oldfinal 01 – PAPAGEORGE, MATT – Due: May 6 2008, 4:00 am
1
Question 1, chap 16, sect 3.
part 1 of 1
10 points
While waiting for Stan Speedy to arrive on
a late passenger train, Kathy Kool notices
beats occurring as a result of two trains blow
ing their whistles simultaneously.
One train
is at rest and the other is approaching her
at a speed of 13
.
9 km
/
h.
Assume that both
whistles have the same frequency and that the
speed of sound is 344 m
/
s.
If Kathy hears beat frequency at 7
.
39 Hz,
what is the frequency of the whistles?
Correct answer: 651
.
011
Hz (tolerance
±
1
%).
Explanation:
The frequency of sound waves produced
by the moving whistle at Kathy’s location is
given, according to Doppler E
ff
ect, by
f
=
v
sound
v
sound

v
train
f
=
344 m
/
s
344 m
/
s

3
.
86111 m
/
s
f
= (1
.
01135)
f .
The beat frequency equals the di
ff
erence in
frequency between the two sources.
f
b
=

f

f

= (1
.
01135

1)
f .
Therefore,
f
=
f
b
(1
.
01135

1)
=
7
.
39 Hz
(1
.
01135

1)
= 651
.
011 Hz
.
Question 2, chap 16, sect 4.
part 1 of 1
10 points
An elastic string of mass 5
.
2 g is stretched
to a length of 1
.
791 m by the tension force
44 N. The string is fixed at both ends and has
fundamental frequency
f
1
. When the tension
force increases to 624
.
8 N the string stretches
to a length 8
.
68635 m and its fundamental
frequency becomes
f
2
.
Calculate the ratio
f
2
f
1
.
Correct answer: 1
.
71109
(tolerance
±
1 %).
Explanation:
Let :
L
1
= 1
.
791 m
,
F
1
= 44 N
,
L
2
= 8
.
68635 m
,
and
F
2
= 624
.
8 N
.
The fundamental mode of a string has
wavelength
λ
= 2
L
and therefore frequency
f
=
v
2
L
,
where
v
=
F
μ
=
F
m
L
=
F L
m
is the speed of waves on the string.
Alto
gether,
f
=
1
2
L
F L
m
=
1
2
F
m L
.
Thus the ratio is
f
2
f
1
=
1
2
F
2
m L
2
1
2
F
1
m L
1
=
F
2
L
1
F
1
L
2
=
(624
.
8 N) (1
.
791 m)
(44 N) (8
.
68635 m)
=
1
.
71109
.
Note:
You don’t need the value of the string’s
mass since it cancels out of the frequency
ratio.
Question 3, chap 17, sect 3.
part 1 of 1
10 points
A variablelength air column is placed just
below a vibrating wire that is fixed at the
both ends. The length of air column open at
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oldfinal 01 – PAPAGEORGE, MATT – Due: May 6 2008, 4:00 am
2
one end is gradually increased from zero until
the first position of resonance is observed at
24
.
9 cm.
The wire is 127
.
2 cm long and is
vibrating in its third harmonic.
Find the speed of transverse waves in the
wire if the speed of sound in air is 340 m
/
s.
Correct answer: 289
.
478 m
/
s (tolerance
±
1
%).
Explanation:
Given :
L
= 24
.
9 cm
,
L
wire
= 127
.
2 cm
,
and
v
sound
= 340 m
/
s
.
The length of the air column when the first
resonance is heard is
L
=
λ
air
4
,
where
λ
air
is the wavelength of the sound in
air. The frequency of sound wave, and hence
the vibrating wire producing sound, is
f
=
v
sound
λ
air
=
v
sound
4
L
.
When the wire vibrates in its third harmonic,
its wavelength is related to its length as fol
lows;
L
wire
=
3
2
λ
wire
,
where
λ
wire
is the wavelength of waves trav
eling in the wire. Therefore,
λ
wire
=
2
3
L
wire
,
and the speed of transverse waves in the wire
is
v
wire
=
λ
wire
f
=
2
3
L
wire
v
sound
4
L
=
L
wire
v
sound
6
L
=
(127
.
2 cm) (340 m
/
s)
6 (24
.
9 cm)
= 289
.
478 m
/
s
.
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 Friction, Correct Answer, Orders of magnitude, kg, m/s, MATT –

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