# Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch16

(Parte 1 de 4)

Chapter 16

1. (a) The angular wave number is 1223.49m. 1.80m

(b) The speed of the wave is ()()1.80m110rads 31.5m s.

2. The distance d between the beetle and the scorpion is related to the transverse speed tv and longitudinal speed vl as t td v t v t= = l l where t and tl are the arrival times of the wave in the transverse and longitudinal directions, respectively. With 50m/stv= and 150m/sv=l, we have t v

3. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one-fourth of a period. The period is T = 4(0.170 s) = 0.680 s.

(b) The frequency is the reciprocal of the period:

(c) A sinusoidal wave travels one wavelength in one period:

4. (a) The speed of the wave is the distance divided by the required time. Thus,

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853seats 21.87 seats/s 2 seats/s

(b) The width w is equal to the distance the wave has moved during the average time required by a spectator to stand and then sit. Thus,

5. Let y1 = 2.0 m (corresponding to time t1) and y2 = –2.0 m (corresponding to time t2). Then we find

kx + 600t2 + φ = sin−1(–2.0/6.0) . Subtracting equations gives

6. Setting x = 0 in u = −ω ym cos(k x − ω t + φ) (see Eq. 16-21 or Eq. 16-28) gives as the function being plotted in the graph. We note that it has a positive “slope” (referring to its t-derivative) at t = 0:

= = − ym ω² sin(−ω t + φ) > 0 at t = 0.

This implies that – sinφ > 0 and consequently that φ is in either the third or fourth quadrant. The graph shows (at t = 0) u = −4 m/s, and (at some later t) umax = 5 m/s. We note that umax = ym ω. Therefore, u = − umax cos(− ω t + φ)|t = 0 ⇒ φ = cos−1( ) = ± 0.6435 rad

(bear in mind that cosθ = cos(−θ )), and we must choose φ = −0.64 rad (since this is about −37° and is in fourth quadrant). Of course, this answer added to 2npi is still a valid answer (where n is any integer), so that, for example, φ = −0.64 + 2pi = 5.64 rad is also an acceptable result.

7. Using v = fλ, we find the length of one cycle of the wave is

From f = 1/T, we find the time for one cycle of oscillation is T = 1/50 = 2.0 × 10–3 s = 2.0 ms.

(a) A cycle is equivalent to 2pi radians, so that pi/3 rad corresponds to one-sixth of a cycle. The corresponding length, therefore, is λ/6 = 700/6 = 117 m.

(b) The interval 1.0 ms is half of T and thus corresponds to half of one cycle, or half of 2pi rad. Thus, the phase difference is (1/2)2pi = pi rad.

8. (a) The amplitude is ym = 6.0 cm. (b) We find λ from 2pi/λ = 0.020pi: λ = 1.0×102 cm.

(c) Solving 2pif = ω = 4.0pi, we obtain f = 2.0 Hz.

(d) The wave speed is v = λf = (100 cm) (2.0 Hz) = 2.0×102 cm/s.

(e) The wave propagates in the –x direction, since the argument of the trig function is kx + ωt instead of kx – ωt (as in Eq. 16-2).

(f) The maximum transverse speed (found from the time derivative of y) is

(g) y(3.5 cm, 0.26 s) = (6.0 cm) sin[0.020pi(3.5) + 4.0pi(0.26)] = –2.0 cm. 9. (a) Recalling from Ch. 12 the simple harmonic motion relation um = ymω, we have

Since ω = 2pif, we obtain f = 64 Hz. (b) Using v = fλ, we find λ = 80/64 = 1.26 m 1.3 m≈.

(c) The amplitude of the transverse displacement is 24.0 cm4.010m.my− ==×

(d) The wave number is k = 2pi/λ = 5.0 rad/m. (e) The angular frequency, as obtained in part (a), is 216/0.0404.010rad/s.ω==×

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(f) The function describing the wave can be written as ( )0.040sin 5 400y x t φ= − + where distances are in meters and time is in seconds. We adjust the phase constant φ to satisfy the condition y = 0.040 at x = t = 0. Therefore, sin φ = 1, for which the “simplest” root is φ = pi/2. Consequently, the answer is

(g) The sign in front of ω is minus. 10. With length in centimeters and time in seconds, we have

Squaring this and adding it to the square of 15piy, we have

 u = = 15pi .

so that

Therefore, where y = 12, u must be ± 135pi. Consequently, the speed there is 424 cm/s = 4.24 m/s.

