20.1. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the right every second.Visualize: Please refer to Figure Ex20.1. The snapshot graph shows the wave at all points on the x-axis at t = 0 s.You can see that nothing is happening at x = 6 m at this instant of time because the wave has not yet reached thispoint. The leading edge of the wave is still 1 m away from x = 6 m. Because the wave is traveling at 1 m/s, it willtake 1 s for the leading edge to reach x = 6 m. Thus, the history graph for x = 6 m will be zero until t = 1 s. The firstpart of the wave causes an upward displacement of the medium. The rising portion of the wave is 2 m wide, so itwill take 2 s to pass the x = 6 m point. The constant part of the wave, whose width is 2 m, will take 2 seconds topass x = 6 m and during this time the displacement of the medium will be a constant (∆y = 1 cm). The trailing edgeof the pulse arrives at t = 5 s at x = 6 m. The displacement now becomes zero and stays zero for all later times.

20.2. Model: This is a wave traveling at constant speed. The pulse moves 1 m to the left every second.Visualize: Please refer to Figure Ex20.2. This snapshot graph shows the wave at all points on the x-axis at t = 2 s.You can see that the leading edge of the wave at t = 2 s is precisely at x = 0 m. That is, in the first 2 seconds, thedisplacement is zero at x = 0 m. The first part of the wave causes a downward displacement of the medium, soimmediately after t = 2 s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is3 m wide and takes 3 s to pass x = 0 m. The positive portion begins to pass through x = 0 m at t = 5 s and until t = 8 sthe displacement of the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zerofor all later times.

20.3. Model: This is a wave traveling at constant speed to the right at 1 m/s.Visualize: Please refer to Figure Ex20.3. This is the history graph of a wave at x = 0 m. The graph shows that thex = 0 m point of the medium first sees the negative portion of the pulse wave at t = 1 s. Thus, the snapshot graph ofthis wave at t = 1 s must have the leading negative portion of the wave at x = 0 m.

20.4. Model: This is a wave traveling at constant speed to the left at 1 m/s.Visualize: Please refer to Figure Ex20.4. This is the history graph of a wave at x = 2 m. Because the wave ismoving to the left at 1 m/s, the wave passes the x = 2 m position a distance of 1 m in 1 s. Because the flat part ofthe history graph takes 2 s to pass the x = 2 m position, its width is 2 m. Similarly, the width of the linearlyincreasing part of the history graph is 2 m. The center of the flat part of the history graph corresponds to both t = 0 sand x = 2 m.

20.5. Visualize:

Figure Ex20.5 shows a snapshot graph at t = 0 s of a longitudinal wave. This diagram shows a row of particles withan inter-particle separation of 1.0 cm at equilibrium. Because the longitudinal wave has a positive amplitude of 0.5 cmbetween x = 3 cm and x = 8 cm, the particles at x = 3, 4, 5, 6, 7 and 8 cm are displaced to the right by 0.5 cm.

20.6. Visualize:

We first draw the particles of the medium in the equilibrium positions, with an inter-particle spacing of 1.0 cm. Justunderneath, the positions of the particles as a longitudinal wave is passing through are shown at time t = 0 s. It isclear that relative to the equilibrium the particle positions are displaced negatively on the left side and positively onthe right side. For example, the particles at x = 0 cm and x = 1 cm are at equilibrium, the particle at x = 2 cm isdisplaced left by 0.5 cm, the particle at x = 3 cm is displaced left by 1.0 cm, the particle at x = 4 cm is displaced leftby 0.5 cm, and the particle at x = 5 cm is undisplaced. The behavior of particles for x > 5 cm is opposite of that forx < 5 cm.

20.7. Model: The wave is a traveling wave on a stretched string.Solve: The wave speed on a stretched string with linear density µ is v Tstring S= µ . The wave speed if the

tension is doubled will be

′ = = = ( ) =vT

vstringS

string 200 m / s 283 m / s2

2 2µ

20.8. Model: The wave is a traveling wave on a stretched string.Solve: The wave speed on a stretched string with linear density µ is

vT

stringS 150 m / s

75 N= ⇒ =µ µ

⇒ = × −µ 3 333 10 3. kg / m

For a wave speed of 180 m/s, the required tension will be

T vS string kg / m 180 m / s 108 N= = ×( )( ) =−µ 2 3 23 333 10.

20.9. Model: The wave is a traveling wave on a stretched string.Solve: The wave speed on a string whose radius is R, length is L, and mass density is ρ is v Tstring S= µ with

µ ρ ρπ ρπ= = = =m

L

V

L

R L

LR

22

If the string radius doubles, then

′ =( )

= = =vT

R

vstring

S string

2 2

280 m / s

2140 m / s

ρπ 2

20.10. Solve: (a) The wave number is

k = = =2 2πλ

π2.0 m

3.14 rad / m

(b) The wave speed is

v f= =

=

=λ λ ω

π π22 0

30

29 55( . ) . m

rad / s m / s

20.11. Solve: (a) The wavelength is

λ π π= = =2 2

k 1.5 rad / m4.19 m

(b) The frequency is

fv= = =λ

200 m / s

4.19 m47.7 Hz

20.12. Model: The wave is a traveling wave.Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 3.5 cm, k = 2.7 rad/m, ω = 124 rad/s,and φ0 = 0 rad . The frequency is

f = = =ωπ π2 2

124 rad / s19.7 Hz

(b) The wavelength is

λ π π= = =2 2

k 2.7 rad / m2.33 m

(c) The wave speed v f= =λ 45.9 m / s .

20.13. Model: The wave is a traveling wave.Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 5.2 cm, k = 5.5 rad/m, ω = 72rad/s, and φ0 = 0 rad . The frequency is

f = = =ωπ π2 2

72 rad / s11.5 Hz

(b) The wavelength is

λ π π= = =2 2

k 5.5 rad / m1.14 m

(c) The wave speed v f= =λ 13.1 m / s .

20.14. Model: The wave is a traveling wave.Solve:

(a) The wave equation at t = 0 s is D = (2.0 cm) sin(2πx). This graph is shown as a dotted line. The equation at

t = 18 s is D = (2.0 cm) sin(2πx –π/2 rad). Its graph is a dashed line.

