Transcript Chapter 15 Periodic Motion

Lecture Outline
Chapter 15
Periodic Motion
© 2015 Pearson Education, Inc.
Slide 1-1
Chapter 15: Periodic Motion
Chapter Goal: Study the kinematics and dynamics
of periodic motion, i.e., motion that repeats itself at
regular intervals.
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Slide 1-2
Section 15.1: Periodic motion and energy
• Any motion that repeats itself at
regular time intervals is called
periodic motion.
• The figure shows the periodic
motion of a spring-cart system.
• Hooke’s law: If a spring is stretched (or
compressed) by a small distance x – x0 from
its unstretched length x0, the x component of
the force it exerts on the load is
(Fby spring on load)x = –k(x – x0)
where k is called the spring constant of the
spring.
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Slide 1-3
Checkpoint 15.1
(a) List the forces exerted on the spring-cart
system of Figure 15.1 right after it is released, and
draw a free-body diagram for each object in the
system. (b) Which of these forces do work on the
system as it oscillates? (c) As the hand pushes on the
cart and compresses the spring, is the work done by
the cart on the spring positive, negative, or zero?
(d) How does the work
done by the cart on the
spring compare with the
work done by the spring
on the cart?
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Fsc
Fsc = -k(x – x0 )
x – x0
Slide 1-4
Section 15.1: Periodic motion and energy
• The time interval it takes to complete a full
cycle of the motion is the period T.
• The inverse of the period is called the
frequency: f = 1/T
and has units of Hz = 1/sec.
• The angular frequency is
ω = 2πf
and has the same unit (s–1) as frequency.
• The object’s maximum displacement from
the equilibrium position is called the
amplitude A.
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Slide 1-5
Section 15.1: Periodic motion and energy
• In practice, periodic motion
in mechanical systems will
die out due to energy
dissipation.
• If we ignore these damping
effects, we find that
• Periodic motion is
characterized by a
continuous conversion
between potential and
kinetic energy in a
closed system.
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Slide 1-6
Section 15.1: Periodic motion and energy
• The figure shows examples of oscillating systems.
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Slide 1-7
Section 15.2: Simple harmonic motion
http://www.animatedscience.co.uk/blog/wp-content/uploads/focus_waves/auxcircle.html
• Simple harmonic motion is closely related to circular motion.
• The figure shows the shadow of a ball projected onto a screen.
• As the ball moves in circular motion with constant rotational
speed , the shadow moves with simple harmonic motion.
• The ball sweeps out at an angle  = t in time t.
• Then the position of the ball’s shadow is described by Asin(t),
where A is the radius of the circle.
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Slide 1-8
Section 15.2: Simple harmonic motion
• As illustrated in the figure, the correspondence
between circular motion and simple harmonic motion
can be demonstrated experimentally.
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Slide 1-9
Section 15.6: Simple harmonic motion and
springs
• Consider the spring-cart
system shown. Let x0 = 0.
• The force exerted by the
spring on the cart is
http://www.compadre.org/Physlets/waves/ex16_3.cfm
FSCC x = -kx
• Using SFx = max = -kx,
we can find the equation
of motion for the car to
be
d 2x
k
=- x
2
m
dt
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Slide 1-10
Section 15.6: Simple harmonic motion and
springs
d 2x
k
a = 2 = - x = -w 2 x define
m
dt
w =+
k
m
• The motion of the cart is given by x(t) = A sin (wt + fi )
dx Ad sin(w t + fi )
ux º =
= w A cos(w t + fi ) (simple harmonic motion)
dt
dt
du x d 2 x w A d cos(w t + fi )
ax º
= 2 =
= -w 2 A sin(w t + fi ) (simple harmonic motion)
dt
dt
dt
• The figure shows four different
solutions that satisfy the equation of
motion of the spring-cart system.
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Slide 1-11
Section 15.5: Energy of a simple harmonic
oscillator
SFx = max = -kx,
d 2x
k
a = 2 = - x = -w 2 x
m
dt
(
x(t) = A sin w t + fi
)
ax = -w 2 A sin(w t + fi ) (simple harmonic motion)
ax = -w 2 x(t) (simple harmonic motion)
• Comparing equations for x(t) and a(t), we can write
ax = –ω2x (simple harmonic motion)
• Using Newton’s 2nd law, SFx = max , we get
2
SFx = -mw x (simple harmonic motion)
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Slide 1-12
Checkpoint 15.2
15.2 (a) In Figure 15.2e, the cart’s
displacement from the equilibrium position is
maximum. Is the x component of the cart’s
acceleration at that instant positive, negative,
or zero? (b) At which instant(s) in Figure 15.2
is the magnitude of the cart’s acceleration
greatest? At which instant(s) is it smallest?
