and the rightwave has moved to the left by 1 m. This is because the distance covered by the wave pulse in 1 s is 1 m. Thesnapshot graphs at t = 2 s, 3 s, and 4 s are a superposition of the left and the right moving waves. The overlappingparts of the two waves are shown by the dotted lines.

21.2. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:

The snapshot graph at t = 1 s differs from the graph t = 0 s in that the left wave has moved to the right by 1 m andthe right wave has moved to the left by 1 m. This is because the distance covered by each wave in 1 s is 1 m. Thesnapshot graphs at t = 2 s, 3 s, and 4 s are a superposition of the left and the right moving waves. The overlappingparts of the two waves are shown by the dotted lines.

21.3. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:

The graph at t = 2 s differs from the graph at t = 0 s in that the left wave has moved to the right by 2 m and the rightwave has moved to the left by 2 m. This is because the distance covered by the wave pulse in 2 s is 2 m. Thesnapshot graphs at t = 4 s and t = 6 s are a superposition of the left and the right moving waves. The overlappingparts of the two are shown by the dotted lines.

21.4. Model: The principle of superposition comes into play whenever the waves overlap.Visualize:

The graph at t = 2 s differs from the graph at t = 0 s in that the left wave has moved to the right by 2 m and the rightwave has moved to the left by 2 m. This is because the distance covered by the wave pulse in 2 s is 2 m. Thesnapshot graphs at t = 4 s and t = 6 s are a superposition of the left and the right moving waves. The overlappingparts of the two waves are shown by the dotted lines.

21.5. Model: The principle of superposition comes into play whenever the waves overlap.Solve: (a) As graphically illustrated in the figure below, the snapshot graph of Figure Ex 21.5b was taken at t = 4 s.

(b)

21.6. Model: A wave pulse reflected from the string-wall boundary is inverted and its amplitude is unchanged.Visualize:

The graph at t = 2 s differs from the graph at t = 0 s in that both waves have moved to the right by 2 m. This isbecause the distance covered by the wave pulse in 2 s is 2 m. The shorter pulse wave encounters the boundary wallat 2.0 s and is inverted on reflection. This reflected pulse wave overlaps with the broader pulse wave, as shown inthe snapshot graph at t = 4 s. At t = 6 s, only half of the broad pulse is reflected and hence inverted; the shorterpulse wave continues to move to the left with a speed of 1 m/s. Finally, at t = 8 s both the reflected pulse waves areinverted and they are both moving to the left.

21.7. Model: Reflections at both ends of the string cause the formation of a standing wave.Solve: Figure Ex21.7 indicates three full wavelengths on the 2.0-m-long string. Thus, the wavelength of thestanding wave is λ = ( ) =1

3 2 0 0 667. . m m. The frequency of the standing wave is

fv= = =λ

40 m / s

0.667 m60 Hz

21.8. Model: Reflections at the string boundaries cause a standing wave on the string.Solve: Figure Ex21.8 indicates one and a half full wavelengths on the string. Hence λ = =2

3 60 40( ) . cm cm Thus

v f= = ( )( ) =λ 0.40 m Hz 40 m / s100

21.9. Model: Reflections at the string boundaries cause a standing wave on the string.Visualize: Please refer to Figure Ex21.9.Solve: (a) When the frequency is doubled ′ =( )f f2 0 , the wavelength is halved ′ =( )λ λ1

2 0 . This halving of the

wavelength will increase the number of antinodes to six.(b) Increasing the tension by a factor of 4 means

vT

vT T

v= ⇒ ′ = ′ = =µ µ µ

42

For the string to continue to oscillate as a standing wave with three antinodes means ′ =λ λ0 . Hence,

′ = ⇒ ′ ′ = ⇒ ′ = ⇒ ′ =v v f f f f f f2 2 2 20 0 0 0 0 0λ λ λ λ

That is, the new frequency is twice the original frequency.

21.10. Model: A string fixed at both ends supports standing waves.Solve: (a) A standing wave can exist on the string only if its wavelength is

λm

L

mm= =2

1 , 2, 3, K

The three longest wavelengths for standing waves will therefore correspond to m = 1, 2, and 3. Thus,

λ λ λ1 2 3

2

1

2

2

2= ( ) = = ( ) = = ( ) =

2.40 m4.80 m

2.40 m2.40 m

2.40 m

31.60 m

(b) Because the wave speed on the string is unchanged from one m value to the other,

f f ff

2 2 3 3 32 2

3

λ λ λλ

= ⇒ = = ( )( ) =50 Hz 2.40 m

1.60 m75 Hz

21.11. Model: A string fixed at both ends supports standing waves.Solve: (a) We have fa = 24 Hz = mf1 where f1 is the fundamental frequency that corresponds to m = 1. The nextsuccessive frequency is fb = 36 Hz = (m + 1) f1. Thus,

f

f

m f

mf

m

mb

a

36 Hz

24 Hz=

+( ) = + = =1 1

1 51

1

. ⇒ + = ⇒ = ⇒m m m1 21.5 f1 = =24 Hz

212 Hz

The wave speed is

v fL

f= = = ( )( ) =λ1 1 1

2

112 242.0 m Hz m / s

(b) The frequency of the third harmonic is 36 Hz. For m = 3, the wavelength is

λm

L

m= = ( ) =2 2 2

3

1 m

3 m

21.12. Model: A string fixed at both ends forms standing waves.Solve: (a) The wavelength of the third harmonic is calculated as follows:

λ λm

L

m

L= ⇒ = = =2 2

33

2.42 m

30.807 m

(b) The speed of waves on the string is v = λ3f3 = (0.807 m)(180 Hz) = 145.3 m/s. The speed is also given byv T= S / µ , so the tension is

T vm

LvS

0.004 kg

1.21 m145.3 m / s N= = = ( ) =µ 2 2 2 69 7.

21.13. Model: A string fixed at both ends forms standing waves.Solve: (a) Three antinodes means the string is vibrating as the m = 3 standing wave. The frequency is f3 = 3f1, so thefundamental frequency is f1 = 1

3 (420 Hz) = 140 Hz. The fifth harmonic will have the frequency f5 = 5f1 = 700 Hz.(b) The wavelength of the fundamental mode is λ1 = 2L = 1.20 m. The wave speed on the string is v = λ1f1 =(1.20 m)(140 Hz) = 168 m/s. Alternatively, the wavelength of the n = 3 mode is λ3 = 1

3 (2L) = 0.40 m, from which v =λ3f3 = (0.40 m)(420 Hz) = 168 m/s. The wave speed on the string is given by

vT= S

µ⇒ = = ( )( ) =T vS 0.0020 kg / m 168 m / s 56.4 Nµ 2 2

Assess: You must remember to use the linear density in SI units of kg/m. Also, the speed is the same for allmodes, but you must use a matching λ and f to calculate the speed.

21.14. Model: The laser light forms a standing wave inside the cavity.Solve: The wavelength of the laser beam is

λ λ µm

L

m= ⇒ = ( ) =2 2

100 00010 6100 000, ,

.0.5300 m

m

The frequency is

fc

100 000100 000

8

6133 0 10

10 6 102 83 10,

,

.

..= = ×

×= ×−λ

m / s

m Hz

21.15. Model: We have an open-open tube that forms standing sound waves.Visualize: Please refer to Figure Ex21.15.Solve: The gas molecules at the ends of the tube exhibit maximum displacement, making antinodes at the ends.There is another antinode in the middle of the tube. Thus, this is the m = 2 mode and the wavelength of the standingwave is equal to the length of the tube, that is, λ = 0.80 m. Since the frequency f = 500 Hz, the speed of sound inthis case is v = fλ = (500 Hz)(0.80 m) = 400 m/s.Assess: The experiment yields a reasonable value for the speed of sound.

21.16. Solve: (a) For the open-open tube, the two open ends exhibit antinodes of a standing wave. The possiblewavelengths for this case are

λm

L

mm= =2

1 2, 3, , K

The three longest wavelengths are

λ λ λ1 2 3

2

1

2 2= ( ) = = ( ) = = ( ) =

1.21 m2.42 m

1.21 m

21.21 m

1.21 m

30.807 m

(b) In the case of an open-closed tube,

λm

L

mm= =4

1 3, 5, , K

The three longest wavelengths are

λ λ λ1 2 3

4 4 4= ( ) = = ( ) = = ( ) =

1.21 m

14.84 m

1.21 m

31.61 m

1.21 m

50.968 m

21.17. Model: An organ pipe has a “sounding” hole where compressed air is blown across the edge of the pipe.This is one end of an open-open tube with the other end at the true “end” of the pipe.Solve: For an open-open tube, the fundamental frequency is f1 = 16.4 Hz. We have

λ λ1

1

1

2

1 2

1

2

1

20 5= ⇒ = =

=

=L

Lv

fsound 343 m / s

16.4 Hz1 m.

Assess: The length of the organ pipe is ≈ 34.5 feet. That is actually somewhat of an overestimate since theantinodes of real tubes are slightly outside the tube. The actual length in a real organ is about 32 feet, and this is thetallest pipe in the so called “32 foot rank” of pipes.