1. (a) The amplitude ym is half of the 6.0 m vertical range shown in the figure, i.e., 3.0 m.my=

(b) The speed of the wave is v = d/t = 15 m/s, where d = 0.060 m and t = 0.0040 s. The angular wave number is k = 2pi/λ where λ = 0.40 m. Thus, k = = 16 rad/m . (c) The angular frequency is found from

(d) We choose the minus sign (between kx and ωt) in the argument of the sine function because the wave is shown traveling to the right [in the +x direction] – see section 16-5). Therefore, with SI units understood, we obtain

12. The slope that they are plotting is the physical slope of sinusoidal waveshape (not to be confused with the more abstract “slope” of its time development; the physical slope is an x-derivative whereas the more abstract “slope” would be the t-derivative). Thus, where the figure shows a maximum slope equal to 0.2 (with no unit), it refers to the maximum of the following function:

= = ym k cos(k x − ω t) .

The problem additionally gives t = 0, which we can substitute into the above expression if desired. In any case, the maximum of the above expression is ym k , where

 Therefore, setting ym k equal to 0.20 allows us to solve for the amplitude ym We find

13. From Eq. 16-10, a general expression for a sinusoidal wave traveling along the +x direction is

(a) The figure shows that at x = 0, (0,)sin()mytytωφ=−+is a positive sine function, i.e.,

(0,)sin.mytytω=+ Therefore, the phase constant must be φpi=. At t =0, we then have

which is a negative sine function. A plot of y(x,0) is depicted on the right.

(b) From the figure we see that the amplitude is ym = 4.0 cm. (c) The angular wave number is given by k = 2pi/λ = pi/10 = 0.31 rad/cm.

(d) The angular frequency is ω = 2pi/T = pi/5 = 0.63 rad/s.

(e) As found in part (a), the phase is φpi=. (f) The sign is minus since the wave is traveling in the +x direction.

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(g) Since the frequency is f = 1/T = 0.10 s, the speed of the wave is v = fλ = 2.0 cm/s. (h) From the results above, the wave may be expressed as x t x ty x t pi pi pi pipi

Taking the derivative of y with respect to t, we find

which yields u(0,5.0) = –2.5 cm/s.

14. From v=τµ, we have new newnew old old old 2.v

15. The wave speed v is given by v = τµ, where τ is the tension in the rope and µ is the linear mass density of the rope. The linear mass density is the mass per unit length of rope:

16. The volume of a cylinder of height l is V = pir2l= pid2l/4. The strings are long, narrow cylinders, one of diameter d1 and the other of diameter d2 (and corresponding linear densities µ1 and µ2). The mass is the (regular) density multiplied by the volume: m = ρV, so that the mass-per-unit length is

and their ratio is

d d d d µ piρ µ piρ

Therefore, the ratio of diameters is

17. (a) The amplitude of the wave is ym=0.120 m.

(b) The wave speed is given by v = τµ, where τ is the tension in the string and µ is the linear mass density of the string, so the wavelength is λ = v/f = τµ/f and the angular wave number is

(c) The frequency is f = 100 Hz, so the angular frequency is ω = 2pif = 2pi(100 Hz) = 628 rad/s.

(d) We may write the string displacement in the form y = ym sin(kx + ωt). The plus sign is used since the wave is traveling in the negative x direction. In summary, the wave can be expressed as

19. (a) The wave speed is given by v = λ/T = ω/k, where λ is the wavelength, T is the period, ω is the angular frequency (2pi/T), and k is the angular wave number (2pi/λ). The displacement has the form y = ym sin(kx + ωt), so k = 2.0 m–1 and ω = 30 rad/s. Thus

(b) Since the wave speed is given by v = τµ, where τ is the tension in the string and µ is the linear mass density of the string, the tension is

20. (a) Comparing with Eq. 16-2, we see that k = 20/m and ω = 600/s. Therefore, the speed of the wave is (see Eq. 16-13) v = ω/k = 30 m/s.

