Periodic Motion
133
(b)
Eq.(13.18):
0
0
arctan(
/
)
x
v
x
φ
ω
=
−
/
(300 N/m)/2.00 kg
12.25 rad/s
k m
ω
=
=
=
(
4.00 m/s)
arctan
arctan( 1.633)
58.5
(12.25 rad/s)(0.200 m)
φ
⎛
⎞
−
=
−
=
+
=
°
⎜
⎟
⎝
⎠
(or 1.02 rad)
(c)
cos(
)
x
A
t
ω
φ
=
+
gives
(0.383 m)cos([12.2rad/s]
1.02 rad)
x
t
=
+
E
VALUATE
:
At
0
t
=
the block is displaced 0.200 m from equilibrium but is moving, so
0.200 m.
A
>
According
to Eq.(13.15), a phase angle
φ
in the range 0
90
φ
<
<
°
gives
0
0.
x
v
<
13.13.
I
DENTIFY
:
For SHM,
2
2
(2
)
x
a
ω
x
π
f
x
= −
= −
. Apply Eqs.(13.13), (13.15) and (13.16), with
A
and
φ
from Eqs.(13.18)
and (13.19).
S
ET
U
P
:
1.1 cm
x
=
,
0
15 cm/s
x
v
= −
.
2
f
ω
π
=
, with
2.5 Hz
f
=
.
E
XECUTE
:
(a)
(
)
2
2
2
2
(2.5 Hz)
(1.1
10
m)
2.71 m/s
.
x
a
π
−
= −
×
= −
(b)
From Eq. (13.19) the amplitude is 1.46 cm, and from Eq. (13.18) the phase angle is 0.715 rad. The angular
frequency is 2
15.7 rad/s,
π
f
=
so
(1.46 cm) cos ((15.7 rad/s)
0.715 rad)
x
t
=
+
,
(
22.9 cm s) sin ((15.7 rad/s)
0.715 rad)
x
v
t
= −
+
and
2
(
359 cm/s
) cos ((15.7 rad/s)
0.715 rad) .
x
a
t
= −
+
E
VALUATE
:
We can verify that our equations for
x
,
x
v
and
x
a
give the specified values at
0
t
=
.
13.14.
I
DENTIFY
and
S
ET
U
P
:
Calculate
x
using Eq.(13.13). Use
T
to calculate
ω
and
0
x
to calculate
.
φ
E
XECUTE
:
0
x
=
at
0
t
=
implies that
/2 rad
φ
π
= ±
Thus
cos(
/2).
x
A
t
ω
π
=
±
2
/
T
π ω
=
so
2
/
2
/1.20 s
5.236 rad/s
T
ω
π
π
=
=
=
(0.600 m)cos([5.236 rad/s][0.480 s]
/2)=
0.353 m.
x
π
=
±
∓
The distance of the object from the equilibrium position is 0.353 m.
E
VALUATE
:
The problem doesn’t specify whether the object is moving in the
x
+
or
x
−
direction at
0.
t
=
13.15.
I
DENTIFY
:
Apply
2
m
T
k
π
=
. Use the information about the empty chair to calculate
k
.
S
ET
U
P
:
When
42.5 kg
m
=
,
1.30 s
T
=
.
E
XECUTE
:
Empty chair:
2
m
T
π
k
=
gives
2
2
2
2
4
4
(42.5 kg)
993 N/m
(1.30 s)
π
m
π
k
T
=
=
=
With person in chair:
2
m
T
k
π
=
gives
2
2
2
2
(2.54 s)
(993 N/m)
162 kg
4
4
T k
m
π
π
=
=
=
and
person
162 kg
42.5 kg
120 kg
m
=
−
=
.
E
VALUATE
:
For the same spring, when the mass increases, the period increases.
13.16.
I
DENTIFY
and
S
ET
U
P
:
Use Eq.(13.12) for
T
and Eq.(13.4) to relate
x
a
and
k
.
E
XECUTE
:
2
/ ,
T
m k
π
=
0.400 kg
m
=
Use
2
2.70 m/s
x
a
= −
to calculate
k
:
x
kx
ma
−
=
gives
2
(0.400 kg)(
2.70 m/s
)
3.60 N/m
0.300 m
x
ma
k
x
−
= −
= −
= +
2
/
2.09 s
T
m k
π
=
=
E
VALUATE
:
a
x
is negative when
x
is positive.
/
x
ma x
has units of N/m and
/
m k
has units of s.
13.17.
I
DENTIFY
:
2
m
T
k
π
=
.
x
k
a
x
m
= −
so
max
k
a
A
m
=
.
F
kx
= −
.
S
ET
U
P
:
x
a
is proportional to
x
so
x
a
goes through one cycle when the displacement goes through one cycle. From
the graph, one cycle of
a
x
extends from
0.10 s
t
=
to
0.30 s,
t
=
so the period is
0.20 s.
T
=
2.50 N/cm
250 N/m.
k
=
=
From the graph the maximum acceleration is
2
12.0 m/s
.
E
XECUTE
:
(a)
2
m
T
k
π
=
gives
2
2
0.20 s
(250 N/m)
0.253 kg
2
2
T
m
k
π
π
⎛
⎞
⎛
⎞
=
=
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
(b)
2
max
(0.253 kg)(12.0 m/s
)
0.0121 m
1.21 cm
250 N/m
ma
A
k
=
=
=
=
(c)
max
(250 N/m)(0.0121 m)
3.03 N
F
kA
=
=
=
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134
Chapter 13
E
VALUATE
:
We can also calculate the maximum force from the maximum acceleration:
2
max
max
(0.253 kg)(12.0 m/s
)
3.04 N,
F
ma
=
=
=
which agrees with our previous results.
13.18.
I
DENTIFY
:
The general expression for
( )
x
v
t
is
( )
sin(
)
x
v
t
A
t
ω
ω
φ
= −
+
. We can determine
ω
and
A
by comparing
the equation in the problem to the general form.
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 Spring '06
 Buchler
 Physics, Energy, Kinetic Energy, m/s

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