21.18. Solve: For the open-open tube, the fundamental frequency of the standing wave is f1 = 1500 Hz whenthe tube is filled with helium gas at 0°C. Using λm L m= 2 ,

fv

L1 heliumhelium 970 m / s= =λ1 2

Similarly, when the tube is filled with air,

fv

L

f

ff1 air

air 1 air

1 helium1 air

331 m / s 331 m / s

970 m / s

331 m / s

970 m / s1500 Hz 512 Hz= = ⇒ = ⇒ =

( ) =

λ1 2

Assess: Note that the length of the tube is one-half the wavelength whether the tube is filled with helium or air.

21.19. Model: A string fixed at both ends forms standing waves.Solve: A simple string sounds the fundamental frequency f1 = v/2L. Initially, when the string is of length LA = 30cm, the note has the frequency f1A = v/2LA. For a different length, f1B = v/2LB. Taking the ratio of each side of thesetwo equations gives

f

f

v L

v L

L

LL

f

fL1A

1B

A

B

B

AB

1A

1BA= = ⇒ =/

/

2

2

We know that the second frequency is desired to be f1B = 523 Hz. The string length must be

LB

440 Hz

523 Hz30 cm 25.2 cm= ( ) =

The question is not how long the string must be, but where must the violinist place his finger. The full string is 30cm long, so the violinist must place his finger 4.8 cm from the end.Assess: A fingering distance of 4.8 cm from the end is reasonable.

21.20. Model: Reflections at the string boundaries cause a standing wave on a stretched string.Solve: Because the vibrating section of the string is 1.9 m long, the two ends of this vibrating wire are fixed, andthe string is vibrating in the fundamental harmonic. The wavelength is

λ λm

L

mL= ⇒ = = ( ) =2

2 21 1.90 m 3.80 m

The wave speed along the string is v = f1λ1 = (27.5 Hz)(3.80 m) = 104.5 m/s. The tension in the wire can be foundas follows:

vT

T v v= ⇒ = =

=

( ) =S

S

mass

length

0.400 kg

2.00 m104.5 m / s 2180 N

µµ 2 2 2

21.21. Model: The interference of two waves depends on the difference between the phases ∆φ( ) of the two

waves.Solve: (a) Because the speakers are in phase, ∆φ0 = 0 rad . Let d represent the path-length difference. Using m =0 for the smallest d and the condition for destructive interference, we get

∆ ∆ ∆φ πλ

φ π= + = +( )2 2012

xm rad m = 0, 1, 2, 3 …

⇒ + =2 0πλ

φ π∆ ∆x rad ⇒ + =2 0π

λπd

rad rad ⇒ = =

=

=d

v

f

λ2

1

2

1

2

343 m / s

686 Hz0.25 m

(b) When the speakers are out of phase, ∆φ π0 = . Using m = 1 for the smallest d and the condition for constructiveinterference, we get

∆ ∆ ∆φ πλ

φ π= + =2 20

xm m = 0, 1, 2, 3, …

⇒ + =2 2πλ

π πd ⇒ = =

=

=d

v

f

λ2

1

2

1

2

343 m / s

686 Hz0.25 m

21.22. Model: Interference occurs according to the difference between the phases ∆φ( ) of the two waves.Solve: (a) A separation of 20 cm between the speakers leads to maximum intensity on the x-axis, but a separationof 60 cm leads to zero intensity. That is, the waves are in phase when (∆x)1 = 20 cm but out of phase when (∆x)2 =60 cm. Thus,

∆ ∆x x( ) − ( ) = ⇒ = −( ) =2 1 22

λ λ 60 cm 20 cm 80 cm

(b) If the distance between the speakers continues to increase, the intensity will again be a maximum when theseparation between the speakers that produced a maximum has increased by one wavelength. That is, when theseparation between the speakers is 20 cm + 80 cm = 100 cm.

21.23. Model: We assume that the speakers are identical and that they are emitting in phase.Solve: Since you don’t hear anything, the separation between the two speakers corresponds to the condition ofdestructive interference. With ∆φ0 0= rad, Equation 21.23 becomes

2 2 12π

λπd

m= +( ) rad ⇒ = +( )d m 12 λ ⇒ =d

λ λ λ2

, , 3

2 5

2

Since the wavelength is

λ = = =v

f

340 m / s

170 Hz2.0 m

three possible values for d are 1.0 m, 3.0 m, and 5.0 m.

21.24. Model: Reflection is maximized if the two reflected waves interfere constructively.Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32:

λ λC

C m

nm= ⇒ = =×( )( )

( )( )=

−2

2

600 10 1

2 1 39216

9nd

md

m

n .

where we have used m = 1 to calculate the thinnest film.Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicablebecause nair < nfilm < nglass.

21.25. Model: Reflection is maximized if the two reflected waves interfere constructively.Solve: The film thickness that causes constructive interference at wavelength λ is given by Equation 21.32:

λ λC

C m

nm= ⇒ = =×( )( )

( )( ) =−

2

2

500 10 1

2 1 25200

9nd

md

m

n .

where we have used m = 1 to calculate the thinnest film.Assess: The film thickness is much less than the wavelength of visible light. The above formula is applicablebecause nair < noil < nwater.

21.26. Visualize: Please refer to Figure Ex21.26.Solve: (a) The circular wave fronts emitted by the two sources show that the two sources are in phase. This isbecause the wave fronts of each source have moved the same distance from their sources.(b) Let us label the top source as 1 and the bottom source as 2. Since the sources are in phase, ∆φ0 0= rad. For thepoint P, r1 = 3λ and r2 = 4λ. Thus, ∆r = r2 − r1 = 4λ − 3λ = λ. The phase difference is

∆ ∆φ πλ

π λλ

π= = ( ) =2 22

r

This corresponds to constructive interference.For the point Q, r1

72= λ and r2 = 2λ. The phase difference is

∆ ∆φ πλ

π λλ

π= = ( ) =2 23

32r

This corresponds to destructive interference.For the point R, r1

52= λ and r2

72= λ . The phase difference is

∆φπ λλ

π= ( ) =2

2

This corresponds to constructive interference.r1 r2 ∆r C/D

P 3λ 4λ λ CQ 7

2 λ 2λ 32 λ D

R 52 λ 7

2 λ λ C

21.27. Visualize: Please refer to Figure Ex21.27.Solve: (a) The circular wave fronts emitted by the two sources indicate the sources are out of phase. This isbecause the wave fronts of each source have not moved the same distance from their sources.(b) Let us label the top source as 1 and the bottom source as 2. The phase difference between the sources is∆φ π0 = . For the point P, r1 = 2λ and r2 = 3λ. The phase difference is

∆ ∆ ∆φ πλ

φπ λ λ

λπ π= + =

−( ) + =2 2 3 230

r

This corresponds to destructive interference.For the point Q, r1 = 3λ and r2

32= λ . The phase difference is

∆φπ λ

λπ π= ( ) + =

24

32

This corresponds to constructive interference.For the point R, r1

52= λ and r2 = 3λ. The phase difference is

∆φπ λ

λπ π= ( ) + =

22

12

This corresponds to constructive interference.r1 r2 ∆r C/D

P 2λ 3λ λ DQ 3λ 3

2 λ 32 λ C

R 52 λ 3λ 1

2 λ C

Assess: Note that it is not r1 or r2 that matter, but the difference ∆r between them.

21.28. Model: The two speakers are identical, and so they are emitting circular waves in phase. The overlap ofthese waves causes interference.Visualize:

Solve: From the geometry of the figure,

r r2 12 2 2 2= + ( ) = ( ) + ( ) =2.0 m 4.0 m 2.0 m 4.472 m

So, ∆r r r= − = − =2 1 4.472 m 4.0 m 0.472 m . The phase difference between the sources is ∆φ0 0= rad and thewavelength of the sound waves is

λ = = =v

f

340 m / s

1800 Hz0.1889 m

Thus, the phase difference of the waves at the point 4.0 m in front of one source is

∆ ∆ ∆φ πλ

φ π π π= + = ( ) + =22

50

r 0.472 m

0.1889 m0 rad rad = 2.5(2 rad)

This is a half-integer multiple of 2π rad, so the interference is perfect destructive.

21.29. Model: The two radio antennas are emitting out-of-phase, circular waves. The overlap of these wavescauses interference.Visualize:

Solve: From the geometry of the figure, r1 = 800 m and

r22 2= ( ) + ( ) =800 m 600 m 1000 m

So, ∆r = r2 − r1 = 200 m and ∆φ π0 = rad . The wavelength of the waves is

λ = = ××

=c

f

3 10

3 0 10

8

6

m / s

Hz100 m

.