(b) The wave speed is

vk

= = =ω ππ

4

2

rad / s

rad / m2.0 m / s

In agreement, you can see that the peak of the graph has moved forward 0.5 m in 14 s.

20.15. Visualize: Please refer to Figure Ex20.15.Solve: The amplitude of the wave is the maximum displacement which is 4.0 cm. The wavelength is the distancebetween two consecutive peaks, which gives λ = − =14 m 2 m 12 m . The frequency of the wave is

fv= = =λ

24 m / s

12 m2.0 Hz

20.16. Visualize: Please refer to Figure Ex20.16.Solve: The amplitude of the wave is the maximum displacement which is 6.0 cm. The period of the wave is0.60 s, so the frequency f T= = =1 1 0 60. s 1.67 Hz . The wavelength is

λ = = =v

f

2 m / s

1.667 Hz1.2 m

20.17. Solve: According to Equation 20.28, the phase difference between two points on a wave is

∆ ∆φ φ φ πλ

πλ

= − = = −( )2 1 2 122r

r r ⇒ −( ) = −( )32π πλ

rad 0 rad 80 cm 20 cm ⇒ =λ 40 cm

20.18. Solve: According to Equation 20.28, the phase difference between two points on a wave is

∆φ φ φ πλ

= − = −( )2 1 2 1

2r r

If φ π1 = rad at r1 = 4.0 m, we can determine φ2 at any r value at the same instant using this equation. At r2 = 3.5 m,

φ φ πλ

π π π2 1 2 1

2 2

2= + −( ) = + −( ) =r r rad

2.0 m3.5 m 4.0 m rad

At r2 = 4.5 m, φ π232= rad.

20.19. Visualize:

Solve: For a sinusoidal wave, the phase difference between two points on the wave is given by Equation 20.28:

∆∆

φ φ φ πλ

πλ

λ πφ

= − = −( ) = −( ) ⇒ = ( )2 1 2 1

2 240 30

210r r m m m

∆φ π= 2 for two points on adjacent wavefronts and ∆φ π= 4 for two points separated by 2λ. Thus, λ = 10 mwhen ∆φ π= 2 , and λ = 5 m when ∆φ π= 4 . The crests corresponding to these two wavelengths are shown in thefigure. One can see that a crest of the wave passes the 40 m-listener and the 30 m-listener simultaneously. Thelowest two possible frequencies will occur for the largest two possible wavelengths, which are 10 m and 5 m. Thus,the lowest frequency is

fv

1 = = =λ

340 m / s

10 m34 Hz

The next highest frequency is f2 = 68 Hz .

20.20. Visualize:

Solve: (a) Because the same wavefront simultaneously reaches listeners at x = −7.0 m and x = +3.0 m,

∆φ πλ

= = −( ) ⇒ =02

2 1 2 1 rad r r r r

Thus, the source is at x = −2 m, so that it is equidistant from the two listeners.

(b) The third person is also 5 m away from the source. Her y-coordinate is thus y = 5 m 2 m 4.58 m( ) − ( ) =2 2 .

20.21. Model: Assume the speed of sound is the 20° C value 343 m/s.Visualize:

Solve: The time elapsed from the stage to you in the middle row is the distance divided by the speed of sound:13 m

343 m / s0.0379 s=

The time elapsed for the same sound to get reflected from the back wall and reach your ear is26 m 13 m

343 m / s0.114 s

+ =

Difference in the two times is thus 0.114 s – 0.038 s = 0.076 s

20.22. Solve: Two pulses of sound are detected because one pulse travels through the metal to the microphonewhile the other travels through the air to the microphone. The time interval for the sound pulse traveling throughthe air is

∆ ∆t

x

vairair

4.0 m

343 m / s0.01166 s 11.66 ms= = = =

Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphonebefore the pulse traveling through the air. Because the pulses are separated in time by 11.0 ms, the pulse travelingthrough the metal takes ∆tmetal = 0.66 ms to travel the 4.0 m to the microphone. Thus, the speed of sound in the metal is

vx

tmetalmetal

4.0 m

0.00066 s6060 m / s= = =∆

∆

20.23. Visualize:

Solve: The explosive’s sound travels down the lake and into the granite, and then it is reflected by the oil surface.The echo time is thus equal to

t t t t t

d dd

echo water down granite down granite up water up

granite granitegranite0.94 s

500 m

1480 m / s 6000 m / s 6000 m / s

500 m

1480 m / s793 m

= + + +

= + + + ⇒ =

20.24. Solve: (a) In aluminum, the speed of sound is 6420 m/s. The wavelength is thus equal to

λ = =×

= × =−v

f

6420 m / s

2.0 10 Hz m 3.21 mm6 3 21 10 3.

(b) The speed of an electromagnetic wave is c. The frequency would be

fc= = ×

×= ×−λ

3 0 10

3 21 109 35 10

8

310.

..

m / s

m Hz

20.25. Solve: (a) The frequency is

fv= = =air 343 m / s

0.20 m1715 Hz

λ(b) The frequency is

fc= = × = × =λ

3 0 101 50 10

89.

. m / s

0.20 m Hz 1.50 GHz

(c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be

λ = =×

= × =−v

fwater

9

1480 m / s

1.50 10 Hz m 987 nm9 87 10 7.

20.26. Model: Light is an electromagnetic wave that travels with a speed of 3 × 108 m/s.Solve: (a) The frequency of the blue light is

fc

blue

m / s

m Hz= = ×

×= ×−λ

3 0 10

450 106 67 10

8

914.

.

(b) The frequency of the red light is

fred 9

m / s

650 10 m Hz= ×

×= ×−

3 0 104 62 10

814.

.

(c) Using Equation 20.30 to calculate the index of refraction,

λ λmaterial

vacuum=n

⇒ = = =nλλ

vacuum

material

650 nm

450 nm1 44.

20.27. Model: Microwaves are electromagnetic waves that travel with a speed of 3 × 108 m/s.Solve: (a) The frequency of the microwave is

fc

microwaves

m / s

m Hz 10 GHz= = ×

×= × =−λ

3 0 10

3 0 101 0 10

8

210.

..

