* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download 16-11. From Eq. (16.10), a general expression for a sinusoidal wave
Density of states wikipedia , lookup
Casimir effect wikipedia , lookup
Weightlessness wikipedia , lookup
Electric charge wikipedia , lookup
Field (physics) wikipedia , lookup
Time in physics wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Anti-gravity wikipedia , lookup
Photon polarization wikipedia , lookup
Speed of gravity wikipedia , lookup
Lorentz force wikipedia , lookup
Aharonov–Bohm effect wikipedia , lookup
Centripetal force wikipedia , lookup
Work (physics) wikipedia , lookup
Electrostatics wikipedia , lookup
Classical central-force problem wikipedia , lookup
Wave packet wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
16-11. From Eq. (16.10), a general expression for a sinusoidal wave traveling along the +x direction is y( x, t ) ym sin(kx t ) (a) Figure 16.34 shows that at x = 0, y(0, t ) ym sin(t ) is a positive sine function, i.e., y(0, t ) ym sin t. Therefore, the phase constant must be . At t =0, we then have y( x, 0) ym sin(kx ) ym sin kx which is a negative sine function. A plot of y(x,0) is depicted below. (b) From the figure we see that the amplitude is ym = 4.0 cm. (c) The angular wave number is given by k = 2/ = /10 = 0.31 rad/cm. (d) The angular frequency is = 2/T = /5 = 0.63 rad/s. (e) As found in part (a), the phase is . (f) The sign is minus since the wave is traveling in the +x direction. (g) Since the frequency is f = 1/T = 0.10 s, the speed of the wave is v = f = 2.0 cm/s. (h) From the results above, the wave may be expressed as x t x t y ( x, t ) 4.0sin 4.0sin . 10 5 10 5 Taking the derivative of y with respect to t, we find u ( x, t ) y x t 4.0 cos t t 10 5 which yields u(0,5.0) = –2.5 cm/s. 16-19. (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 = 55 cm, so = (55 cm – 15 cm) = 40 cm = 0.40 m. (c) The wave speed is v / , where is the tension in the string and is the linear mass density of the string. Thus, v 3.6 N 12 m/s. 25 103 kg/m (d) The frequency is f = v/ = (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 = ym = 2fym = 2(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s. (f) The angular wave number is k = 2/ = 2/(0.40 m) = 16 m–1. (g) The angular frequency is = 2f = 2(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 . We wish to select so that 5.0 10–2 sin = 4.0 10–2. 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 = 0.93 rad. (i) The string displacement has the form y (x, t) = ym sin(kx + t + ). 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 y ( x, t ) 5.0 102 m sin (16 m 1 ) x (190s 1 )t 0.93 . 16-20. (a) The general expression for y (x, t) for the wave is y (x, t) = ym sin(kx – t), which, at x = 10 cm, becomes y (x = 10 cm, t) = ym sin[k(10 cm – t)]. Comparing this with the expression given, we find = 4.0 rad/s, or f = /2 = 0.64 Hz. (b) Since k(10 cm) = 1.0, the wave number is k = 0.10/cm. Consequently, the wavelength is = 2/k = 63 cm. (c) The amplitude is ym 5.0 cm. (d) In part (b), we have shown that the angular wave number is k = 0.10/cm. (e) The angular frequency is = 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 into the general expression for y (x, t), with centimeters and seconds understood, we obtain y ( x, t ) 5.0sin (0.10 x 4.0t ). (g) Since v /k / , the tension is 2 k2 (4.0g / cm) (4.0s1 )2 6400g cm/s2 0.064 N. 1 2 (0.10cm ) 16-59. (a) Recalling the discussion in §16-5, we see that the speed of the wave given by a function with argument x – 5.0t (where x is in centimeters and t is in