**UFBA**

# Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch17

(Parte **3** de 4)

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(d) For tube 4, dv=2v /3. 65. (a) The expression for the Doppler shifted frequency is

v vf f v v where f is the unshifted frequency, v is the speed of sound, vD is the speed of the detector (the uncle), and vS is the speed of the source (the locomotive). All speeds are relative to the air. The uncle is at rest with respect to the air, so vD = 0. The speed of the source is vS = 10 m/s. Since the locomotive is moving away from the uncle the frequency decreases and we use the plus sign in the denominator. Thus

vf f v v

(b) The girl is now the detector. Relative to the air she is moving with speed vD = 10.0 m/s toward the source. This tends to increase the frequency and we use the plus sign in the numerator. The source is moving at vS = 10.0 m/s away from the girl. This tends to decrease the frequency and we use the plus sign in the denominator. Thus (v + vD) =

(v + vS) and f′ = f = 500.0 Hz.

(c) Relative to the air the locomotive is moving at vS = 20.0 m/s away from the uncle.

Use the plus sign in the denominator. Relative to the air the uncle is moving at vD = 10.0 m/s toward the locomotive. Use the plus sign in the numerator. Thus

v vf f v v

(d) Relative to the air the locomotive is moving at vS = 20.0 m/s away from the girl and the girl is moving at vD = 20.0 m/s toward the locomotive. Use the plus signs in both the numerator and the denominator. Thus (v + vD) = (v + vS) and f′ = f = 500.0 Hz.

6. We use Eq. 17–47 with f = 500 Hz and v = 343 m/s. We choose signs to produce f′ > f.

(a) The frequency heard in still air is

(b) In a frame of reference where the air seems still, the velocity of the detector is 30.5 – 30.5 = 0, and that of the source is 2(30.5). Therefore,

(c) We again pick a frame of reference where the air seems still. Now, the velocity of the source is 30.5 – 30.5 = 0, and that of the detector is 2(30.5). Consequently,

67. The Doppler shift formula, Eq. 17–47, is valid only when both uS and uD are measured with respect to a stationary medium (i.e., no wind). To modify this formula in the presence of a wind, we switch to a new reference frame in which there is no wind.

(a) When the wind is blowing from the source to the observer with a speed w, we have u′S

= u′D = w in the new reference frame that moves together with the wind. Since the observer is now approaching the source while the source is backing off from the observer, we have, in the new reference frame,

v u v wf f f v u v w

In other words, there is no Doppler shift.

(b) In this case, all we need to do is to reverse the signs in front of both u′D and u′S. The result is that there is still no Doppler shift:

v u v wf f f v u v w

In general, there will always be no Doppler shift as long as there is no relative motion between the observer and the source, regardless of whether a wind is present or not.

68. We note that 1350 km/h is vS = 375 m/s. Then, with θ = 60º, Eq. 17-57 gives v = 3.3×102 m/s.

69. (a) The half angle θ of the Mach cone is given by sin θ = v/vS, where v is the speed of sound and vS is the speed of the plane. Since vS = 1.5v, sin θ = v/1.5v = 1/1.5. This means θ = 42°.

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(b) Let h be the altitude of the plane and suppose the Mach cone intersects Earth's surface a distance d behind the plane. The situation is shown on the diagram below, with P indicating the plane and O indicating the observer. The cone angle is related to h and d by tan θ = h/d, so d = h/tan θ. The

shock wave reaches O in the time the plane takes to fly the distance d:

5000 m11s tan 1.5(331m/s)tan42 d ht

70. The altitude H and the horizontal distance x for the legs of a right triangle, so we have tan tan 1.25 sinpH x v t vt= = =θ θ θ where v is the speed of sound, vp is the speed of the plane and v v

v v

71. (a) Incorporating a term (λ/2) to account for the phase shift upon reflection, then the path difference for the waves (when they come back together) is

leads to d = 0, 2.10 m, … | Since the problem explicitly excludes the d = 0 possibility, |

Setting this equal to the condition needed to destructive interference (λ/2, 3λ/2, 5λ/2 …) then our answer is d = 2.10 m.