(Use the blue position curve to answer these
questions.)
Fsc = -k(x – x0 ) = mac
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Slide 1-13
Section 15.2: Simple harmonic motion
• Investigation of oscillating systems reveal that, when the
amplitude is not too large, the period is independent of the
amplitude.
• An oscillating system that exhibits this property is called
isochronous. SFx = max = -kx, w = + k , x(t) = A sin (wt + fi )
m
T=
2p
w
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Slide 1-14
Section 15.2: Simple harmonic motion
• The x(t) curve of an isochronous oscillation are sinusoidal.
• Periodic motion that yields a sinusoidal x(t) curve is called a simple
harmonic motion (SHM):
• A object executing simple harmonic motion is subject to a
linear restoring force that tends to return the object to its
equilibrium position and is linearly proportional to the
object’s displacement from its equilibrium position.
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Slide 1-15
Section 15.3: Fourier’s theorem
• Fourier’s theorem: Any periodic function with period T can
be written as a sum of sinusoidal functions of frequency fn =
n/T, where n ≥ 1 is an integer.
• f1 = 1/T is called the fundamental frequency or the first
harmonic.
• All other components, called higher harmonics, are integer
multiples of the fundamental frequency:
fn = nf1 = n/T
• According to Fourier’s theorem, any periodic motion can be
treated as a superposition of simple harmonic motions.
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Slide 1-16
Section 15.3: Fourier’s theorem
• Fourier’s theorem: Any periodic function with period T can be written as a sum of
sinusoidal functions of frequency fn = n/T, where n ≥ 1 is an integer.
• By adjusting the amplitude An of each harmonic, we can make the
sum of harmonics fit the original nonsinusoidal periodic function.
http://www.falstad.com/fourier/
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Slide 1-17
Section 15.3: Fourier’s theorem
• The breaking down of a
function into harmonic
components is called
Fourier analysis.
• The resulting sum of
sinusoidal functions is
called a Fourier series.
• The figure shows a
graphical representation
of the Fourier series.
• The plot of An2 as a
function of the frequency
is called the spectrum of
the periodic function.
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Slide 1-18
Section 15.3: Fourier’s theorem
• Fourier analysis has many applications, ranging from
electronic signal processing to chemical analysis and voice
recognition.
• The figure shows the spectrum analyzer of a stereo system,
which displays the frequency content of the sound being
http://www.falstad.com/fourier/
played.
https://en.wikipedia.org/wiki/Fourier_series
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Slide 1-19
Checkpoint 15.6
15.6 (a) What does the spectrum of a single sinusoidal function
of period T look like? (b) As T is increased, what change occurs
in the spectrum?
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Slide 1-20
Lecture Outline
Chapter 15
Periodic Motion
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Slide 1-21
Section 15.4: Restoring forces in simple
harmonic motion
Periodic motion requires a
restoring force that tends to return
the object to the equilibrium
position.
A consequence of the restoring
force about a stable equilibrium
position is
• In the absence of friction, a
small displacement of a
system from a position of
stable equilibrium causes the
U (x)
system to oscillate.
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SFx = max = -kx
d 2x
k
a = 2 = - x = -w 2 x
m
dt
(
x(t) = A sin w t + fi
)
Slide 1-22
Checkpoint 15.7
15.7 (a) Are there any equilibrium positions in the force-versus-
distance graph in Figure 15.13? If so, is the equilibrium stable,
unstable, or neutral? (b) Compare the magnitude of the restoring
force for equal displacements on either side of x0 in Figure 15.13.
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Slide 1-23
Section 15.4: Restoring forces in simple
harmonic motion
• As illustrated in the figure:
• For sufficiently small
displacements away from
the equilibrium position x0,
restoring forces are always
linearly proportional to the
displacement.
• Consequently, for small
displacements, objects
execute simple harmonic
motion about a stable
equilibrium position.