(b) The refractive index of air is 1.0003, so the speed of microwaves in air is v cair = 1 00. ≈ c. The time for themicrowave signal to travel is

tv

= = ××( ) =50 km m

3.0 10 m 1.000.167 ms

air8

50 103

Assess: A small time of 0.167 ms for the microwaves to cover a distance of 50 km shows that the electromagneticwaves travel very fast.

20.28. Model: Radio waves are electromagnetic waves that travel with speed c.Solve: (a) The wavelength is

λ = = × =c

f

3 0 108. m / s

101.3 MHz2.96 m

(b) The speed of sound in air at 20°C is 343 m/s. The frequency is

fv= = =sound 343 m / s

2.96 m116 Hz

λ

20.29. Model: Light is an electromagnetic wave.Solve: (a) The time light takes is

tv c n

= = × = ××( ) = ×

− −−3.0 mm m m

m / s 1.50 s

glass

3 0 10 3 0 10

3 0 101 50 10

3 3

8

11. .

..

(b) The thickness of water is

d v tc

nt= = = × ×( ) =−

waterwater

m / s s 3.38 mm

3 0 10

1 331 50 10

811.

..

20.30. Solve: (a) The speed of light in a material is given by Equation 20.29:

nc

vv

c

n= ⇒ =

matmat

The refractive index is

n v c= ⇒ = = ×( ) = ×λλ

λλ

vac

matsolid

solid

vac

m / s420 nm

670 nm m / s3 0 10 1 88 108 8. .

(b) The frequency is

fv= = × = ×solid

solid

m / s

420 nm Hz

λ1 88 10

4 48 108

14..

20.31. Model: The laser beam is an electromagnetic wave that travels with the speed of light.Solve: The speed of light in the liquid is

vliquid 9

m

1.38 10 s m / s= ×

×= ×

−

−

30 102 174 10

28.

The liquid’s index of refraction is

nc

v= = ×

×=

liquid

3 0 10

2 174 101 38

8

8

.

..

Thus the wavelength of the laser beam in the liquid is

λ λliquid

vac nm459 nm= = =

n

633

1 38.

20.32. Solve: The energy delivered to the eardrum in time t is E = Pt, where P is the power of the wave. Theintensity of the wave is I P a= where a is the area of the ear drum. Putting the above information together, we have

E Pt Ia t I r t= = ( ) = = ×( ) ×( ) ( ) = ×− − −π π2 3 3 2 62 0 10 3 0 10 3 39 10. . . W / m m 60 s J2

20.33. Solve: The energy delivered to an area a in time t is E = Pt, where the power P is related to the intensityI as I P a= . Thus, the energy received by your back is

E = Pt = Iat = (0.80)(1400 W/m2)(0.30 × 0.50 m2)(3600 s) = 6.05 × 105 J

20.34. Solve: If a source of spherical waves radiates uniformly in all directions, the ratio of the intensities atdistances r1 and r2 is

I

I

r

r1

2

= 22

12 ⇒ =

= × −I

I50 m

2 m

2 m

50 m

231 6 10.

⇒ = ×( ) = ( ) ×( ) = ×− − −I I50 m 2 m2 22.0 W / m W / m1 6 10 1 6 10 3 2 103 3 3. . .

Assess: The power generated by the sound source is P = I2m [4π(2 m)2] = (2.0 W/m2)(50.27) = 101 W. This is asignificant amount of power.

20.35. Solve: (a) The intensity of a uniform spherical source of power Psource a distance r away is I P r= source 4 2π .Thus, the intensity at the position of the microphone is

I50 m235 W

50 m W / m=

( )= × −

41 11 102

3

π.

(b) The sound energy impinging on the microphone per second is

P Ia= = ×( ) ×( ) = × = ×− − − −1.11 10 W / m m W 1.11 10 J / s3 2 2 71 0 10 1 11 104 7. .

⇒Energy impinging on the microphone in 1 second =1.11 10 J.7× −

20.36. Model: The frequency of the opera singer’s note is altered by the Doppler effect.Solve: (a) Using 90 km/hr = 25 m/s, the frequency as her convertible approaches the stationary person is

ff

v v+ =−

=−

=0

1 1647

S

600 Hz25 m / s

343 m / s

Hz

(b) The frequency as her convertible recedes from the stationary person is

ff

v v− =+

=+

=0

1 1S

600 Hz25 m / s

343 m / s

559 Hz

20.37. Model: The mother hawk’s frequency is altered by the Doppler effect.Solve: The frequency is f+ as the hawk approaches you is

ff

v v v+ =−

⇒ =−

⇒0

1 1S S900 Hz

800 Hz

343 m / s

vS 38.1 m / s=

Assess: The mother hawk’s speed of 38.1 m/s ≈ 80 mph is reasonable.

20.38. Model: Sound frequency is altered by the Doppler effect. The frequency increases for an observerapproaching the source and decreases for an observer receding from a source.Solve: You need to ride your bicycle away from your friend to lower the frequency of the whistle. The minimumspeed you need to travel is calculated as follows:

fv

vf

v− = −

⇒ = −

( ) ⇒1 10

0020 kHz

343 m / s21 kHz v0 = 16.3 m / s

Assess: A speed of 16.3 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on abicycle.

20.39. Model: Your friend’s frequency is altered by the Doppler effect. The frequency of your friend’s noteincreases as he races towards you (moving source and a stationary observer). The frequency of your note for yourapproaching friend is also higher (stationary source and a moving observer).Solve: (a) The frequency of your friend’s note as heard by you is

ffv

v

+ =−

=−

=0

1 1S

400 Hz25.0 m / s

340 m / s

432 Hz

(b) The frequency heard by your friend of your note is

f fv

v+ = +

= ( ) +

=001 1400 Hz

25.0 m / s

340 m / s429 Hz

20.40. Visualize: The function D(x, t) represents a pulse that travels in the positive x-direction without changingshape.Solve: (a)

(b) The leading edge of the pulse moves forward 3 m each second. Thus, the wave speed is 3.0 m/s.(c) | |x t− 3 is a function of the form D(x – vt), so the pulse moves to the right at v = 3 m/s.

20.41. Visualize: The function D(x, t) represents a pulse that travels in the negative x-direction without changingshape.Solve: (a)

(b) We can see from the graphs that the wave pulse is moving to the left. From the graphs, the pulse moves ∆x =−2 m during each ∆t = 1 s. The wave’s velocity (not speed) is v x t= = −( ) ( ) = −∆ ∆ 2 m 1 s 2 m / s, and the wave’sspeed is 2 m/s.(c) x t+ 2 is a function of the form D(x + vt), so the pulse moves to the left at v = 2 m/s.

