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

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

(Parte 2 de 4)

103. The intensity is increased by a factor of 1.0×103.

(b) The pressure amplitude is proportional to the square root of the intensity so it is increased by a factor of 100032.≈

30. (a) Eq. 17-29 gives the relation between sound level β and intensity I, namely

Thus we find that for a β = 70 dB level we have a high intensity value of Ihigh = 10 µW/m2. (b) Similarly, for β = 50 dB level we have a low intensity value of Ilow = 0.10 µW/m2.

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(c) Eq. 17-27 gives the relation between the displacement amplitude and I. Using the values for density and wave speed, we find sm = 70 nm for the high intensity case.

(d) Similarly, for the low intensity case we have sm = 7.0 nm.

We note that although the intensities differed by a factor of 100, the amplitudes differed by only a factor of 10.

31. We use β = 10 log(I/Io) with Io = 1 × 10–12 W/m2 and Eq. 17–27 with ω = 2pif = 2pi(260 Hz), v = 343 m/s and ρ = 1.21 kg/m3.

()()28.527o11027.610m0.76 m.2mmIIvfssρµ−

32. (a) Since ω = 2pif, Eq. 17-15 leads to

which yields sm = 0.26 nm. The nano prefix represents 10–9. We use the speed of sound and air density values given at the beginning of the exercises and problems section in the textbook.

(b) We can plug into Eq. 17–27 or into its equivalent form, rewritten in terms of the pressure amplitude:

mp I

3. We use β = 10 log (I/Io) with Io = 1 × 10–12 W/m2 and I = P/4pir2 (an assumption we are asked to make in the problem). We estimate r ≈ 0.3 m (distance from knuckle to ear) and find

34. The difference in sound level is given by Eq. 17-37:

iII

Thus, if 5.0 dbβ∆=, then log(/)1/2fiII=, which implies that 10fiII=. On the other hand, the intensity at a distance r from the source is 24 PI rpi =, where P is the power of the source. A fixed P implies that 2 iiffIrIr=. Thus, with 1.2 m,ir= we obtain

Ir r I

pi pi

(b) Let A (= 0.750 cm2) be the cross-sectional area of the microphone. Then the power intercepted by the microphone is

36. Combining Eqs.17-28 and 17-29 we have β = 10 log. Taking differences (for sounds A and B) we find

∆β = 10 log – 10 log = 10 log using well-known properties of logarithms. Thus, we see that ∆β is independent of r and can be evaluated anywhere.

(a) We can solve the above relation (once we know ∆β = 5.0) for the ratio of powers; we find PA /PB ≈ 3.2.

(b) At r = 1000 m it is easily seen (in the graph) that ∆β = 5.0 dB. This is the same ∆β we expect to find, then, at r = 10 m.

37. (a) As discussed on page 408, the average potential energy transport rate is the same as that of the kinetic energy. This implies that the (average) rate for the total energy is

= 2 = 2 ( ¼ ρ A v ω2 sm 2 ) using Eq. 17-4. In this equation, we substitute ρ = 1.21 kg/m3, A = pir2 = pi(0.020 m)2, v = 343 m/s, ω = 3000 rad/s, sm = 12 ×10−9 m, and obtain the answer 3.4 × 10−10 W.

(b) The second string is in a separate tube, so there is no question about the waves superposing. The total rate of energy, then, is just the addition of the two: 2(3.4 × 10−10

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(c) Now we do have superposition, with φ = 0, so the resultant amplitude is twice that of the individual wave which leads to the energy transport rate being four times that of part

(d) In this case φ = 0.4pi, which means (using Eq. 17-39)

This means the energy transport rate is (1.618)2 = 2.618 times that of part (a). We obtain 2.618(3.4 × 10−10 W) = 8.8 × 10−10 W.

(e) The situation is as shown in Fig. 17-14(b). The answer is zero.