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SFx = max = -k(x - xo )
(
x(t) = xo + A sin w t + fi
)
Slide 1-24
Section 15.5: Energy of a simple harmonic
oscillator
• The work done by the forces exerted on the harmonic oscillator as it
moves from the equilibrium position toward the positive x direction is
SFx = max = -kx,
k
w=
m
W=
ò
x
x0
å Fx (x) dx = - ò kx dx
x
x0
• This work causes a change in kinetic energy, given by
x
W = DK = -k ò xdx = -k[ 12 x 2 ]xx = 12 kx02 - 12 kx 2
x0
0
• For a closed system E = K + U = 0, which gives us
-DK = DU = U (x) -U (x0 ) = 12 kx 2 - 12 kx02
= 12 kx 2
xo = 0
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Slide 1-25
Section 15.5: Energy of a simple harmonic
oscillator
Fx = -kx,
x
xo = 0
Objects move to the
minimum potential energy
U (x) = 12 kx 2
x
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W = -U =
ò
x
x0
Fx (x) dx
dU
= Fx
dx
The negative derivative of the
potential energy is the force
Slide 1-26
Section 15.4: Restoring forces in simple
harmonic motion
Periodic motion requires a restoring
force that tends to return the object
to the equilibrium position.
In the absence of friction, a small
displacement of a system from a
position of stable equilibrium
causes the system to oscillate.
dU
= Fx
dx
Fx = -kx,
Fx = +kx,
U (x) = - 12 kx 2
U (x)
U (x) = 12 kx 2
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Slide 1-27
Section 15.7: Restoring torques
Example 15.8 The simple pendulum
as a one-dimensional oscillator.
❶ The bob is displaced a horizontal distance
x from its equilibrium position, (Figure
15.35a). The free-body diagram for the bob
G
(Figure 15.35b), showing FEbG and FEb^
,with
the component vectors of FEbG parallel to and
perpendicular to the string in its displaced
orientation.
The bob is constrained to move on a circular
path.
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Slide 1-28
Section 15.7: Restoring torques
Example 15.8 The simple pendulum as a onedimensional oscillator.
G
Eb^
❸ From the free-body diagram, F
x
= -mg sin q = -mg ,
ℓ
d 2s
x
G
å F = ma = m dt 2 = FEb^ = -mg ℓ
For small angles, x ~ s
d 2x
x
m 2 = -mg
ℓ
dt
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Slide 1-29
Section 15.7: Restoring torques
Example 15.8 The simple pendulum as a onedimensional oscillator.
d 2x
x
❸ In the small-angle approximation, m 2 = -mg
ℓ
dt
mg
Thus, max » x,
d 2x
g
or ax º 2 » - x.
ℓ
dt
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Slide 1-30
Section 15.7: Restoring torques
Example 15.8 The simple pendulum as a onedimensional oscillator.
❸
d x
g
Rewrite in form similar to spring,
»
x.
ℓ
dt 2
d 2x
2
F = -mg sin q
or w = g/ℓ
»
w
x.
x
dt 2
= -mg ,
2
G
Eb^
(
x(t) = A sin w t + fi
ℓ
)
d 2x
k
a = 2 = - x = -w 2 x
m
dt
define w = +
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k
m
SFx = max = -kx
Slide 1-31
Section 15.4: Restoring forces in simple
harmonic motion
w=
• Oscillations arise from interplay between inertia
and a restoring force.
• For springs, strings, and vibrating masses
• The period of an oscillating object increases when
its mass is increased and decreases when the
magnitude of the restoring force is increased.
• For pendulums:
• The period of a pendulum is independent of the
mass of the pendulum.
g/ℓ
(
x(t) = A sin w t + fi
)
w=
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k
m
Slide 1-32
Checkpoint 15.9
15.9 (a) What effect, if any, does
increasing the length ℓ of a simple pendulum
have on its period?
(b) A pendulum is taken to the Moon, where
the acceleration due to gravity is smaller
than that on Earth. How does the period of
the pendulum on the Moon compare with its
period on Earth?
w = g/ℓ
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Slide 1-33
Section 15.4
Clicker Question 4
A child and an adult are on adjacent swings at the
playground. Is the adult able to swing in synchrony
with the child?
1. No, this is impossible because the inertia of the
two are different.
2. Yes, as long as the lengths of the two swings are
adjusted.
3. Yes, as long as the lengths of the swings are the
same.
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Slide 1-34
Section 15.4
Clicker Question 4
A child and an adult are on adjacent swings at the
playground. Is the adult able to swing in synchrony
with the child?
1. No, this is impossible because the inertia of the
two are different.
