bello (rtb473) – hw12 – Demkov – (59910)
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001
(part 1 of 3) 10.0 points
A particle rotates counterclockwise in a circle
of radius 5
.
9 m with a constant angular speed
of 7 rad
/
s
.
At
t
= 0, the particle has an
x
coordinate of 4
.
5 m and
y >
0
.
x
y
(4
.
5 m
, y
)
Figure:
Not drawn to scale.
radius
5
.
9 m
7 rad
/
s
Determine the
x
coordinate of the particle
at
t
= 1
.
14 s
.
Correct answer:

4
.
35108 m.
Explanation:
Let :
x
0
= 4
.
5 m
,
ω
= 7 rad
/
s
,
t
0
= 0 s
,
and
R
= 5
.
9 m
,
Since the amplitude of the particle’s mo
tion
equals
the
radius
of
the
circle
and
ω
= 7 rad
/
s , we have
x
=
A
cos(
ω t
+
φ
)
= (5
.
9 m) cos
bracketleftBig
(7 rad
/
s)
t
+
φ
bracketrightBig
.
We can find
φ
using the initial condition that
x
0
= 4
.
5 m at
t
= 0
(4
.
5 m) = (5
.
9 m) cos(0 +
φ
)
,
which implies
φ
= arccos
x
0
R
= arccos
(4
.
5 m)
(5
.
9 m)
= 40
.
2962
◦
= 0
.
7033 rad
.
Therefore, at time
t
= 1
.
14 s , the
x
coordinate
of the particle is
x
=
R
cos
bracketleftBig
ω t
+
φ
bracketrightBig
= (5
.
9 m) cos
bracketleftBig
(7 rad
/
s) (1
.
14 s)
+ (0
.
7033 rad)
bracketrightBig
=

4
.
35108 m
.
Note:
The angles in the cosine are in radians.
002
(part 2 of 3) 10.0 points
Find the
x
component of the particle’s veloc
ity at
t
= 1
.
14 s.
Correct answer:

27
.
8931 m
/
s.
Explanation:
Differentiating the function
x
(
t
) with re
spect to
t
, we find the
x
component of the
particle’s velocity at any time
t
v
x
=
d x
dt
=

ω A
sin(
ω t
+
φ
)
,
so at
t
= 1
.
14 s
,
the argument of the sine is
φ
2
≡
ω t
+
φ
= (7 rad
/
s) (1
.
14 s) + (0
.
7033 rad)
= 8
.
6833 rad
,
and the
x
component of the velocity of the
particle is
v
x
=

ω R
sin(
φ
2
)
=

(7 rad
/
s) (5
.
9 m) sin(8
.
6833 rad)
=

27
.
8931 m
/
s
.
003
(part 3 of 3) 10.0 points
Find the
x
component of the particle’s accel
eration at
t
= 1
.
14 s.
Correct answer: 213
.
203 m
/
s
2
.
Explanation:
Differentiating the
x
component of the par
ticle’s velocity with respect to
t
, we find the
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bello (rtb473) – hw12 – Demkov – (59910)
2
x
component of the particle’s acceleration at
any time
t
a
x
=
d v
x
dt
=

ω
2
A
cos(
ω t
+
φ
)
,
so at
t
= 1
.
14 s
,
the
x
component of the
particle’s acceleration is
a
x
=

ω
2
R
cos
φ
2
=

(7 rad
/
s)
2
(5
.
9 m) cos(8
.
6833 rad)
=
213
.
203 m
/
s
2
.
004
10.0 points
When a body executes simple harmonic mo
tion, its period is
1.
proportional to its acceleration.
2.
independent of its amplitude.
correct
3.
inversely proportional to its accelera
tion.
4.
the reciprocal of its speed.
5.
proportional to its amplitude.
Explanation:
The period of simple harmonic motion is
the same regardless of the amplitude.
005
(part 1 of 3) 10.0 points
A simple harmonic oscillator takes 17
.
1 s to
undergo five complete vibrations.
Find the period of its motion.
Correct answer: 3
.
42 s.
Explanation:
Let :
t
= 17
.
1 s
and
n
= 5
.
The period is, by definition, how long it
takes the oscillator to complete one full cycle;
i.e.
, to return to where it started.
Since we
are told how long it takes to undergo
five
complete vibrations, the period is
T
=
t
n
=
17
.
1 s
5
=
3
.
42 s
.
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 Spring '07
 Swinney
 mechanics, Correct Answer, Bello

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