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001
(part 1 oF 3) 10.0 points
Consider the oscillation oF a massspring
system, where
x
=
A
cos(
ω t
+
φ
)
.
At the time
t
= 0, the mass
m
is at
x
= 0
(the equilibrium point) and it is moving with
a positive velocity
v
0
.
k
m
v
0
x
= 0
x
±ind the phase angle
φ
.
(Hint:
Consider
x
as the projection oF a
counterclockwise uniForm circular motion.)
1.
φ
= 2
π
2.
φ
=
3
2
π
correct
3.
φ
=
5
4
π
4.
φ
=
π
5.
φ
=
7
4
π
6.
φ
=
1
2
π
7.
φ
=
1
4
π
8.
φ
= 0
9.
φ
=
3
4
π
Explanation:
Basic Concepts:
x
=
A
cos
ω
(
t
+
φ
)
ω
=
r
m
k
A
B
C
D
E
F
G
H
φ
ϖ
x
The SHM can be represented by the
x

projection oF a uniForm circular motion:
x
=
A
cos
ω
(
t
+
φ
)
.
At
t
= 0
,x
= 0. ±rom inspection, it should be
either
C
or
G
. At
C
,
v <
0; while at
G
,
v >
0.
So
G
is the correct choice, or
φ
=
3
2
π
.
002
(part 2 oF 3) 10.0 points
Let the mass be
m
= 5
.
09 kg, spring constant
k
= 567 N
/
m and the initial velocity
v
0
=
1
.
81 m
/
s.
±ind the amplitude
A
.
Correct answer: 0
.
171493 m.
Explanation:
v
=
dx
dt
=

ω A
sin(
ω t
+
φ
)
So the velocity amplitude or the maximum
speed is
v
max
=
ωA
;
i.e.
,
v
0
=
ωA,
so
A
=
v
0
ω
=
v
0
r
m
k
= (1
.
81 m
/
s)
R
(5
.
09 kg)
(567 N
/
m)
= 0
.
171493 m
.
003
(part 3 oF 3) 10.0 points
±ind the total energy oF oscillation at
t
=
T
8
;
i.e.
, at oneeighth oF the period.
homework 32 – Turner – (58185)
1
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2
(Hint: Consider what happens to the total
energy during oscillatory motion.)
1.
E
= 2
m v
2
0
2.
E
=
5
2
mv
2
0
3.
E
=
mv
2
0
4.
E
=
1
4
mv
2
0
5.
E
=
3
2
mv
2
0
6.
E
=
1
2
mv
2
0
correct
7.
E
=
1
2
√
2
m v
2
0
8.
E
=
3
4
mv
2
0
Explanation:
Since the spring force is a conservative force
the total energy is conserved. One may equate
the energy at
t
= 0 where
x
= 0. So
E
=
K
+
U
=
K
max
=
1
2
mv
2
0
.
004
(part 1 of 3) 10.0 points
A block of mass 0
.
2 kg is attached to a
spring of spring constant 24 N
/
m on a fric
tionless track. The block moves in simple har
monic motion with amplitude 0
.
11 m. While
passing through the equilibrium point from
left to right, the block is struck by a bullet,
which stops inside the block.
The velocity of the bullet immediately be
fore it strikes the block is 54 m
/
s and the mass
of the bullet is 4
.
29 g.
24 N
/
m
0
.
2 kg
4
.
29 g
54 m
/
s
Find the speed of the block immediately
before the collision.
Correct answer: 1
.
20499 m
/
s.
Explanation:
Let :
M
= 0
.
2 kg
,
A
= 0
.
11 m
,
and
k
= 24 N
/
m
.
Since the block moves in simple harmonic
motion, we know that
x
=
A
sin(
ω
0
t
+
ϕ
)
,
and since the only force acting on it is the
spring force
F
s
=

k x
, the equation of mo
tion is
F
s
=
M a
=
M
d
2
x
dt
2
=

k x,
so the angular velocity of this initial motion is
ω
0
=
r
k
M
=
R
24 N
/
m
0
.
2 kg
= 10
.
9545 rad
/
s
.
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Mass, Work

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