Thus, the phase difference of the waves at the point (300 m, 800 m) is

∆ ∆ ∆φ πλ

φ π π π π= + = ( ) + =22

50

r 200 m

100 m rad rad = 2.5(2 rad)

This is a half-integer multiple of 2π rad, so the interference is perfect destructive.

21.30. Solve: The beat frequency isf f f f fbeat Hz 200 Hz 203 Hz= − ⇒ = − ⇒ =1 2 1 13

f1 is larger than f2 is because the increased tension increases the wave speed and hence the frequency.

21.31. Solve: The flute player’s initial frequency is either 523 Hz + 4 Hz = 527 Hz or 523 Hz − 4 Hz = 519 Hz.Since she matches the tuning fork’s frequency by lengthening her flute, she is increasing the wavelength of thestanding wave in the flute. A wavelength increase means a decrease of frequency because v = fλ. Thus, her initialfrequency was 527 Hz.

21.32. Model: The principle of superposition applies when the waves overlap.Visualize:

Solve: Because the wave pulses travel along the string at a speed of 100 m/s, they move a distance d = vt =(100 m/s)(0.05 s) = 5 m in 0.05 s. The tail end of the wave pulse moving right was therefore located at x = 6 m atthe earlier time t = 0 s. This helps us draw the wave pulse at t = 0 s (shown as a dashed line). Subtracting this wavesnapshot from the resultant (shown as a solid line) yields the second wave’s snapshot at t = 0 s (shown as a dottedline) which is traveling to the left. Finally, the snapshot graph of the wave pulse moving left at t = 0.05 s is thesame as at t = 0 s (shown as a dotted line) except that it is shifted to the left by 5 m.

21.33. Model: The principle of superposition applies to overlapping waves.Visualize:

Solve: Because the wave pulses travel along the string at a speed of 100 m/s, they move a distance of d = vt =(100 m/s)(0.05 s) = 5 m in 0.05 s. The front of the wave pulse moving left, which is located at x = 1 m at t = 0.05 s,was thus located at x = 6 m at t = 0 s. This helps us draw the snapshot of the wave pulse moving left at t = 0 s (shownas a dashed line). Subtracting this wave snapshot from the resultant at t = 0 s (shown as a solid line) yields the right-traveling wave’s snapshot at t = 0 s (shown as a dotted line). Finally, the snapshot graph of the wave pulse movingright at t = 0.05 s is the same as at t = 0 s (shown as a dotted line) except that it is shifted to the right by 5 m.

21.34. Model: The superposition principle applies to overlapping waves.Solve: (a)

The function a kx tsin cosω is shown as a dotted line, the function a kx tcos sinω is shown as a dashed line, and thesum of the functions

D a kx t a kx t= +sin cos cos sinω ωis shown as a solid line.(b) The superposition of the two standing waves is a traveling wave and it is moving to the left. A dashed line isdrawn as a guide to the eye to illustrate the motion of one of the peaks as a function of time.(c) Using the trigonometric identity

sin sin cos cos sinα β α β α β+( ) = +

the combined wave can be written D a kx t= +( )sin ω . This is the equation of a wave that is moving to the left. Themathematical representation of the wave is thus consistent with the graphical representation.

21.35. Model: The wavelength of the standing wave on a string vibrating at its second-harmonic frequency isequal to the string’s length.Visualize:

Solve: The length of the string L = 2.0 m, so λ = L = 2.0 m. This means the wave number is

k = = =2 2πλ

π π2.0 m

rad m

According to Equation 21.5, the displacement of a medium when two sinusoidal waves superpose to give astanding wave is D x t A x t, cos( ) = ( ) ω , where A x a kx A kx( ) = =2 sin sinmax . The amplitude function gives theamplitude of oscillation from point to point in the medium. For x = 10 cm,

A x =( ) = ( ) ( )( )[ ] =10 cm 2.0 cm rad m 0.10 m 0.62 cmsin π

Similarly, A x =( ) =20 cm 1.18 cm, A x =( ) =30 cm 1.62 cm , A x =( ) =40 cm 1.90 cm, and A x =( ) =50 cm2.00 cm.Assess: Consistent with the above figure, the amplitude of oscillation is a maximum at x = 0.50 m.

21.36. Model: The wavelength of the standing wave on a string is λm L m= 2 , where m = 1, 2, 3, … We assumethat 30 cm is the first place from the left end of the string where A A= max .2

Visualize:

Solve: The amplitude of oscillation on the string is A x A kx( ) = max sin . Since the string is vibrating in the thirdharmonic, the wave number is

kL L

= = ( ) =2 2

2 33

πλ

π π

Substituting into the equation for the amplitude,

1

2

3A A

Lmax max sin= ( )

π0.30 m ⇒ ( )

=sin

3 1

2

πL

0.30 m ⇒ ( ) =30.30 m rad

π πL 6

⇒ =L 5.40 m

21.37. Model: The wavelength of the standing wave on a string vibrating at its fundamental frequency is equalto 2L.Solve: The amplitude of oscillation on the string is A x a kx( ) = 2 sin , where a is the amplitude of the travelingwave and the wave number is

kL L

= = =2 2

2

πλ

π π

Substituting into the above equation,

A x L aL

L=( ) =

14 2

4= 2.0 cm sin

π ⇒ =

1.0 cm a1

2⇒ = =a 2 cm 1.41 cm

21.38. Visualize: Please refer to Figure 21.4.Solve: You can see in Figure 21.4 that the time between two successive instants when the antinodes are atmaximum height is half the period, or 1

2 T. Thus T = =2 0 25 0 50( . ) . , s s and so

fT

v

f= = = ⇒ = = =1 1

500.50 s

2.0 Hz3.0 m / s

2.0 Hz1 mλ .

21.39. Model: The standing wave on a guitar string, vibrating at its fundamental frequency, has a wavelength λequal to twice the length L.Solve: The wave speed on the stretched string is

vT

stringS 200 N

0.001 kg / m447.2 m / s= = =

µ

The wavelength of the wave on the string is λ = = ( ) =2 2L 0.80 m 1.60 m. Thus, the frequency of the wave is

fv

= = =string 447.2 m / s

1.60 m279.5 Hz

λThe wave created by the guitar string travels as a sound wave with a speed of 343 m/s in air. Thus, the wavelengthof the sound wave that reaches your ear is

λairsound 343 m / s

279.5 Hz1.23 m= = =v

f

21.40. Model: The wave on a stretched string with both ends fixed is a standing wave.Solve: We must distinguish between the sound wave in the air and the wave on the string. The listener hears asound wave of wavelength λsound = 40 cm = 0.40 m. Thus, the frequency is

fv= = =sound

sound

343 m / s

0.40 m857.5 Hz

λ

The violin string oscillates at the same frequency, because each oscillation of the string causes one oscillation of theair. But the wavelength of the standing wave on the string is very different because the wave speed on the string isnot the same as the wave speed in air. Bowing a string produces sound at the string’s fundamental frequency, so thewavelength of the string is

λ λ λstring string string0.60 m 0.60 m 857.5 Hz 514.5 m / s= = = ⇒ = = ( )( ) =1 2L v f

The tension is the string is found as follows:

vT

stringS=

µ⇒ = ( ) = ( )( ) =T vS string 0.001 kg / m 514.5 m / s 265 Nµ

2 2

21.41. Model: The wave on a stretched string with both ends fixed is a standing wave. For vibration at itsfundamental frequency, λ = 2L.Solve: The wavelength of the wave reaching your ear is 39.1 cm = 0.391 m. So the frequency of the sound wave is

fv= = =air 344 m / s

0.391 m879.8 Hz

λThis is also the frequency emitted by the wave on the string. Thus,

879.8 Hz 150 N

0.0006 kg / mstring S= = =

v T

λ λ µ λ1 1 ⇒ =λ 0.568 m

⇒ = = =L 12 λ 0.284 m 28.4 cm

21.42. Model: For the stretched wire vibrating at its fundamental frequency, the wavelength of the standingwave is λ1 = 2L.Visualize:

Solve: The wave speed on the steel wire isv f f Lwire 80 Hz m 144 m / s= = ( ) = ( ) ×( ) =λ 2 2 0 90.

and is also equal to TS µ , where

µ = = × = ×−

−mass

length

kg

0.90 m kg / m

5 0 105 555 10

33.

.

The tension TS in the wire equals the weight of the sculpture or Mg. Thus,

vMg

wire =µ

⇒ = =×( )( )

=−

Mv

g

µ wire2

kg / m m / s

9.8 m / s11.8 kg

2 3 25 555 10 144.

