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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
r
m
k
= 2
π
R
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, ±nd 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
p

Δ
E

E
P
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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View Full Document 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 diFerences 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
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Force, Work

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