20.42. Visualize: Please refer to Figure P20.42.Solve: (a) We can see from the graph that the wavelength is λ = 2.0 m. We are given that the wave’s frequency isf = 5.0 Hz. Thus, the wave speed is v = λf = 10.0 m/s.(b) The snapshot graph was made at t = 0 s. Reading the graph at x = 0 m, we see that the displacement is

D x t D A= =( ) = ( ) = =0 0 0 0 m s m s 0.5 mm 12, ,

Thus

D A A0 0 12 0 m s, sin( ) = = φ ⇒ = ( ) =−φ π π

01 1

2 6

5

6sin rad or rad

Note that the value of D(0 m, 0 s) alone gives two possible values of the phase constant. One of the values willcause the displacement to start at 0.5 mm and increase with distance—as the graph shows—while the other willcause the displacement to start at 0.5 mm but decrease with distance. Which is which? The wave equation for t =0 s is

D x t Ax

, sin =( ) = +

0

20

πλ

φ

If x is a point just to the right of the origin and is very small, the angle 2 0π λ φx +( ) is just slightly bigger than the

angle φ0 . Now sin31° > sin30°, but sin151° < sin150°, so the value φ π016= rad is the phase constant for which the

displacement increases as x increases.(c) The equation for a sinusoidal traveling wave can be written as

D x t Ax

ft Ax

ft, sin sin( ) = − +

= −

+

22 20 0

πλ

π φ πλ

φ

Substituting in the values found above,

D x tx

t, sin 1.0 mm2.0 m

5.0 s( ) = ( ) − ( )

+

−26

1π π

20.43. Visualize: Please refer to Figure P20.43.Solve: (a) We see from the history graph that the period T = 0.20 s and the wave speed v = 4.0 m/s. Thus, thewavelength is

λ = = = ( )( ) =v

fvT 4.0 m / s 0.20 s 0.80 m

(b) The phase constant φ0 is obtained as follows:

D A0 m, 0 s( ) = sinφ0 ⇒ − = ( ) ⇒ = −2 10 0 mm 2 mm sin sinφ φ ⇒ = −φ π012 rad

(c) The displacement equation for the wave is

D x t Ax

ftx t

x t, sin . sin. .

sin . mm m s

2.0 mm( ) = − +

= ( ) − −

= ( ) − −( )2

2 2 02

0 80

2

0 20 22 5 100

12

πλ

π φ π π π π π π

where x and t are in m and s, respectively.

20.44. Solve: The time for the sound wave to travel down the tube and back is t = 440 µs since 1 division isequal to 100 µs. So, the speed of the sound wave in the liquid is

v = × =2 25

440

cm

s1140 m / s

µ

20.45. Model: The wave pulse is a traveling wave on a stretched string.Solve: The wave speed on a stretched string with linear density µ is

vT T

m/L

LT

m mmstring

S S S3

2.0 m

50 10 s

2.0 m 20 N0.025 kg 25 g= = = ⇒

×= ( )( ) ⇒ = =−µ

20.46. Model: The wave pulse is a traveling wave on a stretched string.Visualize:

Solve: The pulse travels along the rope with speed

vx

t

T= = = =∆∆

3.0 m

0.150 s20 m / s S

µ

where we used the relationship between the wave’s speed, the rope’s tension, and the linear density µ = m Lrope .

We can find the tension in the rope from a Newtonian analysis. Bob experiences horizontal forces rT and

rfk . He is

being pulled at constant speed, so he is in dynamic equilibrium with rFnet N= 0 . Thus T f n= =k kµ . From the

vertical forces we see that n w m g= = Bob . So we find the rope’s tension to be

T m gS k Bob260 kg 9.8 m / s 117.6 N= = ( )( )( ) =µ 0 20.

Using this in the velocity equation, we can find the rope’s mass:

vT

m L= S

rope

⇒ = = ( )( )( )

= =mLT

vropeS 3.0 m 117.6 N

20 m / s0.882 kg 882 g2 2

20.47. Model: The wave pulse is a traveling wave on a stretched string.Visualize: Please refer to Figure P20.47.Solve: While the tension TS is the same in both the strings, the wave speeds in the two strings are not. We have

vT

11

= S

µ and v

T2

2

= S

µ⇒ = =v v T1

21 2

22µ µ S

Because v L t v L t1 1 1 2 2 2= = and , and because the pulses are to reach the ends of the string simultaneously theabove equation can be simplified to

L

t

L

t12

12

22

22

µ µ= ⇒ = = =L

L1

2

2

1

2µµ

4.0 g / m

2.0 g / m⇒ =L L1 22

Since L1 + L2 = 4 m,

2 2 2 2L L L+ = ⇒ =4 m 1.66 m and L1 2= ( ) =1.66 m 2.34 m

20.48. Solve: ∆t is the time the sound wave takes to travel down to the bottom of the ocean and then up to theocean surface. The depth of the ocean is

2d v t d t= ( ) ⇒ = ( )sound in water 750 m / s∆ ∆

Using this relation and the data from Figure P20.48, we can generate the following table for the ocean depth (d ) atvarious positions (x) of the ship.

x (km) ∆t (s) d (km)0

2040455060

644842

4.53.03.06.03.01.5

20.49. Model: The wave is a sinusoidal traveling wave on a stretched string.Solve: The variables for the first 1.0 m-length of the string will be denoted by the subscript 1. The variables forthe rest of the string will be identified with the subscript 2. The tension is the same throughout the string, so thewave speeds in the string are

vT

11

= S

µ v

T2

2

= S

µBecause µ µ2

14 1= , the wave speeds are

vT

v21

1

42 2= = = ( ) =S 2.0 m / s 4.0 m / s

µThe wavelengths are

λ11= = =v

f

2.0 m / s

5.0 Hz0.40 m λ2

2= = =v

f

4.0 m / s

5.0 Hz0.80 m

20.50. Model: Assume a room temperature of 20°C.Visualize:

Solve: The distance between the source and the left ear (EL) is

d x yL 0.1 m 5.0 m cos45 5.0 m sin45 0.1 m m= + +( ) = ( ) °[ ] + ( ) ° +[ ] =2 2 2 25 0712.