21.43. Model: The stretched string with both ends fixed forms standing waves.Visualize:

Solve: The astronauts have created a stretched string whose vibrating length is L = 2.0 m. The weight of the hangingmass creates a tension T MgS = in the string, where M = 1.0 kg. As a consequence, the wave speed on the string is

vT Mg= =S

µ µwhere µ = (0.0050 kg)/(2.5 m) = 0.0020 kg/m is the linear density. The astronauts then observe standing waves atfrequencies of 64 Hz and 80 Hz. The first is not the fundamental frequency of the string because 80 Hz ≠ 2 × 64Hz. But we can easily show that both are multiples of 16 Hz: 64 4 1 Hz = f and 80 Hz = 5 1f . Both frequencies arealso multiples of 8 Hz. But 8 Hz cannot be the fundamental frequency because, if it were, there would be a standingwave resonance at 9(8 Hz) = 72 Hz. So the fundamental frequency is f1 = 16 Hz. The fundamental wavelength isλ1 = 2L = 4.0 m. Thus, the wave speed on the string is v f= =λ1 1 64.0 m / s . Now we can find g on Planet X:

vMg=µ

⇒ = = ( ) =gM

vµ 2 20.0020 kg / m

1.0 kg64 m / s 8.19 m / s2

21.44. Model: The stretched bungee cord that forms a standing wave with two antinodes is vibrating at thesecond harmonic frequency.Visualize:

Solve: Because the vibrating cord has two antinodes, λ2 = =L 1.80 m . The wave speed on the cord is

v fcord 20 Hz 1.80 m 36 m / s= = ( )( ) =λ

The tension TS in the cord is equal to k∆L, where k is the bungee’s spring constant and ∆L is the 0.60 m the bungeehas been stretched. Thus,

vT k L

cordS= =

µ µ∆ ⇒ = =

( )( )

=kv

L

µ cord20.075 kg

1.20 m

36 m / s

0.60 m135 N / m

2

∆

21.45. Model: The steel wire is under tension and it vibrates with three antinodes.Visualize: Please refer to Figure P21.45.Solve: When the spring is stretched 8.0 cm, the standing wave on the wire has three antinodes. This meansλ3

23= L and the tension TS in the wire is TS = k(0.080 m), where k is the spring constant. For this tension,

vT

wireS=

µ⇒ =f

Tλµ3

S ⇒ = ( )f

L

k3

2

0.08 m

µ

We will let the stretching of the spring be ∆x when the standing wave on the wire displays two antinodes. Thismeans λ2 = L and ′ =T kxS . For the tension ′TS ,

′ = ′ ⇒ = ′v

Tf

Twire

S S

µλ

µ2 ⇒ =fL

k x1 ∆µ

The frequency f is the same in the above two situations because the wire is driven by the same oscillating magneticfield. Now, equating the two frequency equations,

1 3

2L

k x

L

k∆µ µ

= ( )0.080 m ⇒ = =∆x 0.18 m 18 cm

21.46. Model: The microwave forms a standing wave between the two reflectors.Solve: There are reflectors at both ends, so the electromagnetic standing wave acts just like the standing wave ona string that is tied at both ends. The frequencies of the standing waves are

f mv

Lm

c

Lm m mm = = = ×

( )= ×( ) =light m / s

0.10 m Hz 1.5 GHz

2 2

3 0 10

21 5 10

89.

.

where we have noted that electromagnetic waves of all frequencies travel with the speed of light c. The generatorcan produce standing waves at any frequency between 10 GHz and 20 GHz. These are

m fm (GHz) 7 8 910111213

10.512.013.515.016.518.019.5

There are 7 different standing wave frequencies. Even-numbered values of n create a node at the center, and odd-numbered values of n create an antinode at the center. So the frequencies where the midpoint is an antinode are10.5, 13.5, 16.5, and 19.5 GHz.

21.47. Model: The fundamental wavelength of an open-open tube is 2L and that of an open-closed tube is 4L.Solve: We are given that

f1 open-closed = f3 open-open = 3f1 open-open

⇒ =v vair

1 open-closed

air

1 open-openλ λ3 ⇒ =1

4

3

2L Lopen-closed open-open

⇒ = = ( ) =LL

open-closedopen-open 78.0 cm

cm2

12

2

1213 0.

21.48. Model: A tube forms standing waves.Solve: (a) The fundamental frequency cannot be 390 Hz because 520 Hz and 650 Hz are not integer multiples ofit. But we note that the difference between 390 Hz and 520 Hz is 130 Hz as is the difference between 520 Hz and650 Hz. We see that 390 Hz = 3 × 130 Hz = 3f1, 520 Hz = 4f1, and 650 Hz = 5f1. So we are seeing the third, fourth,and fifth harmonics of a tube whose fundamental frequency is 130 Hz. According to Equation 21.17, this is anopen-open tube because fm = mf1 with m = 1, 2, 3, 4, … For an open-closed tube m has only odd values.(b) Knowing f1, we can now find the length of the tube:

Lv

f= =

( )=

2 1

343 m / s

2 130 Hz1.32 m

(c) 520 Hz is the fourth harmonic. This is a sound wave, not a wave on a string, so the wave will have four nodesand will have antinodes at the ends, as shown.

(d) With carbon dioxide, the new fundamental frequency is

fv

L1 2= =

( )=280 m / s

2 1.32 m106 Hz

Thus the frequencies of the n = 3, 4, and 5 modes are f3 = 3f1 = 318 Hz, f4 = 4f1 = 424 Hz, and f5 = 5f1 = 530 Hz.

21.49. Model: Particles of the medium at the nodes of a standing wave have zero displacement.Visualize: Please refer to Figure P21.49.Solve: The cork dust settles at the nodes of the sound wave where there is no motion of the air molecules. Theseparation between the centers of two adjacent piles is 1

2 λ . Thus,

123 cm

382 cm= ⇒ =λ λ

2

Because the piston is driven at a frequency of 400 Hz, the speed of the sound wave in oxygen is

v f= = ( )( ) =λ 400 Hz 0.82 m 328 m / s

Assess: A speed of 328 m/s in oxygen is close to the speed of sound in air, which is 343 m/s at 20°C.

21.50. Model: The nodes of a standing wave are spaced λ/2 apart.Visualize:

Solve: The wavelength of the mth mode of an open-open tube is λm = 2L/m. Or, equivalently, the length of thetube that generates the mth mode is L = m(λ/2). Here λ is the same for all modes because the frequency of thetuning fork is unchanged. Increasing the length of the tube to go from mode m to mode m + 1 requires a lengthchange

∆L = (m + 1)(λ/2) – mλ/2 = λ/2

That is, lengthening the tube by λ/2 adds an additional antinode and creates the next standing wave. This isconsistent with the idea that the nodes of a standing wave are spaced λ/2 apart. This tube is first increased ∆L =56.7 cm – 42.5 cm = 14.2 cm, then by ∆L = 70.9 cm – 56.7 cm = 14.2 cm. Thus λ/2 = 14.2 cm and thus λ = 28.4 cm =0.284 m. Therefore the frequency of the tuning fork is

fv= = =λ

3431208

m / s

0.284 m Hz

21.51. Model: The open-closed tube forms standing waves.Visualize:

Solve: When the air column length L is the proper length for a 580 Hz standing wave, a standing wave resonancewill be created and the sound will be loud. From Equation 21.18, the standing wave frequencies of an open-closedtube are fm = m(v/4L), where v is the speed of sound in air and m is an odd integer: m = 1, 3, 5, … The frequency isfixed at 580 Hz, but as the length L changes, 580 Hz standing waves will occur for different values of m. The lengththat causes the mth standing wave mode to be at 580 Hz is

Lm

= ( )( )( )

343 m / s

4 580 HzWe can place the values of L, and corresponding values of h = 1 m − L, in a table:

m L h = 1 m − L

1 0.148 m 0.852 m = 85.2 cm3 0.444 m 0.556 m = 55.6 cm

5 0.739 m 0.261 m = 26.1 cm7 1.035 m h can’t be negative

So water heights of 26.1 cm, 55.6 cm, and 85.2 cm will cause a standing wave resonance at 580 Hz. The figureshows the m = 3 standing wave at h = 55.6 cm.

