le (txl65) – Vibration and wave – mohanty – (81440)
1
This printout should have 17 questions.
Multiplechoice questions may continue on
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beFore answering.
001
10.0 points
A piston in an automobile engine is in simple
harmonic motion. Its amplitude oF oscilla
tion From equilibrium (centered) position is
±
3
.
46 cm and its mass is 1
.
27 kg.
±ind the maximum velocity oF the piston
when the auto engine is running at the rate oF
3470 rev
/
min.
Correct answer: 12
.
5729 m
/
s.
Explanation:
The simple harmonic motion is described
by
x
=
A
cos
ω t,
where
ω
is the Frequency in rad/s iF
t
is in
seconds.
To fnd
ω
From the given value in rev/min,
Note:
60 rev
/
min = 1 rev
/
s = 1 Hz, so
ω
= 2
π f
= 2
π
(57
.
8334 rev
/
s)
= 363
.
378 rad
/
s
.
Di²erentiating
x
above we fnd the velocity
v
=
d x
dt
=

A ω
sin(
ω t
)
meaning the maximum velocity is
v
max
=
A ω
= (0
.
0346 m) (363
.
378 rad
/
s)
= 12
.
5729 m
/
s
,
since sine has a maximum value oF 1.
002
10.0 points
As your hand moves back and Forth to gener
ate longitudinal pulses in a spiral spring, your
hand completes 2
.
13 backandForth cycles ev
ery 5
.
82 s. The velocity oF the pulse in the
spring is 1
.
03 cm
/
s.
What is the wavelength?
Correct answer: 0
.
0281437 m.
Explanation:
±requence is the number oF cycles generated
per unit time, so
f
=
n
t
The velocity oF the wave is defned by
v
=
f λ
=
n
t
λ
λ
=
v t
n
=
(1
.
03 cm
/
s) (5
.
82 s)
2
.
13
·
1 m
100 cm
=
0
.
0281437 m
.
keywords:
003 (part 1 of 2) 10.0 points
A sinusoidal wave oF the Form
y
=
A
sin(
k x

ω t
)
is traveling along a string in the
x
direc
tion, where
A
= 0
.
3 mm
, k
= 1
.
6 m
−
1
,
ω
= 34 rad
/
s
,
with
x
in meters and
t
in
seconds.
±or this string, the mass per unit length is
given by
μ
= 0
.
01 kg
/
m.
0
.
185 s
0
.
185 s
0
.
3 mm

0
.
3 mm
3
.
93 m
3
.
93 m
0
.
3 mm

0
.
3 mm
Δ
m
What is the maximum value oF the acceler
ation oF the segment oF the string at
x
= 0
.
5
m?
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View Full Documentle (txl65) – Vibration and wave – mohanty – (81440)
2
Correct answer: 0
.
3468 m
/
s
2
.
Explanation:
Let :
A
= 0
.
3 mm = 0
.
0003 m
and
ω
= 34 rad
/
s
.
The maximum acceleration is independent of
x
:
a
max
=
p
d
2
y
dt
2
P
max
=
A ω
2
= (0
.
0003 m) (34 rad
/
s)
2
=
0
.
3468 m
/
s
2
.
004 (part 2 of 2) 10.0 points
For a length segment Δ
x
= 1 cm along the
string, what is the total energy of oscillation?
Correct answer: 5
.
202
×
10
−
9
J.
Explanation:
Let :
μ
= 0
.
01 kg
/
m
and
Δ
x
= 1 cm = 0
.
01 m
.
The total energy is
Δ
E
= Δ
K
+ Δ
U
= Δ
K
max
=
1
2
Δ
m v
2
max
=
1
2
μ
Δ
x
(
ω A
)
2
=
1
2
(0
.
01 kg
/
m) (0
.
01 m)
×
(34 rad
/
s)
2
(0
.
0003 m)
2
=
5
.
202
×
10
−
9
J
.
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 Spring '11
 sdsd
 Energy, Simple Harmonic Motion, Frequency, Correct Answer, Orders of magnitude

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