Similarly dR m= 4 9298. . Thus,

d d dL R 0.1414 m− = =∆For the sound wave with a speed of 343 m/s, the difference in arrival times at your left and right ears is

∆ ∆t

d= = =343 m / s

0.1414 m

343 m / ss412 µ

20.51. Model: The temperature is 20°C for both air and water.Solve: For the sound speed of vair = 343 m/s, the wavelength of the sound wave in air is

λair

343 m / s

256 Hz1.340 m= =

On entering water the frequency does not change, so fwater = fair and fwater/fair = 1. The wave speed in water is vwater =1480 m/s, so

v

vwater

air

1480 m / s

343 m / s= = 4 31.

Finally, the wavelength in water is

λ λλwater

water

water

water

air

1480 m / s

256 Hz5.781 m

5.781 m

1.340 m= = = ⇒ = =v

f4 31.

Assess: This last result is expected because v = fλ and the frequency remains unchanged as the wave enters fromair into water.

20.52. Solve: The difference in the arrival times for the P and S waves is

∆t t td

v

d

v= − = −S P

S P

⇒ = −

120 s4500 m / s 8000 m / s

d1 1 ⇒ = ×d 1 23 106. m = 1230 km

Assess: d is approximately one-fifth of the radius of the earth and is reasonable.

20.53. Solve: The time for the wave to travel from California to the South Pacific is

td

v= = × =8.00 10 m

1480 m / s5405.4 s

6

A time decrease to 5404.4 s implies the speed has changed to vd

t= = 1480 28. m / s.

Since the 4.0 m/s increase in velocity is due to an increase of 1°C, an increase of 0.28 m/s occurs due to atemperature increase of

1°

( ) = °C

4.0 m / s0.28 m / s 0.07 C

Thus, a temperature increase of approximately 0.07°C can be detected by the researchers.

20.54. Model: This is a sinusoidal wave.Solve: (a) The equation is of the form D y t A ky t, sin ( ) = + +( )ω φ0 , so the wave is traveling along the y-axis.

Because it is +ωt rather than −ωt the wave is traveling in the negative y-direction.(b) Sound is a longitudinal wave, meaning that the medium is displaced parallel to the direction of travel. So the airmolecules are oscillating back and forth along the y-axis.(c) The wave number is k = −8.96 m 1 , so the wavelength is

λ π π= = −

2 2

8 96 1k . m= 0.70 m

The angular frequency is ω = −3140 s 1, so the wave’s frequency is

f = =−ω

π π2

3140

2

1 s= 500 Hz

Thus, the wave speed v = λf = (0.70 m)(500 Hz) = 350 m/s. The period T = 1/f = 0.0020 s = 2.0 ms.(d) The interval t = 0 s to t = 4 ms is exactly 2 cycles of the wave. The initial value at y = 1 m is

D y t= =( ) = ( ) +( )1 m, 0 s 0.02 mm sin .8 96 14 π = −0.0063 mm

Assess: The wave is a sound wave with speed v = 350 m/s. This is greater than the room-temperature speed of343 m/s, so the air temperature must be greater than 20°.

20.55. Model: This is a sinusoidal wave.Solve: (a) The displacement of a wave traveling in the positive x-direction with wave speed v must be of the formD(x, t) = D(x− vt). Since the variables x and t in the given wave equation appear together as x + vt, the wave istraveling toward the left, that is, in the −x direction.(b) The speed of the wave is

vk

= = =ω ππ2 0 20

2

. s

rad 2.4 m12.0 m / s

The frequency is

f = =ωπ

ππ2

2 rad 0.20 s

2= 5.0 Hz

The wave number is

k = 2π rad

2.4 m= 2.62 rad / m

(c) The displacement is

D 0.20 m, 0.50 s 3.0 cm0.20 m

2.4 m

0.50 s

0.20 s cm( ) = ( ) + +

= −sin .2 1 1 50π

20.56. Model: This is a sinusoidal wave traveling on a stretched string in the +x direction.Solve: (a) From the displacement equation of the wave, A = 2.0 cm, k = 12.57 rad/m, and ω = 638 rad/s. Usingthe equation for the wave speed in a stretched string,

vT

T vkstring

SS string

32

kg / m638 rad / s

12.57 rad / m= 12.6 N= ⇒ = =

= ×( )

−

µµ µ ω2

235 0 10.

(b) The maximum displacement is the amplitude D x tmax , 2.0 cm( ) = .(c) From Equation 20.17,

v Ay max 638 rad / s m 12.8 m / s= = ( ) ×( ) =−ω 2 0 10 2.

20.57. Solve: The wave number and frequency are calculated as follows:

k vk= = = ⇒ = = ( )( ) =2 2

0 504 4 0 4 16

πλ

π π ω π π rad

m rad / m m / s rad / m rad / s

..

Thus, the displacement equation for the wave is

D y t y t, sin 5.0 cm rad / m rad / s( ) = ( ) ( ) + ( )[ ]4 16π πAssess: The positive sign in the sine function’s argument indicates motion along the −y direction.

20.58. Solve: The angular frequency and wave number are calculated as follows:

ω π π π ω π π= = ( ) = ⇒ = = =2 2 400400

400f k

v200 Hz rad / s

rad / s

m / s rad / m

The displacement equation for the wave is

D x t x t, sin 0.010 mm rad / m rad / s rad( ) = ( ) ( ) − ( ) +[ ]π π π400 12

Assess: Note the negative sign with ωt in the sine function’s argument. This indicates motion along the +x direction.

20.59. Solve: The graphs in Figure 20.11 show that the function f(x – d) looks exactly like the function f(x)except that it is shifted to the right by distance d. Similarly, the function f(x + d) looks exactly like the function f(x)except that it is shifted to the left by distance d. Whatever value f(x) has at x = 0, the function f(x + d) has the samevalue at x = – d where x + d = 0. If d = vt, then a graph of the function at t = 0 is shifted leftward by the amount vt at alater time t. In other words, the graph of the function is moving to the left at speed v. In particular, if the function is D,a function describing the displacement of a medium, then D(x + vt) describes a wave moving to the left at speed v.

