Model: 20.1. 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 this point. 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 will take 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 first part of the wave causes an upward displacement of the medium. The rising portion of the wave is 2 m wide, so it will 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 to pass x = 6 m and during this time the displacement of the medium will be a constant (y = 1 cm). The trailing edge of 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, the displacement is zero at x = 0 m. The first part of the wave causes a downward displacement of the medium, so immediately after t = 2 s the displacement at x = 0 m will be negative. The negative portion of the wave pulse is 3 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 s the displacement of the medium is positive. The displacement at x = 0 m returns to zero at t = 8 s and remains zero for 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 the x = 0 m point of the medium first sees the negative portion of the pulse wave at t = 1 s. Thus, the snapshot graph of this 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 is moving 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 of the history graph takes 2 s to pass the x = 2 m position, its width is 2 m. Similarly, the width of the linearly increasing part of the history graph is 2 m. The center of the flat part of the history graph corresponds to both t = 0 s and 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 with an inter-particle separation of 1.0 cm at equilibrium. Because the longitudinal wave has a positive amplitude of 0.5 cm between 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. Just underneath, the positions of the particles as a longitudinal wave is passing through are shown at time t = 0 s. It is clear that relative to the equilibrium the particle positions are displaced negatively on the left side and positively on the right side. For example, the particles at x = 0 cm and x = 1 cm are at equilibrium, the particle at x = 2 cm is displaced 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 left by 0.5 cm, and the particle at x = 5 cm is undisplaced. The behavior of particles for x > 5 cm is opposite of that for x < 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 is vstring = TS . The wave speed if the tension is doubled will be
vstring = 2TS = 2 vstring = 2 (200 m / s) = 283 m / s
20.8. Model: The wave is a traveling wave on a stretched string.
Solve:
The wave speed on a stretched string with linear density is
vstring =
TS 150 m / s =
75 N = 3.333 10 -3 kg / m
2
For a wave speed of 180 m/s, the required tension will be
2 TS = vstring = (3.333 10 -3 kg / m )(180 m / s) = 108 N
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 is vstring = TS with
=
If the string radius doubles, then
m V R 2 L = = = R 2 L L L
vstring =
vstring 280 m / s TS = = 140 m / s 2 = 2 2 (2 R)
20.10. Solve: (a) The wave number is
k=
(b) The wave speed is
2 2 = = 3.14 rad / m 2.0 m
30 rad / s v = f = = (2.0 m ) = 9.55 m / s 2 2
20.11. Solve: (a) The wavelength is
=
2 2 = = 4.19 m k 1.5 rad / m
(b) The frequency is
f = v 200 m / s = = 47.7 Hz 4.19 m
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, = 124 rad/s, and 0 = 0 rad . The frequency is
f =
(b) The wavelength is
124 rad / s = = 19.7 Hz 2 2
2 2 = = 2.33 m k 2.7 rad / m
=
(c) The wave speed v = f = 45.9 m / s .
Solve: (a) A comparison of the wave equation with Equation 20.14 yields: A = 5.2 cm, k = 5.5 rad/m, = 72 rad/s, and 0 = 0 rad . The frequency is
20.13. Model: The wave is a traveling wave.
f =
(b) The wavelength is
72 rad / s = = 11.5 Hz 2 2
2 2 = = 1.14 m k 5.5 rad / m
=
(c) The wave speed v = f = 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(2x). This graph is shown as a dotted line. The equation at t = 1 s is D = (2.0 cm) sin(2x /2 rad). Its graph is a dashed line. 8 (b) The wave speed is
v=
4 rad / s = = 2.0 m / s k 2 rad / m
1 4
In agreement, you can see that the peak of the graph has moved forward 0.5 m in
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 distance between two consecutive peaks, which gives = 14 m - 2 m = 12 m . The frequency of the wave is
f =
v 24 m / s = = 2.0 Hz 12 m
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 is 0.60 s, so the frequency f = 1 T = 1 0.60 s = 1.67 Hz . The wavelength is
=
v 2m/s = = 1.2 m f 1.667 Hz
20.17. Solve: According to Equation 20.28, the phase difference between two points on a wave is
= 2 - 1 = 2 r 2 2 = (r2 - r1 ) (3 rad - 0 rad) = (80 cm - 20 cm) = 40 cm
20.18. Solve: According to Equation 20.28, the phase difference between two points on a wave is
= 2 - 1 = 2 (r2 - r1 )
If 1 = rad at r1 = 4.0 m, we can determine 2 at any r value at the same instant using this equation. At r2 = 3.5 m,
2 = 1 +
3 At r2 = 4.5 m, 2 = 2 rad.
2 2 (r2 - r1 ) = rad + 2.0 m (3.5 m - 4.0 m) = 2 rad
20.19.
