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Chapter 15
443
P15.5
(a)
At
t
=
0,
x
=
0 and
v
is positive (to the right). Therefore, this situation corresponds to
xA
t
=
sin
ω
and
vv
t
i
=
cos
Since
f
=
150
.
H
z
,
ωπ
π
==
23
0
0
f
.
Also,
A
=
200
c
m
, so that
xt
=
300
.s
i
n
.
cm
af
(b)
A
i
max
..
.
== =
=
=
200300
600
188
cm s
. cm s
The particle has this speed at
t
=
0 and next at
t
T
2
1
3
s
(c)
aA
max
.
=
=
2
2
2
2 00 3 00
18 0
178
cm s
cm s
22
This positive value of acceleration first occurs at
tT
3
4
0500
s
(d)
Since
T
=
2
3
s and
A
=
c
m
, the particle will travel 8.00 cm in this time.
Hence, in
100
3
2
s
=
F
H
G
I
K
J
T
, the particle will travel
8 00
4 00
12 0
.
cm
cm
cm
+=
.
P15.6
The proposed solution
x
t
v
t
i
i
=+
F
H
G
I
K
J
cos
sin
implies velocity
v
dx
dt
v
t
ii
−
+
ωω
sin
cos
and acceleration
a
dv
dt
v
t
x
t
v
tx
i
i
−
−
=
−
+
F
H
G
I
K
J
F
H
G
I
K
J
=−
2
cos
sin
cos
sin
(a)
The acceleration being a negative constant times position means we do have SHM, and its
angular frequency is
. At
t
=
0 the equations reduce to
xx
i
=
and
i
=
so they satisfy all
the requirements.
(b)
v a
x
x
tv
t
x
t
v
t
i
i
i
i
2
2
2
−=
−
+
−
−
−
+
F
H
G
I
K
J
F
H
G
I
K
J
sin
cos
cos
sin
cos
sin
bg
ej
va
x
x
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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