seconds) must be 5.0 cm/s . (b) In part (c), we show several “snapshots” of the wave: the one on the left is as shown in Figure 16–45 (at t = 0), the middle one is at t = 1.0 s, and the rightmost one is at t 2.0 s . It is clear that the wave is traveling to the right (the +x direction). (c) The third picture in the sequence below shows the pulse at 2.0 s. The horizontal scale (and, presumably, the vertical one also) is in centimeters. (d) The leading edge of the pulse reaches x = 10 cm at t = (10 – 4.0)/5 = 1.2 s. The particle (say, of the string that carries the pulse) at that location reaches a maximum displacement h = 2 cm at t = (10 – 3.0)/5 = 1.4 s. Finally, the trailing edge of the pulse departs from x = 10 cm at t = (10 – 1.0)/5 = 1.8 s. Thus, we find for h(t) at x = 10 cm (with the horizontal axis, t, in seconds): 16-50. The nodes are located from vanishing of the spatial factor sin 5x = 0 for which the solutions are 5x 0, , 2,3, 1 2 3 x 0, , , , 5 5 5 (a) The smallest value of x which corresponds to a node is x = 0. (b) The second smallest value of x which corresponds to a node is x = 0.20 m. (c) The third smallest value of x which corresponds to a node is x = 0.40 m. (d) Every point (except at a node) is in simple harmonic motion of frequency f = /2 = 40/2 = 20 Hz. Therefore, the period of oscillation is T = 1/f = 0.050 s. (e) Comparing the given function with Eq. 16–58 through Eq. 16–60, we obtain y1 0.020sin(5x 40t ) and y2 0.020sin(5x 40t ) for the two traveling waves. Thus, we infer from these that the speed is v = /k = 40/5 = 8.0 m/s. (f) And we see the amplitude is ym = 0.020 m. (g) The derivative of the given function with respect to time is u y (0.040) (40)sin(5x )sin(40t ) t which vanishes (for all x) at times such as sin(40t) = 0. Thus, 40t 0, , 2,3, t 0, 1 2 3 , , , 40 40 40 Thus, the first time in which all points on the string have zero transverse velocity is when t = 0 s. (h) The second time in which all points on the string have zero transverse velocity is when t = 1/40 s = 0.025 s. (i) The third time in which all points on the string have zero transverse velocity is when t = 2/40 s = 0.050 s. 51. From the x = 0 plot (and the requirement of an anti-node at x = 0), we infer a standing wave function of the form y ( x, t ) (0.04) cos(kx) sin(t ), where 2 / T rad/s , with length in meters and time in seconds. The parameter k is determined by the existence of the node at x = 0.10 (presumably the first node that one encounters as one moves from the origin in the positive x direction). This implies k(0.10) = /2 so that k = 5 rad/m. (a) With the parameters determined as discussed above and t = 0.50 s, we find y (0.20 m, 0.50 s) 0.04 cos(kx) sin( t ) 0.040 m . (b) The above equation yields y (0.30 m, 0.50 s) 0.04 cos(kx) sin( t ) 0 . (c) We take the derivative with respect to time and obtain, at t = 0.50 s and x = 0.20 m, u dy 0.04 cos kx cos t 0 . dt d) The above equation yields u = –0.13 m/s at t = 1.0 s. (e) The sketch of this function at t = 0.50 s for 0 x 0.40 m is shown below: 17-12. 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 t + )|t=0 = smcos(kxA + ), where xA = 2.00 m. Regarding “air molecule B” we have sB = + 1 3 sm = sm cos( kxB t + )|t=0 = sm cos( kxB + ). These statements lead to the following conditions: kxA + kxB + cos1(1/3) = 1.231 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 = 2 we find = 0.357 m, which means f = v/ = 343/0.357 = 960 Hz. Another way to complete this problem (once k is found) is to use kv = and then the fact that = f. 17-54. 