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 1 de 4)

Chapter 17

1. (a) When the speed is constant, we have v = d/t where v = 343 m/s is assumed. Therefore, with t = 15/2 s being the time for sound to travel to the far wall we obtain d =

(343 m/s) × (15/2 s) which yields a distance of 2.6 km.

(b) Just as the 12factor in part (a) was 1/(n + 1) for n = 1 reflection, so also can we write

for multiple reflections (with d in meters). For d = 25.7 m, we find n = 199 2.010≈×.

2. The time it takes for a soldier in the rear end of the column to switch from the left to the right foot to stride forward is t = 1 min/120 = 1/120 min = 0.50 s. This is also the time for the sound of the music to reach from the musicians (who are in the front) to the rear end of the column. Thus the length of the column is

3. (a) The time for the sound to travel from the kicker to a spectator is given by d/v, where d is the distance and v is the speed of sound. The time for light to travel the same distance is given by d/c, where c is the speed of light. The delay between seeing and hearing the kick is ∆t = (d/v) – (d/c). The speed of light is so much greater than the speed of sound that the delay can be approximated by ∆t = d/v. This means d = v ∆t. The distance from the kicker to spectator A is dA = v ∆tA = (343 m/s)(0.23 s) = 79 m.

(b) The distance from the kicker to spectator B is dB = v ∆tB = (343 m/s)(0.12 s) = 41 m.

(c) Lines from the kicker to each spectator and from one spectator to the other form a right triangle with the line joining the spectators as the hypotenuse, so the distance between the spectators is

CHAPTER 17

From /vBρ= we find

5. Let tf be the time for the stone to fall to the water and ts be the time for the sound of the splash to travel from the water to the top of the well. Then, the total time elapsed from dropping the stone to hearing the splash is t = tf + ts. If d is the depth of the well, then the kinematics of free fall gives

The sound travels at a constant speed vs, so d = vsts, or ts = d/vs. Thus the total time is

2//stdgdv=+. This equation is to be solved for d. Rewrite it as 2//sdgtdv=− and square both sides to obtain

Now multiply by g2sv and rearrange to get gd2 – 2vs(gt + vs)d + g2svt2 = 0. This is a quadratic equation for d. Its solutions are s s s s sv gt v v gt v g v t d

The physical solution must yield d = 0 for t = 0, so we take the solution with the negative sign in front of the square root. Once values are substituted the result d = 40.7 m is obtained.

6. Using Eqs. 16-13 and 17-3, the speed of sound can be expressed as

where (/)/BdpdVV=−. Since ,andVλρ are not changed appreciably, the frequency ratio becomes

i i i i f v B dp dV

Thus, we have s i i

i s s dV dp B f

7. If d is the distance from the location of the earthquake to the seismograph and vs is the speed of the S waves then the time for these waves to reach the seismograph is ts. = d/vs. Similarly, the time for P waves to reach the seismograph is tp = d/vp. The time delay is

s p

p s v v t d v v

We note that values for the speeds were substituted as given, in km/s, but that the value for the time delay was converted from minutes to seconds.

8. Let l be the length of the rod. Then the time of travel for sound in air (speed vs) will be /sstv=l. And the time of travel for compressional waves in the rod (speed vr) will be

/rrtv=l. In these terms, the problem tells us that

t t v v

Thus, with vs = 343 m/s and vr = 15vs = 5145 m/s, we find 44ml=. 9. (a) Using λ = v/f, where v is the speed of sound in air and f is the frequency, we find

(b) Now, λ = v/f, where v is the speed of sound in tissue. The frequency is the same for air and tissue. Thus

10. (a) The amplitude of a sinusoidal wave is the numerical coefficient of the sine (or cosine) function: pm = 1.50 Pa.

(b) We identify k = 0.9pi and ω = 315pi (in SI units), which leads to f = ω/2pi = 158 Hz.

CHAPTER 17

(c) We also obtain λ = 2pi/k = 2.2 m.

(d) The speed of the wave is v = ω/k = 350 m/s.

1. Without loss of generality we take x = 0, and let t = 0 be when s = 0. This means the phase is φ = −pi/2 and the function is s = (6.0 nm)sin(ωt) at x = 0. Noting that ω = 3000 rad/s, we note that at t = sin−1(1/3)/ω = 0.13 ms the displacement is s = +2.0 nm.

Doubling that time (so that we consider the excursion from –2.0 nm to +2.0 nm) we conclude that the time required is 2(0.13 ms) = 0.23 ms.

12. The key idea here is that the time delay t∆ is due to the distance d that each wavefront must travel to reach your left ear (L) after it reaches your right ear (R).

(a) From the figure, we find sind Dt

(b) Since the speed of sound in water is now wv, with 90θ=°, we have sin90

w w w

D Dt

(c) The apparent angle can be found by substituting /wDv for t∆:

sin w

D Dt

Solving for θ with 1482m/swv= (see Table 17-1), we obtain

13. (a) Consider a string of pulses returning to the stage. A pulse which came back just before the previous one has traveled an extra distance of 2w, taking an extra amount of time ∆t = 2w/v. The frequency of the pulse is therefore

(b) Since f ∝ 1/w, the frequency would be higher if w were smaller.