Visualize:
Solve:
For a sinusoidal wave, the phase difference between two points on the wave is given by Equation 20.28:
= 2 - 1 =
2 2 2 (r2 - r1 ) = (40 m - 30 m) = (10 m)
= 2 for two points on adjacent wavefronts and = 4 for two points separated by 2. Thus, = 10 m when = 2 , and = 5 m when = 4 . The crests corresponding to these two wavelengths are shown in the figure. One can see that a crest of the wave passes the 40 m-listener and the 30 m-listener simultaneously. The lowest two possible frequencies will occur for the largest two possible wavelengths, which are 10 m and 5 m. Thus, the lowest frequency is
f1 =
The next highest frequency is f2 = 68 Hz .
v 340 m / s = = 34 Hz 10 m
20.20.
Visualize:
Solve:
(a) Because the same wavefront simultaneously reaches listeners at x = -7.0 m and x = +3.0 m,
= 0 rad =
2 (r2 - r1 ) r2 = r1
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 - (2 m)2 = 4.58 m .
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 = 0.0379 s 343 m / s The time elapsed for the same sound to get reflected from the back wall and reach your ear is 26 m + 13 m = 0.114 s 343 m / 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 microphone
while the other travels through the air to the microphone. The time interval for the sound pulse traveling through the air is x 4.0 m = = 0.01166 s = 11.66 ms tair = vair 343 m / s Sound travels faster through solids than gases, so the pulse traveling through the metal will reach the microphone before the pulse traveling through the air. Because the pulses are separated in time by 11.0 ms, the pulse traveling through 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
vmetal =
x 4.0 m = = 6060 m / s t metal 0.00066 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 echo = t water down + t granite down + t granite up + t water up 0.94 s = dgranite dgranite 500 m 500 m + + + dgranite = 793 m 1480 m / s 6000 m / s 6000 m / s 1480 m / s
20.24. Solve: (a) In aluminum, the speed of sound is 6420 m/s. The wavelength is thus equal to
=
v 6420 m / s = = 3.21 10 -3 m = 3.21 mm f 2.0 10 6 Hz
(b) The speed of an electromagnetic wave is c. The frequency would be
f =
c 3.0 10 8 m / s = = 9.35 1010 Hz 3.21 10 -3 m
20.25. Solve: (a) The frequency is
f =
(b) The frequency is
vair 343 m / s = = 1715 Hz 0.20 m
f =
c 3.0 10 8 m / s = = 1.50 10 9 Hz = 1.50 GHz 0.20 m
vwater 1480 m / s = = 9.87 10 -7 m = 987 nm 1.50 10 9 Hz f
(c) The speed of a sound wave in water is vwater = 1480 m/s. The wavelength of the sound wave would be
=
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
fblue =
(b) The frequency of the red light is
c 3.0 10 8 m / s = = 6.67 1014 Hz 450 10 -9 m 3.0 10 8 m / s = 4.62 1014 Hz 650 10 -9 m
fred =
(c) Using Equation 20.30 to calculate the index of refraction,
material =
vacuum 650 nm n = vacuum = = 1.44 material 450 nm n
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
fmicrowaves =
c 3.0 10 8 m / s = = 1.0 1010 Hz = 10 GHz 3.0 10 -2 m
(b) The refractive index of air is 1.0003, so the speed of microwaves in air is vair = c 1.00 c. The time for the microwave signal to travel is
t=
50 km 50 10 3 m = = 0.167 ms vair (3.0 10 8 m 1.00)
Assess: A small time of 0.167 ms for the microwaves to cover a distance of 50 km shows that the electromagnetic waves travel very fast.