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 v u2 f1 f1 (1000 Hz) v u1 5470 70 3 1.02 10 Hz. 5470 50 v u1 v u1 v u 2 v u1 f '1 f1 f ' r f r v u2 v u2 v u1 v u 2 (b) 1000 5470 70 5470 50 1.044 10 3 Hz 5470 505470 70 17-57. As a result of the Doppler effect, the frequency of the reflected sound as heard by the bat is v ubat v v / 40 4 4 fr f (3.9 10 Hz) 4.110 Hz. v v / 40 v ubat 21-66. (a) A force diagram for one of the balls is shown below. The force of gravity mg acts downward, the electrical force Fe of the other ball acts to the left, and the tension in the thread acts along the thread, at the angle to the vertical. The ball is in equilibrium, so its acceleration is zero. The y component of Newton’s second law yields T cos – mg = 0 and the x component yields T sin – Fe = 0. We solve the first equation for T and obtain T = mg/cos. We substitute the result into the second to obtain mg tan – Fe = 0. Examination of the geometry of Figure 21-43 leads to tan x2 . bg L2 x 2 2 If L is much larger than x (which is the case if is very small), we may neglect x/2 in the denominator and write tan x/2L. This is equivalent to approximating tan by sin. The magnitude of the electrical force of one ball on the other is Fe q2 4 0 x 2 by Eq. 21-4. When these two expressions are used in the equation mg tan = Fe, we obtain F G H mgx 1 q2 q2 L x 2 L 4 0 x 2 2 0mg IJ . K 1/ 3 (b) We solve x3 = 2kq2L/mg) for the charge (using Eq. 21-5): mgx3 q 2kL 0.010 kg 9.8m s2 0.050 m 2 8.99 109 N m2 C2 1.20 m 3 2.4 108 C. Thus, the magnitude is | q | 2.4 108 C. 22-12. By symmetry we see the contributions from the two charges q1 = q2 = +e cancel each other, and we simply use Eq. 22-3 to compute magnitude of the field due to q3 = +2e. (a) The magnitude of the net electric field is | Enet | 19 1 2e 1 2e 1 4e ) 9 4(1.60 10 (8.99 10 ) 160 N/C. 2 2 6 2 4 0 r 4 0 (a / 2) 2 4 0 a (6.00 10 ) (b) This field points at 45.0°, counterclockwise from the x axis. 22-19. Consider the figure below. (a) The magnitude of the net electric field at point P is 1 q Enet 2 E1 sin 2 2 2 4 0 d / 2 r d /2 d / 2 2 r2 1 qd 4 0 d / 2 2 r 2 3/ 2 For r d , we write [(d/2)2 + r2]3/2 r3 so the expression above reduces to | Enet | 1 qd . 4 0 r 3 (b) From the figure, it is clear that the net electric field at point P points in the j direction, or 90 from the +x axis. 22-20. Referring to Eq. 22-6, we use the binomial expansion (see Appendix E) but keeping higher order terms than are shown in Eq. 22-7: q d 3 d2 1 d3 d 3 d2 1 d3 1 + + + + … 1 z 4 z2 2 z3 z + 4 z2 2 z3 + … 4o z2 E = = qd 2o z3 + q d3 4o z5 +… Therefore, in the terminology of the problem, Enext = q d3/ 40z5. 22-46. Due to the fact that the electron is negatively charged, then (as a consequence of Eq. 22-28 and Newton’s second law) the field E pointing in the +y direction (which we will call “upward”) leads to a downward acceleration. This is exactly like a projectile motion problem as treated in Chapter 4 (but with g replaced with a = eE/m = 8.78 1011 m/s2). Thus, Eq. 4-21 gives t= x 3.00 m = 1.53 x 107 m/s vo cos 40º = 1.96 106 s. This leads (using Eq. 4-23) to vy = vo sin 40º a t = 4.34 105 m/s . Since the x component of velocity does not change, then the final velocity is v = (1.53 106 m/s) i (4.34 105 m/s) j 2 23-2. We use E A , where A Aj 140 . m j . ^ ^ . b g 2 (a) 6.00 N C ˆi 1.40 m ˆj 0. 