38. (a) Using Eq. 17–39 with v = 343 m/s and n = 1, we find f = nv/2L = 86 Hz for the fundamental frequency in a nasal passage of length L = 2.0 m (subject to various assumptions about the nature of the passage as a “bent tube open at both ends”).

(b) The sound would be perceptible as sound (as opposed to just a general vibration) of very low frequency.

(c) Smaller L implies larger f by the formula cited above. Thus, the female's sound is of higher pitch (frequency).

39. (a) From Eq. 17–53, we have

(b) The frequency of the wave on the string is the same as the frequency of the sound wave it produces during its vibration. Consequently, the wavelength in air is

40. The distance between nodes referred to in the problem means that λ/2 = 3.8 cm, or λ = 0.076 m. Therefore, the frequency is

41. (a) We note that 1.2 = 6/5. This suggests that both even and odd harmonics are present, which means the pipe is open at both ends (see Eq. 17-39).

(b) Here we observe 1.4 = 7/5. This suggests that only odd harmonics are present, which means the pipe is open at only one end (see Eq. 17-41).

42. At the beginning of the exercises and problems section in the textbook, we are told to assume vsound = 343 m/s unless told otherwise. The second harmonic of pipe A is found from Eq. 17–39 with n = 2 and L = LA, and the third harmonic of pipe B is found from Eq. 17–41 with n = 3 and L = LB. Since these frequencies are equal, we have

(a) Since the fundamental frequency for pipe A is 300 Hz, we immediately know that the second harmonic has f = 2(300 Hz) = 600 Hz. Using this, Eq. 17–39 gives

(b) The length of pipe B is 340.429m.BALL==

43. (a) When the string (fixed at both ends) is vibrating at its lowest resonant frequency, exactly one-half of a wavelength fits between the ends. Thus, λ = 2L. We obtain v = fλ = 2Lf = 2(0.220 m)(920 Hz) = 405 m/s.

(b) The wave speed is given by /,vτµ= where τ is the tension in the string and µ is the linear mass density of the string. If M is the mass of the (uniform) string, then µ = M/L. Thus,

(c) The wavelength is λ = 2L = 2(0.220 m) = 0.440 m.

(d) The frequency of the sound wave in air is the same as the frequency of oscillation of the string. The wavelength is different because the wave speed is different. If va is the speed of sound in air the wavelength in air is

4. The frequency is f = 686 Hz and the speed of sound is vsound = 343 m/s. If L is the length of the air-column, then using Eq. 17–41, the water height is (in unit of meters)

where n = 1, 3, 5,… with only one end closed. (a) There are 4 values of n (n = 1,3,5,7) which satisfies h > 0.

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(b) The smallest water height for resonance to occur corresponds to n = 7 with 0.125 mh=.

(c) The second smallest water height corresponds to n = 5 with h = 0.375 m.

45. (a) Since the pipe is open at both ends there are displacement antinodes at both ends and an integer number of half-wavelengths fit into the length of the pipe. If L is the pipe length and λ is the wavelength then λ = 2L/n, where n is an integer. If v is the speed of sound then the resonant frequencies are given by f = v/λ = nv/2L. Now L = 0.457 m, so

To find the resonant frequencies that lie between 1000 Hz and 2000 Hz, first set f = 1000 Hz and solve for n, then set f = 2000 Hz and again solve for n. The results are 2.6 and 5.32, which imply that n = 3, 4, and 5 are the appropriate values of n. Thus, there are 3 frequencies.

(b) The lowest frequency at which resonance occurs is (n = 3) f = 3(376.4 Hz) = 1129 Hz. (c) The second lowest frequency at which resonance occurs is (n = 4) f = 4(376.4 Hz) = 1506 Hz.

46. (a) Since the difference between consecutive harmonics is equal to the fundamental frequency (see section 17-6) then f1 = (390 – 325) Hz = 65 Hz. The next harmonic after 195 Hz is therefore (195 + 65) Hz = 260 Hz.

(b) Since fn = nf1 then n = 260/65 = 4.

