Chapter 15
1. (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
=
ϖ
x
m
, where
ϖ
is the
angular frequency. Since
ϖ
= 2
π
f
, where
f
is the frequency,
(
29
(
29
3
= 2
= 2
120 Hz
1.0
10
m
= 0.75 m/s.
m
m
v
fx
π
π

×
(c) The maximum acceleration is
(
29
(
29
(
29
(
29
2
2
2
3
2
2
=
=
2
=
2
120 Hz
1.0
10
m
= 5.7
10
m/s .
m
m
m
a
x
f
x
ϖ
π
π

×
×
2. (a) The acceleration amplitude is related to the maximum force by Newton’s second
law:
F
max
=
ma
m
. The textbook notes (in the discussion immediately after Eq. 157) that
the acceleration amplitude is
a
m
=
ϖ
2
x
m
, where
ϖ
is the angular frequency (
ϖ
= 2
π
f
since
there are 2
π
radians in one cycle). The frequency is the reciprocal of the period:
f
= 1/
T
=
1/0.20 = 5.0 Hz, so the angular frequency is
ϖ
= 10
π
(understood to be valid to two
significant figures). Therefore,
=
= 0.12
10
0.085
= 10
.
2
2
F
m
x
m
max
kg
rad / s
m
N
ϖ
b
gb
gb
g
π
(b) Using Eq. 1512, we obtain
(
29
(
29
2
2
0.12kg
10
rad/s
1.2 10 N/m.
k
k
m
ϖ
π
=
⇒
=
=
×
3. (a) The angular frequency
ϖ
is given by
ϖ
= 2
π
f
= 2
π
/
T
, where
f
is the frequency and
T
is the period. The relationship
f
= 1/
T
was used to obtain the last form. Thus
ϖ
= 2
π
/(1.00 × 10
–
5
s) = 6.28 × 10
5
rad/s.
(b) The maximum speed
v
m
and maximum displacement
x
m
are related by
v
m
=
ϖ
x
m
, so
=
=
1.00
10
6.28
10
= 1.59
10
.
3
5
3
x
v
m
m
ϖ
×
×
×

m / s
rad / s
m
657
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CHAPTER 15
4. The textbook notes (in the discussion immediately after Eq. 157) that the acceleration
amplitude is
a
m
=
ϖ
2
x
m
, where
ϖ
is the angular frequency (
ϖ
= 2
π
f
since there are 2
π
radians in one cycle). Therefore, in this circumstance, we obtain
= 2
6.60
0.0220
= 37.8
2
2
a
m
π
Hz
m
m / s
b
g
c
hb
g
.
5. (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
ϖ
is
ϖ
= 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
ϖ
=
k m
. We solve for
k
:
k
=
m
ϖ
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
=
ϖ
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.
6. (a) The problem describes the time taken to execute one cycle of the motion. The
period is
T
= 0.75 s.
(b) Frequency is simply the reciprocal of the period:
f
= 1/
T
≈
1.3 Hz, where the SI unit
abbreviation Hz stands for Hertz, which means a cyclepersecond.
(c) Since 2
π
radians are equivalent to a cycle, the angular frequency
ϖ
(in radiansper
second) is related to frequency
f
by
ϖ
= 2
π
f
so that
ϖ
≈
8.4 rad/s.
7. (a) During simple harmonic motion, the speed is (momentarily) zero when the object is
at a “turning point” (that is, when
x
= +
x
m
or
x
= –
x
m
). Consider that it starts at
x
= +
x
m
and we are told that
t
= 0.25 second elapses until the object reaches
x
= –
x
m
. To execute a
full cycle of the motion (which takes a period
T
to complete), the object which started at
x
= +
x
m
must return to
x
= +
x
m
(which, by symmetry, will occur 0.25 second
after
it was
at
x
= –
x
m
). Thus,
T
= 2
t
= 0.50 s.
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 Spring '08
 IASHVILI
 Energy, Kinetic Energy, Potential Energy, Angular frequency, xm

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