2 (b) 2.00 N C ˆj 1.40 m ˆj 3.92 N m 2 C. 2 (c) 3.00 N C ˆi 400 N C kˆ 1.40 m ˆj 0 . (d) The total flux of a uniform field through a closed surface is always zero. 23-3. We use E dA and note that the side length of the cube is (3.0 m–1.0 m) = 2.0 m. z (a) On the top face of the cube y = 2.0 m and dA dA ĵ . Therefore, we have 2 E 4iˆ 3 2.0 2 ˆj 4iˆ 18jˆ . Thus the flux is top E dA top 4iˆ 18jˆ dA ˆj 18 top dA 18 2.0 N m2 C 72 N m2 C. (b) On the bottom face of the cube y = 0 and 2 dA dA j . Therefore, we have b gej c h E 4 i 3 02 2 j 4 i 6j . Thus, the flux is bottom E dA bottom 4iˆ 6jˆ dA ˆj 6 bottom dA 6 2.0 N m2 C 24 N m 2 C. 2 (c) On the left face of the cube dA dA î . So Eˆ dA left left 4iˆ E ˆj dA ˆi 4 y bottom dA 4 2.0 N m2 C 16 N m2 C. 2 (d) On the back face of the cube dA dA k̂ . But since E has no z component E dA 0 . Thus, = 0. (e) We now have to add the flux through all six faces. One can easily verify that the flux through the front face is zero, while that through the right face is the opposite of that through the left one, or +16 N·m2/C. Thus the net flux through the cube is = (–72 + 24 – 16 + 0 + 0 + 16) N·m2/C = – 48 N·m2/C. 23-7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the shape of a cube, of edge length d, with a proton of charge q 1.6 1019 C situated at the inside center of the cube. The cube has six faces, and we expect an equal amount of flux through each face. The total amount of flux is net = q/0, and we conclude that the flux through the square is one-sixth of that. Thus, = q/60 = 3.01 10–9 Nm2/C. 23-33. In the region between sheets 1 and 2, the net field is E1 – E2 + E3 = 2.0 105 N/C . In the region between sheets 2 and 3, the net field is at its greatest value: E1 + E2 + E3 = 6.0 105 N/C . The net field vanishes in the region to the right of sheet 3, where E1 + E2 = E3 . We note the implication that 3 is negative (and is the largest surface-density, in magnitude). These three conditions are sufficient for finding the fields: E1 = 1.0 105 N/C , E2 = 2.0 105 N/C , From Eq. 23-13, we infer (from these values of E) E3 = 3.0 105 N/C . |3| |2| 3.0 x 105 N/C = 2.0 x 105 N/C = 1.5 . Recalling our observation, above, about 3, we conclude 3 2 = –1.5 . 23-36. The field due to the sheet is E = 2 . The force (in magnitude) on the electron (due to that field) is F = eE, and assuming it’s the only force then the acceleration is a= e 2o m = slope of the graph ( = 2.0 105 m/s divided by 7.0 1012 s) . Thus we obtain = 2.9 106 C/m2. 23-50. The field is zero for 0 r a as a result of Eq. 23-16. Thus, (a) E = 0 at r = 0, (b) E = 0 at r = a/2.00, and (c) E = 0 at r = a. For a r b the enclosed charge qenc (for a r b) is related to the volume by qenc F4r G H3 3 IJ K 4a 3 . 3 Therefore, the electric field is F G H IJ K 1 qenc 4r 3 4a 3 r 3 a3 E 4 0 r 2 4 0r 2 3 3 3 0 r 2 for a r b. (d) For r =1.50a, we have E (1.50a)3 a3 a 2.375 (1.84 109 )(0.100) 2.375 7.32 N/C. 3 0 (1.50a) 2 3 0 2.25 3(8.85 1012 ) 2.25 (e) For r = b=2.00a, the electric field is E (2.00a)3 a3 a 7 (1.84 109 )(0.100) 7 12.1 N/C. 3 0 (2.00a)2 3 0 4 3(8.85 1012 ) 4 (f) For r b we have E qtotal / 4 r 2 or E b3 a . 3 0 r 2 Thus, for r = 3.00b = 6.00a, the electric field is E (2.00a)3 a3 a 7 (1.84 109 )(0.100) 7 1.35 N/C. 3 0 (6.00a) 2 3 0 36 3(8.85 1012 ) 4 23-59. None of the constant terms will result in a nonzero contribution to the flux (see Eq. 23-4 and Eq. 23-7), so we focus on the x dependent term only: ^ Enon-constant = (4.00y2 ) i (in SI units) . ^ The face of the cube located at y = 4.00 has area A = 4.00 m2 (and it “faces” the + j direction) and has a “contribution” to the flux equal to Enon-constant A = (4)(42)(4) = –256 (in SI units). The face of the cube located at y = 2.00 m has the same area A (and this one “faces” the –j direction) and a contribution to the flux: Enon-constant A = (4)(22)(4) = (in