obaidi (aao476) – OH12 – Bradley – (191)
1
This
printout
should
have
17
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
This is Online Homework No. 12. It is due
07:00 AM on Thursday 20 November (UAE
Time), unless you are otherwise notified.
001
10.0 points
An oscillator is described by
1
2
3
4
5
6
7
8
9 10 11 12
1
2
x
(t)
t
(sec)
What is the angular frequency
ω
?
1.
3.1 rad/s
2.
2.0 rad/s
3.
6.2 rad/s
4.
1.6 rad/s
correct
5.
4.0 rad/s
Explanation:
From the graph,
T
= 4 s.
ω
=
2
π
T
=
2
π
4 sec
=
π
2
1
/
sec =
1
.
5708 1
/
s
.
002
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
3
correct
2.
T
= 6
π
3.
T
=
3
2
π
4.
T
=
9
2
π
5.
T
=
2
π
9
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
.
003
(part 1 of 2) 5.0 points
A piston in an automobile engine is in simple
harmonic motion. Its amplitude of oscillation
from the equilibrium (centered) position is
±
7
.
05 cm and its mass is 1
.
242 kg.
Find the maximum velocity of the piston
when the auto engine is running at the rate of
5380 rev
/
min.
Correct answer: 39
.
7192 m
/
s.
Explanation:
Let :
A
= 7
.
05 cm
,
m
= 1
.
242 kg
,
and
f
= 5380 rev
/
min
.
ω
= 2
π f
= 2
π
(5380 rev
/
min)
1 min
60 s
= 563
.
392 rad
/
s
.
The simple harmonic motion is described
by
x
=
A
cos
ω t ,
where
ω
is the frequency in rad/s if
t
is in
seconds.
The velocity is
v
=
d x
dt
=

A ω
sin(
ω t
)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
obaidi (aao476) – OH12 – Bradley – (191)
2
meaning the maximum velocity is
v
max
=
A ω
= (0
.
0705 m) (563
.
392 rad
/
s)
=
39
.
7192 m
/
s
,
since sine has a maximum value of 1.
004
(part 2 of 2) 5.0 points
Find the maximum acceleration of the piston
when the auto engine is running at this rate.
Correct answer: 22
.
3775 km
/
s
2
.
Explanation:
The acceleration is
a
=
d
2
x
dt
2
=
d v
dt
=

A ω
2
cos(
ω t
)
,
so the maximum acceleration is
a
max
=
A ω
2
= (0
.
0705 m) (563
.
392 rad
/
s)
2
=
22
.
3775 km
/
s
2
,
since cosine has a maximum value of 1.
005
10.0 points
A body oscillates with simple harmonic mo
tion along the
x
axis. Its displacement varies
with time according to the equation
A
=
A
0
sin
parenleftBig
ω t
+
π
3
parenrightBig
,
where
ω
=
π
radians per second,
t
is in sec
onds, and
A
0
= 5
.
9 m.
What is the phase of the motion at
t
=
4
.
4 s?
Correct answer: 14
.
8702 rad.
Explanation:
Let :
t
= 4
.
4 s
and
ω
=
π .
x
=
A
0
sin(
ω t
+
φ
)
The phase is the angle in the argument of the
sine function, and from the problem state
ment we see it is
φ
=
π t
+
π
3
= (
π
rad
/
s) (4
.
4 s) +
π
3
=
14
.
8702 rad
.
006
(part 1 of 4) 3.0 points
A 12
.
4 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: 89
.
8027 s
−
1
.
Explanation:
Let :
M
= 12
.
4 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 .
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 curtise
 Correct Answer, OH12 – Bradley

Click to edit the document details