(b) Setting this equal to the condition needed to constructive interference (λ, 2λ, 3λ …) leads to d = 1.47 m, … Our answer is d = 1.47 m.

72. When the source is stationary (with respect to the air) then Eq. 17-47 gives

where vd is the speed of the detector (assumed to be moving away from the source, in the way we’ve written it, above). The difference between the approach and the recession is

1 1 2d d dv v vf f f f v v v which, after setting (f′′′−)/f =1/2, leads to an equation which can be solved for the ratio vd /v. The result is 1/4. Thus, vd /v = 0.250.

73. (a) Adapting Eq. 17-39 to the notation of this chapter, we have sm′ = 2 sm cos(φ/2) = 2(12 nm) cos(pi/6) = 20.78 nm. Thus, the amplitude of the resultant wave is roughly 21 nm.

(b) The wavelength (λ = 35 cm) does not change as a result of the superposition.

(c) Recalling Eq. 17-47 (and the accompanying discussion) from the previous chapter, we conclude that the standing wave amplitude is 2(12 nm) = 24 nm when they are traveling in opposite directions.

(d) Again, the wavelength (λ = 35 cm) does not change as a result of the superposition.

74. (a) The separation distance between points A and B is one-quarter of a wavelength; therefore, λ = 4(0.15 m) = 0.60 m. The frequency, then, is

(b) The separation distance between points C and D is one-half of a wavelength; therefore, λ = 2(0.15 m) = 0.30 m. The frequency, then, is f = v/λ = (343 m/s)/(0.30 m) = 14 Hz (or approximately 1.14 kHz).

75. Any phase changes associated with the reflections themselves are rendered inconsequential by the fact that there are an even number of reflections. The additional path length traveled by wave A consists of the vertical legs in the zig-zag path: 2L. To be

(minimally) out of phase means, therefore, that 2L = λ/2 (corresponding to a half-cycle, or

180°, phase difference). Thus, L = λ/4, or L/λ = 1/4 = 0.25.

76. Since they are approaching each other, the sound produced (of emitted frequency f) by the flatcar-trumpet received by an observer on the ground will be of higher pitch f ′. In these terms, we are told f ′ – f = 4.0 Hz, and consequently that f ‘/ f = 4/440 = 1.0091.

With vS designating the speed of the flatcar and v = 343 m/s being the speed of sound, the Doppler equation leads to f v v f v v

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7. The siren is between you and the cliff, moving away from you and towards the cliff.

Both “detectors” (you and the cliff) are stationary, so vD = 0 in Eq. 17–47 (and see the discussion in the textbook immediately after that equation regarding the selection of ± signs). The source is the siren with vS = 10 m/s. The problem asks us to use v = 330 m/s for the speed of sound.

(a) With f = 1000 Hz, the frequency fy you hear becomes

vf f v v

(b) The frequency heard by an observer at the cliff (and thus the frequency of the sound reflected by the cliff, ultimately reaching your ears at some distance from the cliff) is

vf f v v

(c) The beat frequency is fc – fy = 60 beats/s (which, due to specific features of the human ear, is too large to be perceptible).

78. Let r stand for the ratio of the source speed to the speed of sound. Then, Eq. 17-5 (plus the fact that frequency is inversely proportional to wavelength) leads to

2 = |

Solving, we find r = 1/3. Thus, vs/v = 0.3.

79. The source being isotropic means Asphere = 4pir2 is used in the intensity definition I = P/A, which further implies

I P r r I P r r pi

(b) Using Eq. 17–27 with I1 = 9.60 × 10–4 W/m2, ω = 2pi(2000 Hz), v = 343 m/s and ρ = 1.21 kg/m3, we obtain

(c) Eq. 17-15 gives the pressure amplitude:

0.893 Pa.mmpvsρω∆==

80. When φ = 0 it is clear that the superposition wave has amplitude 2∆pm. For the other cases, it is useful to write

The factor in front of the sine function gives the amplitude ∆pr. Thus, / 2cos( / 2).r mp p φ∆ ∆ =

(b) Using that value of P in Eq. 17–28 with a new value for r, we obtain

Alternatively, a ratio I′ /I = (r/r′ )2 could have been used. (c) Using Eq. 17–29 with I = 0.0080 W/m2, we have

82. We use /vBρ= to find the bulk modulus B:

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83. Let the frequencies of sound heard by the person from the left and right forks be fl and fr, respectively.

