432
CHAPTE R 15 OSCI LL ATIONS
on the left side. Thus,
ϭ 5.0 s.
(Answer)
Because T ϭ 0.34 s, this is about 15 periods of oscillation.
(c) How long does it take for the mechanical energy to drop
to one-half its initial value?
KEY IDEA
From Eq. 15-44, the mechanical energy at time t is kx
2
m e
Ϫbt/m
.
Calculations: The mechanical energy has the value
kx
2
m at t ϭ 0. Thus, we must find the value of t for which
.
If we divide both sides of this equation by kx
2
m and solve for
t as we did above, we find
(Answer)
This is exactly half the time we calculated in (b), or about
7.5 periods of oscillation. Figure 15-17 was drawn to illustrate this sample problem.
t ϭ
Ϫm ln
1
2
b
ϭ
Ϫ(0.25 kg)(ln
1
2 )
0.070 kg/s
ϭ 2.5 s.
1
2
1
2 kx
2
m e
Ϫbt/m ϭ
1
2 (
1
2 kx
2
m )
1
2
1
2
t ϭ
Ϫ2m ln
1
2
b
ϭ
Ϫ(2)(0.25 kg)(ln
1
2 )
0.070 kg/s
Sample Problem 15.06 Damped harmonic oscillator, time to decay, energy
For the damped oscillator of Fig. 15-16, m ϭ 250 g, k ϭ
85 N/m, and b ϭ 70 g/s.
(a) What is the period of the motion?
KEY IDEA
Because b Ӷ
ϭ4.6 kg/s, the period is approximately
that of the undamped oscillator.
Calculation: From Eq. 15-13, we then have
(Answer)
(b) How long does it take for the amplitude of the damped
oscillations to drop to half its initial value?
KEY IDEA
The amplitude at time t is displayed in Eq. 15-42 as x m e
Ϫbt/2m
.
Calculations: The amplitude has the value x m at t ϭ 0. Thus,
we must find the value of t for which
x m e
Ϫbt/2m ϭ x m .
Canceling x m and taking the natural logarithm of the equation that remains, we have ln on the right side and
ln(e
Ϫbt/2m
) ϭ Ϫbt/2m
1
2
1
2
T ϭ 2p A
m
k
ϭ 2p A
0.25 kg
85 N/m
ϭ 0.34 s.
1km
Additional examples, video, and practice available at WileyPLUS
15-6 FORCED OSCILLATIONS AND RESONANCE
After reading this module, you should be able to . . .
15.43 Distinguish between natural angular frequency v and
driving angular frequency v d .
15.44 For a forced oscillator, sketch a graph of the oscillation
amplitude versus the ratio v d /v of driving angular frequency to natural angular frequency, identify the approximate location of resonance, and indicate the effect of increasing the damping constant.
15.45 For a given natural angular frequency v, identify the approximate driving angular frequency v d that gives resonance.
● If an external driving force with angular frequency v d acts
on an oscillating system with natural angular frequency v, the
system oscillates with angular frequency v d .
● The velocity amplitude v m of the system is greatest when
v d ϭ v,
a condition called resonance. The amplitude x m of the system
is (approximately) greatest under the same condition.
Learning Objectives
Key Ideas
Forced Oscillations and Resonance
A person swinging in a swing without anyone pushing it is an example of free
oscillation. However, if someone pushes the swing periodically, the swing has
CHAPTE R 15 OSCI LL ATIONS
on the left side. Thus,
ϭ 5.0 s.
(Answer)
Because T ϭ 0.34 s, this is about 15 periods of oscillation.
(c) How long does it take for the mechanical energy to drop
to one-half its initial value?
KEY IDEA
From Eq. 15-44, the mechanical energy at time t is kx
2
m e
Ϫbt/m
.
Calculations: The mechanical energy has the value
kx
2
m at t ϭ 0. Thus, we must find the value of t for which
.
If we divide both sides of this equation by kx
2
m and solve for
t as we did above, we find
(Answer)
This is exactly half the time we calculated in (b), or about
7.5 periods of oscillation. Figure 15-17 was drawn to illustrate this sample problem.
t ϭ
Ϫm ln
1
2
b
ϭ
Ϫ(0.25 kg)(ln
1
2 )
0.070 kg/s
ϭ 2.5 s.
1
2
1
2 kx
2
m e
Ϫbt/m ϭ
1
2 (
1
2 kx
2
m )
1
2
1
2
t ϭ
Ϫ2m ln
1
2
b
ϭ
Ϫ(2)(0.25 kg)(ln
1
2 )
0.070 kg/s
Sample Problem 15.06 Damped harmonic oscillator, time to decay, energy
For the damped oscillator of Fig. 15-16, m ϭ 250 g, k ϭ
85 N/m, and b ϭ 70 g/s.
(a) What is the period of the motion?
KEY IDEA
Because b Ӷ
ϭ4.6 kg/s, the period is approximately
that of the undamped oscillator.
Calculation: From Eq. 15-13, we then have
(Answer)
(b) How long does it take for the amplitude of the damped
oscillations to drop to half its initial value?
KEY IDEA
The amplitude at time t is displayed in Eq. 15-42 as x m e
Ϫbt/2m
.
Calculations: The amplitude has the value x m at t ϭ 0. Thus,
we must find the value of t for which
x m e
Ϫbt/2m ϭ x m .
Canceling x m and taking the natural logarithm of the equation that remains, we have ln on the right side and
ln(e
Ϫbt/2m
) ϭ Ϫbt/2m
1
2
1
2
T ϭ 2p A
m
k
ϭ 2p A
0.25 kg
85 N/m
ϭ 0.34 s.
1km
Additional examples, video, and practice available at WileyPLUS
15-6 FORCED OSCILLATIONS AND RESONANCE
After reading this module, you should be able to . . .
15.43 Distinguish between natural angular frequency v and
driving angular frequency v d .
15.44 For a forced oscillator, sketch a graph of the oscillation
amplitude versus the ratio v d /v of driving angular frequency to natural angular frequency, identify the approximate location of resonance, and indicate the effect of increasing the damping constant.
15.45 For a given natural angular frequency v, identify the approximate driving angular frequency v d that gives resonance.
● If an external driving force with angular frequency v d acts
on an oscillating system with natural angular frequency v, the
system oscillates with angular frequency v d .
● The velocity amplitude v m of the system is greatest when
v d ϭ v,
a condition called resonance. The amplitude x m of the system
is (approximately) greatest under the same condition.
Learning Objectives
Key Ideas
Forced Oscillations and Resonance
A person swinging in a swing without anyone pushing it is an example of free
oscillation. However, if someone pushes the swing periodically, the swing has