homework 11 – JACOBS, AMBER – Due: Apr 22 2008, 1:00 am
1
Question 1, chap 15, sect 6.
part 1 of 3
10 points
A pendulum that moves through its equilib
rium position once every 1.000 s is sometimes
called a “seconds pendulum.”
a) What is the period of any seconds pen
dulum?
1.
4.000 s
2.
None of these
3.
2.000 s
correct
4.
0.250 s
5.
1.000 s
6.
0.500 s
Explanation:
Basic Concept:
A pendulum passes through its equilibrium
position twice each cycle.
Solution:
T
= 2(1
.
000 s)
= 2
.
000 s
Question 2, chap 15, sect 6.
part 2 of 3
10 points
In Cambridge, England, a seconds pendu
lum is 0.9942 m long.
b) What is the freefall acceleration in Cam
bridge?
1.
None of these
2.
9
.
81236 m
/
s
2
correct
3.
9
.
81341 m
/
s
2
4.
9
.
79552 m
/
s
2
5.
9
.
81 m
/
s
2
6.
9
.
79657 m
/
s
2
Explanation:
Basic Concept:
T
= 2
π
radicalBigg
L
g
Solution:
parenleftbigg
T
2
π
parenrightbigg
2
=
L
g
g
1
=
L
1
parenleftbigg
2
π
T
parenrightbigg
2
= (0
.
9942 m)
parenleftbigg
2
π
2 s
parenrightbigg
2
= 9
.
81236 m
/
s
2
Question 3, chap 15, sect 6.
part 3 of 3
10 points
In Tokyo, Japan, a seconds pendulum is
0.9926 m long.
c)
What
is
the
freefall
acceleration
in
Tokyo?
1.
None of these
2.
9
.
81 m
/
s
2
3.
9
.
79552 m
/
s
2
4.
9
.
79657 m
/
s
2
correct
5.
9
.
81341 m
/
s
2
6.
9
.
81236 m
/
s
2
Explanation:
Solution:
g
2
=
L
2
parenleftbigg
2
π
T
parenrightbigg
2
= (0
.
9926 m)
parenleftbigg
2
π
2 s
parenrightbigg
2
= 9
.
79657 m
/
s
2
Question 4, chap 15, sect 7.
part 1 of 3
10 points
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homework 11 – JACOBS, AMBER – Due: Apr 22 2008, 1:00 am
2
A particle rotates counterclockwise in a cir
cle of radius 6
.
6 m with a constant angular
speed of 18 rad
/
s
.
At
t
= 0, the particle has
an
x
coordinate of 5
.
3 m and
y >
0
.
x
y
(5
.
3 m
, y
)
Figure:
Not drawn to scale.
radius
6
.
6 m
18 rad
/
s
Determine the
x
coordinate of the particle
at
t
= 1
.
71 s
.
Correct answer:
6
.
59998
m (tolerance
±
1
%).
Explanation:
LET :
x
0
= 5
.
3 m
,
ω
= 18 rad
/
s
,
t
0
= 0 s
,
and
R
= 6
.
6 m
,
Since the amplitude of the particle’s mo
tion
equals
the
radius
of
the
circle
and
ω
= 18 rad
/
s , we have
x
=
A
cos(
ω t
+
φ
)
= (6
.
6 m) cos
bracketleftBig
(18 rad
/
s)
t
+
φ
bracketrightBig
.
We can find
φ
using the initial condition that
x
0
= 5
.
3 m at
t
= 0
(5
.
3 m) = (6
.
6 m) cos(0 +
φ
)
,
which implies
φ
= arccos
x
0
R
= arccos
(5
.
3 m)
(6
.
6 m)
= 36
.
5796
◦
= 0
.
638433 rad
.
Therefore, at time
t
= 1
.
71 s , the
x
coordinate
of the particle is
x
=
R
cos
bracketleftBig
ω t
+
φ
bracketrightBig
= (6
.
6 m) cos
bracketleftBig
(18 rad
/
s) (1
.
71 s)
+ (0
.
638433 rad)
bracketrightBig
=
6
.
59998 m
.
Note:
The angles in the cosine are in radians.
Question 5, chap 15, sect 7.
part 2 of 3
10 points
Find the
x
component of the particle’s ve
locity at
t
= 1
.
71 s.
Correct answer:

0
.
297822 m
/
s (tolerance
±
1 %).
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
.
71 s
,
the argument of the sine is
φ
2
≡
ω t
+
φ
= (18 rad
/
s) (1
.
71 s) + (0
.
638433 rad)
= 31
.
4184 rad
,
and the
x
component of the velocity of the
particle is
v
x
=

ω R
sin(
φ
2
)
=

(18 rad
/
s) (6
.
6 m) sin(31
.
4184 rad)
=

0
.
297822 m
/
s
.
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 Spring '07
 KOPP
 Physics, Work, Frequency, Correct Answer

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