§
13 Problem 7 (
a
): If the frequency is
f
= 6
.
00 Hz = 6
.
00 oscillations per second, then the period is
T
= 1
/f
=
0
.
167 s.
(
b
): The angular frequency is
ω
= 2
πf
= 37
.
7
/
s.
(
c
): We have
ω
=
radicalbigg
k
m
⇒
m
=
k
ω
2
=
120 N
/
m
(37
.
7
/
s)
2
= 0
.
0844 kg
.
§
13 Problem 26 (
a
): The total energy at this point is
E
=
1
2
mv
2
+
1
2
kx
2
=
1
2
(0
.
150 kg)(0
.
300 m
/
s)
2
+
1
2
(300 N
/
m)(0
.
0120 m)
2
= 0
.
0284 J
.
(
b
): The amplitude
A
is the maximum value of
x
, at which point all the energy is potential, so
1
2
kA
2
=
E
⇒
A
=
radicalbigg
2
E
k
= 0
.
0137 m
.
(
c
): When the object is at maximimum speed all the energy is kinetic, so
1
2
mv
2
max
=
E
⇒
v
max
=
2
E
m
= 0
.
615 m
/
s
.
§
13 Problem 36: The moment of inertia is
I
=
1
2
mR
2
= 4
.
84
×
10

7
kg m
2
.
For a torsional pendulum we have
T
= 2
π
radicalbigg
I
κ
⇒
κ
=
4
π
2
I
T
2
= 1
.
91
×
10

5
N m
.
§
13 Problem 47: The period of oscillation is
T
= 1
.
36 s. We have
T
= 2
π
radicalBigg
L
g
⇒
g
=
4
π
2
L
T
2
= 10
.
7 m
/
s
2
.
§
13 Problem 55: We treat both problems as physical pendula, and use
T
= 2
π
radicalBigg
I
Mgd
,
where
d
is the distance from the pivot to the center of mass (which is
L
for both pendula) and
I
is the
moment of inertia of the pendulum about the pivot. For the small bob we have
I
=
ML
2
, while for the large
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 Summer '08
 Chamberlin
 Force, Work

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