Niu (qn269) – HW10 – Tsoi – (58020)
1
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001
10.0 points
A 0
.
462 kg mass is attached to a spring and
undergoes simple harmonic motion with a pe
riod oF 0
.
66 s. The total energy oF the system
is 4
.
2 J.
±ind the amplitude oF the motion.
Correct answer: 0
.
447902 m.
Explanation:
Let :
m
= 0
.
462 kg
,
T
= 0
.
66 s
,
and
E
= 4
.
2 J
.
The period is
T
= 2
π
r
m
k
k
=
4
π
2
m
T
2
=
4
π
2
(0
.
462 kg)
(0
.
66 s)
2
= 41
.
871 N
/
m
,
and the energy is
E
=
1
2
k A
2
A
=
r
2
E
k
=
R
2 (4
.
2 J)
41
.
871 N
/
m
=
0
.
447902 m
.
002
10.0 points
IF almost any system in stable equilibrium is
slightly disturbed, it will then exhibit simple
harmonic motion because
1.
the potential energy oF a system near a
state oF static equilibrium is proportional to
the cube oF the displacement From the equi
librium position.
2.
momentum and mechanical energy are
both conserved.
3.
mechanical energy is conserved.
4.
the Force on a system in stable equilibrium
is zero.
5.
the momentum oF a system in stable equi
librium is zero.
6.
the potential energy oF a system near a
state oF static equilibrium is linearly propor
tional to the displacement From the equilib
rium position.
7.
the potential energy oF a system near
a state oF static equilibrium is proportional
to the square oF the displacement From the
equilibrium position.
correct
8.
momentum is conserved.
9.
the kinetic energy on a system in stable
equilibrium is zero.
10.
the Force on a system in unstable equilib
rium is zero.
Explanation:
±or simple harmonic motion, the restoring
Force obeys Hooke’s law (
F
∝
x
−
x
0
, the
displacement From the equilibrium position),
so the potential energy
U
∝
(
x
−
x
0
)
2
.
003
10.0 points
A particle oscillates harmonically
x
=
A
sin(
ωt
+
φ
0
)
,
with amplitude 13 m, angular Frequency
π
s
−
1
, and initial phase
π
3
radians.
Every
now and then, the particle’s kinetic energy
and potential energy happen to be equal to
each other (
K
=
U
).
When does this equality happen For the frst
time aFter
t
= 0?
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View Full DocumentNiu (qn269) – HW10 – Tsoi – (58020)
2
1.
0.4167 s
correct
2.
0.9967 s
3.
0.7615 s
4.
0.6547 s
5.
0.5884 s
6.
0.8623 s
7.
0.5267 s
8.
0.2238 s
9.
0.3467 s
10.
0.1294 s
Explanation:
Let :
A
= 13 m
,
ω
=
π
s
−
1
,
and
φ
0
=
π
3
.
x
=
A
cos(
ω t
+
φ
0
)
v
=
−
A π
sin(
ω t
+
φ
0
)
and
ω
=
r
k
m
.
K
=
U
1
2
mv
2
=
1
2
k x
2
A
2
ω
2
sin
2
(
ω t
+
φ
0
) =
k
m
A
2
cos
2
(
ω t
+
φ
0
)
ω
R
sin
2
(
ω t
+
φ
0
) =
r
k
m
R
cos
2
(
ω t
+
φ
0
)
v
v
v
sin
p
π t
+
π
3
Pv
v
v
=
v
v
v
cos
p
π t
+
π
3
Pv
v
v
.
Since

sin
x

=

cos
x

when
x
=
(2
n
+ 1)
π
4
, n
= 0
,
±
1
,
±
2
, ... ,
(2
n
+ 1)
π
4
=
π t
+
π
3
2
n
+ 1
4
=
t
+
1
3
t
=
2
n
+ 1
4
−
1
3
.
Since
t >
0 for
n
≥
1, the Frst positive time is
when
t
= 1:
t
= 0
.
416667
004
10.0 points
If almost any system in stable equilibrium is
slightly disturbed, it will then exhibit simple
harmonic motion because
1.
the force on a system in stable equilibrium
is zero.
2.
the change in potential energy is propor
tional to the square of the displacement from
the equilibrium position.
correct
3.
momentum is conserved.
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 Spring '09
 Staff
 Physics, Energy, Force, Friction, Mass, Simple Harmonic Motion

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