20.28. Model: Radio waves are electromagnetic waves that travel with speed c.
Solve: (a) The wavelength is
=
c 3.0 10 8 m / s = = 2.96 m f 101.3 MHz
(b) The speed of sound in air at 20C is 343 m/s. The frequency is
f =
vsound 343 m / s = = 116 Hz 2.96 m
20.29. Model: Light is an electromagnetic wave.
Solve: (a) The time light takes is
t=
3.0 mm 3.0 10 -3 m 3.0 10 -3 m = = = 1.50 10 -11 s vglass cn 3.0 10 8 m / s) 1.50 (
c nwater 3.0 10 8 m / s (1.50 10 -11 s) = 3.38 mm 1.33
(b) The thickness of water is
d = vwater t =
t=
20.30. Solve: (a) The speed of light in a material is given by Equation 20.29:
n=
The refractive index is
c c vmat = vmat n
n=
(b) The frequency is
vac 420 nm vsolid = c solid = (3.0 10 8 m / s) = 1.88 10 8 m / s 670 nm mat vac
vsolid 1.88 10 8 m / s = = 4.48 1014 Hz 420 nm solid
f =
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 =
The liquid's index of refraction is
30 10 -2 m = 2.174 10 8 m / s 1.38 10 -9 s
n=
c vliquid
=
3.0 10 8 = 1.38 2.174 10 8
Thus the wavelength of the laser beam in the liquid is
liquid =
vac 633 nm = = 459 nm n 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. The
intensity 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 = Ir 2 t = (2.0 10 -3 W / m 2 ) (3.0 10 -3 m ) (60 s) = 3.39 10 -6 J
2
20.33. Solve: The energy delivered to an area a in time t is E = Pt, where the power P is related to the intensity
I 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 at
distances r1 and r2 is
I I1 r22 2m = 50 m = = 1.6 10 -3 I2 r12 I2 m 50 m
2
I50 m = I2 m (1.6 10 -3 ) = (2.0 W / m 2 )(1.6 10 -3 ) = 3.2 10 -3 W / m 2
Assess: The power generated by the sound source is P = I2m [4(2 m)2] = (2.0 W/m2)(50.27) = 101 W. This is a significant amount of power.
20.35. Solve: (a) The intensity of a uniform spherical source of power Psource a distance r away is I = Psource 4r 2 .
Thus, the intensity at the position of the microphone is
I50 m =
35 W -3 W / m2 2 = 1.11 10 4 (50 m )
(b) The sound energy impinging on the microphone per second is
P = Ia = (1.11 10 -3 W / m 2 )(1.0 10 -4 m 2 ) = 1.11 10 -7 W = 1.11 10 -7 J / s
Energy impinging on the microphone in 1 second = 1.11 10 -7 J.
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
f+ =
f0 600 Hz = = 647 Hz 1 - vS v 1 - 25 m / s 343 m / s f0 600 Hz = = 559 Hz 1 + vS v 1 + 25 m / s 343 m / s
(b) The frequency as her convertible recedes from the stationary person is
f- =
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
f+ =
f0 800 Hz 900 Hz = vS = 38.1 m / s vS 1 - vS v 1- 343 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 observer
approaching 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 minimum speed you need to travel is calculated as follows:
v v0 f- = 1 - 0 f0 20 kHz = 1 - (21 kHz) v0 = 16.3 m / s 343 m / s v
Assess: A speed of 16.3 m/s corresponds to approximately 35 mph. This is a possible but very fast speed on a bicycle.
20.39. Model: Your friend's frequency is altered by the Doppler effect. The frequency of your friend's note
increases as he races towards you (moving source and a stationary observer). The frequency of your note for your approaching friend is also higher (stationary source and a moving observer). Solve: (a) The frequency of your friend's note as heard by you is
f+ =
f0 400 Hz = = 432 Hz vS 25.0 m / s 1- 1- 340 m / s v
(b) The frequency heard by your friend of your note is
v 25.0 m / s f+ = f0 1 + 0 = ( 400 Hz)1 + = 429 Hz 340 m / s v
20.40. Visualize: The function D(x, t) represents a pulse that travels in the positive x-direction without changing
shape.
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 - 3t | 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 changing
shape.