21.52. Model: A stretched wire, which is fixed at both ends, forms a standing wave whose fundamentalfrequency f1 wire is the same as the fundamental frequency f1 open-closed of the open-closed tube. The two frequencies

are the same because the oscillations in the wire drive oscillations of the air in the tube.Visualize:

Solve: The fundamental frequency in the wire is

fv

L L

T1 wire

wire

wire wire

S

1.0 m

440 N

0.0010 kg / 0.050 m= = =

( ) ( )2

1

2

1

µ= 469 Hz

The fundamental frequency in the open-closed tube is

fv

L L1 open-closedair

tube tube

Hz 340 m / s= = =469

4 4⇒ =

( )= =Ltube

340 m / s

4 469 Hz0.181 m 18.1 cm

21.53. Model: A stretched wire, which is fixed at both ends, creates a standing wave whose fundamentalfrequency is f1 wire . The second vibrational mode of an open-closed tube is f3 open-closed . These two frequencies areequal because the wire’s vibrations generate the sound wave in the open-closed tube.Visualize:

Solve: The frequency in the tube is

fv

L3 open-closedair

tube

340 m / s

4 0.85 cm300 Hz= = ( )

( ) =3

4

3

⇒ = = =fv

L L

T1 wire

wire

wire wire

S300 Hz 2

1

2 µ

⇒ = ( ) ( )T LS wire300 Hz 2 22 µ = (300 Hz)2 (2 × 0.25 m)2(0.020 kg/m) = 450 N

21.54. Model: A stretched wire, which is fixed at both ends, creates a standing wave whose fundamentalfrequency is f v L1 wire wire wire= 2 . A standing wave in an open-closed tube exhibits an antinode at the open end and anode at the closed end.Visualize:

Solve: The wave velocity along the wire is

v fT

wire 1 wireS= =λ

µ1 ⇒ = = =×( ) ( )f

T

L

T1 wire

S

wire

S

m

440 N

0.001 kg 0.50 m

1 1

2

1

2 0 501λ µ µ .= 469 Hz

The nodes of a standing wave are spaced λ 2 apart. From the above figure, it is clear that 36 cm is half the wavelength ofthe wave set up in the air column of the tube. That is, λair 0.36 m 0.72 m= ( ) =2 . Because it is the vibrating wire thatcreates sound waves, f fair 1 wire 469 Hz= = . The speed of sound in air is

v fsound in air air air 469 Hz 0.72 m 338 m / s= ( )( ) = ( )( ) =λ

Assess: This value for the speed of sound in air is close to the value of 343 m/s at 20°C.

21.55. Model: In a rod in which a longitudinal standing wave can be created, the standing wave is equivalentto a sound standing wave in an open-open tube. Both ends of the rod are antinodes, and the rod is vibrating in thefundamental mode.Visualize: Please refer to Figure P21.55.Solve: Since the rod is in the fundamental mode, λ1 = 2L = 2(2.0 m) = 4.0 m. Using the speed of sound inaluminum, the frequency is

fv

11

= = =Al 6420 m / s

4.0 m1605 Hz

λ

21.56. Visualize: Please refer to Figure P21.56.Solve: (a) The left end of the tube is closed because the air molecules show zero displacement (i.e., a node). Theright end, on the other hand, is open because the air molecules show full displacement relative to their equilibriumposition (i.e., an antinode).(b) The snapshot picture shows a third harmonic. Thus,

LL= ⇒ = = ( ) =3

4

4

3

433

λ λ 1.5 m

32.0 m

(c)

21.57. Model: A standing wave in an open-closed tube must have a node at the closed end of the tube and anantinode at the open end.Visualize:

Solve: We first draw a series of pictures showing all the possible standing waves. By examination, we see that thefirst standing wave mode is 1

4 of a wavelength, so the tube’s length is L = 14 λ . The next mode is 3

4 of awavelength. The tube’s length hasn’t changed, so in this mode L = 3

4 λ . The next mode is now slightly more than awavelength. That is, L = 5

4 λ . The next mode is 74 of a wavelength, so L = 7

4 λ . We see that there is a pattern. Thelength of the tube and the possible standing wave wavelengths are related by

Lm= λ4

m = 1, 3, 5, 7, … = odd integers

Solving for λ, we find that the wavelengths and frequencies of standing waves in an open-closed tube are

λ

λ

m

mm

L

m

fv

mv

L

m=

= =

= =

4

4

1 3 5, , , odd integersK

21.58. Model: Constructive or destructive interference occurs according to the phases of the two waves.Visualize:

Solve: (a) To go from destructive to constructive interference requires moving the speaker ∆x = 12 λ, equivalent

to a phase change of π rad. Since ∆x = 40 cm, we find λ = 80 cm.(b) Destructive interference at ∆x = 10 cm requires

2 210

0 0πλ

φ π φ π∆ ∆ ∆x + =

+ = cm

80 cm rad rad ⇒ =∆φ π

0

3

4 rad

(c) When side by side, with ∆x = 0, the phase difference is ∆φ = ∆φ0 = 3π/4 rad. The amplitude of the superpositionof the two waves is

a a a=

=2

22

3

8cos cos

∆φ π = 0.765a

21.59. Model: The amplitude is determined by the interference of the two waves.Solve: For interference in one dimension, where the speakers are separated by a distance ∆x, the amplitude of the netwave is A a= ( )2 1

2cos ∆φ , where a is the amplitude of each wave and ∆ ∆ ∆φ π λ φ= +2 0x is the phase differencebetween the two waves. The speakers are emitting identical waves so they have identical phase constants and ∆φ0 0= .Thus,

A a ax= =

1 5 2. cos

πλ∆ ⇒ =

−∆xλπ

cos.1 1 5

2

The wavelength of a 1000 Hz tone is λ = =v fsound 0.343 m . Thus the separation must be

∆x = ( ) = =−0.343 m0.0789 m 7.89 cm

πcos .1 0 75

It is essential to note that the argument of the arccosine is in radians, not in degrees.

21.60. Model: Constructive or destructive interference occurs according to the phases of the two waves.Solve: The phase difference between the sound waves from the two speakers is

∆ ∆ ∆φ πλ

φ= +2 0

x

With no delay between the two signals, ∆φ0 0= rad and

∆φ π π π= ( ) = ( )

=2

2 42.0 m

2.0 m340 Hz

340 m / s rad

v f

According to Equation 21.22, this corresponds to constructive interference. A delay of 1.47 ms corresponds to aninherent phase difference of

∆φ π π π π0

22 2=

( ) = ( )( ) = ( )( ) =

Tf1.47 ms rad 1.47 ms rad 1.47 ms 340 Hz rad rad

The phase difference ∆φ between the signals is then

∆ ∆ ∆φ πλ

φ π π π=

+ = + =2 4 50

x rad rad rad

Thus, the interference along the x-axis will be perfect destructive.

21.61. Model: Interference occurs according to the difference between the phases of the two waves.Visualize:

Solve: (a) The phase difference between the sound waves from the two speakers is

∆ ∆ ∆φ πλ

φ= +2 0

x

We have a maximum intensity when ∆x = 0.50 m and ∆x = 0.90 m. This means

2 2 2 2 10 0πλ

φ π πλ

φ π0.50 m rad

0.90 m rad

( ) + =

+ = +( )∆ ∆m m

Taking the difference of the above two equations,

2 2πλ

π λ0.40 m0.40 m

= ⇒ = ⇒ = = =f

vsound 340 m / s

0.40 m850 Hz

λ(b) Using again the equations that correspond to constructive interference,

2 20π φ π0.50 m

0.40 m rad

+ =∆ m ⇒ = − = −∆φ φ φ π

0 20 10 2 rad

We have taken m = 1 in the last equation. This is because we always specify phase constants in the range –π radto π rad (or 0 rad to 2π rad). m = 1 gives − 1

2 π rad (or equivalently, m = 2 will give 32 π rad).

21.62. Model: Reflection is maximized for constructive interference of the two reflected waves, but minimizedfor destructive interference.Solve: (a) Constructive interference of the reflected waves occurs for wavelengths given by Equation 21.32:

λm

nd

m m m= = ( )( ) = ( )2 2 1 42. 500 nm 1420 nm

Thus, 1420 nm,λ1 = λ212= ( ) =1420 nm 710 nm, λ3 = 473 nm, λ4 = 355 nm,KOnly the wavelength of 473 nm

is in the visible range.(b) For destructive interference of the reflected waves,

λ =−

= ( )( )−

=−

2 212

12

12

nd

m m m

1.42 500 nm 1420 nm

Thus, 1420 nm nm,λ1 2 2840= × = λ223= ( ) =1420 nm 947 nm, λ3 = 568 nm, λ4 = 406 nm, K The wavelengths

of 406 nm and 568 nm are in the visible range.(c) Beyond the limits 430 nm and 690 nm the eye’s sensitivity drops to about 1 percent of its maximum value. Thereflected light is enhanced in blue (473 nm). The transmitted light at mostly 568 nm will be yellowish green.