20.60. Solve: A sinusoidal traveling wave is represented as D x t A kx t, sin ( ) = −( )ω . Replacing t with t + Tand using the relationship ω = 2π/T between the angular frequency and period,

D x t T A kx t T, sin +( ) = − +( )( )ω = − −( )A kx t Tsin ω ω = − −( )A kx tsin ω π2

= −( ) ( ) − −( ) ( )[ ]A kx t kx tsin cos cos sinω π ω π2 2 = −( )A kx tsin ω = D x t, ( )

20.61. Solve: According to Equation 20.28, the phase difference between two points on a wave is ∆φ =2 2 1π λr r−( ) . For the first point and second point,

r

r

12 2 2

22 2 2

= −( ) + −( ) + −( ) =

= − −( ) + −( ) + −( ) =

1.0 cm 0 cm 3.0 cm 0 cm 2.0 cm 0 cm 3.742 cm

1.0 cm 0 cm 1.5 cm 0 cm 2.5 cm 0 cm 3.082 cm

The wavelength is

λ = = =v

f

346 m / s

13,100 Hz0.02641 m = 2.641 cm

⇒ =−( ) = × ° = °∆φ π π π

π2

2 290

3.742 cm 3.082 cm

2.641 cm rad = rad

180

rad

20.62. Model: We have a sinusoidal traveling wave on a stretched string.Solve: (a) The wave speed on a string and the wavelength are calculated as follows:

vT v

f= = = ⇒ = = =S 20 N

0.002 kg / m100.0 m / s

100.0 m / s

100 Hz1.0 m

µλ

(b) The amplitude is determined by the oscillator at the end of the string and is A = 1.0 mm. The phase constant canbe obtained from Equation 20.15 as follows:

D A0 m, 0 s( ) = sinφ0 ⇒ − = ( )1.0 mm 1.0 mm sinφ0 ⇒ = −φ π0 2

rad

(c) The wave (as distinct from the oscillator) is described by D x t A kx t, sin ( ) = − +( )ω φ0 . In this equation thewave number and angular frequency are

k = = =2 22

πλ

π π1.0 m

rad / m ω π π= = ( )( ) =vk 100.0 m / s rad / m rad / s2 200

Thus, the wave’s displacement equation is

D x t x t, sin 1.0 mm rad / m rad / s rad( ) = ( ) ( ) − ( ) −[ ]2 200 12π π π

(d) The displacement is

D 0.50 m, 0.015 s 1.0 mm rad / s 0.50 m rad / s 0.015 s( ) = ( ) ( )( ) − ( )( ) −[ ]sin 2 200 12π π π = −1.00 mm

20.63. Model: We have a sinusoidal traveling wave going in the +y-direction. We will assume a sound speedof 343 m/s.Solve: The wave’s frequency is 686 Hz. Thus the wavelength is λ = = =v f/ 0.500 m 50.0 cm . Since one crestis located at y = 12.5 cm at t = 0 s, the other crests will be spaced every 50 cm. The crests will be at 62.5 cm, 112.5cm, and so on, as well as at −37.5 cm, −87.5 cm, and so on. This is sufficient information to draw the snapshotgraph at t = 0 s. Note that the y-axis is drawn horizontally in this representation of the wave, even though the waveis physically moving along the y-axis in a vertical direction.

(b) The displacement at y = 0 m and t = 0 s is

D A0 0 0 0 00 01 m s m or , sin sin( ) = = ⇒ = ( ) =−φ φ π rad

How do we distinguish between these two mathematically possible answers? Consider the rest of the y-axis at t =0 s. The displacement is

D y t Ay

Ay

, sin sin cm

=( ) = +

= +

0

2 2

500 0

πλ

φ π φ

We know that the displacement at y = 12.5 cm is a crest at t = 0 s (D = A), so the sine must equal to 1 at t = 0 s.Using the displacement equation at y = 12.5 cm yields:

D t A A12.5 cm, s cm

cm=( ) = ( ) +

= +

0

2 12 5

50 20 0sin.

sinπ φ π φ

φ0 0= correctly gives D = A, whereas φ π0 = rad gives D = −A. So the phase constant is φ0 0= rad.(c) Using the known values of λ, f, and φ0 , we can write the wave equation as

D y t Ay

ft A y t, sin sin 12.57 m 4310 s1 1( ) = − +

= ( ) − ( )[ ]− −2

2 0

πλ

π φ

(d) At t = 0 s, one crest is located at y = 12.5 cm and the others are spaced every 50 cm. This leads to the wavefront diagram shown below. The wave fronts are numbered.(e) The period of the wave is

Tf

= = =11 458

1

686 Hz ms.

0.729 ms is exactly 12 T —half a period. During half a period, the wave crests each move forward half a

wavelength, or 25 cm.(f) The phase of the wave is the full value of the argument of the sine function at a particular point in space andtime. At y = −12 5. cm, the phase is

φ ω φ

π

= − + = ( ) − ( )= ( ) −( ) − ( )( ) = − = −

− −

− −

ky t y t0

32

12.57 m 4310 s

12.57 m 0.125 m 4310 s 0.000729 s 4.70 rad rad

1 1

1 1

This phase corresponds to a wave crest. In the wave front graph at t = 0.729 ms, you can see a crest at y = −12.5 cm.Similarly, the phase at y = +12 5. cm is φ = −1 57. rad = − 1

2 π rad. This is a trough.

20.64. Model: We have a wave traveling to the right on a string.Visualize:

Solve: The snapshot of the wave as it travels to the right for an infinitesimally small time ∆t shows that thevelocity at point 1 is downward, at point 3 is upward, and at point 2 is zero. Furthermore, the speed at points 1 and3 is the maximum speed given by Equation 20.17: v v A1 3= = ω . The frequency of the wave is

ω π πλ

π π ω π= = = ( ) = ⇒ = ( ) ×( ) =−2 22

300 300 2 0 10 2fv

A45 m / s

0.30 m rad / s rad / s m 18.85 m / s.

Thus, v1 = −18.85 m/s, v2 = 0 m/s, and v3 = +18.85 m/s.