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's speed is 2 m/s. (c) x + 2t 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 = 2.0 m. We are given that the wave's frequency is f = 5.0 Hz. Thus, the wave speed is v = f = 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 = 0 m, t = 0 s) = D(0 m, 0 s) = 0.5 mm = 1 A 2
Thus
D(0 m, 0 s) = 1 A = A sin 0 0 = sin -1 ( 1 ) = 2 2
5 rad or rad 6 6
Note that the value of D(0 m, 0 s) alone gives two possible values of the phase constant. One of the values will cause the displacement to start at 0.5 mm and increase with distance--as the graph shows--while the other will cause the displacement to start at 0.5 mm but decrease with distance. Which is which? The wave equation for t = 0 s is
2x D( x, t = 0) = A sin + 0
If x is a point just to the right of the origin and is very small, the angle (2x + 0 ) is just slightly bigger than the angle 0 . Now sin31 > sin30, but sin151 < sin150, so the value 0 = 1 rad is the phase constant for which the 6 displacement increases as x increases. (c) The equation for a sinusoidal traveling wave can be written as
2x x D( x, t ) = A sin - 2ft + 0 = A sin 2 - ft + 0
Substituting in the values found above,
x D( x, t ) = (1.0 mm ) sin 2 - 5.0 s -1 )t + 2.0 m ( 6
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, the wavelength is
=
v = vT = ( 4.0 m / s)(0.20 s) = 0.80 m f
(b) The phase constant 0 is obtained as follows:
D(0 m, 0 s) = A sin 0 -2 mm = (2 mm ) sin 0 sin 0 = -1 0 = - 1 rad 2
(c) The displacement equation for the wave is
2x 2x 2t D( x, t ) = A sin - 2ft + 0 = (2.0 mm ) sin - - = (2.0 mm ) sin(2.5x - 10t - 1 ) 2 0.80 m 0.20 s 2
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 s since 1 division is
equal to 100 s. So, the speed of the sound wave in the liquid is
v=
2 25 cm = 1140 m / s 440 s
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 is
vstring =
TS =
TS = m/L
LTS 2.0 m = 50 10 -3 s m
(2.0 m)(20 N)
m
m = 0.025 kg = 25 g
20.46. Model: The wave pulse is a traveling wave on a stretched string.
Visualize:
Solve:
The pulse travels along the rope with speed
v=
x 3.0 m = = 20 m / s = t 0.150 s
TS
where we used the relationship between the wave's speed, the rope's tension, and the linear density = mrope L . r r We can find the tension in the rope from a Newtonian analysis. Bob experiences horizontal forces T and fk . He is r being pulled at constant speed, so he is in dynamic equilibrium with Fnet = 0 N . Thus T = fk = k n . From the vertical forces we see that n = w = mBob g . So we find the rope's tension to be
TS = k mBob g = (0.20)(60 kg)(9.8 m / s 2 ) = 117.6 N
Using this in the velocity equation, we can find the rope's mass:
v=
LT TS (3.0 m)(117.6 N) mrope = 2S = = 0.882 kg = 882 g mrope L v (20 m / s)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
v1 =
TS 1
and
v2 =
TS 2 v12 1 = v2 2 = TS 2
Because v1 = L1 t1 and v2 = L2 t2 , and because the pulses are to reach the ends of the string simultaneously the above equation can be simplified to
2 L L1 1 L2 2 = 22 1 = L2 t2 t
2 = 1
4.0 g / m = 2 L1 = 2 L2 2.0 g / m
and
Since L1 + L2 = 4 m,
2 L2 + L2 = 4 m L2 = 1.66 m
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 the
ocean surface. The depth of the ocean is
2d = (vsound in water )t d = (750 m / s)t
Using this relation and the data from Figure P20.48, we can generate the following table for the ocean depth (d ) at various positions (x) of the ship. x (km) 0 20 40 45 50 60 t (s) 6 4 4 8 4 2 d (km) 4.5 3.0 3.0 6.0 3.0 1.5
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 The 1. variables for the rest of the string will be identified with the subscript 2. The tension is the same throughout the string, so the wave speeds in the string are
v1 =
Because 2 = 1 1 , the wave speeds are 4
TS 1
v2 =
TS 2
v2 =
The wavelengths are
4TS = 2v1 = 2(2.0 m / s) = 4.0 m / s 1
1 =
v1 2.0 m / s = = 0.40 m f 5.0 Hz
2 =
v2 4.0 m / s = = 0.80 m f 5.0 Hz
20.50. Model: Assume a room temperature of 20C.
Visualize:
Solve:
The distance between the source and the left ear (EL) is
dL =
x 2 + ( y + 0.1 m ) =
2
[(5.0 m)cos45] + [(5.0 m)sin45
2
+ 0.1 m ] = 5.0712 m
2
Similarly d R = 4.9298 m . Thus,
dL - dR = d = 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 d 0.1414 m t = = = 412 s 343 m / s 343 m / s
For the sound speed of vair = 343 m/s, the wavelength of the sound wave in air is 343 m / s air = = 1.340 m 256 Hz 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 vwater 1480 m / s = = 4.31 343 m / s vair Solve: Finally, the wavelength in water is v 1480 m / s 5.781 m water = water = = 5.781 m water = = 4.31 256 Hz 1.340 m air fwater Assess: This last result is expected because v = f and the frequency remains unchanged as the wave enters from air into water.