14. (a) The period is T = 2.0 ms (or 0.0020 s) and the amplitude is ∆pm = 8.0 mPa (which is equivalent to 0.0080 N/m2). From Eq. 17-15 we get sm = = = 6.1 × 10−9 m . where ρ = 1.21 kg/m3 and v = 343 m/s.

(b) The angular wave number is k = ω/v = 2pi/vT = 9.2 rad/m.

(c) The angular frequency is ω = 2pi/T = 3142 rad/s 3.110 rad/s≈×.

The results may be summarized as s(x, t) = (6.1 nm) cos[(9.2 m−1)x – (3.1 × 103 s−1)t].

(d) Using similar reasoning, but with the new values for density (ρ′ = 1.35 kg/m3) and speed (v′= 320 m/s), we obtain m m p ps

v v Tρ ω ρ pi

(e) The angular wave number is k = ω/v’ = 2pi/v’T = 9.8 rad/m.

(f) The angular frequency is ω = 2pi/T = 3142 rad/s 3.110 rad/s≈×.

The new displacement function is s(x, t) = (5.9 nm) cos[(9.8 m−1)x – (3.1 × 103 s−1)t].

15. The problem says “At one instant..” and we choose that instant (without loss of generality) to be t = 0. Thus, the displacement of “air molecule A” at that instant is sA = +sm = smcos(kxA − ωt + φ)|t=0 = smcos(kxA + φ), where xA = 2.0 m. Regarding “air molecule B” we have sB = + sm = sm cos( kxB − ωt + φ )|t=0 = sm cos( kxB + φ ). These statements lead to the following conditions:

where xB = 2.07 m. Subtracting these equations leads to

k(xB − xA) = 1.231 ⇒k = 17.6 rad/m.

Using the fact that k = 2pi/λ we find λ = 0.357 m, which means f = v/λ = 343/0.357 = 960 Hz.

CHAPTER 17

Another way to complete this problem (once k is found) is to use kv = ω and then the fact that ω = 2pif.

16. Let the separation between the point and the two sources (labeled 1 and 2) be x1 and x2, respectively. Then the phase difference is

(330m/s)/540Hz x x x xft ft pi piφ φ φ pi pi

17. (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 note that at all points (at large distance from the origin) along the x axis there will be quiet ones; one way to see this is to note that the path-length difference (for the waves traveling from their respective sources) divided by wavelength gives the (dimensionless) value 3.5, implying a half-wavelength (180º) phase difference (destructive interference) between the waves. To distinguish the destructive interference along the +x axis from the destructive interference along the –x axis, we label one with +3.5 and the other –3.5. This labeling is useful in that it suggests that the complete enumeration of the quiet directions in the upper-half plane (including the x axis) is: –3.5, –2.5, –1.5, –0.5, +0.5, +1.5, +2.5, +3.5. Similarly, the complete enumeration of the loud directions in the upper-half plane is: –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 discussion about the “quiet” directions was started in part (a). The number of values in the list: –3.5, –2.5, –1.5, –0.5, +0.5, +1.5, +2.5, +3.5 along with –2.5, –1.5, –0.5, +0.5, +1.5, +2.5 (for the lower-half plane) is 14. There are 14 “quiet” directions.

18. At the location of the detector, the phase difference between the wave which traveled straight down the tube and the other one which took the semi-circular detour is

For r = rmin we have ∆φ = pi, which is the smallest phase difference for a destructive interference to occur. Thus,

min 40.0cm 17.5cm.

pi − pi −

19. Let L1 be the distance from the closer speaker to the listener. The distance from the other speaker to the listener is 2 21LLd=+, where d is the distance between the speakers. The phase difference at the listener is φ = 2pi(L2 – L1)/λ, where λ is the wavelength.

For a minimum in intensity at the listener, φ = (2n + 1)pi, where n is an integer. Thus, λ = 2(L2 – L1)/(2n + 1). The frequency is

Now 20,0/343 = 58.3, so 2n + 1 must range from 0 to 57 for the frequency to be in the audible range. This means n ranges from 0 to 28.

(a) The lowest frequency that gives minimum signal is (n = 0) min,1343 Hz.f=

(b) The second lowest frequency is (n = 1) min,2min,1[2(1)1]343 Hz1029 Hz3.f=+== Thus, the factor is 3.

(c) The third lowest frequency is (n=2) min,3min,1[2(2)1]343 Hz1715 Hz5.f=+== Thus, the factor is 5.

For a maximum in intensity at the listener, φ = 2npi, where n is any positive integer. Thus

Since 20,0/686 = 29.2, n must be in the range from 1 to 29 for the frequency to be audible.

(d) The lowest frequency that gives maximum signal is (n =1) max,1686 Hz.f=

(e) The second lowest frequency is (n = 2) max,2max,12(686 Hz)1372 Hz2.f=== Thus, the factor is 2.

(f) The third lowest frequency is (n = 3) max,3max,13(686 Hz)2058 Hz3.f=== Thus, the factor is 3.

