Unformatted text preview: 14 Oscillations
This loudspeaker cone generates
sound waves by oscillating back
and forth at audio frequencies.
Looking Ahead The goal of Chapter 14 is to understand systems that oscillate with simple harmonic motion.
Simple Harmonic Motion
Springs
Pendulums
The most basic
oscillation, with
sinusoidal motion,
is called simple
harmonic motion.
Simple harmonic motion occurs when
there is a linear restoring force. The
simplest example is
a mass on a spring.
You will learn how to
determine the period
of oscillation.
A mass swinging at the end of a string or
rod is a pendulum. Its motion is another
example of simple harmonic motion.
Oscillation
The oscillating cart
is an example of
simple harmonic
motion. You’ll learn
how to use the
mass and the spring
constant to deter
mine the frequency
of oscillation.
In this chapter you will learn to:
■ Represent
simple harmonic motion
both graphically and mathematically.
■ Understand the dynamics of oscillating systems.
■ Recognize the similarities among
many types of oscillating systems.
Simple harmonic motion has a very
close connection to uniform circular
motion. You’ll learn that an edgeon
view of uniform circular motion is none
other than simple harmonic motion.
Looking Back
Section 4.5 Uniform circular motion
The “bounce” at the
bottom of a bungee
jump is an exhilarating
example of a mass
oscillating on a spring.
Looking Back
Section 10.4 Restoring forces
Damping and Resonance
Energy of Oscillations
If there’s drag or other dissipation, then
the oscillation “runs down.” This is
called a damped oscillation.
If there is no friction or other dissipation, then the mechanical energy of an
oscillator is conserved. Conservation of
energy will be an important tool.
All kinetic
The system oscil
lates between all
kinetic energy and
all potential energy
A
0
All potential
Looking Back
Section 10.5 Elastic potential energy
Section 10.6 Energy diagrams
Illustrated Chapter Previews
give an overview of the
upcoming ideas for each
chapter, setting them in context,
explaining their utility, and tying
them to existing knowledge
(through Looking Back
references). These previews
build on the cognitive
psychology concept of an
“advance organizer.”
The period of a pendu
lum is determined by
the length of the string;
neither the mass nor
the amplitude matters.
Consequently, the pen
dulum was the basis of
time keeping for many
centuries.
A
x
The amplitude of
a damped oscil
lation undergoes
exponential
decay.
A
x
0
A
Oscillations can increase in amplitude,
sometimes dramatically, when driven at
their natural oscillation frequency. This
is called resonance.
t
378 c h a p t e r 14 . Oscillations
FiguRE 14.1 Examples of positionversus
time graphs for oscillating systems.
The oscillation takes
place around an
equilibrium position.
Position
T
t
14.1 Simple Harmonic Motion
Objects or systems of objects that undergo oscillatory motion—a repetitive motion
back and forth around an equilibrium position—are called oscillators. FiguRE 14.1
shows positionversustime graphs for three different oscillating systems. Although
the shapes of the graphs are different, all these oscillators have two things in common:
1. The oscillation takes place about an equilibrium position, and
2. The motion is periodic, repeating at regular intervals of time.
Position
The motion is periodic.
One cycle takes time T.
T
t
Position
This oscillation
is sinusoidal.
T
The time to complete one full cycle, or one oscillation, is called the period of the
motion. Period is represented by the symbol T.
A closely related piece of information is the number of cycles, or oscillations, com1
pleted per second. If the period is 10 s, then the oscillator can complete 10 cycles
1
in one second. Conversely, an oscillation period of 10 s allows only 10 of a cycle to be
completed per second. In general, T seconds per cycle implies that 1/T cycles will be
completed each second. The number of cycles per second is called the frequency f of
the oscillation. The relationship between frequency and period is
t
f=
1
T
or
T=
1
f
(14.1)
The units of frequency are hertz, abbreviated Hz, named in honor of the German
physicist Heinrich Hertz, who produced the first artificially generated radio waves in
1887. By definition,
TABLE 14.1
Units of frequency
Frequency
Period
103 Hz = 1 kilohertz = 1 kHz
1 ms
106 Hz = 1 megahertz = 1 MHz
1 ms
Figures are carefully
1 ns
109 Hz = 1 gigahertz = 1 GHz
streamlined in detail and
color so students focus on the
physics—for instance, the object
of interest in mechanics.
FiguRE 14.2 A prototype simple
Explicit instruction as
annotations directly on figures
helps students to interpret
figures and graphs.
harmonicmotion experiment.