21.63. Model: Maximum constructive interference between the two reflected waves maximizes reflectedintensity. Perfect destructive interference, on the other hand, causes the reflected intensity to be zero.Solve: Constructive interference occurs for wavelengths given by Equation 21.32. The condition for constructiveinterference for blue light is

λ λC

480 nm

2 1.39= ⇒ = =

( )2

2

nd

md

nm m

The condition for destructive interference for red light is

λ λD

640 nm

2 1.39=

−( ) ⇒ = −( ) =( )

−( )2

212

12

12

nd

md

nm m

Equating the two equations for d,480 nm 640 nm( )

( )= ( )

( )−( )

2 1 39 2 1 3912. .

m m ⇒ = ⇒ =( )

× =m d2 2 345480 nm

2 1.39 nm

21.64. Solve: (a) The intensity of reflected light from the uncoated glass is I Ca02= , where a is the amplitude

of the reflected light. We will assume that the amplitude of the reflected light from both the bottom and the top ofthe coated film is a. The interference of the two reflected waves determines the amplitude of the resultant wavewhich is given by

A ax= ( ) = +2 2 2 0cos ∆ ∆ ∆ ∆φ φ π

λφ where

coat

With ∆φ0 0= rad, ∆x d= 2 , and λ λcoat air= n , we have

∆φ πλ

πλ

πλ λ

= ( ) + = = ( )( ) =2 2

04 4 1 39d

n

dn

air

rad92 nm 1607 nm.

⇒ =A a2 cos803.5 nm

λ⇒ = =

I CA C aλ λ

2 2 24 cos803.5 nm ⇒ =

I

Iλ

λ0

24cos803.5 nm

(b) The values of I Iλ 0( ) at λ = 400, 450, 500, 550, 600, 650, and 700 nm are 0.719, 0.182, 0.005, 0.048, 0.211,0.431, and 0.674 respectively.(c)

21.65. Model: Reflection is minimized when the two reflected waves interfere destructively.Solve: Equation 21.2 gives the condition for perfect destructive interference between the two waves:

∆ ∆ ∆φ πλ

φ π= + = +( )2 2012

xm rad

The wavelength of the sound is

λ = = =v

f

343 m / s

1200 Hz0.2858 m

Let d be the separation between the mesh and the wall. Substituting ∆φ0 = 0 rad , ∆x d= 2 , m = 0, and the abovevalue for the wavelength,

2 2π πd( ) + =0.2858 m

0 rad rad ⇒ = = =d0.2858 m

0.0715 m 7.15 cm4

21.66. Model: A light wave that reflects from a boundary at which the index of refraction increases has a phaseshift of π rad.Solve: (a) Because nfilm > nair, the wave reflected from the outer surface of the film (called 1) is inverted due to thephase shift of π rad. The second reflected wave does not go through any phase shift of π rad because the index ofrefraction decreases at the boundary where this wave is reflected, which is on the inside of the soap film. We canwrite for the phases

φ φ π1 1 10= + +kx rad φ φ2 2 20= + +kx 0 rad

⇒ = − = −( ) + −( ) −∆φ φ φ φ φ π2 1 2 1 20 10k x x rad = + −k x∆ ∆φ π0 rad = −k x∆ π rad

∆φ0 = 0 rad because the sources are identical. For constructive interference,

∆φ π= ⇒2m rad k x m∆ − =π π rad rad2 ⇒

( ) = +( )22 2 1

πλ

πfilm

radd m

⇒ = =+

λ λfilm

C

n

d

m

212

⇒ =+

=+

λC

2 2 6612

12

nd

m

d

m

. m = 0, 1, 2, 3, …

(b) For m = 0 the wavelength for constructive interference is

λC

390 nm2075 nm= ( )( )

( ) =2 66

12

.

For m = 1 and 2, λC 692 nm ~ red= ( ) and λC 415 nm ~ violet= ( ) . Red and violet together give a purplish color.

21.67. Model: The two radio antennas are sources of in-phase, circular waves. The overlap of these waves causesinterference.Visualize:

Solve: Maxima occur along lines such that the path difference to the two antennas is ∆r m= λ . The 750 MHz =7.50 × 108 Hz wave has a wavelength λ = =c f 0.40 m . Thus, the antenna spacing d = 2.0 m is exactly 5λ. Themaximum possible intensity is on the line connecting the antennas, where ∆r d= = 5λ . So this is a line ofmaximum intensity. Similarly, the line that bisects the two antennas is the ∆r = 0 line of maximum intensity. Inbetween, in each of the four quadrants, are four lines of maximum intensity with ∆r = λ, 2λ, 3λ, and 4λ. Althoughwe have drawn a fairly accurate picture, you do not need to know precisely where these lines are located to knowthat you have to cross them if you walk all the way around the antennas. Thus, you will cross 20 lines where ∆r =mλ and will detect 20 maxima.

21.68. Model: The changing sound intensity is due to the interference of two overlapped sound waves.Visualize: Please refer to Figure P21.68.Solve: Minimum intensity implies destructive interference. Destructive interference occurs where the path lengthdifference for the two waves is ∆r m= +( )1

2 λ . We have assumed ∆φ0 0= rad for two speakers playing “exactlythe same” tone. The wavelength of the sound is λ = = ( ) =v fsound 343 m / s 686 Hz 0.500 m. Consider a point thatis a distance x in front of the top speaker. Let r1 be the distance from the top speaker to the point and r2 the distancefrom the bottom speaker to the point. We have

r x1 = r x22 23= + ( ) m

Destructive interference occurs at distances x such that

∆r x x m= + − = +( )2 129 m2 λ

To solve for x, isolate the square root on one side of the equation and then square:

x x m x m x m2 12

2 2 12

12

2 29 2+ = + +( )[ ] = + +( ) + +( ) m λ λ λ ⇒ =− +( )

+( )xm

m

9

2

12

2 2

12

m λλ

Evaluating x for different values of m:m x (m)

0123

17.885.622.981.79

Because you start at x = 2.5 m and walk away from the speakers, you will only hear minima for values x > 2.5 m.Thus, minima will occur at distances of 2.98 m, 5.62 m, and 17.88 m.

21.69. Model: The changing sound intensity is due to the interference of two overlapped sound waves.Visualize: The listener moving relative to the speakers changes the phase difference between the waves.

Solve: (a) Initially when you are at P, equidistant from the speakers, you hear a sound of maximum intensity.This implies that the two speakers are in phase (∆φ0 = 0). However, on moving to Q you hear a minimum of soundintensity implying that the path length difference from the two speakers to Q is λ / .2 Thus,

12 1

2 21

2 2λ = = ( ) + ( ) − = ( ) + ( ) − =∆r r r5.0 m 12.0 m 5.0 m 12.0 m 1.0 m

⇒ =λ 2.0 m ⇒ = = =fv

λ340 m / s

2.0 m170 Hz

(b) At Q, the condition for perfect destructive interference is

∆∆φ π

λπ= ( ) + = −( )2

2 12

rm0 rad rad ⇒ = −( )2

2 12

π π∆r

v fm rad

⇒ = −( ) = −( )f m

v

rm1

212∆

340 m / s

1.0 m

For m = 1, 2, and 3, f1 = 170 Hz , f2 = 510 Hz, and f3 = 850 Hz .

21.70. Model: The amplitude is determined by the interference of the two waves.Visualize:

Solve: The amplitude of the sound wave is A a= ( )2 12cos ∆φ . With ∆φ0 = 0 rad , the phase difference between

the waves is

∆ ∆ ∆ ∆φ φ φ πλ

π π= − = = ⇒ =

2 1 2 2 2

r rA a

r

2.0 m 2.0 mcos

At the coordinates (0.0 m, 0.0 m), ∆r = 0 m, so A = 2a. At the coordinates (0.0 m, 0.5 m),

∆r A a a= ( ) + ( ) − ( ) + ( ) = ⇒ = ( ) =3.0 m 2.5 m 3.0 m 1.5 m 0.551 m0.551 m

2.0 m2 2 2 2 2 1 30cos .

π

At the coordinates (0.0 m, 1.0 m),

∆r A a a= ( ) + ( ) − ( ) + ( ) = ⇒ = ( )

=3.0 m 3.0 m 3.0 m 1.0 m 1.08 m1.08 m

2.0 m2 2 2 2 2 0 25cos .

π

At the coordinates (0.0 m, 1.5 m), ∆r A a= =1.568 m and 1 56. . At the coordinates (0.0 m, 2.0 m), ∆r = 2.0 mand A a= 2 .

