Download 16-11. From Eq. (16.10), a general expression for a sinusoidal wave

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 103 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 = 2fym = 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  = 2f = 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  102 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) (4.0s1 )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 5x = 0 for
which the solutions are
5x  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(5x  40t ) and y2  0.020sin(5x  40t )
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(5x )sin(40t )
t
which vanishes (for all x) at times such as sin(40t) = 0. Thus,
40t  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 + cos1(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  505470  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) 
  4.110 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 108 C.
Thus, the magnitude is | q | 2.4 108 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 + … 
4o z2 
 
E =
=
qd
2o z3
+
q d3
4o z5
+…
Therefore, in the terminology of the problem, Enext = q d3/ 40z5.
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  106 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  Aj  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  6j . 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 1019 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/60 =
3.01 10–9 Nm2/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
2o m
= slope of the graph ( = 2.0  105 m/s divided by 7.0  1012
s) .
Thus we obtain  = 2.9 106 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
F4r
 G
H3
3
IJ
K
4a 3
.

3
Therefore, the electric field is
F
G
H
IJ
K
1 qenc

4r 3 4a 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 109 )(0.100) 2.375


 7.32 N/C.
3 0 (1.50a) 2
3 0 2.25
3(8.85 1012 )
2.25
(e) For r = b=2.00a, the electric field is
E
 (2.00a)3  a3  a 7 (1.84 109 )(0.100) 7


 12.1 N/C.
3 0 (2.00a)2
3 0 4
3(8.85 1012 ) 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 109 )(0.100) 7


 1.35 N/C.
3 0 (6.00a) 2
3 0 36
3(8.85 1012 ) 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  109
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 1015 )
 1 1 1 1






 5.62 104 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  2.11017 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 1017 N) î . Thus, the magnitude of the force is
F  1.6 1017 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 1012 )(0.130)  5 5  101 
2

 

  1.03 10 V.
12
 0 
10
8.85

10
10


