alameri (aoa434) – Hwk12 – Stokes – (19102)
1
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19
questions.
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This is ONLINE HOMEWORK No 12. It
is due before 6.00 am on Tuesday 28th April,
Abu Dhabi Time (9.00 pm Monday in Texas)
001
10.0 points
The equation of motion of a simple harmonic
oscillator is
d
2
x
dt
2
=
−
9
x ,
where
x
is displacement and
t
is time.
What is the period of oscillation?
1.
T
=
2
π
9
2.
T
=
3
2
π
3.
T
=
9
2
π
4.
T
=
2
π
3
correct
5.
T
= 6
π
Explanation:
d
2
x
dt
2
=
−
ω
2
x ,
where
ω
is the angular frequency, so the pe
riod of oscillation is
T
=
2
π
ω
=
2
π
√
9
=
2
π
3
.
002
(part 1 of 3) 10.0 points
A 668 g mass is connected to a light spring
of force constant 2 N
/
m that is free to oscillate
on a horizontal, frictionless track. The mass
is displaced 8 cm from the equilibrium point
and released from rest.
2 N
/
m
668 g
8 cm
x
= 0
x
Find the period of the motion.
Correct answer: 3
.
63122 s.
Explanation:
Let :
m
= 668 g = 0
.
668 kg
and
k
= 2 N
/
m
.
This situation corresponds to the special
case
x
(
t
) =
A
cos
ωt ,
so the frequency is
ω
=
radicalbigg
k
m
=
radicalBigg
2 N
/
m
0
.
668 kg
= 1
.
73032 s
−
1
and the period is
T
=
2
π
ω
=
2
π
1
.
73032 s
−
1
=
3
.
63122 s
.
003
(part 2 of 3) 10.0 points
What is the maximum speed of the mass?
Correct answer: 0
.
138426 m
/
s.
Explanation:
Let :
A
= 8 cm
.
The velocity as a function of time is
v
(
t
) =
−
ω A
sin(
ω t
)
,
so the maximum speed of the mass is
v
max
=
ω A
= (1
.
73032 s
−
1
) (0
.
08 m)
=
0
.
138426 m
/
s
.
004
(part 3 of 3) 10.0 points
What is the maximum acceleration of the
mass?
Correct answer: 0
.
239521 m
/
s
2
.
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alameri (aoa434) – Hwk12 – Stokes – (19102)
2
Explanation:
The acceleration as a function of time is
a
(
t
) =
−
ω
2
A
cos(
ω t
)
,
so the maximum acceleration of the mass is
a
max
=
ω
2
A
= (1
.
73032 s
−
1
)
2
(0
.
08 m)
=
0
.
239521 m
/
s
2
.
005
(part 1 of 4) 10.0 points
A 15
.
1 kg mass is suspended on a 1
×
10
5
N
/
m
spring. The mass oscillates up and down from
the equilibrium position
y
eq
= 0 according to
y
(
t
) =
A
sin(
ωt
+
φ
0
)
.
Find the angular frequency of the oscillat
ing mass.
Correct answer: 81
.
3788 s
−
1
.
Explanation:
Let :
M
= 15
.
1 kg
and
k
= 1
×
10
5
N
/
m
.
When the mass moves out of equilibrium,
it suffers a net restoring force
F
net
y
=
F
spring
−
Mg
=
−
k
(
y
−
y
eq
) =
−
ky ,
and accelerates back towards the equilibrium
position at the rate
a
y
=
F
net
y
M
=
−
k
M
y .
Therefore, the mass oscillates harmonically
with angular frequency
ω
=
radicalbigg
−
a
y
y
=
radicalbigg
k
M
=
radicalBigg
1
×
10
5
N
/
m
15
.
1 kg
=
81
.
3788 s
−
1
.
006
(part 2 of 4) 10.0 points
At time
t
0
= 0 the mass is at 19
.
2 cm and
moving upward at velocity +21
.
1 m
/
s.
Find the amplitude of the oscillating mass.
Correct answer: 32
.
2631 cm.
Explanation:
Let :
y
0
= 19
.
2 cm
and
v
0
= 21
.
1 m
/
s
.
The mass oscillates according to the SHM
equation
y
(
t
) =
A
sin(
ωt
+
φ
0
)
,
so its velocity is
v
y
(
t
) =
d y
dt
=
A ω
cos(
ωt
+
φ
0
)
.
At time
t
0
= 0, we have
y
0
≡
y
(
t
= 0) =
A
sin
φ
0
and
v
0
=
v
y
(
t
= 0) =
A ω
cos
φ
0
,
so
A
sin
φ
0
=
y
0
and
A
cos
φ
0
=
v
0
ω
.
(1)
Consequently,
A
2
= (
A
sin
φ
0
)
2
+ (
A
cos
φ
0
)
2
= (
y
0
)
2
+
parenleftBig
v
0
ω
parenrightBig
2
A
=
radicalbigg
(
y
0
)
2
+
parenleftBig
v
0
ω
parenrightBig
2
=
radicalBigg
(19
.
2 cm)
2
+
parenleftbigg
21
.
1 m
/
s
81
.
3788 s
−
1
parenrightbigg
2
=
32
.
2631 cm
.
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 Spring '09
 NAJAFZADEH
 mechanics, Friction, Simple Harmonic Motion, Work, Correct Answer, k2

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