20.65. Model: This is a wave traveling to the left at a constant speed of 50 cm/s.Solve: The particles at positions between x = 2 cm and x = 7 cm have a speed of 10 cm/s, and the particles betweenx = 7 cm and x = 9 cm have a speed of −25 cm/s. That is, at the time the snapshot of the velocity is shown, the particlesof the medium have upward motion for 2 cm ≤ x ≤ 7 cm, but downward motion for 7 cm ≤ x ≤ 9 cm. The width of thefront section of the wave pulse is 7 cm – 2 cm = 5 cm and the width of the rear section is 9 cm – 7 cm = 2 cm. With awave speed of 50 cm/s, the time taken by the front section to pass through a particular point is 5 cm 50 cm / s 0.1 s=and the time taken by the rear section of the wave to pass through a point is 2 cm 50 cm / s 0.04 s= . Thus the wavecauses the upward moving particles to go through a displacement of A = ( )( ) =10 cm / s 0.1 s 1.0 cm. The downwardmoving particles have a maximum displacement of −( )( ) = −25 cm / s 0.04 s 1.0 cm .

20.66. Model: The wave is traveling on a stretched string.Solve: The wave speed on the string is

vT= = =S 50 N

0.005 kg / m100 m / s

µThe velocity of the particle on the string, however, is given by Equation 20.17. The maximum speed is calculatedas follows:

v A kx ty = − − +( )ω ω φcos 0 ⇒ = = =v A fAv

Ay max ω π πλ

2 2 =

( ) =2π 100 m / s

2.0 m0.030 m 9.42 m / s

20.67. Model: A sinusoidal wave is traveling along a stretched string.Solve: From Equation 20.17 and Equation 20.20, vy max = ωA and ay max = ω2A. These two equations can be combinedto give

ω ωπ

= = = ⇒ = =a

vfy

y

max

max

200 m / s

2.0 m / s100 rad / s 15.9 Hz

2

2⇒ = = =A

vy max .ω

2.0 m / s

100 rad / s cm2 0

20.68. Solve: Because the sun radiates waves uniformly in all directions, the intensity I of the sun’s rays whenthey impinge upon the earth is

IP

rI

P

r= ⇒ = = ×

×( )=sun

earthsun

earth2

2 W

m1 20 W / m

4 4

4 10

4 1 496 1042

26

11 2π π π .

With rsun-Venus = 1.082 × 1011 m and rsun-Mars = 2.279 × 1011 m, the intensities of electromagnetic waves at theseplanets are I IVenus

2Mars

22 20 W / m and 610 W / m= =7 .

20.69. Solve: (a) At a distance r from the bulb, the 2 watts of visible light have spread out to cover the surfaceof a sphere of radius r. The surface area of a sphere is a = 4πr2. Thus, the intensity at a distance of 2 m is

IP

a

P

r= = =

( )=

4 42 2π π2.0 W

2.0 m0.040 W / m2

Note that the presence of the wall has nothing to do with the intensity. The wall allows you to see the light, but thelight wave has the same intensity at all points 2 m from the bulb whether it is striking a surface or moving throughempty space.(b) Unlike the light from a light bulb, a laser beam does not spread out. We ignore the small diffraction spread ofthe beam. The laser beam creates a dot of light on the wall that is 2 mm in diameter. The full 2 watts of light isconcentrated in this dot of area a r= = ( ) =π π2 20.001 m 3 14 10 6. × − m2 . The intensity is

IP

a= =

×=−

2 W

3.14 10 m637,000 W / m6 2

2

Although the power of the light source is the same in both cases, the laser produces light on the wall whoseintensity is about 16 million times that of the light bulb.

20.70. Model: The radio wave is an electromagnetic wave.Solve: At a distance r, the 25 kW power station spreads out waves to cover the surface of a sphere of radius r.The surface area of a sphere is 4πr2. Thus, the intensity of the radio waves is

IP

r= = ×

×( )= × −source 2 W

m W / m

4

25 10

4 10 101 99 102

3

3 25

π π.

20.71. Model: We have a traveling wave radiated by the tornado siren.Solve: (a) The power of the source is calculated as follows:

IP

r

P50 m

2 source source0.10 W / m50 m

= = =( )4 42 2π π

⇒ = ( ) ( ) = ( )Psource20.10 W / m 50 m W4 10002π π

The intensity at 1000 m is

IP

1000 msource

1000 m

W

1000 m250 W m=

( )= ( )

( )=

4

1000

42 22

ππ

πµ

(b) The maximum distance is calculated as follows:

IP

r rr= ⇒ × = ( ) ⇒ =−source 2 W / m

W15.8 km

41 0 10

1000

426

2ππ

π.

20.72. Solve: (a) The peak power of the light pulse is

PE

tpeak 8

mJ

10 ns

0.500 J

1.0 10 s W= = =

×= ×−

∆∆

5005 0 107.

(b) The average power is

PE

avgtotal

1 s

mJ

1 s

5 J

1 s5 W= = × = =10 500

The laser delivers pulses of very high power. But the laser spends most of its time “off,” so the average power isvery much less than the peak power.(c) The intensity is

IP

alaser 11 22 W

5.0 m

W

7.85 10 m W / m= = ×

( )= ×

×= ×−

5 0 10 5 0 106 4 10

7

2

717. .

.π µ

(d) The ratio is

I

Ilaser

sun

2

2

W / m

W / m= ×

×= ×6 4 10

1 1 105 8 10

17

314.

..

20.73. Model: The bat’s chirping frequency is altered by the Doppler effect. The frequency is increased as thebat approaches and it decreases as the bat recedes away.Solve: The bat must fly away from you, so that the chirp frequency observed by you is less than 25 kHz. FromEquation 20.38,

ff

v v− =+

0

1 S /⇒ =

+

20,000 Hz25000 Hz

343 m / sS1

v⇒ =vS 85.8 m / s

Assess: This is a rather large speed: 85.8 m/s ≈ 180 mph. This is not possible for a bat.

20.74. Model: The sound generator’s frequency is altered by the Doppler effect. The frequency increases asthe generator approaches the student, and it decreases as the generator recedes from the student.Solve: The generator’s speed is

v r r fS 1.0 m rev / s 10.47 m / s= = ( ) = ( )

=ω π π2 2

100

60The frequency of the approaching generator is

ff

v v+ =−

=−

=0

1 1S

600 Hz10.47 m / s

343 m / s

619 Hz

Doppler effect for the receding generator, on the other hand, is

ff

v v− =+

=+

=0

1 S

600 Hz

110.47 m / s

343 m / s

582 Hz

Thus, the highest and the lowest frequencies heard by the student are 619 Hz and 582 Hz.