(c) Only odd harmonics are present in tube B so the difference between consecutive harmonics is equal to twice the fundamental frequency in this case (consider taking differences of Eq. 17-41 for various values of n). Therefore, f1 = (1320 – 1080) Hz = 120 Hz. The next harmonic after 600 Hz is consequently [600 + 2(120)] Hz = 840 Hz.

(d) Since fn = nf1 (for n odd) then n = 840/120 = 7.

47. The string is fixed at both ends so the resonant wavelengths are given by λ = 2L/n, where L is the length of the string and n is an integer. The resonant frequencies are given by f = v/λ = nv/2L, where v is the wave speed on the string. Now /v=τµ, where τ is the tension in the string and µ is the linear mass density of the string. Thus

(/2)/fnL=τµ. Suppose the lower frequency is associated with n = n1 and the higher frequency is associated with n = n1 + 1. There are no resonant frequencies between so you know that the integers associated with the given frequencies differ by 1. Thus 1(/2)/fnL=τµ and

48. (a) Using Eq. 17–39 with n = 1 (for the fundamental mode of vibration) and 343 m/s for the speed of sound, we obtain sound

tube

(b) For the wire (using Eq. 17–53) we have wire wire wire1 2 2 nvf where µ = mwire/Lwire. Recognizing that f = f ′ (both the wire and the air in the tube vibrate at the same frequency), we solve this for the tension τ:

2 2 2 3wire wire wirewire

49. The top of the water is a displacement node and the top of the well is a displacement anti-node. At the lowest resonant frequency exactly one-fourth of a wavelength fits into the depth of the well. If d is the depth and λ is the wavelength then λ = 4d. The frequency is f = v/λ = v/4d, where v is the speed of sound. The speed of sound is given by

/,vBρ= where B is the bulk modulus and ρ is the density of air in the well. Thus

Bd f ρ

50. We observe that “third lowest … frequency” corresponds to harmonic number nA = 3 for pipe A which is open at both ends. Also, “second lowest … frequency” corresponds to harmonic number nB = 3 for pipe B which is closed at one end.

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(a) Since the frequency of B matches the frequency of A, using Eqs. 17-39 and 17-41, we have

which implies /2(1.20 m)/20.60 mBALL===. Using Eq. 17-40, the corresponding wavelength is

The change from node to anti-node requires a distance of λ/4 so that every increment of

0.20 m along the x axis involves a switch between node and anti-node. Since the closed end is a node, the next node appears at x = 0.40 m So there are 2 nodes. The situation corresponds to that illustrated in Fig. 17-15(b) with 3n=.

(b) The smallest value of x where a node is present is x = 0. (c) The second smallest value of x where a node is present is x = 0.40m.

(d) Using v = 343 m/s, we find f3 = v/λ = 429 Hz. Now, we find the fundamental resonant frequency by dividing by the harmonic number, f1 = f3/3 = 143 Hz.

51. Let the period be T. Then the beat frequency is 1/440Hz4.00beats/s.T−=

Therefore, T = 2.25 × 10–3 s. The string that is “too tightly stretched” has the higher tension and thus the higher (fundamental) frequency.

52. Since the beat frequency equals the difference between the frequencies of the two tuning forks, the frequency of the first fork is either 381 Hz or 387 Hz. When mass is added to this fork its frequency decreases (recall, for example, that the frequency of a mass-spring oscillator is proportional to 1/m). Since the beat frequency also decreases the frequency of the first fork must be greater than the frequency of the second. It must be 387 Hz.

53. Each wire is vibrating in its fundamental mode so the wavelength is twice the length of the wire (λ = 2L) and the frequency is where /vτµ= is the wave speed for the wire, τ is the tension in the wire, and µ is the linear mass density of the wire. Suppose the tension in one wire is τ and the oscillation frequency of that wire is f1. The tension in the other wire is τ + ∆τ and its frequency is f2.