2. Yes, as long as the lengths of the two swings are
adjusted.
3. Yes, as long as the lengths of the swings are the
same.
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Slide 1-35
Chapter 15: Self-Quiz #3
1. In the cart-spring system shown in the figure, is the
cart’s speed greater at A or at B?
2. Once released from the stretched position shown,
the cart speeds up as it moves left toward the
equilibrium position, so its speed is greater at A than
at B.
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Slide 1-36
Chapter 15: Self-Quiz #3
1. At which of these two positions is the restoring force
acting on the cart greater?
2. The restoring force is proportional to the
displacement from the equilibrium position, so it is
greater at B than at A.
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Slide 1-37
Section 15.6: Simple harmonic motion and
springs
Example 15.5 Vertical oscillations
A block of mass m is suspended
from a spring of spring constant k.
(a) How far below the end of the
relaxed spring at x0 is the
equilibrium position xeq of the
suspended block (Figure 15.29a)?
k
m
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Slide 1-38
Section 15.6: Simple harmonic motion and
springs
Example 15.5 Vertical oscillations
When the suspended block is in
translational equilibrium at xeq (which
lies below x0), the vector sum of these
forces must be zero. With the block at
xeq, the spring is stretched such that the
end attached to the block is also at xeq.
SFx = F
c
sb x
+F
G
Eb x
= –k(xeq – x0 ) + mg = 0,
k(xeq - x0 ) = mg
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Slide 1-39
Section 15.6: Simple harmonic motion and
springs
Vertical oscillations
(b) Is the frequency f with which the
block oscillates about this
equilibrium position xeq greater than,
smaller than, or equal to that of an
identical system that oscillates
horizontally about x0 on a surface for
which friction can be ignored
(Figure 15.29b)?
k
m
SFx = max = -k(x - xo )
( )
x(t) = x0 + A sin wt
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k
w=
m
k
m
Slide 1-40
Section 15.6: Simple harmonic motion and
springs
❶ When the block is below xeq so the
spring is stretched more than when the
block is at xeq, causing the magnitude of Fsbc
to increase above the equilibrium value, and
so the vector sum of the forces exerted on
the block points upward.
SFx = Fsbc x + FEbG x = –k(x – x0 ) + mg = ma
(x – x0 ) = (x – xeq ) + (xeq – x0 )
SFx = –k(x – xeq ) - k(xeq – x0 ) + mg = ma
SFx = –k(x – xeq ) = ma
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( )
x(t) = xeq + A sin wt
w=
k
m
Slide 1-41
Section 15.6: Simple harmonic motion and
springs
Example 15.5 Vertical oscillations
EVALUATE RESULT The two
oscillations take place about different
equilibrium positions (x0 for the
horizontal case, xeq for the vertical
case), but the effect of the combined
gravitational and elastic forces in the
vertical arrangement is the same as that
of just the elastic force in the
horizontal arrangement because the
force exerted by the spring is linear in
the displacement.
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k
m
k
m
Slide 1-42
Section 15.6
Clicker Question 6
If you know the mass of an object hanging from a
spring in an oscillating system, what else do you need
to know to determine the period of the motion?
Answer all that apply.
1.
2.
3.
4.
The spring constant
The initial velocity
The angular frequency
The amplitude
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Slide 1-43
Section 15.6
Clicker Question 6
If you know the mass of an object hanging from a
spring in an oscillating system, what else do you need
to know to determine the period of the motion?
Answer all that apply.
1.
2.
3.
4.
The spring constant
The initial velocity
The angular frequency
The amplitude
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Slide 1-44
Section 15.7
Clicker Question 7
What approximation is necessary in order for a pendulum
to execute simple harmonic motion? Answer all that apply.
1. No approximations are necessary; a pendulum always
executes simple harmonic motion.
2. The mass must be very large.
3. The mass must be very small.
4. The length must be very large.
5. The length must be very small.
6. The amplitude of the angular displacement must be
much less than 1 radian.
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Slide 1-45
Section 15.7
Clicker Question 7
What approximation is necessary in order for a pendulum
to execute simple harmonic motion? Answer all that apply.
1. No approximations are necessary; a pendulum always
executes simple harmonic motion.
2. The mass must be very large.
3. The mass must be very small.
4. The length must be very large.
5. The length must be very small.
6. The amplitude of the angular displacement must be
much less than 1 radian.