(a)
Oscillation
The point on the
object that is
measured
Air track
(b)
0
A
Turning
point
A
A
A
t
x
x is measured from
the equilibrium
position where
the object would
be at rest.
The motion
is sinusoidal,
indicating
SHM.
The motion is symmetrical about the
equilibrium position. Maximum distance
to the left and to the right is A.
1 Hz K 1 cycle per second = 1 s 1
We will frequently deal with very rapid oscillations and make use of the units shown
in Table 14.1.
Uppercase and lowercase letters are important. 1 MHz is 1 megahertz =
106 Hz, but 1 mHz is 1 millihertz = 103 Hz!
NOTE
ExAMPLE 14.1
Frequency and period of a loudspeaker cone
What is the oscillation period of a loudspeaker cone that vibrates back and forth 5000 times
per second?
SOLvE The oscillation frequency is f = 5000 cycles/s = 5000 Hz = 5.0 kHz. The period
is the inverse of the frequency; hence
T=
1
1
=
= 2.0 * 104 s = 200 ms
f
5000 Hz
A system can oscillate in many ways, but we will be especially interested in
the smooth sinusoidal oscillation (i.e., like a sine or cosine) of the third graph in
Figure 14.1. This sinusoidal oscillation, the most basic of all oscillatory motions, is
called simple harmonic motion, often abbreviated SHM. Let’s look at a graphical
description before we dive into the mathematics of simple harmonic motion.
FiguRE 14.2a shows an airtrack glider attached to a spring. If the glider is pulled out
a few centimeters and released, it will oscillate back and forth on the nearly frictionless air track. FiguRE 14.2b shows actual results from an experiment in which a computer was used to measure the glider’s position 20 times every second. This is a
positionversustime graph that has been rotated 90 from its usual orientation in order
for the xaxis to match the motion of the glider.
The object’s maximum displacement from equilibrium is called the amplitude A
of the motion. The object’s position oscillates between x =  A and x = + A. When
using a graph, notice that the amplitude is the distance from the axis to the maximum,
not the distance from the minimum to the maximum.
14.1 . Simple Harmonic Motion
shows the data with the graph axes in their “normal” positions. You
can see that the amplitude in this experiment was A = 0.17 m, or 17 cm. You can
also measure the period to be T = 1.60 s. Thus the oscillation frequency was
f = 1/T = 0.625 Hz.
FiguRE 14.3b is a velocityversustime graph that the computer produced by using
x/ t to find the slope of the position graph at each point. The velocity graph is also
sinusoidal, oscillating between  vmax (maximum speed to the left) and + vmax (maximum speed to the right). As the figure shows,
FiguRE 14.3a
■
■
The instantaneous velocity is zero at the points where x = { A. These are the turn
ing points in the motion.
The maximum speed vmax is reached as the object passes through the equilibrium
position at x = 0 m. The velocity is positive as the object moves to the right but
negative as it moves to the left.
Position and velocity graphs
of the experimental data.
FiguRE 14.3
(a) The speed is zero
when x
A.
0
0.1
0.2
x
0
12
v
vmax
v
t (s)
6.0
vmax
0
0
2.0
4.0
t (s)
6.0
The positionversustime
graph for simple harmonic motion.
FiguRE 14.4
(14.2)
where the notation x(t) indicates that the position x is a function of time t. Because
cos(2p) = cos(0), it’s easy to see that the position at time t = T is the same as the position at t = 0. In other words, this is a cosine function with period T. Be sure to convince
yourself that this function agrees with the five special points shown in Figure 14.4.
NOTE The argument of the cosine function is in radians. That will be true throughout this chapter. It’s especially important to remember to set your calculator to
radian mode before working oscillation problems. Leaving it in degree mode will
lead to errors.
We can write Equation 14.2 in two alternative forms. Because the oscillation frequency is f = 1/T, we can write
1. Starts at x A
2. Passes through
x
x 0 at t 1 T
4
5. Returns to
x A at t
(14.5)
Equations 14.2, 14.3, and 14.5 are equivalent ways to write the position of an object
moving in simple harmonic motion.
