salazar (ggs332) – HW11 – TSOI – (58160)
1
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beFore answering.
001
10.0 points
A body oscillates with simple harmonic mo
tion along the
x
axis. Its displacement varies
with time according to the equation
x
(
t
) =
A
sin(
ω t
+
φ
)
.
IF
A
= 3 m,
ω
= 3
.
572 rad
/
s, and
φ
=
1
.
0472 rad, what is the acceleration oF the
body at
t
= 4 s? Note: The argument oF the
sine Function is in radians rather than degrees.
Correct answer:
−
13
.
9404 m
/
s
2
.
Explanation:
Let :
A
= 3 m
,
ω
= 3
.
572 rad
/
s
,
φ
= 1
.
0472 rad
,
and
t
= 4 s
.
x
=
A
sin(
ω t
+
φ
)
v
=
dx
dt
=
ω A
cos(
ω t
+
φ
)
a
=
dv
dt
=
−
ω
2
A
sin(
ω t
+
φ
)
=
−
ω
2
A
sin(
ω t
+
φ
)
=
−
(3
.
572 rad
/
s)
2
(3 m)
×
sin[(3
.
572 rad
/
s)(4 s) + 1
.
0472 rad]
=
−
13
.
9404 m
/
s
2
.
002
10.0 points
A mass attached to a spring executes sim
ple harmonic motion in a horizontal plane
with an amplitude oF 3
.
64 m.
At a point
1
.
3468 m away From the equilibrium, the mass
has speed 0
.
886 m
/
s.
What is the period oF oscillation oF the
mass? Consider equations For
x
(
t
) and
v
(
t
)
and use sin
2
+ cos
2
= 1 to calculate
ω
.
Correct answer: 23
.
9816 s.
Explanation:
Let
A
= 3
.
64 m
,
x
= 1
.
3468 m
,
and
v
= 0
.
886 m
/
s
.
The simplest solution uses the equation For
simple harmonic motion
x
(
t
) =
A
sin(
ω t
)
,
and its time derivative
v
(
t
) =
dt
=
Aω
cos(
ω t
)
.
±or any angle (such as the phase angle
ω t
)
sin
2
(
ω t
) + cos
2
(
ω t
) = 1
,
so
p
x
A
P
2
+
p
v
A ω
P
2
= 1
,
and
p
v
ω
P
2
=
A
2
−
x
2
.
Consequently,
ω
=
v
√
A
2
−
x
2
=
0
.
886 m
/
s
r
(3
.
64 m)
2
−
(1
.
3468 m)
2
= 0
.
262 s
−
1
and
T
=
2
π
ω
=
2
π
0
.
262 s
−
1
=
23
.
9816 s
.
003
(part 1 oF 2) 10.0 points
A 1
.
91 kg object on a Frictionless horizontal
track is attached to the end oF a horizontal
spring whose Force constant is 4
.
59 N
/
m. The
object is displaced 4
.
4 m to the right From its
equilibrium position and then released, which
initiates simple harmonic motion.
What is the magnitude oF the Force acting
on the object 3
.
67 s aFter it is released?
Correct answer: 16
.
7374 N.
Explanation:
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View Full Documentsalazar (ggs332) – HW11 – TSOI – (58160)
2
Let :
A
= 4
.
4 m
,
k
= 4
.
59 N
/
m
,
m
= 1
.
91 kg
,
and
t
= 3
.
67 s
.
The block exhibits simple harmonic motion,
so the displacement is
x
=
A
cos(
ω t
)
where the angular frequency is
ω
=
r
k
m
.
Thus the force is
F
=
−
k x
=
−
k A
cos
p
r
k
m
t
P
=
−
(4
.
59 N
/
m) (4
.
4 m) cos
b
R
4
.
59 N
/
m
1
.
91 kg
(3
.
67 s)
B
=
−
16
.
7374 N
a force of
16
.
7374 N
directed to the left.
004
(part 2 of 2) 10.0 points
How many times does the object oscillate in
3
.
67 s?
Correct answer: 0
.
905474 turn.
Explanation:
The angular frequency is
w
=
r
k
m
and the period of oscillation is
T
=
2
π
ω
= 2
π
r
m
k
,
so the number of oscillations made in 3
.
67 s is
N
=
Δ
t
T
=
Δ
t
2
π
r
k
m
=
3
.
67 s
2
π
R
4
.
59 N
/
m
1
.
91 kg
=
0
.
905474 turn
.
005
10.0 points
A uniform plank of mass
m
is pivoted at
one end.
A spring of force constant
k
is
attached to the center of the plank, as shown
in the Fgure.
The height of the pivot has
been adjusted so that the plank will be in
equilibrium when it is horizontally oriented.
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 Spring '10
 TSOI
 Physics, Energy, Force, Simple Harmonic Motion, Correct Answer, Salazar

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