**UFBA**

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

(Parte **4** de 4)

9. (a) The period is the reciprocal of the frequency:

(b) Using v = 343 m/s, we find λ = v/f = 3.8 m.

100. (a) The problem asks for the source frequency f. We use Eq. 17–47 with great care (regarding its ± sign conventions).

Therefore, with f ′ = 950 Hz, we obtain f = 880 Hz.

(b) We now have

so that with f = 880 Hz, we find f ′ = 824 Hz.

101. (a) The blood is moving towards the right (towards the detector), because the Doppler shift in frequency is an increase: ∆f > 0.

(b) The reception of the ultrasound by the blood and the subsequent remitting of the signal by the blood back toward the detector is a two-step process which may be compactly written as v vf f f

where bloodcos.xvvθ= If we write the ratio of frequencies as R = (f + ∆f)/f, then the solution of the above equation for the speed of the blood is

R v v

(c) We interpret the question as asking how ∆f (still taken to be positive, since the detector is in the “forward” direction) changes as the detection angle θ changes. Since larger θ means smaller horizontal component of velocity vx then we expect ∆f to decrease towards zero as θ is increased towards 90°.

CHAPTER 17

102. Pipe A (which can only support odd harmonics – see Eq. 17-41) has length LA. Pipe B (which supports both odd and even harmonics [any value of n] – see Eq. 17-39) has length LB = 4LA . Taking ratios of these equations leads to the condition:

= |

Solving for nB we have nB = 2nodd.

(a) Thus, the smallest value of nB at which a harmonic frequency of B matches that of A is nB = 2(1)=2.

(b) The second smallest value of nB at which a harmonic frequency of B matches that of A is nB = 2(3)=6.

(c) The third smallest value of nB at which a harmonic frequency of B matches that of A is nB = 2(5)=10.

103. The points and the least-squares fit is shown in the graph that follows.

The graph has frequency in Hertz along the vertical axis and 1/L in inverse meters along the horizontal axis. The function found by the least squares fit procedure is f = 276(1/L) + 0.037. We shall assume that this fits either the model of an open organ pipe (mathematically similar to a string fixed at both ends) or that of a pipe closed at one end.

(a) In a tube with two open ends, f = v/2L. If the least-squares slope of 276 fits the first model, then a value of

(b) In a tube with only one open end, f = v/4L, and we find v = 4(276 m/s) = 1106 m/s 31.110 m/s≈×which is more “in the ballpark” of the 1400 m/s value cited in the problem.

(c) This suggests that the acoustic resonance involved in this situation is more closely related to the n = 1 case of Figure 17-15(b) than to Figure 17-14.

104. (a) Since the source is moving toward the wall, the frequency of the sound as received at the wall is

vf f v v

(b) Since the person is moving with a speed u toward the reflected sound with frequency f ′, the frequency registered at the source is

343m/sr v uf f v

105. Using Eq. 17-47 with great care (regarding its ± sign conventions), we have

106. (a) Let P be the power output of the source. This is the rate at which energy crosses the surface of any sphere centered at the source and is therefore equal to the product of the intensity I at the sphere surface and the area of the sphere. For a sphere of radius r, P

= 4pir2 I and I = P/4pir2. The intensity is proportional to the square of the displacement amplitude sm. If we write 2mICs=, where C is a constant of proportionality, then 2/4mCsPr=pi. Thus,

The displacement amplitude is proportional to the reciprocal of the distance from the source. We take the wave to be sinusoidal. It travels radially outward from the source, with points on a sphere of radius r in phase. If ω is the angular frequency and k is the angular wave number then the time dependence is sin(kr – ωt). Letting /4,bPC=pi the displacement wave is then given by

(b) Since s and r both have dimensions of length and the trigonometric function is dimensionless, the dimensions of b must be length squared.

107. (a) The problem is asking at how many angles will there be “loud” resultant waves, and at how many will there be “quiet” ones? We consider the resultant wave (at large distance from the origin) along the +x axis; we note that the path-length difference (for the waves traveling from their respective sources) divided by wavelength gives the (dimensionless) value n = 3.2, implying a sort of intermediate condition between

CHAPTER 17 constructive interference (which would follow if, say, n = 3) and destructive interference (such as the n = 3.5 situation found in the solution to the previous problem) between the waves. To distinguish this resultant along the +x axis from the similar one along the –x axis, we label one with n = +3.2 and the other n = –3.2. This labeling facilitates the complete enumeration of the loud directions in the upper-half plane: n = –3, –2, –1, 0, +1, +2, +3. Counting also the “other” –3, –2, –1, 0, +1, +2, +3 values for the lower-half plane, then we conclude there are a total of 7 + 7 = 14 “loud” directions.

(b) The labeling also helps us enumerate the quiet directions. In the upper-half plane we find: n = –2.5, –1.5, –0.5, +0.5, +1.5, +2.5. This is duplicated in the lower half plane, so the total number of quiet directions is 6 + 6 = 12.

108. The source being isotropic means Asphere = 4pir2 is used in the intensity definition I = P/A. Since intensity is proportional to the square of the amplitude (see Eq. 17–27), this further implies sI P r r I s P r r pi

(b) Using the notation A instead of sm for the amplitude, we find

4.0mA A

109. (a) In regions where the speed is constant, it is equal to distance divided by time. Thus, we conclude that the time difference is

L d d Lt

where the first term is the travel time through bone and rock and the last term is the expected travel time purely through rock. Solving for d and simplifying, we obtain

(b) If we estimate d ≈ 10 cm (as the lower limit of a range that goes up to a diameter of 20 cm), then the above expression (with the numerical values given in the problem) leads to

∆t = 0.8 µs (as the lower limit of a range that goes up to a time difference of 1.6 µs).

110. (a) We expect the center of the star to be a displacement node. The star has spherical symmetry and the waves are spherical. If matter at the center moved it would move equally in all directions and this is not possible.

(b) We assume the oscillation is at the lowest resonance frequency. Then, exactly one-fourth of a wavelength fits the star radius. If λ is the wavelength and R is the star radius then λ = 4R. The frequency is f = v/λ = v/4R, where v is the speed of sound in the star. The period is T = 1/f = 4R/v.

(c) The speed of sound is /vB=ρ, where B is the bulk modulus and ρ is the density of stellar material. The radius is R = 9.0 × 10–3Rs, where Rs is the radius of the Sun (6.96 × 108 m). Thus

1.3 10 Pa T R B

1. We find the difference in the two applications of the Doppler formula:

which leads to24.810 Hzf=×.

112. (a) We proceed by dividing the (velocity) equation involving the new (fundamental) frequency f ′ by the equation when the frequency f is 440 Hz to obtain

/ | / |

f f f f λ τ µ τ λ τ µ τ where we are making an assumption that the mass-per-unit-length of the string does not change significantly. Thus, with τ′=1.2τ, we have/4401.2,f′=which gives

(b) In this case, neither tension nor mass-per-unit-length change, so the wave speed v is unchanged. Hence, using Eq. 17–38 with 1n=,

(Parte **4** de 4)