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Chapter 15 – Student Solutions Manual
3. (a) The amplitude is half the range of the displacement, or
x
m
= 1.0 mm.
(b) The maximum speed
v
m
is related to the amplitude
x
m
by
v
m
=
ω
x
m
, where
is the
angular frequency. Since
= 2
π
f
, where
f
is the frequency,
( )( )
3
= 2
= 2
120 Hz 1.0 10 m = 0.75 m/s.
mm
vf
x
ππ
−
×
(c) The maximum acceleration is
()
(
)
( ) ( )
2
2
23
=
= 2
= 2
120 Hz
1.0 10 m = 5.7 10 m/s .
m
ax
f
x
ωπ
π
−
××
2
2
7. (a) The motion repeats every 0.500 s so the period must be
T
= 0.500 s.
(b) The frequency is the reciprocal of the period:
f
= 1/
T
= 1/(0.500 s) = 2.00 Hz.
(c) The angular frequency
is
= 2
π
f
= 2
π
(2.00 Hz) = 12.6 rad/s.
(d) The angular frequency is related to the spring constant
k
and the mass
m
by
=
km
. We solve for
k
and obtain
k
=
m
2
= (0.500 kg)(12.6 rad/s)
2
= 79.0 N/m.
(e) Let
x
m
be the amplitude. The maximum speed is
v
m
=
x
m
= (12.6 rad/s)(0.350 m) = 4.40 m/s.
(f) The maximum force is exerted when the displacement is a maximum and its
magnitude is given by
F
m
=
kx
m
= (79.0 N/m)(0.350 m) = 27.6 N.
9. The magnitude of the maximum acceleration is given by
a
m
=
2
x
m
, where
is the
angular frequency and
x
m
is the amplitude.
(a) The angular frequency for which the maximum acceleration is
g
is given by
=
gx
m
/
, and the corresponding frequency is given by
2
6
11
9
.
8
m
/
s
498 Hz.
2
2
2
1.0 10 m
m
g
f
x
−
==
=
=
×
(b) For frequencies greater than 498 Hz, the acceleration exceeds
g
for some part of the
motion.
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View Full Document 17. The maximum force that can be exerted by the surface must be less than
μ
s
F
N
or else
the block will not follow the surface in its motion. Here, μ
s
is the coefficient of static
friction and
F
N
is the normal force exerted by the surface on the block. Since the block
does not accelerate vertically, we know that
F
N
=
mg
, where
m
is the mass of the block. If
the block follows the table and moves in simple harmonic motion, the magnitude of the
maximum force exerted on it is given by
F
=
ma
m
=
m
ω
2
x
m
=
m
(2
π
f
)
2
x
m
,
where
a
m
is the magnitude of the maximum acceleration,
is the angular frequency, and
f
is the frequency. The relationship
= 2
π
f
was used to obtain the last form. We
substitute
F
=
m
(2
π
f
)
2
x
m
and
F
N
=
mg
into
F
<
μ
s
F
N
to obtain
m
(2
π
f
)
2
x
m
<
μ
s
mg
. The
largest amplitude for which the block does not slip is
=
2
=
0.50 9.8
22
.
0
0 031
2
2
2
x
g
f
m
s
ππ
b
g
b
gc
h
b
g
m/s
Hz
×
=
..
m
A larger amplitude requires a larger force at the end points of the motion. The surface
cannot supply the larger force and the block slips.
19. (a) Let
=
2
2
1
x
At
T
cos
π
F
H
G
I
K
J
be the coordinate as a function of time for particle 1 and
=
2
2
+
6
2
x
T
cos
F
H
G
I
K
J
be the coordinate as a function of time for particle 2. Here
T
is the period. Note that since
the range of the motion is
A
, the amplitudes are both
A
/2. The arguments of the cosine
functions are in radians. Particle 1 is at one end of its path (
x
1
=
A
/2) when
t
= 0. Particle
2 is at
A
/2 when 2
π
t
/
T
+
π
/6 = 0 or
t
= –
T
/12. That is, particle 1 lags particle 2 by one
twelfth a period. We want the coordinates of the particles 0.50 s later; that is, at
t
= 0.50 s,
1
20
.
5
0
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This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.
 Spring '08
 drty
 mechanics

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