SI units). Thus, the net flux is = 256 + 64 = 192 N·m/C2. According to Gauss’s law, we therefore have qenc = = 1.70 109 C. ^ 24-6. (a) By Eq. 24-18, the change in potential is the negative of the “area” under the curve. Thus, using the area-of-a-triangle formula, we have V 10 z x 2 0 1 E ds 2 20 2 bgb g which yields V = 30 V. z (b) For any region within 0 x 3 m, E ds is positive, but for any region for which x > 3 m it is negative. Therefore, V = Vmax occurs at x = 3 m. V 10 z x 3 0 1 E ds 3 20 2 bgb g which yields Vmax = 40 V. (c) In view of our result in part (b), we see that now (to find V = 0) we are looking for some X > 3 m such that the “area” from x = 3 m to x = X is 40 V. Using the formula for a triangle (3 < x < 4) and a rectangle (4 < x < X), we require bgb gb gb g 1 1 20 X 4 20 40 . 2 Therefore, X = 5.5 m. 24-8. We connect A to the origin with a line along the y axis, along which there is no change of potential (Eq. 24-18: E ds 0 ). Then, we connect the origin to B with a z line along the x axis, along which the change in potential is V z x 4 0 4 I zx dx 4.00FG H2 J K E ds 4.00 4 2 0 which yields VB – VA = –32.0 V. 24-15. A charge –5q is a distance 2d from P, a charge –5q is a distance d from P, and two charges +5q are each a distance d from P, so the electric potential at P is (in SI units) V q (8.99 109 )(5.00 1015 ) 1 1 1 1 5.62 104 V. 2 4 0 2d d d d 8 0 d 2(4.00 10 ) q The zero of the electric potential was taken to be at infinity. 24-64. (a)When the electron is released, its energy is K + U = 3.0 eV 6.0 eV (the latter value is inferred from the graph along with the fact that U = qV and q = e). Because of the minus sign (of the charge) it is convenient to imagine the graph multiplied by a minus sign so that it represents potential energy in eV. Thus, the 2 V value shown at x = 0 would become –2 eV, and the 6 V value at x = 4.5 cm becomes –6 eV, and so on. The total energy (3.0 eV) is constant and can then be represented on our (imagined) graph as a horizontal line at 3.0 V. This intersects the potential energy plot at a point we recognize as the turning point. Interpolating in the region between 1.0 cm and 4.0 cm, we find the turning point is at x = 1.75 cm 1.8 cm. (b) There is no turning point towards the right, so the speed there is nonzero. Noting that the kinetic energy at x = 7.0 cm is 3.0 eV (5.0 eV) = 2.0 eV, we find the speed using energy conservation: v= 2(2.0 eV) m = 2(2.0 eV)(1.6 x 10-19 J/eV) = 8.4 105 m/s . 9.11 x 10-31 kg (c) The electric field at any point P is the (negative of the) slope of the voltage graph evaluated at P. Once we know the electric field, the force on the electron follows immediately from F = q E , where q = e for the electron. In the region just to the left of x = 4.0 cm, the field is E = (133 V/m) î and the magnitude of the force is F 2.11017 N . (d) The force points in the +x direction. (e) In the region just to the right of x = 5.0 cm, the field is E = +100 V/m î and the force is F = ( –1.6 x 1017 N) î . Thus, the magnitude of the force is F 1.6 1017 N . (f) The minus sign indicates that F points in the –x direction. 65. We treat the system as a superposition of a disk of surface charge density and radius R and a smaller, oppositely charged, disk of surface charge density – and radius r. For each of these, Eq 24-37 applies (for z > 0) V 2 0 ez 2 j R2 z 2 0 ez 2 j r2 z . This expression does vanish as r , as the problem requires. Substituting r = 0.200R and z = 2.00R and simplifying, we obtain V R 5 5 101 (6.20 1012 )(0.130) 5 5 101 2 1.03 10 V. 12 0 10 8.85 10 10