(a) If the speeds of both forks are u, then fl,r = fv/(v ± u) and r l fuvf f f fv

v u v u v u

(b) If the speed of the listener is u, then fl,r = f(v ± u)/v and

343m/sl r uf f f f v

84. The rule: if you divide the time (in seconds) by 3, then you get (approximately) the straight-line distance d. We note that the speed of sound we are to use is given at the beginning of the problem section in the textbook, and that the speed of light is very much larger than the speed of sound. The proof of our rule is as follows:

sound light sound sound .

343m/s 0.343km/s d d dt t t t v

Cross-multiplying yields (approximately) (0.3 km/s)t = d which (since 1/3 ≈ 0.3) demonstrates why the rule works fairly well.

85. (a) The intensity is given by 2212,mIvsρω= where ρ is the density of the medium, v is the speed of sound, ω is the angular frequency, and sm is the displacement amplitude. The displacement and pressure amplitudes are related by ∆pm = ρvωsm, so sm = ∆pm/ρvω and I

= (∆pm)2/2ρv. For waves of the same frequency the ratio of the intensity for propagation in water to the intensity for propagation in air is

,w mw a a a ma w w

I p v where the subscript a denotes air and the subscript w denotes water. Since Ia = Iw, mw w w

ma a a p v p v

The speeds of sound are given in Table 17-1 and the densities are given in Table 15-1.

(b) Now, ∆pmw = ∆pma, so w a a

a w w

I v

I v

(b) Since mmpsI∆∝∝, we have

(c) The displacement amplitude ratio is 1212//71.msI==

87. (a) When the right side of the instrument is pulled out a distance d the path length for sound waves increases by 2d. Since the interference pattern changes from a minimum to the next maximum, this distance must be half a wavelength of the sound. So 2d = λ/2, where λ is the wavelength. Thus λ = 4d and, if v is the speed of sound, the frequency is

(b) The displacement amplitude is proportional to the square root of the intensity (see Eq.

17–27). Write mICs=, where I is the intensity, sm is the displacement amplitude, and C is a constant of proportionality. At the minimum, interference is destructive and the displacement amplitude is the difference in the amplitudes of the individual waves: sm = sSAD – sSBD, where the subscripts indicate the paths of the waves. At the maximum, the waves interfere constructively and the displacement amplitude is the sum of the amplitudes of the individual waves: sm = sSAD + sSBD. Solve

100()SADSBDCss=− and 900()SADSBDCss=− for sSAD and sSBD. Adding the equations give

SADs = ( 100 900 /2 20/ ,C C+ = while subtracting them yields

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Thus, the ratio of the amplitudes is sSAD/sSBD = 2.

(c) Any energy losses, such as might be caused by frictional forces of the walls on the air in the tubes, result in a decrease in the displacement amplitude. Those losses are greater on path B since it is longer than path A.

89. The round-trip time is t = 2L/v where we estimate from the chart that the time between clicks is 3 ms. Thus, with v = 1372 m/s, we find 122.1mLvt==.

90. The wave is written as (,)cos()msxtskxtω=±.

(a) The amplitude ms is equal to the maximum displacement: 0.30 cmms=.

(b) Since λ = 24 cm, the angular wave number is 12/0.26 cmkpiλ− = = .

(c) The angular frequency is 2(25 Hz)1.610 rad/sfωpipi===×.

(d) The speed of the wave is v = λf = (24 cm)(25 Hz) = 6.0 × 102 cm/s.

