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CHAPTER 17
WAVES—I
473
C
HAPTER
17
Answer to Checkpoint Questions
.
(
a
) ; (
b
) ; (
c
)
.
(
a
) , , ; (
b
) , then and tie
.
a
.
:
and
:
tie, then
:
,
:
.
(
a
) ; (
b
) ; (
c
)
.
(
a
)
Hz; (
b
)
Hz
Answer to Questions
.
d
.
b, a, c
.
tie of A and B, then C, D
.
(
a
) ; (
b
) ; (
c
)
.
intermediate (closer to fully destructive interference)
.
(
a
) all tie; (
b
) intermediate (closer to fully constructive interference)
.
a and d tie, then b and c tie
.
(
a
)
and
; (
b
)
,
, and
; or
,
, and
.
(
a
) ; (
b
) antinode; (
c
) longer; (
d
) lower
.
(
a
) decrease; (
b
) disappears
.
(
a
) integer multiples of ; (
b
) node; (
c
) node
.
d
.
string A
.
(
a
) increase, decrease; (
b
) same, increase; (
c
) same, decrease
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CHAPTER 17
WAVES—I
.
decrease
E
(
a
) The frequency is
f
=
v=
= (
m/s)
=
(
:
m) =
Hz
:
(
b
) The period is
T
=
=f
=
=
Hz =
:
s.
E
(
a
)
k
=
=
:
m
=
:
m
:
(
b
)
v
=
f
=
!
=
(
:
m)(
rad/s)
=
:
m/s
:
E
(
a
) For visible light
f
min
=
c
max
=
:
m/s
m
=
:
Hz
and
f
max
=
c
min
=
:
m/s
m
=
:
Hz
:
(
b
) For radio waves
min
=
c
max
=
:
m/s
Hz
=
:
m
and
max
=
c
min
=
:
m/s
:
Hz
=
:
m
:
(
c
) For X rays
f
min
=
c
max
=
:
m/s
:
m
=
:
Hz
y
(cm)
x
(cm)
0
80
160
(
t
=0)
2.0
2.0
CHAPTER 17
WAVES—I
475
and
f
max
=
c
min
=
:
m/s
:
m
=
:
Hz
:
E
(
a
)
T
= (
:
s) =
:
s.
(
b
)
f
=
=T
=
=
:
s =
:
Hz
:
(
c
)
v
=
f
= (
:
m)(
:
Hz) =
:
m/s
:
E
(
a
)
y
(
x; t
) =
y
m
sin(
kx
+
!t
) =
y
m
sin
f
x
v
+
t
= (
:
m) sin
(
Hz)
x
m/s
+
t
=
:
m sin[
(
:
x
+
t
)]
;
where
x
is in meters and
t
is in seconds.
E
(
a
)
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2.0
y
(cm)
0
80
160
(
t
=0.05s)
2.0
2.0
y
(cm)
0
80
160
(
t
=0.10s)
x
(cm)
x
(cm)
476
CHAPTER 17
WAVES—I
(
b
)
(
c
) The wave speed is
v
= (
x=
t
)
peak
=
cm
=
:
s =
cm/s
:
The wave travels in
the negative
x
direction.
E
You can easily verify these alternative forms by using the following relations:
k
=
=
,
=
vT
=
v=f
, and
!
=
f
=
=T
=
kv
.
E
(
a
) Compare the wave
f
(
x; t
) =
h
(
x
t
) with the general form
f
(
x; t
) =
f
(
x
vt
) to
obtain
v
= cm/s
:
(
b
) It travels in the positive
x
direction.
h
(
x
)
2
11
13
14
x
t
=2s
h
(
t
)
2
t
x
=10cm
1.2
1.4
1.8
CHAPTER 17
WAVES—I
477
(
c
)
E
(
a
) Use Eq.

:
u
max
=
@y
@t
max
= [
!y
m
cos(
kx
!t
)]
max
=
fy
m
:
(
b
)
a
y
=
@u
@t
=
@
@t
[
!y
m
cos(
kx
!t
)] =
!
y
m
sin(
kx
!t
)
:
Thus
a
y;
max
= [
!
y
m
sin(
kx
!t
)]
max
=
f
y
m
:
E
(
a
)
A
=
:
mm
; f
=
s
=
=
Hz
; v
= (
s
)
=
(
m
) =
m/s
;
and
=
v=f
= (
m/s)
=
(
Hz) =
:
m
:
(
b
)
u
max
=
!A
=
fA
=
(
Hz)(
:
mm) =
:
m/s
:
E
(
a
) Let the displacements of the wave at (
y; t
) be
z
(
y; t
). Then
z
(
y; t
) =
z
m
sin(
ky
!t
),
where
z
m
=
:
mm,
k
=
cm
, and
!
=
=T
=
=
:
s =
s
. Thus
z
(
y; t
) = (
:
mm) sin[(
cm
)
y
(
s
)
t
]
:
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.
0
1
20
y(cm)
x
(cm)
t
=0
4.0
478
CHAPTER 17
WAVES—I
(
b
)
u
z;
max
=
!z
m
= (
=
:
s)(
:
mm) =
mm/s
:
P
(
a
)
A
=
:
cm
:
(
b
) Solve
from
=
=
:
:
=
cm
:
(
c
) Solve
f
=
:
:
f
=
:
Hz
:
(
d
)
v
=
f
= (
cm)(
:
Hz) =
cm/s
:
(
e
) The wave propagates in the negative
x
direction.
(
f
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.
 Fall '07
 GROUPTEST

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