21.71. Model: The two radio transmitters are sources of out-of-phase, circular waves. The overlap of thesewaves causes interference.Visualize:

Please refer to Figure 21.29b.Solve: The phase difference of the waves at point P is given by

∆ ∆ ∆φ πλ

φ= +2 0

r

∆r = ( ) + ( ) − ( ) + ( ) =3000 m 85 m 3000 m 35 m 0.99976 m2 2 2 2

The intensity at P is a maximum. Using m = 1 for the first maximum, and ∆φ π0 = rad since the transmitters are outof phase, the condition for constructive interference is ∆φ π π= =2 2m . Thus,

2 2π πλ

π radr

rad= +∆ ⇒ = = ( )λ 2 2∆r 0.99976 m ⇒ = = ×( )

=fc

λ3 10

2

8 m / s

0.99976 m150 MHz

21.72. Model: The two radio antennas are sources of in-phase waves. The overlap of these waves causesinterference.Visualize:

Please refer to Figure 21.29a.Solve: (a) The phase difference of the two waves at point P is given by

∆ ∆ ∆φ πλ

φ= +2 0

r ∆r = ( ) + ( ) − ( ) + ( ) =800 m 650 m 800 m 550 m 59.96 m2 2 2 2

The wavelength of the radio wave is

λ = = ××

=c

f

3 0 10

3 0 10

8

6

.

.

m / s

m100 m

Since the sources are identical, ∆φ0 0= rad . The phase difference at P due to the two waves is

∆φ π π=

+ =2 1 20

59.96 m

100 m0 rad rad.

(b) Since ∆φ π π= = ( )1 20 0 6 2. . , which is neither m2π nor m +( )12 2π , the interference at P is somewhere in

between maximum constructive and perfect destructive.(c) At a point 10 m further north we have

∆r = ( ) + ( ) − ( ) + ( ) =800 m 660 m 800 m 560 m 60.58 m2 2 2 2

⇒ =

+ = = ( )∆φ π π π2 1 21 2

60.58 m

100 m0 rad rad 0.605.

Because the phase difference is increasing as you move north, you are moving from a destructive interferencecondition ∆φ π= +( )m 1

2 2 with m = 0 toward a constructive interference condition ∆φ π= ( )m 2 with m = 1. Thesignal strength will therefore increase.

21.73. Model: The amplitude is determined by the interference of the two waves.Visualize: Please refer to Figure P21.73.Solve: (a) We have three identical loudspeakers as sources. ∆r between speakers 1 and 2 is 1.0 m and λ = 2.0 m.Thus ∆r = 1

2 λ, which gives perfect destructive interference for in-phase sources. That is, the interference of thewaves from loudspeakers 1 and 2 is perfectly destructive, leaving only the contribution due to speaker 3. Thus theamplitude is a.(b) If loudspeaker 2 is moved away by one-half of a wavelength or 1.0 m, then all three waves will reach you inphase. The amplitude of the superposed waves will therefore be maximum and equal to A = 3a.(c) The maximum intensity is I CA Camax = =2 29 . The ratio of the intensity to the intensity of a single speaker is

I

I

Ca

Camax

single speaker

= =99

2

2

21.74. Model: The superposition of two slightly different frequencies gives rise to beats.Solve: The third harmonic of note A and the second harmonic of note E are

f f f f3A 1A 2E 1E440 Hz 1320 Hz 659 Hz 1318 Hz= = ( ) = = = ( ) =3 3 2 2

⇒ − = − =f f3A 2E 1320 Hz 1318 Hz 2 Hz

(b) The beat frequency between the first harmonics is

f1E − f1A = 659 Hz − 440 Hz = 219 Hz

The beat frequency between the second harmonics is

f2E − f2A = 1318 Hz − 880 Hz = 438 Hz

The beat frequency between f3A and f2E is 2 Hz. It therefore emerges that the tuner looks for a beat frequency of 2 Hz.(c) If the beat frequency is 4 Hz, then the second harmonic frequency of the E string is

f f2E 1E1320 Hz 4 Hz 1316 Hz 1316 Hz 658 Hz= − = ⇒ = ( ) =12

Note that the second harmonic frequency of the E string could also be

f f2E 1E1320 Hz 4 Hz 1324 Hz 662 Hz= + = ⇒ =

This higher frequency can be ruled out because the tuner started with low tension in the E string and we know that

v fT

f Tstring = = ⇒ ∝λµ

21.75. Model: The superposition of two slightly different frequencies creates beats.Solve: (a) The wavelength of the sound initially created by the flutist is

λ = =342 m / s

440 Hz0.77727 m

When the speed of sound inside her flute has increased due to the warming up of the air, the new frequency of theA note is

′ = =f346 m / s

0.77727 m445 Hz

Thus the flutist will hear beats at the following frequency:

′ − = − =f f 445 Hz 440 Hz 5 beats/s

Note that the wavelength of the A note is determined by the length of the flute rather than the temperature of air orthe increased sound speed.(b) The initial length of the flute is L = = ( ) =1

212λ 0.77727 m 0.3886 m . The new length to eliminate beats needs

to be

′ = ′ = ′

=

=L

v

f

λ2

1

2

1

2

346 m / s

440 Hz 0.3932 m

Thus, she will have to extend the “tuning joint” of her flute by

0.3932 m 0.3886 m 0.0046 m 4.6 mm− = =

21.76. Model: The superposition of two slightly different frequencies creates beats.Solve: Let λ1 = 780.54510 nm and λ λ2 1> . This means f f2 1< and

∆f f fc c= − = − = ×1 2

1 2

698 5 10λ λ

. Hz

⇒ =−

=× − × ×

=

−λλ2

19 6 8

1

1

1

1 780 54510 10 98 5 10 3 00 10

780 54530

( / ) ( / ) / ( . ) ( . ) / ( . )

.

∆f c m Hz m / s

nm

Assess: A small difference in wavelengths, λ λ2 1−( ) = =0.00020 nm 0.20 pm , can yield beats at a relatively

high frequency of 98.5 MHz.

21.77. Model: The frequency of the loudspeaker’s sound in the back of the pick-up truck is Doppler shifted.As the truck moves away from you, its frequency is decreased.Solve: Because you hear 8 beats per second as the truck drives away from you, the frequency of the sound fromthe speaker in the pick-up truck is f− = − =400 Hz 8 Hz 392 Hz. This frequency is

ff

v v− =+

0

1 S

⇒ + = =1 1 020408vS

343 m / s

400 Hz

392 Hz. ⇒ =vS 7.0 m / s

That is, the velocity of the source vS and hence the pick-up truck is 7.0 m/s.

21.78. Solve: (a) Yvette’s speed is the width of the room divided by time. This means

vn

tY = ( )12 λ

⇒ =n

t

v2 Y

λNote that 1

2 λ is the distance between two consecutive antinodes, and n is the number of such half wavelengths thatfill the entire width of the room.(b) Yvette observes a higher frequency f+ of the source she is moving toward and a lower frequency f− of the sourceshe is receding from. If v is the speed of sound and f is the sound wave’s frequency, we have

f fv

vf f

v

v+ −= +

= −

1 1Y Y

The expression for the beat frequency is

f f fv

vf

v

vf

v

v

v v

v

v+ −− = +

− −

= = =1 1 2 2

2Y Y Y Y Y

λ λ(c) The answers to part (a) and (b) are the same. Even though you and Yvette have different perspectives, youshould agree as to how many modulations per second she hears.

21.79. Model: A stretched string under tension supports standing waves.Solve: (a) The wave speed on a stretched string is

vT

f fT

string = = ⇒ =µ

λλ µ1

The wavelength λ cannot change if the length of the string does not change. So,

df

dTT

T T

T

Tf= ( ) = =

=−1 1 1

2

1 1

2

1 1

212

1 2

λ µ λ µ λ µ/ ⇒ =∆ ∆f

f

T

T2

(b) Since there are 5 beats per second,

∆ ∆f

f T

T= = 5 Hz

2⇒ = = = =∆T

T f

10 Hz 10 Hz

500 Hz0 020 2 0. . %

That is, an increase of 2.0% in the tension of one of the strings will cause 5 beats per second.

21.80. Model: The microphone will detect a loud sound only if there is a standing wave resonance in the tube.The sound frequency does not change, but changing the length of the tube can create a standing wave.Visualize: Please refer to Figure CP21.80.Solve: The standing wave condition is

f mv

Lm= = =280 Hz

21 2 3, , , K

where L is the total length of the tube. When the slide is extended a distance s, the tube has two straight sides, eachof length s + 80 cm, plus a semicircular bend of length 1

2 2πr( ) . The radius is r = ( ) =12 10 cm 5.0 cm . The tube’s

total length is

L s s s= +( ) + ×( ) = + = +2 2 5 0 2 21280 cm cm 175.7 cm 1.757 mπ .

A standing wave resonance will be created if

L s mv

fm m= +[ ] = =

( )=

1 757 2

20 6125. .

343 m / s

2 280 Hz

⇒ = −s m0 3063. 0.8785 meters

We can tabulate the different extensions s that correspond to standing wave modes m = 1, m = 2, m = 3, and so on.m s

123456

−0.572 m−0.266 m

0.040 m = 4.0 cm0.347 m = 34.7 cm0.653 m = 65.3 cm

1.959 m

Physically, the extension must be greater than 0 cm and less than 80 cm. Thus, the three slide extensions that createa standing wave resonance at 280 Hz are 4.0 cm, 34.7 cm, and 65.3 cm.