20. (a) To be out of phase (and thus result in destructive interference if they superpose) means their path difference must be λ/2 (or 3λ/2 or 5λ/2 or …). Here we see their path difference is L, so we must have (in the least possibility) L = λ/2, or q =L/λ = 0.5.

(b) As noted above, the next possibility is L = 3λ/2, or q =L/λ = 1.5.

CHAPTER 17

21. Building on the theory developed in §17 – 5, we set /1/2, 1,2,...Lnnλ∆=−= in order to have destructive interference. Since v = fλ, we can write this in terms of frequency:

min, where we have used v = 343 m/s (note the remarks made in the textbook at the beginning of the exercises and problems section) and ∆L = (19.5 – 18.3 ) m = 1.2 m.

(a) The lowest frequency that gives destructive interference is (n = 1) min,1(1/2)(286 Hz)143 Hz.f=−= (b) The second lowest frequency that gives destructive interference is (n = 2)

So the factor is 3. (c) The third lowest frequency that gives destructive interference is (n = 3)

Now we set 12/L∆=λ (even numbers) — which can be written more simply as “(all integers n = 1, 2,…)” — in order to establish constructive interference. Thus,

(d) The lowest frequency that gives constructive interference is (n =1)max,1(286 Hz).f= (e) The second lowest frequency that gives constructive interference is (n = 2)

(f) The third lowest frequency that gives constructive interference is (n = 3)

2. (a) The problem indicates that we should ignore the decrease in sound amplitude which means that all waves passing through point P have equal amplitude. Their superposition at P if d = λ/4 results in a net effect of zero there since there are four sources (so the first and third are λ/2 apart and thus interfere destructively; similarly for the second and fourth sources).

(b) Their superposition at P if d = λ/2 also results in a net effect of zero there since there are an even number of sources (so the first and second being λ/2 apart will interfere destructively; similarly for the waves from the third and fourth sources).

(c) If d = λ then the waves from the first and second sources will arrive at P in phase; similar observations apply to the second and third, and to the third and fourth sources. Thus, four waves interfere constructively there with net amplitude equal to 4sm.

23. (a) If point P is infinitely far away, then the small distance d between the two sources is of no consequence (they seem effectively to be the same distance away from P). Thus, there is no perceived phase difference.

(b) Since the sources oscillate in phase, then the situation described in part (a) produces fully constructive interference.

(c) For finite values of x, the difference in source positions becomes significant. The path lengths for waves to travel from S1 and S2 become now different. We interpret the question as asking for the behavior of the absolute value of the phase difference |∆φ|, in which case any change from zero (the answer for part (a)) is certainly an increase.

The path length difference for waves traveling from S1 and S2 is

2 2 for 0.d x x x∆ = + − >l The phase difference in “cycles” (in absolute value) is therefore

Thus, in terms of λ, the phase difference is identical to the path length difference:

rearranging, and solving, we find 2

In general, if lx=Dl for some multiplier ξ > 0, we find

CHAPTER 17

where we have used d = 16.0 m and λ = 2.0 m. (d) For 0.50l=Dl, or 0.50ξ=, we have 0.50) m127.5 m128 mx=(64.0/0.50−=≈.

Note that since whole cycle phase differences are equivalent (as far as the wave superposition goes) to zero phase difference, then the ξ = 1, 2 cases give constructive interference. A shift of a half-cycle brings “troughs” of one wave in superposition with

“crests” of the other, thereby canceling the waves; therefore, the 351 2,,ξ= cases produce destructive interference.

24. (a) Since intensity is power divided by area, and for an isotropic source the area may be written A = 4pir2 (the area of a sphere), then we have

(b) This calculation may be done exactly as shown in part (a) (but with r = 2.5 m instead of r = 1.0 m), or it may be done by setting up a ratio. We illustrate the latter approach. Thus,

I P r r

I P r r

25. The intensity is the rate of energy flow per unit area perpendicular to the flow. The rate at which energy flow across every sphere centered at the source is the same, regardless of the sphere radius, and is the same as the power output of the source. If P is the power output and I is the intensity a distance r from the source, then P = IA = 4pir2I, where A (= 4pir2) is the surface area of a sphere of radius r. Thus

26. Sample Problem 17-5 shows that a decibel difference ∆β is directly related to an intensity ratio (which we write as /I=′R). Thus,

27. The intensity is given by 2212,mIvs=ρω where ρ is the density of air, v is the speed of sound in air, ω is the angular frequency, and sm is the displacement amplitude for the sound wave. Replace ω with 2pif and solve for sm:

pi pi

28. (a) The intensity is given by I = P/4pir2 when the source is “point-like.” Therefore, at r

(b) The sound level there is

29. (a) Let I1 be the original intensity and I2 be the final intensity. The original sound level is β1 = (10 dB) log(I1/I0) and the final sound level is β2 = (10 dB) log(I2/I0), where I0 is the reference intensity. Since β2 = β1 + 30 dB which yields

Divide by 10 dB and use log(I2/I0) – log(I1/I0) = log(I2/I1) to obtain log(I2/I1) = 3. Now use each side as an exponent of 10 and recognize that 21log 2110/I=. The result is I2/I1 =

(Parte 1 de 4)

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