20.51. Model: The temperature is 20C for both air and water.
20.52. Solve: The difference in the arrival times for the P and S waves is
t = tS - t P =
1 1 d d 6 120 s = d - - d = 1.23 10 m = 1230 km 4500 m / s 8000 m / s vS v P
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
t= d 8.00 10 6 m = = 5405.4 s v 1480 m / s
d = 1480.28 m / s. t
A time decrease to 5404.4 s implies the speed has changed to v =
Since the 4.0 m/s increase in velocity is due to an increase of 1C, an increase of 0.28 m/s occurs due to a temperature increase of
1C (0.28 m / s) = 0.07C 4.0 m / s
Thus, a temperature increase of approximately 0.07C 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 sin( ky + t + 0 ) , so the wave is traveling along the y-axis. Because it is +t rather than -t 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 air molecules are oscillating back and forth along the y-axis. (c) The wave number is k = 8.96 m -1 , so the wavelength is 2 2 = = = 0.70 m 8.96 m -1 k The angular frequency is = 3140 s -1 , so the wave's frequency is 3140 s -1 f = = = 500 Hz 2 2 Thus, the wave speed v = f = (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 = 1 m, t = 0 s) = (0.02 mm ) sin(8.96 + 1 ) = -0.0063 mm 4
Assess: The wave is a sound wave with speed v = 350 m/s. This is greater than the room-temperature speed of 343 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 form D(x, t) = D(x- vt). Since the variables x and t in the given wave equation appear together as x + vt, the wave is traveling toward the left, that is, in the -x direction. (b) The speed of the wave is 2 0.20 s v= = = 12.0 m / s k 2 rad 2.4 m The frequency is
f =
The wave number is
2 rad 0.20 s = = 5.0 Hz 2 2
k= 2 rad = 2.62 rad / m 2.4 m
(c) The displacement is
0.20 m 0.50 s D(0.20 m, 0.50 s) = (3.0 cm ) sin 2 + + 1 = -1.50 cm 2.4 m 0.20 s
20.56. Model: This is a sinusoidal wave traveling on a stretched string in the +x direction.
2 2
Solve: (a) From the displacement equation of the wave, A = 2.0 cm, k = 12.57 rad/m, and = 638 rad/s. Using the equation for the wave speed in a stretched string,
vstring =
TS 638 rad / s 2 TS = vstring = = (5.0 10 -3 kg / m 3 ) = 12.6 N k 12.57 rad / m
(b) The maximum displacement is the amplitude Dmax ( x, t ) = 2.0 cm . (c) From Equation 20.17,
v y max = A = (638 rad / s)(2.0 10 -2 m ) = 12.8 m / s
20.57. Solve: The wave number and frequency are calculated as follows:
k= 2 2 rad = = 4 rad / m = vk = ( 4.0 m / s)( 4 rad / m ) = 16 rad / s 0.50 m
Thus, the displacement equation for the wave is
D( y, t ) = (5.0 cm ) sin[( 4 rad / m ) y + (16 rad / s)t ]
Assess: 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:
= 2f = 2 (200 Hz) = 400 rad / s k =
The displacement equation for the wave is
400 rad / s = = rad / m v 400 m / s
D( x, t ) = (0.010 mm ) sin[( rad / m ) x - ( 400 rad / s)t + 1 rad ] 2
Assess: Note the negative sign with t 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 same value 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 a later 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 sin(kx - t ) . Replacing t with t + T
and using the relationship = 2/T between the angular frequency and period,
D( x, t + T ) = A sin( kx - (t + T )) = A sin( kx - t - T ) = A sin( kx - t - 2 ) = A[sin( kx - t ) cos(2 ) - cos( kx - t ) sin(2 )] = A sin( kx - t ) = D( x, t )
20.61. Solve: According to Equation 20.28, the phase difference between two points on a wave is =
2 (r2 - r1 ) . For the first point and second point,
r1 = r2 =
The wavelength is
(1.0 cm - 0 cm)2 + (3.0 cm - 0 cm)2 + (2.0 cm - 0 cm)2 = 3.742 cm (-1.0 cm - 0 cm)2 + (1.5 cm - 0 cm)2 + (2.5 cm - 0 cm)2 = 3.082 cm
=
=
v 346 m / s = = 0.02641 m = 2.641 cm f 13,100 Hz
2 (3.742 cm - 3.082 cm ) 180 = rad = rad = 90 rad 2 2 2.641 cm
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:
v=
TS =
20 N v 100.0 m / s = 100.0 m / s = = = 1.0 m 100 Hz 0.002 kg / m f
(b) The amplitude is determined by the oscillator at the end of the string and is A = 1.0 mm. The phase constant can be obtained from Equation 20.15 as follows: D(0 m, 0 s) = A sin 0 -1.0 mm = (1.0 mm ) sin 0 0 = - rad 2 (c) The wave (as distinct from the oscillator) is described by D( x, t ) = A sin( kx - t + 0 ) . In this equation the wave number and angular frequency are