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Slide 1-46
Section 15.8: Damped oscillations
Section Goals
You will learn to
• Integrate the concept of friction to oscillatory motion
and represent the combined effects graphically.
• Model damped harmonic motion mathematically.
• Define the time constant for damped harmonic
motion.
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Slide 1-47
Section 15.8: Damped oscillations
• Mechanical oscillators
always involve some friction
that causes the energy of the
oscillator to convert to
thermal energy.
• This will cause the oscillator
to slow down.
• Such a system is said to
execute a damped
oscillation.
• The figure shows examples
of two damped oscillations.
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Slide 1-48
Section 15.8: Damped oscillations
• The drag force exerted by air and liquids at slow
d
speeds can be modeled as Fac = -bu , where b is called
the damping coefficient.
• In presence of a drag, the equation of motion becomes
d 2x
dx
m 2 + b + kx = 0
dt
dt
• The solution of this equation takes the forms
x(t) = Ae
wd =
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- bt/2m
sin(w d t + fi )
æ b ö
k
b
2
= w -ç
2
m 4m
è 2m ÷ø
2
2
Slide 1-49
Section 15.8: Damped oscillations
• The figure shows oscillations for various values of
the damping coefficient b.
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Slide 1-50
Section 15.8: Damped oscillations
• The ratio m/b has units of time and is called the time
constant:   m/b.
• Amplitude of the damped oscillation at any given
time is given by
xmax(t) = Ae–t/2.
• The mechanical energy of the oscillator can be
expressed as
2
E(t) = 12 mw 2 xmax
= ( 12 mw 2 A2 )e-t/t = E0 e-t/t
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Slide 1-51
Checkpoint 15.18
15.18 A tuning fork that sounds the tone musicians call middle
C oscillates at frequency f = 262 Hz. If the amplitude of the fork’s
oscillation decreases by a factor of 3 in 4.0 s, what are (a) the time
constant of the oscillation and (b) the quality factor?
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Slide 1-52
Section 15.8
Clicker Question 8
During a time interval of one time constant, by what
factor does the mechanical energy of an oscillating
system decrease?
1.
2.
3.
4.
5.
6.
7.
By a factor of 1/e (0.3679)
By a factor of ln(2) (0.6931)
By a factor of (2)1/2
By a factor of 2
By a factor of e
(2.718)
By a factor of pi
By a factor of 10
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Slide 1-53
Section 15.8
Clicker Question 8
During a time interval of one time constant, by what
factor does the mechanical energy of an oscillating
system decrease?
1.
2.
3.
4.
5.
6.
7.
By a factor of 1/e (0.3679)
By a factor of ln(2) (0.6931)
By a factor of (2)1/2
By a factor of 2
By a factor of e
(2.718)
By a factor of pi
By a factor of 10
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Slide 1-54
Chapter 15: Summary
Concepts: Fundamental characteristics of
periodic motion
• Periodic motion is any motion that repeats itself at
regular time intervals. Oscillation (or vibration) is
back-and-forth periodic motion.
• The period T is the minimum time interval in which
periodic motion repeats, and the amplitude A of the
motion is the magnitude of the maximum
displacement of the moving object from its
equilibrium position.
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Slide 1-55
Chapter 15: Summary
Quantitative Tools: Fundamental
characteristics of periodic motion
• The frequency f of periodic motion is the number of
cycles per second and is defined as
1
f º .
T
• The SI unit of frequency is the hertz (Hz):
1 Hz  1 s–1.
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Slide 1-56
Chapter 15: Summary
Concepts: Simple harmonic motion
• Simple harmonic motion is periodic motion in which
the displacement of a system from its equilibrium
position varies sinusoidally with time. A system
moving in this way is called a simple harmonic
oscillator.
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Slide 1-57
Chapter 15: Summary
Concepts: Simple harmonic motion
• A restoring force that is linearly proportional to
displacement tends to return a simple harmonic
oscillator to its equilibrium position. For small
displacements, restoring forces are generally
proportional to the displacement and therefore cause
objects to execute simple harmonic motion about any
stable equilibrium position.
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Slide 1-58
Chapter 15: Summary
Concepts: Simple harmonic motion
• A phasor is a rotating arrow whose component on a
vertical axis traces out simple harmonic motion. The
reference circle is the circle traced out by the tip of
the phasor, and the length of the phasor is equal to the
amplitude A of the simple harmonic motion.