21.81. Model: The stretched wire is vibrating at its second harmonic frequency.Visualize: Let l be the full length of the wire, and L be the vibrating length of the wire. That is, L l= ( )1

2 .

Solve: The wave speed on a stretched wire is

vT

fwires= =

µλ

The frequency f = 100 Hz and the wavelength λ = 12 l because it is a second harmonic wave. The tension Ts =

(1.25 kg)g because the hanging mass is in static equilibrium and µ = 1.00 × 10−3 kg/m. Substituting in these values,

1.25 kg

kg / m100 Hz 100 Hz

kg / m

1.25 kg2.00 m s1 2( )

×( ) = ( ) ⇒ = ( )

×( )( ) = ( )−

−− −g l

gl

l1 00 10 2 2

1 00 103

22 3

2

.

.

To find l we can use the equation for the time period of a simple pendulum:

Tl

gl

Tg g g= ⇒ = = ( ) = ( )2

4 4

2

2

2

2ππ π

314 s / 1000.250 s2

Substituting this expression for l into the equation for g, we get

g g g g

g g g

= ( )( ) ⇒ ( ) − =

⇒ ( ) −[ ] = ⇒ =

− − −

−

2 000 0

1 0

2 2 2 1 2

1

. m s 0.025 s 0.125 m s

0.125 m s 8.00 m / s

1 2 2

2 2

Assess: A value of 8.0 m/s2 is reasonable for the information given in the problem.

21.82. Solve: (a) Because the frequency of the standing wave on the copper wire is the same as the frequencyon the aluminum wire,

f fCu Al= ⇒ =v vCu

Cu

Al

Cuλ λ

Let nCu be the number of half-wavelength antinodes on the copper wire and nAl be the number of half-wavelengthantinodes on the aluminum wire. Thus,

nnCu

CuCu

Cu20.22 m

0.44 mλ λ

= ⇒ = nnAl

AlAl

Al20.60 m

1.20 mλ λ

= ⇒ =

⇒ ( ) = ( )v

n

v

nCu

Cu

Al

Al0.44 m 1.20 m⇒ =

n

n

v

vCu

Al

Al

Cu

0.44 m

1.20 m

We can find vCu and vAl by using the following equations for a stretched wire:

vT

vT

CuCu

AlAl

= =µ µ

The linear densities are calculated as follows:

µ ρ ρ π

π

CuCu Cu Cu Cu

3 3

0.22 m 0.22 m 0.22 m0.22 m

8920 kg / m m kg / m

= = =

( )

= ( ) ×( ) = ×− −

m Vr 2

4 2 35 0 10 7 006 10. .

µ ρ ρ π

π

AlAl Al Al Al

3

0.60 m 0.60 m 0.60 m0.60 m

2700 kg / m m kg / m

= = =

( )

= ( ) ×( ) = ×− −

m Vr 2

3 4 2 35 0 10 2 121 10. .

⇒ =×

=−vCu 3 3

20 N

7.006 10 kg / m53.43 m / s vAl

20 N

kg / m97.12 m / s=

×=−2 121 10 3 3.

Going back to the n nCu Al equation, we have

n

nCu

Al

97.12 m / s

53.43 m / s

0.44 m

1.20 m=

= =0 666

2

3. ⇒ = =n nCu Al and 2 3

Substituting into the expressions for wavelength and frequency,

λCuCu

0.44 m 0.44 m

20.22 m= = =

n⇒ = = =f

vCu

Cu

Cu

53.43 m / s

0.22 m243 Hz

λ

λAlAl

1.20 m 1.20 m

30.40 m= = =

n⇒ = = =f

vAl

Al

Al

97.12 m / s

0.40 m243 Hz

λ

(b) At this frequency of 243 Hz, there are 3 antinodes on the aluminum wire.

21.83. Model: The frequency is Doppler shifted to higher values for a detector moving toward the source. Thefrequency is also shifted to higher values for a source moving towards the detector.Visualize:

Solve: (a) We will derive the formula in two steps. First, the object acts like a moving detector and “observes” afrequency that is given by ′ = +( )f f v v0 0 01 . Second, as this moving object reflects (or acts as a “source” of

ultrasound waves), the frequency fecho as observed by the original source So is f f v vecho = ′ −( )−0 0

11 . Combining

these two equations gives

ff

v v

f v v

v v

v v

v vfecho = ′

−=

+( )−

= +−

0

0

0 0

0

0

001

1

1

(b) If v0 << v, then

f fv

v

v

vecho = +

−

−

00 0

1

1 1 = +

+ +

f

v

v

v

v00 01 1 K = + +

f

v

v001

2K

⇒ = − ≈f f fv

vfbeat echo 0

00

2

(c) Using part (b) for the beat frequency,

65 Hz 1540 m / s

Hz=

×( )22 40 100 6v. ⇒ =v0 2.09 cm / s

(d) Assuming the heart rate is 90 beats per minute the angular frequency is

ω π π= = ( ) =2 2f 1.5 beats / s 9.425 rad / s

Using v v A0 = =max ω ,

Av= = =0

ω2.09 cm / s

9.425 rad / s2.2 mm

21.84. Solve: (a) The wavelengths of the standing wave modes are

λm

L

m= 2

m = 1, 2, 3, ….

⇒ = ( ) = = ( ) = = ( ) =λ λ λ1 2 3

2

1

2

2

2

3

10.0 m20.0 m

10.0 m10.0 m

10.0 m6.67 m

The depth of the pool is 5.0 m. Clearly the standing waves with λ2 and λ3 are “deep water waves” because the 20 mdepth is larger than one-quarter of the wavelength. The wave with λ1 barely qualifies to be a deep water standingwave.

(b) The wave speed for the first standing wave mode is

vg

11

2 2= =

( )( )=λ

π π9.8 m / s 20.0 m

5.59 m / s2

Likewise, v v2 3= =3.95 m / s and 3.22 m / s .(c) We have

vg

fmm m= =λ

πλ

2 ⇒ = =f

g mg

Lmm2 4πλ π

Please note that m is the mode and not the mass.(d) The period of oscillation for the first standing wave mode is calculated as follows

f T1 1

1

4=

( )( )( )

= ⇒ =9.8 m / s

10.0 m0.279 Hz 3.58 s

2

π

Likewise, 2.53 s and 2.07 sT T2 3= = .

21.85. Model: The overlap of the waves causes interference.Visualize: Please refer to Figure CP21.85.Solve: (a) The waves traveling to the left are

D ax t

T1 2= − −

sin πλ

D ax L t

T2 202= −−( ) −

+

sin π

λφ

The phase difference between the waves on the left side of the antenna is thus

∆φ φ φ πλ

φ πλL = − = −

−( ) −

+ − − −

2 1 202 2

x L t

T

x t

T= +2 20π

λφL

On the right side of the antennas, where x x L1 2= + , the two waves are

D ax t

T1 2= −

sin πλ

D ax L t

T2 202=−( ) −

+

sin π

λφ

Thus, the phase difference between the waves on the right is

∆φ φ φ πλ

φ πλR = − =

−( ) −

+ − −

2 1 202 2

x L t

T

x t

T= − +2

20

πλ

φL

We want to have destructive interference on the country side or on the left and constructive interference on theright. This requires

∆φ φ πλ

πL = + = +( )2012

22

Lm ∆φ φ π

λπR = − =20

22

Ln

These are two simultaneous equations, and we can satisfy them both if L and φ20 are properly chosen. Subtractingthe second equation from the first to eliminate φ20 ,

42 1

2

πλ

πLm n= + −( ) ⇒ = + −( )L m n1

2 2

λ

The smallest value of L that works is for n = m, in which case L = 14 λ .

(b) From the ∆φR equation,

φ πλ

φπ λ

λφ π π20 20

14

20

2 2

22− = − ( ) = −

=L

n ⇒ = +φ π π20 22 n rad

Adding integer multiples of 2π to the phase constant doesn’t really change the wave, so the physically significantphase constant is for n = 0, namely φ π20

12= rad.

(c) We have φ π π2012

14 2= = ( ) rad . If the wave from antenna 2 was delayed by one full period T, it would shift the

wave by one full cycle. We would describe this by a phase constant of 2π rad. So a phase constant of 14 2π( ) rad can

be achieved by delaying the wave by ∆t T= 14 .

(d) A wave with frequency f = = ×1000 kHz 1.00 10 Hz6 has a period T = × =−1 00 10 6. s 1.00 sµ and wavelengthλ = =c f/ 300 m . So this broadcast scheme will work if the antennas are spaced L = 75 m apart and if the broadcastfrom antenna 2 is delayed by ∆t = 0.25 µs = 250 ns.