k=
2 2 = = 2 rad / m 1.0 m
= vk = (100.0 m / s)(2 rad / m ) = 200 rad / s
Thus, the wave's displacement equation is
D( x, t ) = (1.0 mm ) sin[(2 rad / m ) x - (200 rad / s)t - 1 rad ] 2
(d) The displacement is
D(0.50 m, 0.015 s) = (1.0 mm ) sin[(2 rad / s)(0.50 m ) - (200 rad / s)(0.015 s) - 1 ] = -1.00 mm 2
20.63. Model: We have a sinusoidal traveling wave going in the +y-direction. We will assume a sound speed
of 343 m/s. Solve: The wave's frequency is 686 Hz. Thus the wavelength is = v / f = 0.500 m = 50.0 cm . Since one crest is 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, 112.5 cm, and so on, as well as at -37.5 cm, -87.5 cm, and so on. This is sufficient information to draw the snapshot graph at t = 0 s. Note that the y-axis is drawn horizontally in this representation of the wave, even though the wave is physically moving along the y-axis in a vertical direction.
(b) The displacement at y = 0 m and t = 0 s is
D(0 m, 0 s) = A sin 0 = 0 m 0 = sin -1 (0) = 0 or 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 2y 2y D( y, t = 0) = A sin + 0 = A sin + 0 50 cm 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:
2 (12.5 cm ) D(12.5 cm, t = 0 s) = A sin + 0 = A sin + 0 2 50 cm
0 = 0 correctly gives D = A, whereas 0 = rad gives D = -A. So the phase constant is 0 = 0 rad. (c) Using the known values of , f, and 0 , we can write the wave equation as
2y D( y, t ) = A sin - 2ft + 0 = A sin (12.57 m -1 ) y - ( 4310 s -1 )t
[
]
(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 wave front diagram shown below. The wave fronts are numbered. (e) The period of the wave is
T=
1 1 = = 1.458 ms f 686 Hz
0.729 ms is exactly 1 T --half a period. During half a period, the wave crests each move forward half a 2 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 and time. At y = -12.5 cm, the phase is
= ky - t + 0 = (12.57 m -1 ) y - ( 4310 s -1 )t
3 = (12.57 m -1 )( -0.125 m ) - ( 4310 s -1 )(0.000729 s) = -4.70 rad = - 2 rad
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 = -1.57 rad = - 1 rad. This is a trough. 2
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 the velocity at point 1 is downward, at point 3 is upward, and at point 2 is zero. Furthermore, the speed at points 1 and 3 is the maximum speed given by Equation 20.17: v1 = v3 = A . The frequency of the wave is
v 2 ( 45 m / s) = = 300 rad / s A = (300 rad / s)(2.0 10 -2 m ) = 18.85 m / s 0.30 m Thus, v1 = -18.85 m/s, v2 = 0 m/s, and v3 = +18.85 m/s.
= 2f = 2
Solve: The particles at positions between x = 2 cm and x = 7 cm have a speed of 10 cm/s, and the particles between x = 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 particles of the medium have upward motion for 2 cm x 7 cm, but downward motion for 7 cm x 9 cm. The width of the front 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 a wave 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 wave causes the upward moving particles to go through a displacement of A = (10 cm / s)(0.1 s) = 1.0 cm . The downward moving particles have a maximum displacement of ( -25 cm / s)(0.04 s) = -1.0 cm .
20.65. Model: This is a wave traveling to the left at a constant speed of 50 cm/s.
20.66. Model: The wave is traveling on a stretched string.
Solve: The wave speed on the string is
v=
TS =
50 N = 100 m / s 0.005 kg / m
The velocity of the particle on the string, however, is given by Equation 20.17. The maximum speed is calculated as follows: v 100 m / s v y = -A cos( kx - t + 0 ) v y max = A = 2fA = 2 A = 2 (0.030 m) = 9.42 m / s 2.0 m
Solve: From Equation 20.17 and Equation 20.20, vy max = A and ay max = 2A. These two equations can be combined to give a y max 200 m / s 2 v 2.0 m / s = = = 100 rad / s f = = 15.9 Hz A = y max = = 2.0 cm v y max 2.0 m / s 2 100 rad / s
20.67. Model: A sinusoidal wave is traveling along a stretched string.
20.68. Solve: Because the sun radiates waves uniformly in all directions, the intensity I of the sun's rays when they impinge upon the earth is
I= Psun P 4 10 26 W 2 I earth = sun = 2 = 1420 W / m 2 2 4r 4rearth 4 (1.496 1011 m )
With rsun-Venus = 1.082 10 11 m and rsun-Mars = 2.279 10 11 m, the intensities of electromagnetic waves at these planets are I Venus = 2720 W / m 2 and I Mars = 610 W / m 2 .