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(b) From Eq. 16–26, we find

21. (a) We read the amplitude from the graph. It is about 5.0 cm.

(b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm and again with the same slope at about x = 5 cm, so λ = (5 cm – 15 cm) = 40 cm = 0.40 m.

(c) The wave speed is /,v=τµ where τ is the tension in the string and µ is the linear mass density of the string. Thus,

(d) The frequency is f = v/λ = (12 m/s)/(0.40 m) = 30 Hz and the period is

T = 1/f = 1/(30 Hz) = 0.033 s. (e) The maximum string speed is um = ωym = 2pifym = 2pi(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s. (f) The angular wave number is k = 2pi/λ = 2pi/(0.40 m) = 16 m–1.

(g) The angular frequency is ω = 2pif = 2pi(30 Hz) = 1.9×102 rad/s

(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 × 10–2 m. The formula for the displacement gives y(0, 0) = ym sin φ. We wish to select φ so that

The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select φ = 0.93 rad.

(i) The string displacement has the form y (x, t) = ym sin(kx + ωt + φ). A plus sign appears in the argument of the trigonometric function because the wave is moving in the negative x direction. Using the results obtained above, the expression for the displacement is

2. (a) The general expression for y (x, t) for the wave is y (x, t) = ym sin(kx – ωt), which, at x = 10 cm, becomes y (x = 10 cm, t) = ym sin[k(10 cm – ωt)]. Comparing this with the expression given, we find ω = 4.0 rad/s, or f = ω/2pi = 0.64 Hz.

(b) Since k(10 cm) = 1.0, the wave number is k = 0.10/cm. Consequently, the wavelength is λ = 2pi/k = 63 cm.

(c) The amplitude is 5.0 cm.my= (d) In part (b), we have shown that the angular wave number is k = 0.10/cm.

(e) The angular frequency is ω = 4.0 rad/s. (f) The sign is minus since the wave is traveling in the +x direction.

Summarizing the results obtained above by substituting the values of k and ω into the general expression for y (x, t), with centimeters and seconds understood, we obtain

(g) Since //,vk==ωτµ the tension is

23. The pulses have the same speed v. Suppose one pulse starts from the left end of the wire at time t = 0. Its coordinate at time t is x1 = vt. The other pulse starts from the right end, at x = L, where L is the length of the wire, at time t = 30 ms. If this time is denoted by t0 then the coordinate of this wave at time t is x2 = L – v(t – t0). They meet when x1 = x2, or, what is the same, when vt = L – v(t – t0). We solve for the time they meet: t = (L + vt0)/2v and the coordinate of the meeting point is x = vt = (L + vt0)/2. Now, we calculate the wave speed:

Here τ is the tension in the wire and L/m is the linear mass density of the wire. The coordinate of the meeting point is

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This is the distance from the left end of the wire. The distance from the right end is L – x = (10.0 m – 7.37 m ) = 2.63 m.

24. (a) The tension in each string is given by τ = Mg/2. Thus, the wave speed in string 1 is

(b) And the wave speed in string 2 is

(d) And we solve for the second mass: M2 = M – M1 = (500 g – 187.5 g) ≈ 313 g.

25. (a) The wave speed at any point on the rope is given by v = τµ, where τ is the tension at that point and µ is the linear mass density. Because the rope is hanging the tension varies from point to point. Consider a point on the rope a distance y from the bottom end. The forces acting on it are the weight of the rope below it, pulling down, and the tension, pulling up. Since the rope is in equilibrium, these forces balance. The weight of the rope below is given by µgy, so the tension is τ = µgy. The wave speed is

(b) The time dt for the wave to move past a length dy, a distance y from the bottom end, is dddtyvygy== and the total time for the wave to move the entire length of the rope is

L y y Lt

26. Using Eq. 16–3 for the average power and Eq. 16–26 for the speed of the wave, we solve for f = ω/2pi:

P f

y µ τ µ −

27. We note from the graph (and from the fact that we are dealing with a cosine-squared, see Eq. 16-30) that the wave frequency is f = = 500 Hz, and that the wavelength λ = 0.20 m. We also note from the graph that the maximum value of dK/dt is 10 W. Setting this equal to the maximum value of Eq. 16-29 (where we just set that cosine term equal to 1) we find µ v ω2 ym2 = 10 with SI units understood. Substituting in µ = 0.002 kg/m, ω = 2pif and v = f λ , we solve for the wave amplitude:

ym = = 0.0032 m .