(e) Since the direction of propagation is x−, the sign is plus, i.e., (,)cos()msxtskxtω=+.

91. The source being a “point source” means Asphere = 4pir2 is used in the intensity definition I = P/A, which further implies

I P r r I P r r pi

From the discussion in §17-5, we know that the intensity ratio between “barely audible” and the “painful threshold” is 10–12 = I2/I1. Thus, with r2 = 10000 m, we find

92. (a) The time it takes for sound to travel in air is ta = L/v, while it takes tm = L/vm for the sound to travel in the metal. Thus,

m m

(b) Using the values indicated (see Table 17-1), we obtain tL v v

93. (a) We observe that “third lowest … frequency” corresponds to harmonic number n = 5 for such a system. Using Eq. 17–41, we have

(b) As noted, n = 5; therefore, f1 = 750/5 = 150 Hz.

94. We note that waves 1 and 3 differ in phase by pi radians (so they cancel upon

superposition) | Consequently, there is no resultant wave. |

superposition). Waves 2 and 4 also differ in phase by pi radians (and also cancel upon

95. Since they oscillate out of phase, then their waves will cancel (producing a node) at a point exactly midway between them (the midpoint of the system, where we choose x = 0). We note that Figure 17-14, and the n = 3 case of Figure 17-15(a) have this property (of a node at the midpoint). The distance ∆x between nodes is λ/2, where λ = v/f and f = 300

Hz and v = 343 m/s. Thus, ∆x = v/2f = 0.572 m.

Therefore, nodes are found at the following positions: (0.572m), 0,1,2,...xnxnn=∆==±

(a) The shortest distance from the midpoint where nodes are found is ∆x =0. (b) The second shortest distance from the midpoint where nodes are found is ∆x=0.572 m.

(c) The third shortest distance from the midpoint where nodes are found is 2∆x =1.14 m.

96. (a) With f = 686 Hz and v = 343 m/s, then the “separation between adjacent wavefronts” is λ = v/f = 0.50 m.

(b) This is one of the effects which are part of the Doppler phenomena. Here, the wavelength shift (relative to its “true” value in part (a)) equals the source speed sv (with appropriate ± sign) relative to the speed of sound v:

In front of the source, the shift in wavelength is –(0.50 m)(110 m/s)/(343 m/s) = –0.16 m, and the wavefront separation is 0.50 m – 0.16 m = 0.34 m.

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(c) Behind the source, the shift in wavelength is +(0.50 m)(110 m/s)/(343 m/s) = +0.16 m, and the wavefront separation is 0.50 m + 0.16 m = 0.6 m.

97. We use I ∝ r–2 appropriate for an isotropic source. We have

r D d

D dI where d = 50.0 m. We solve for

98. (a) Using m = 7.3 × 107 kg, the initial gravitational potential energy is 113.910 JUmgy==×, where h = 550 m. Assuming this converts primarily into kinetic energy during the fall, then K = 3.9 × 1011 J just before impact with the ground. Using instead the mass estimate m = 1.7 × 108 kg, we arrive at K = 9.2 × 1011 J.

(b) The process of converting this kinetic energy into other forms of energy (during the impact with the ground) is assumed to take ∆t = 0.50 s (and in the average sense, we take the “power” P to be wave-energy/∆t). With 20% of the energy going into creating a seismic wave, the intensity of the body wave is estimated to be

hemisphere 2

K tPI A r using r = 200 × 103 m and the smaller value for K from part (a). Using instead the larger estimate for K, we obtain I = 1.5 W/m2.

(c) The surface area of a cylinder of “height” d is 2pird, so the intensity of the surface wave is

cylinder

K tPI A rd using d = 5.0 m, r = 200 × 103 m and the smaller value for K from part (a). Using instead the larger estimate for K, we obtain I = 58 kW/m2.

(d) Although several factors are involved in determining which seismic waves are most likely to be detected, we observe that on the basis of the above findings we should expect the more intense waves (the surface waves) to be more readily detected.

(Parte **3** de 4)