Homework 9 Chapt. 14 solutions
(1)
Cosine Wave
The graph shows the position
of an oscillating object as a function of time
. The equation of
the graph is
,
Part A
What is A in the equation?
Part B
What is
in the equation?
Part C
What is
in the equation?
From the figure we see when t = N, x = M
Using the cos wave function M = M cos(2
π
/T * (N) +
Φ
)
→
cos(2
π
/T * (N) +
Φ
)=1
(2
π
/T * (N) +
Φ
) = 0,
Φ
= 2
π
N /T
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14.14.
Model:
The oscillating mass is in simple harmonic motion.
Solve:
(a)
The amplitude
A
=
2.0 cm.
(b)
The period is calculated as follows:
2
2
10 rad s
0.628 s
10 rad s
T
T
π
π
ω
=
=
⇒
=
=
(c)
The spring constant is calculated as follows:
(
)
(
)
2
2
0.050 kg
10 rad s
5.0 N m
k
k
m
m
ω
ω
=
⇒
=
=
=
(d)
The phase constant
1
0
4
rad.
φ
π
= 
(e)
The initial conditions are obtained from the equations
( )
(
)
(
)
( )
(
)
(
)
1
1
4
4
2.0 cm
cos 10
and
20.0 cm s sin 10
x
x t
t
v
t
t
π
π
=

= 

At
t
=
0 s, these equations become
(
)
(
)
(
)
(
)
1
1
0
0
4
4
2.0 cm
cos
1.414 cm and
20.0 cm s sin
14.14 cm s
x
x
v
π
π
=

=
= 

=
In other words, the mass is at
+
1.414 cm and moving to the right with a velocity of 14.14 cm/s.
(f)
The maximum speed is
(
)
(
)
max
2.0 cm
10 rad s
20.0 cm s
v
A
ω
=
=
=
.
(g)
The total energy
(
)(
)
2
2
3
1
1
2
2
5.0 N m
0.02 m
1.0
10
J
E
kA

=
=
=
×
.
(h)
At
t
=
0.40 s, the velocity is
(
)
(
)(
)
1
0
4
20.0 cm s sin
10 rad s
0.40 s
1.46 cm s
x
v
π
= 

=
14.19 Model:
Assume a small angle of oscillation so there is simple harmonic motion.
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 Spring '10
 ZHOU
 Force, Simple Harmonic Motion, 0.02 m, 0.050 kg, 0.090 M

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