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Slide 1-59
Chapter 15: Summary
Quantitative Tools: Simple harmonic motion
• The angular frequency  of oscillation is equal to the rotational
speed of a rotating phasor whose vertical component oscillates with
a frequency f. Angular frequency is measured in s–1 and is related to
the frequency (measured in Hz) by
 = 2f.
• For a simple harmonic oscillator of amplitude A, the displacement x
as a function of time is
x(t) = A sin(t + i),
where the sine argument is the phase (t) of the periodic motion,
(t) = t + i,
and i is the initial phase at t = 0.
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Slide 1-60
Chapter 15: Summary
Quantitative Tools: Simple harmonic motion
• The x components of the velocity and acceleration of
a simple harmonic oscillator are
dx
ux º
= w Acos(w t + fi )
dt
2
d x
ax º 2 = -w 2 Asin(w t + fi ).
dt
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Slide 1-61
Chapter 15: Summary
Quantitative Tools: Simple harmonic motion
• Any object that undergoes simple harmonic motion
obeys the simple harmonic oscillator equation:
d 2x
2
=
w
x,
2
dt
where x is the object’s displacement from its
equilibrium position.
• The mechanical energy E of an object of mass m
undergoing simple harmonic motion is
E = 12 mw 2 A2 .
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Slide 1-62
Chapter 15: Summary
Concepts: Fourier series
• Fourier’s theorem says that any periodic function
with period T can be written as a sum of sinusoidal
simple harmonic functions of frequency fn = n/T,
where n is an integer. The n = 1 term is the
fundamental frequency or first harmonic, and the
other components are higher harmonics.
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Slide 1-63
Chapter 15: Summary
Quantitative Tools: Oscillating springs
• For an object attached to a light spring of spring constant k, the
simple harmonic oscillator equation takes the form
d 2x
k
=
x,
2
m
dt
and the angular frequency of the oscillation is
k
w =+
.
m
• The motion of the object is described by
æ k
ö
x(t) = A sin ç
t + fi ÷ .
è m
ø
© 2015 Pearson Education, Inc.
Slide 1-64
Chapter 15: Summary
Concepts: Rotational oscillations
• A horizontal disk suspended at its center by a thin
fiber forms a type of torsional oscillator.
• A pendulum is any object that swings about a pivot.
A simple pendulum consists of a small object (the
bob) attached to a very light wire or rod.
© 2015 Pearson Education, Inc.
Slide 1-65
Chapter 15: Summary
Quantitative Tools: Rotational oscillations
• If a torsional oscillator of rotational inertia I is
twisted through a small angle from its equilibrium
position 0 to position , the restoring torque  is
 = –( – 0),
where  is the torsional constant. When 0 = 0, the
simple harmonic oscillator equation for the torsional
oscillator is
2
dJ
k
= - J.
2
I
dt
© 2015 Pearson Education, Inc.
Slide 1-66
Chapter 15: Summary
Quantitative Tools: Rotational oscillations
• The rotational position  of the torsional oscillator at
instant t is given by
 = max sin(t + i),
where max is the maximum rotational displacement
and
w=
© 2015 Pearson Education, Inc.
k
I
.
Slide 1-67
Chapter 15: Summary
Quantitative Tools: Rotational oscillations
• For small rotational displacements, the simple harmonic
oscillator equation of a pendulum is
m cm g
d 2J
=J
2
I
dt
and its angular frequency is
w=
m
cm
g
,
I
where cm is the distance from the center of mass of the
pendulum to the pivot.
• The period of a simple pendulum is T = 2p
.
g
© 2015 Pearson Education, Inc.
Slide 1-68
Chapter 15: Summary
Concepts: Damped oscillations
• In damped oscillation, the amplitude decreases over
time due to energy dissipation. The cause of the
dissipation is a damping force due to friction, air
drag, or water drag.
• A damped oscillator that has a large quality factor Q
keeps oscillating for many periods.
© 2015 Pearson Education, Inc.
Slide 1-69
Chapter 15: Summary
Quantitative Tools: Damped oscillations
• At low speeds, the damping force Faod tends to be proportional
to the velocity of the object:
Faod = -bu ,
where b, the damping coefficient, has units of kilograms per
second.
• For small damping, the position x(t) of a damped spring is
x(t) = Ae–bt/2m sin(dt + i)
and its angular frequency d is
wd =
© 2015 Pearson Education, Inc.
2
æ b ö
k
b
2
= w -ç
.
2
÷
m 4m
è 2m ø
2
Slide 1-70