homework 13 – ALIBHAI, ZAHID – Due: Apr 25 2007, 4:00 am
1
Question 1
part 1 of 3
10 points
A 2
.
2 kg object oscillates with an initial am
plitude of 101 cm on a spring of force constant
98
.
8 N
/
m.
Find the period.
Correct answer: 0
.
93759 s (tolerance
±
1 %).
Explanation:
Let :
m
= 2
.
2 kg
and
k
= 98
.
8 N
/
m
.
The period of the oscillator is
T
= 2
π
radicalbigg
m
k
= 2
π
radicalBigg
2
.
2 kg
98
.
8 N
/
m
=
0
.
93759 s
.
Question 2
part 2 of 3
10 points
Find the total initial energy.
Correct answer: 50
.
3929 J (tolerance
±
1 %).
Explanation:
Let :
A
= 101 cm = 1
.
01 m
.
The initial energy is
E
0
=
1
2
k A
2
=
1
2
(98
.
8 N
/
m)(1
.
01 m)
2
=
50
.
3929 J
.
Question 3
part 3 of 3
10 points
If the energy decreases by 1 percent per
period, find the damping constant
b
.
Correct answer: 0
.
0234644 kg
/
s (tolerance
±
1 %).
Explanation:
Let :

Δ
E

E
= 0
.
01
,
m
= 2
.
2 kg
,
and
T
= 0
.
93759 s
.
Relating the fractional rate at which the en
ergy decreases,
Q
=
2
π
parenleftbigg

Δ
E

E
parenrightbigg
cycle
=
2
π
0
.
01
.
Express the
Q
value in terms of
b
:
Q
=
ω
0
m
b
=
2
π
0
.
01
b
=
0
.
01
ω
0
m
2
π
=
0
.
01 (2
π
)
m
2
π T
=
0
.
01
m
T
=
0
.
01 (2
.
2 kg)
0
.
93759 s
=
0
.
0234644 kg
/
s
.
Question 4
part 1 of 1
10 points
What beat frequencies are possible with
tuning forks of frequencies 256, 259, and 261
Hz?
1.
There are 3 possible beat frequencies, 515
Hz, 520 Hz, and 517 Hz.
2.
There are 3 possible beat frequencies, 256
Hz, 259 Hz, and 261 Hz.
3.
There are 4 possible beat frequencies, 2
Hz, 3 Hz, 5 Hz, and 7 Hz.
4.
There are 3 possible beat frequencies, 2
Hz, 3 Hz, and 5 Hz.
correct
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homework 13 – ALIBHAI, ZAHID – Due: Apr 25 2007, 4:00 am
2
5.
There are 2 possible beat frequencies, 2
Hz and 3 Hz.
Explanation:
The beat frequencies are the differences in
the fork frequencies:
261 Hz

259 Hz = 2 Hz
,
261 Hz

256 Hz = 5 Hz
,
259 Hz

256 Hz = 3 Hz
.
Question 5
part 1 of 1
10 points
A wave pulse on a string is described by the
equation
y
1
=
A
(
B x

C t
)
2
+
D
.
A second wave pulse on the same string is
described by
y
2
=

A
(
B x
+
C t

E
)
2
+
D
,
where
x
is in meters and
t
in seconds, and
A
= 5
.
7 m,
B
= 3
.
58 m
−
1
,
C
= 6 s
−
1
,
D
=
4
.
59, and
E
= 7
.
08.
At what time will the two waves exactly
cancel everywhere?
Correct answer: 0
.
59 s (tolerance
±
1 %).
Explanation:
Basic Concepts:
Superposition principle:
y
=
y
1
(
x, t
) +
y
2
(
x, t
)
Solution:
Since
y
=
y
1
+
y
2
,
y
will be zero when
y
1
+
y
2
= 0
.
This implies, after some algebra, that
B x

C t
=
±
(
B x
+
C t

E
)
(3
.
58 m
−
1
)
x

(6 s
−
1
)
t
=
±
[(3
.
58 m
−
1
)
x
+ (6 s
−
1
)
t

7
.
08]
.
(The
±
arises from taking a square root.)
When
solving
the
“+”
equation,
we
get
2
C t
=
E
, therefore
t
=
E
2
C
=
(7
.
08)
2 (6 s
−
1
)
=
0
.
59 s
.
Because
x
cancels out, our solution tells us
y
is zero
for all
positions
x
at this particular
time
t
.
Question 6
part 1 of 1
10 points
Two wires are made of the same material
but the second wire has twice the diameter
and twice the length of the first wire. When
the two wires are stretched, and the tension
in the second wire is also twice the tension in
the first wire, the fundamental frequency of
the first wire is 920 Hz.
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 Spring '08
 Turner
 Force, Work, Wavelength, Correct Answer, superposed waves

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