bhappu (nkb284) – Sample Midterm 4 – forin – (58140)
1
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Multiplechoice questions may continue on
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be±ore answering.
001
10.0 points
A mass attached to a spring executes sim
ple harmonic motion in a horizontal plane
with an amplitude o± 2
.
09 m.
At a point
1
.
8601 m away ±rom the equilibrium, the mass
has speed 2
.
01 m
/
s.
What is the period o± oscillation o± the
mass? Consider equations ±or
x
(
t
) and
v
(
t
)
and use sin
2
+ cos
2
= 1 to calculate
ω
.
Correct answer: 2
.
97891 s.
Explanation:
Let
A
=2
.
09 m
,
x
=1
.
8601 m
,
and
v
.
01 m
/
s
.
The simplest solution uses the equation ±or
simple harmonic motion
x
(
t
)=
A
sin(
ω t
)
,
and its time derivative
v
(
t
dx
dt
=
Aω
cos(
)
.
²or any angle (such as the phase angle
)
sin
2
(
) + cos
2
(
) = 1
,
so
±
x
A
²
2
+
±
v
A ω
²
2
,
and
±
v
ω
²
2
=
A
2

x
2
.
Consequently,
ω
=
v
√
A
2

x
2
=
2
.
01 m
/
s
³
(2
.
09 m)
2

(1
.
8601 m)
2
.
10922 s

1
and
T
=
2
π
ω
=
2
π
2
.
10922 s

1
=
2
.
97891 s
.
002
10.0 points
A block o± unknown mass is attached to a
spring o± spring constant 7
.
9N
/
m and under
goes simple harmonic motion with an ampli
tude o± 9
.
2 cm. When the mass is hal±way
between its equilibrium position and the end
point, its speed is measured to be 23 cm
/
s.
Calculate the mass o± the block.
Correct answer: 0
.
948 kg.
Explanation:
Let :
k
=7
.
/
m
,
A
=9
.
2 cm
,
and
v
= 23 cm
/
s
.
I± the maximum displacement (amplitude) is
A
, the hal±way displacement is
A
2
. By energy
conservation,
K
i
+
U
i
=
F
f
+
U
f
0+
1
2
kA
2
=
1
2
mv
2
+
1
2
k
´
A
2
µ
2
2
=
2
+
1
4
2
m
=
3
2
4
v
2
=
3 (7
.
/
m) (0
.
092 m)
2
4 (0
.
23 m
/
s)
2
=
0
.
948 kg
.
003
10.0 points
A hoop (
i.e.
, a ring) o± radius 2
.
1 m and
mass 14 kg is suspended ±rom a pivot on the
perimeter o± the hoop as shown in the Fgure.
The acceleration o± gravity is 9
.
8m
/
s
2
.
x
y
θ
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View Full Documentbhappu (nkb284) – Sample Midterm 4 – forin – (58140)
2
Find the angular ±requency
ω
o± small os
cillations.
Correct answer: 1
.
52753 rad
/
s.
Explanation:
Basic Concepts
T
=2
π
±
I
mg d
Solution:
From the parallel axis theorem,
the moment o± inertia o± the hoop about the
pivot is given by
I
=
I
c
+
Md
2
=
mR
2
+
2
2
Hence the angular ±requency is
ω
=
²
I
=
²
mg R
2
2
=
²
g
2
R
=1
.
52753 rad
/
s
.
004
10.0 points
A0
.
12 kg mass is attached to a spring and
undergoes simple harmonic motion with a pe
riod o± 0
.
279 s. The total energy o± the system
is 3
.
9 J.
Find the amplitude o± the motion.
Correct answer: 0
.
357998 m.
Explanation:
Let :
m
=0
.
12 kg
,
T
.
279 s
,
and
E
=3
.
9J
.
T
π
²
m
k
k
=
4
π
2
m
T
2
=
4
π
2
(0
.
12 kg)
(0
.
279 s)
2
= 60
.
8601 N
/
m
,
so the total energy is
E
=
1
2
kA
2
A
=
²
2
E
k
=
±
2 (3
.
9 J)
60
.
8601 N
/
m
=
0
.
357998 m
.
005
10.0 points
Earthquakes produce two kinds o± seismic
waves: he longitudinal
primary
waves (called
P
waves) and the transverse
secondary
waves
(called
S
waves). Both S waves and P waves
travel through Earth’s crust and mantle, but
at di²erent speeds; the P waves are always
±aster than the S waves, but their exact speeds
depend on depth and location. For the pur
pose o± this exercise, we assume the P wave’s
speed to be 8790 m
/
s while the S waves travel
at a slower speed o± 4080 m
/
s.
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 Spring '10
 SHIH
 Mass, Wavelength, Standing wave, Swave

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