CHAPTER 15 SOLUTIONS
Problem
3. The vibration frequency of a hydrogen chloride molecule is
8.66
!
10
13
Hz.
How long does it take the molecule to
complete one oscillation?
Solution
T
=
1
=
f
=
1
=
(8.66
!
10
Hz)
=
1.15
!
10
"
14
s
=
11.5 fs
(Equation 151).
Problem
5. Determine the amplitude, angular frequency, and phase constant for each of the simple harmonic motions shown in
Fig. 1533.
Solution
The amplitude is the maximum displacement, read along the
x
axis (ordinate) in Fig. 1533. The angular frequency is
2
!
times the reciprocal of the period, which is the time interval between corresponding points read along the
t
axis (abscissa).
The phase constant can be determined from the intercept and slope (displacement and velocity) at
t
=
0.
One sees that
(a)
A
'
20 cm,
ω
' 2
π
=4 s '
1
2
s
"
1
,
and
φ
' 0; (b)
A
' 30 cm,
' 2
=3.2 s ' 2s
−
1
, and
'
−
90
°
'
!
1
2
"
;
(c)
A
'
40 cm,
' 2
=(2
×
2 s) '
1
2
s
"
1
,
and
' cos
−
1
(27=40) ' 48
°
'
1
4
.
FIGURE 1533 Problem 5 Solution.
Problem
7. An astronaut in an orbiting spacecraft is “weighed” by being strapped to a spring of constant
k
=
400 N/m,
and set into
simple harmonic motion. If the oscillation period is
2.5 s,
what is the astronaut’s mass?
Solution
We suppose that the other end of the spring is fastened to the spacecraft, whose mass is much greater than the astronaut’s,
and that the orbiting system is approximately inertial. Then Equation 158c gives
m
=
k
=
2
=
k
(
T
=
2
)
2
=
(400 N/m)(2.5 s
=
2
)
2
=
63.3 kg.
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CHAPTER 15
Problem
9. A simple model of a carbon dioxide (CO
2
) molecule consists of three mass points (the atoms) connected by two
springs (electrical forces), as suggested in Fig. 1534. One way this system can oscillate is if the carbon atom stays fixed
and the two oxygens move symmetrically on either side of it. If the frequency of this oscillation is
4.0
!
10
13
Hz,
what
is the effective spring constant? The mass of an oxygen atom is
16 u.
FIGURE 1534 Problem 9.
Solution
With the carbon atom end of either “spring” fixed, the frequency of either oxygen atom is
!
=
2
"
f
=
=
m
.
Therefore
k
=
(2
"
4
"
10
Hz)
2
(16
"
1.66
"
10
#
27
kg)
=
1.68
"
10
3
N/m.
Problem
13. A mass
m
slides along a frictionless horizontal surface at speed
v
0
.
It strikes a spring of constant
k
attached to a rigid
wall, as shown in Fig. 1535. After a completely elastic encounter with the spring, the mass heads back in the direction it
came from. In terms of
k
,
m
, and
v
0
,
determine (a) how long the mass is in contact with the spring and (b) the maximum
compression of the spring.
FIGURE 1535 Problem 13.
Solution
(a) While the mass is in contact with the spring, the net horizontal force on it is just the spring force, so it undergoes half a
cycle of simple harmonic motion before leaving the spring with speed
v
0
to the left. This takes time equal to half a period
1
2
T
=
=
=
=
k
.
(b)
v
0
is the maximum speed, which is related to the maximum compression of the spring (the
amplitude) by
v
0
=
A
.
Thus
A
=
v
0
=
=
v
0
m
=
k
.
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 Winter '07
 Hicks
 Physics, Energy, Force, Mass, Simple Harmonic Motion

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