20.75. Solve: We will closely follow the details of section 20.7 in the textbook. Figure 20.29 shows that thewave crests are stretched out behind the source. The wavelength detected by Pablo is λ− = 1

3 d , where d is thedistance the wave has moved plus the distance the source has moved at time t = 3T . These distances are∆x vt vTwave = = 3 and ∆x v t v Tsource S S= = 3 . The wavelength of the wave emitted by a receding source is thus

λ− = = + = + = +( )d x x vT v Tv v T

3 3

3 3

3

∆ ∆wave source SS

The frequency detected in Pablo’s direction is thus

fv v

v v T

f

v v−−

= =+( ) =

+λ s S

0

1

20.76. Model: We are looking at the Doppler effect for the light of an approaching source.Solve: (a) The time is

t = ××

= =54 10

3 10

6

5

km

km / s180 s 3.0 min

(b) Using Equation 20.40, the observed wavelength is

λ λ= −+

= −+

( )1

1

1 0 1

1 0 10

v c

v c

c c

c cs

s

540 nm.

. = (0.9045)(540 nm) = 488 nm

Assess: 488 nm is slightly blue shifted from green.

20.77. Model: We are looking at the Doppler effect for the light of a receding source.Visualize:

Note that the daredevil’s tail lights are receding away from your rocket’s light detector with a relative speed of 0.2c.Solve: Using Equation 20.40, the observed wavelength is

λ λ= +−

= +−

( )1

1

1 0 2

1 0 20

v c

v c

c c

c cS

S

650 nm.

. = 796 nm

This wavelength is in the infrared region.

20.78. Model: The Doppler effect for light of a receding source yields an increased wavelength.Solve: Because the measured wavelengths are 5% longer, that is, λ = 1.05λ0, the distant galaxy is receding awayfrom the earth. Using Equation 20.40,

λ λ λ= = +−

⇒ ( ) = +−

⇒ = = ×1 051

11 05

1

11 47 100 0

2 7. . .v c

v c

v c

v cv cs

s

s

ss 0.049 m / s

20.79. Model: The Doppler effect for light of an approaching source leads to a decreased wavelength.Solve: The red wavelength (λ0 = 650 nm) is Doppler shifted to green (λ = 540 nm) due to the approaching lightsource. In relativity theory, the distinction between the motion of the source and the motion of the observerdisappears. What matters is the relative approaching or receding motion between the source and the observer. Thus,we can use Equation 20.40 as follows:

λ λ= −+

⇒ = ( ) −+

⇒ = × = ×

0

4

1

1

1

1

5 5 10

v c

v c

v c

v c

v

s

s

s

s

s8

540 nm 650 nm

km / s 2.0 10 km / hr.

The fine will be

2.0 10 km / hr 50 km / hr $

1 km / hr $ million8× −( )

=1200

The police officer knew his physics.

20.80. Model: The wave pulse is a traveling wave on a stretched string. The two masses hanging from the steelwire are in static equilibrium.Visualize: Please refer to Figure CP20.80 in your textbook.

Solve: The wave speed along the wire is

vwire

4.0 m

0.024 s166.7 m / s= =

Using Equation 20.2,

vT T

Twire m / s = 0.060 kg 8.0 m

208.4 N= = ( ) ⇒ =166 7 1 11.

µ

Because point 1 is in static equilibrium, with r rFnet = 0,

F T T TT

F T w w mg T m

x

y

net

net 2

N

N272.1 N

9.8 m / s17.8 kg

( ) = − ° ⇒ =°

=

( ) = ° − = ⇒ = = ° ⇒ = ( ) °=

1 2 21

2 2

4040

2721

40 0 4040

coscos

sin sinsin

20.81. Visualize: Please refer to Figure CP20.81.Solve: The wave speeds along the two metal wires are

vT

vT

11

22

= = = = = =µ µ

2250 N

0.009 kg / m500 m / s

2250 N

0.025 kg / m300 m / s

The wavelengths along the two wires are

λ λ11

221

3

1

5= = = = = =v

f

v

f

500 m / s

1500 Hz m

300 m / s

1500 Hz m

Thus, the number of wavelengths over two sections of the wire are

1.0 m 1.0 m

m

1.0 m 1.0 m

m13

15λ λ1 2

3 5= ( ) = = ( ) =

The number of complete cycles of the wave in the 2.0-m-long wire is 8.

20.82. Model: The wave pulse is a traveling wave on a stretched wire.Visualize:

Solve: (a) At a distance y above the lower end of the rope, the point P is in static equilibrium. The upwardtension in the rope must balance the weight of the rope that hangs below this point. Thus, at this point

T w Mg y g= = = ( )µwhere µ = m/L is the linear density of the entire rope. Using Equation 20.2, we get

vT yg

gy= = =µ

µµ

(b) The time to travel a distance dy at y, where the wave speed is v gy= , is

dtdy

v

dy

gy= =

Finding the time for a pulse to travel the length of the rope requires integrating from one end of the rope to the other:

∆t dtdy

gy gy

gL

T LL

= = =

=∫ ∫

0 00

12

2 ⇒ =∆tL

g2

20.83. Visualize:

Solve: (a) Using the graph, the refractive index n as a function of distance x can be mathematically expressed as

n nn n

Lx= + −

12 1

At position x, the light speed is v c n= / . The time for the light to travel a distance dx at x is

dtdx

v

n

cdx

cn

n n

Lx dx= = = + −

11

2 1

To find the total time for the light to cover a thickness L of a glass we integrate as follows:

T dtc

nn n

Lx dx

T L

= = + −

∫ ∫

0

12 1

0

1 = +−( )

∫ ∫n

cdx

n n

cLx dx

L L1

0

2 1

0

= + −

= +

n

cL

n n

cL

L n n

cL1 2 1

21 2

2 2

(b) Substituting the given values into this equation,

T =+( )

×( ) × = × −1 50 1 60

2 3 0 100 010

8

. .

..

m / s m 5.17 10 s11