20.69. Solve: (a) At a distance r from the bulb, the 2 watts of visible light have spread out to cover the surface of a sphere of radius r. The surface area of a sphere is a = 4r2. Thus, the intensity at a distance of 2 m is
I= P P 2.0 W 2 = = 2 = 0.040 W / m a 4r 2 4 (2.0 m )
Note that the presence of the wall has nothing to do with the intensity. The wall allows you to see the light, but the light wave has the same intensity at all points 2 m from the bulb whether it is striking a surface or moving through empty space. (b) Unlike the light from a light bulb, a laser beam does not spread out. We ignore the small diffraction spread of the beam. The laser beam creates a dot of light on the wall that is 2 mm in diameter. The full 2 watts of light is 2 concentrated in this dot of area a = r 2 = (0.001 m ) = 3.14 10 -6 m 2 . The intensity is
P 2W = = 637,000 W / m 2 a 3.14 10 -6 m 2 Although the power of the light source is the same in both cases, the laser produces light on the wall whose intensity is about 16 million times that of the light bulb. I=
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 4r2. Thus, the intensity of the radio waves is P 25 10 3 W -5 I = source = W / m2 2 = 1.99 10 4r 2 4 10 10 3 m
(
)
20.71. Model: We have a traveling wave radiated by the tornado siren.
Solve: (a) The power of the source is calculated as follows: P Psource 2 2 I50 m = 0.10 W / m 2 = source = source = 0.10 W / m 4 (50 m ) = (1000 ) W 2 P 2 4r 4 (50 m )
(
)
The intensity at 1000 m is
I1000 m =
Psource (1000 ) W 2 2 = 2 = 250 W m 4 (1000 m ) 4 (1000 m )
(b) The maximum distance is calculated as follows: P (1000 ) W I = source 1.0 10 -6 W / m 2 = r = 15.8 km 2 4r 4r 2
20.72. Solve: (a) The peak power of the light pulse is
Ppeak =
(b) The average power is
E 500 mJ 0.500 J = = = 5.0 10 7 W t 10 ns 1.0 10 -8 s
Etotal 10 500 mJ 5 J = = =5W 1s 1s 1s The laser delivers pulses of very high power. But the laser spends most of its time "off," so the average power is very much less than the peak power. (c) The intensity is P 5.0 10 7 W 5.0 10 7 W I laser = = = 6.4 1017 W / m 2 2 = a (5.0 m ) 7.85 10 -11 m 2 Pavg =
(d) The ratio is
I laser 6.4 1017 W / m 2 = = 5.8 1014 1.1 10 3 W / m 2 Isun
20.73. Model: The bat's chirping frequency is altered by the Doppler effect. The frequency is increased as the
bat 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. From Equation 20.38, f0 25000 Hz 20,000 Hz = f- = vS = 85.8 m / s vS 1 + vS / v 1+ 343 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 as
the generator approaches the student, and it decreases as the generator recedes from the student. Solve: The generator's speed is 100 vS = r = r(2f ) = (1.0 m )2 rev / s = 10.47 m / s 60 The frequency of the approaching generator is f0 600 Hz f+ = = = 619 Hz 1 - vS v 1 - 10.47 m / s 343 m / s Doppler effect for the receding generator, on the other hand, is f0 600 Hz f- = = = 582 Hz 10.47 m / s 1 + vS v 1 + 343 m / s Thus, the highest and the lowest frequencies heard by the student are 619 Hz and 582 Hz.