You want to calculate ∆τ/τ for f1 = 600 Hz and f2 = 606 Hz. Now, 1(1/2)/fL=τµ and

54. (a) The number of different ways of picking up a pair of tuning forks out of a set of five is 5!/(2!3!) = 10. For each of the pairs selected, there will be one beat frequency. If these frequencies are all different from each other, we get the maximum possible number of 10.

(b) First, we note that the minimum number occurs when the frequencies of these forks, labeled 1 through 5, increase in equal increments: fn = f1 + n∆f, where n = 2, 3, 4, 5. Now, there are only 4 different beat frequencies: fbeat = n∆f, where n = 1, 2, 3, 4.

5. In the general Doppler shift equation, the trooper’s speed is the source speed and the speeder’s speed is the detector’s speed. The Doppler effect formula, Eq. 17–47, and its accompanying rule for choosing ± signs, are discussed in §17-10. Using that notation, we have v = 343 m/s, and f = 500 Hz. Thus,

56. The Doppler effect formula, Eq. 17–47, and its accompanying rule for choosing ± signs, are discussed in §17-10. Using that notation, we have v = 343 m/s, vD = 2.4 m/s, f ′ = 1590 Hz and f = 1600 Hz. Thus,

v v f f v v v v v v f

57. We use vS = rω (with r = 0.600 m and ω = 15.0 rad/s) for the linear speed during circular motion, and Eq. 17–47 for the Doppler effect (where f = 540 Hz, and v = 343 m/s for the speed of sound).

(a) The lowest frequency is

vf f v v

(b) The highest frequency is

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0555 HzS vf f v v

58. We are combining two effects: the reception of a moving object (the truck of speed u = 45.0 m/s) of waves emitted by a stationary object (the motion detector), and the subsequent emission of those waves by the moving object (the truck) which are picked up by the stationary detector. This could be figured in two steps, but is more compactly computed in one step as shown here:

final initial 343m/s 45m/s(0.150MHz)0.195MHz.

343m/s 45m/s v uf f v u

59. In this case, the intruder is moving away from the source with a speed u satisfying u/v = 1. The Doppler shift (with u = –0.950 m/s) leads to

60. We use Eq. 17–47 with f = 1200 Hz and v = 329 m/s.

(a) In this case, vD = 65.8 m/s and vS = 29.9 m/s, and we choose signs so that f ′ is larger than f:

(b) The wavelength is λ = v/f ′ = 0.208 m.

(c) The wave (of frequency f ′) “emitted” by the moving reflector (now treated as a

“source,” so vS = 65.8 m/s) is returned to the detector (now treated as a detector, so vD = 29.9 m/s) and registered as a new frequency f ′′:

(d) This has wavelength /vf′′ = 0.152 m.

61. We denote the speed of the French submarine by u1 and that of the U.S. sub by u2. (a) The frequency as detected by the U.S. sub is

(b) If the French sub were stationary, the frequency of the reflected wave would be fr = f1(v+u2)/(v – u2). Since the French sub is moving towards the reflected signal with speed u1, then r r v u v u v uf f f v v v u

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

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

ratio vs/v. The result is– 2 = 0.236. Thus, vs/v = 0.236.

which, after setting (f′′′−)/f = 1/2, leads to an equation which can be solved for the

63. As a result of the Doppler effect, the frequency of the reflected sound as heard by the bat is

4 4batbat

64. The “third harmonic” refers to a resonant frequency f3 = 3 f1, where f1 is the fundamental lowest resonant frequency. When the source is stationary, with respect to the air, then Eq. 17-47 gives

where dv is the speed of the detector (assumed to be moving away from the source, in the way we’ve written it, above). The problem, then, wants us to find dv such that f′ = f1 when the emitted frequency is f = f3. That is, we require 1 – dv/v = 1/3. Clearly, the solution to this is dv/v = 2/3 (independent of length and whether one or both ends are open [the latter point being due to the fact that the odd harmonics occur in both systems]). Thus,

(Parte 2 de 4)

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