(,)sin()myxtykxtω=−, we see that 14.0 mk−=and 7.0 rad/sω=. The speed of the wave is angular wave number120 mk−= and angular frequency4.0 rad/sω=. Thus, the speed of the wave is angular wave number 130 mk−= and angular frequency 6.0 rad/sω=. Thus, the speed of the wave is

31. The displacement of the string is given by

where φ = pi/2. The amplitude is

32. (a) Let the phase difference be φ. Then from Eq. 16–52, 2ym cos(φ/2) = 1.50ym, which gives

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(b) Converting to radians, we have φ = 1.45 rad.

(c) In terms of wavelength (the length of each cycle, where each cycle corresponds to 2pi rad), this is equivalent to 1.45 rad/2pi = 0.230 wavelength.

3. (a) The amplitude of the second wave is 9.0 mmmy=, as stated in the problem.

 number is k = 2pi/0.40 = 16 rad/m

(b) The figure indicates that λ = 40 cm = 0.40 m, which implies that the angular wave

(c) The figure (along with information in the problem) indicates that the speed of each wave is v = dx/t = (56.0 cm)/(8.0 ms) = 70 m/s. This, in turn, implies that the angular frequency is

(d) The figure depicts two traveling waves (both going in the –x direction) of equal amplitude ym. The amplitude of their resultant wave, as shown in the figure, is y′m = 4.0 m. Eq. 16-52 applies:

 y′m = 2 ym cos( φ2) ⇒ φ2 = 2 cos−1(2.0/9.0) = 2.69 rad.

(e) In making the plus-or-minus sign choice in y = ym sin(k x ± ω t + φ), we recall the discussion in section 16-5, where it shown that sinusoidal waves traveling in the –x direction are of the form y = ym sin(k x + ω t + φ). Here, φ should be thought of as the phase difference between the two waves (that is, φ1 = 0 for wave 1 and φ2 = 2.69 rad for wave 2).

In summary, the waves have the forms (with SI units understood):

y1 = (0.00900)sin(16 x +10 t) and y2 = (0.00900)sin(16 x + 10 t + 2.7 ) .

34. (a) We use Eq. 16-26 and Eq. 16-3 with µ = 0.00200 kg/m and ym = 0.00300 m. These give v = = 775 m/s and

Pavg = µv ω2ym2 = 10 W.

(b) In this situation, the waves are two separate string (no superposition occurs). The answer is clearly twice that of part (a); P = 20 W.

(c) Now they are on the same string. If they are interfering constructively (as in Fig.

16-16(a)) then the amplitude ym is doubled which means its square ym2 increases by a factor of 4. Thus, the answer now is four times that of part (a); P = 40 W.

(d) Eq. 16-52 indicates in this case that the amplitude (for their superposition) is

2 ymcos(0.2pi) = 1.618 times the original amplitude ym. Squared, this results in an increase in the power by a factor of 2.618. Thus, P = 26 W in this case.

(e) Now the situation depicted in Fig. 16-16(b) applies, so P = 0.

35. The phasor diagram is shown below: y1m and y2m represent the original waves and ym represents the resultant wave. The phasors corresponding to the two constituent waves make an angle of 90° with each other, so the triangle is a right triangle. The Pythagorean theorem gives

Thus ym = 5.0 cm.

36. (a) As shown in Figure 16-16(b) in the textbook, the least-amplitude resultant wave is obtained when the phase difference is pi rad.

(b) In this case, the amplitude is (8.0 m – 5.0 m) = 3.0 m.

(Parte 1 de 4)