wave crests are stretched out behind the source. The wavelength detected by Pablo is - = 1 d , where d is the 3 distance the wave has moved plus the distance the source has moved at time t = 3T . These distances are x wave = vt = 3vT and xsource = vSt = 3vS T . The wavelength of the wave emitted by a receding source is thus
20.75. Solve: We will closely follow the details of section 20.7 in the textbook. Figure 20.29 shows that the
d x wave + xsource 3vT + 3vS T = = = (v + vS )T 3 3 3 The frequency detected in Pablo's direction is thus f0 v v f- = = = - (v + vs )T 1 + vS v
- =
20.76. Model: We are looking at the Doppler effect for the light of an approaching source.
Solve: (a) The time is
54 10 6 km = 180 s = 3.0 min 3 10 5 km / s (b) Using Equation 20.40, the observed wavelength is t=
=
1 - vs c 1 - 0.1c c 0 = (540 nm) = (0.9045)(540 nm) = 488 nm 1 + vs c 1 + 0.1c c
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
=
1 + vS c 1 + 0.2c c 0 = (650 nm) = 796 nm 1 - vS c 1 - 0.2c c
This wavelength is in the infrared region.
Solve: Because the measured wavelengths are 5% longer, that is, = 1.050, the distant galaxy is receding away from the earth. Using Equation 20.40,
20.78. Model: The Doppler effect for light of a receding source yields an increased wavelength.
= 1.05 0 =
1 + vs c 1 + vs c 2 0 (1.05) = vs = 0.049 c = 1.47 10 7 m / s 1 - vs c 1 - vs c
Solve: The red wavelength (0 = 650 nm) is Doppler shifted to green ( = 540 nm) due to the approaching light source. In relativity theory, the distinction between the motion of the source and the motion of the observer disappears. What matters is the relative approaching or receding motion between the source and the observer. Thus, we can use Equation 20.40 as follows:
20.79. Model: The Doppler effect for light of an approaching source leads to a decreased wavelength.
= 0
1 - vs c 1 - vs c 540 nm = (650 nm ) 1 + vs c 1 + vs c
vs = 5.5 10 4 km / s = 2.0 10 8 km / hr
The fine will be
(2.0 10
The police officer knew his physics.
8
1$ km / hr - 50 km / hr ) = $200 million 1 km / hr
20.80. Model: The wave pulse is a traveling wave on a stretched string. The two masses hanging from the steel
wire are in static equilibrium. Visualize: Please refer to Figure CP20.80 in your textbook.
Solve:
The wave speed along the wire is
vwire =
Using Equation 20.2,
4.0 m = 166.7 m / s 0.024 s
T1 T1 = T1 = 208.4 N (0.060 kg 8.0 m) r r Because point 1 is in static equilibrium, with Fnet = 0, vwire = 166.7 m / s =
1 ( Fnet ) x = T1 - T2 cos 40 T2 = cos 40 = 2721 N
T
( Fnet ) y = T2 sin 40 - w = 0 N w = mg = T2 sin 40 m = (
272.1 N ) sin 40 = 17.8 kg 9.8 m / s 2
20.81. Visualize: Please refer to Figure CP20.81.
Solve: The wave speeds along the two metal wires are
v1 =
T = 1
2250 N = 500 m / s 0.009 kg / m
v1 500 m / s 1 = = m f 1500 Hz 3
1.0 m 1.0 m = 1 =3 1 ( 3 m)
v2 =
T = 2
2250 N = 300 m / s 0.025 kg / m
The wavelengths along the two wires are
1 =
2 =
v2 300 m / s 1 = = m f 1500 Hz 5
Thus, the number of wavelengths over two sections of the wire are
1.0 m 1.0 m = 1 =5 2 ( 5 m)
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 upward tension in the rope must balance the weight of the rope that hangs below this point. Thus, at this point T = w = Mg = ( y)g where = m/L is the linear density of the entire rope. Using Equation 20.2, we get
v=
T =
yg = gy
(b) The time to travel a distance dy at y, where the wave speed is v = gy , is
dt =
dy dy = v 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 = dt =
0
T
0
L
L dy 1 2 = 2 y 0 = g gy g
L t = 2
L g
20.83.
Visualize:
(a) Using the graph, the refractive index n as a function of distance x can be mathematically expressed as n - n1 n = n1 + 2 x L At position x, the light speed is v = c / n. The time for the light to travel a distance dx at x is n - n1 dx n 1 dt = n1 + 2 = dx = x dx v c c L To find the total time for the light to cover a thickness L of a glass we integrate as follows:
Solve:
T = dt =
0
T
L L (n - n1 ) L x dx = n1 L + n2 - n1 L2 = n1 + n2 L n - n1 n 1 n1 + 2 x dx = 1 dx + 2 cL 2 2c c L c 0 cL c 0 0
(b) Substituting the given values into this equation, (1.50 + 1.60) T= 0.010 m = 5.17 10 -11 s 2 3.0 10 8 m / s
(
)