T
A
0
2T
T
A
4. Passes through x
3. Reaches x
A at t
0 at t
1
2T
(14.3)
Recall from Chapter 4 that a particle in circular motion has an angular velocity v that
is related to the period by v = 2p/T, where v is in rad/s. Now that we’ve defined the
frequency f, you can see that v and f are related by
2p
v (in rad/s) =
= 2pf (in Hz)
(14.4)
T
In this context, v is called the angular frequency. The position can be written in terms
of v as
x ( t ) = A cos vt
4.0
0.7
redraws the positionversustime graph of Figure 14.3a as a smooth curve.
Although these are empirical data (we don’t yet have any “theory” of oscillation) the
graph for this particular motion is clearly a cosine function. The object’s position is
x ( t ) = A cos (2pft )
0.17 m
A
2.0
(b)
vx (m/s)
Kinematics of Simple Harmonic Motion
2pt
T
0.17 m
1.60 s
0.1
0.7
x ( t ) = A cos
A
T
0.2
A mass oscillating on a spring is the prototype of simple harmonic motion. Our
analysis, in which we answer these questions, will be of a springmass system. Even
so, most of what we learn will be applicable to other types of SHM.
FiguRE 14.4
The speed is maximum
as the object passes
through x 0.
x
x (m)
We can ask three important questions about this oscillating system:
1. How is the maximum speed vmax related to the amplitude A?
2. How are the period and frequency related to the object’s mass m, the spring
constant k, and the amplitude A?
3. Is the sinusoidal oscillation a consequence of Newton’s laws?
379
Derivatives of sine and
cosine functions
TABLE 14.2
d
1 a sin ( bt + c )2 = + ab cos ( bt + c )
dt
d
1 a cos ( bt + c )2 =  ab sin ( bt + c )
dt
t
3
4T
NOTE paragraphs throughout
guide students away from
known preconceptions and
around common sticking
points and highlight many
math and vocabularyrelated
issues that have been proven
to cause difficulties.
380 c h a p t e r 14 . Oscillations
Position and velocity graphs
for simple harmonic motion.
FiguRE 14.5
Position x
T
A
vx ( t ) =  vmax sin
t
0
2T
T
A
Velocity vx
T
vmax
NOTE vmax
vx ( t ) =
t
0
vmax
2T
T
v (t)
2pt
=  vmax sin(2pft ) =  vmax sin vt
T
(14.6)
is the maximum speed and thus is a positive number.
12
We deduced Equation 14.6 from the experimental results, but we could equally well
find it from the position function of Equation 14.2. After all, velocity is the time derivative of position. Table 14.2 on the previous page reminds you of the derivatives of the
sine and cosine functions. Using the derivative of the position function, we find
A cos vt
x (t)
12
Just as the position graph was clearly a cosine function, the velocity graph shown
in FiguRE 14.5 is clearly an “upsidedown” sine function with the same period T. The
velocity vx, which is a function of time, can be written
dx
2pA
2pt
=sin
=  2pfA sin(2pft ) =  vA sin vt (14.7)
dt
T
T
Comparing Equation 14.7, the mathematical definition of velocity, to Equation 14.6,
the empirical description, we see that the maximum speed of an oscillation is
vmax sin vt
vmax =
2pA
= 2pfA = vA
T
(14.8)
Equation 14.8 answers the first question we posed above, which was how the maximum
speed vmax is related to the amplitude A. Not surprisingly, the object has a greater maximum speed if you stretch the spring farther and give the oscillation a larger amplitude.
ExAMPLE 14.2
Worked examples follow a
consistent problemsolving
strategy and include careful
explanations of the underlying,
and often unstated, reasoning.
A system in simple harmonic motion
An airtrack glider is attached to a spring, pulled 20.0 cm to the
right, and released at t = 0 s. It makes 15 oscillations in 10.0 s.
a. What is the period of oscillation?
b. What is the object’s maximum speed?
c. What are the position and velocity at t = 0.800 s?
MODEL
SOLvE
15 oscillations
= 1.50 oscillations/s = 1.50 Hz
f=
10.0 s
Thus the period is T = 1/f = 0.667 s.
b. The oscillation amplitude is A = 0.200 m. Thus
vmax =
ExAMPLE 14.3
2pA 2p(0.200 m)
=
= 1.88 m/s
T
0.667 s
Finding the time
A mass oscillating in simple harmonic motion starts at x = A and
has period T. At what time, as a fraction of T, does the object first
pass through x = 1 A?
2
Figure 14.4 showed that the object passes through the
equilibrium position x = 0 at t = 1 T. This is onequarter of the
4
total distance in onequarter of a period. You might expect it to
take 1 T to reach 1 A, but this is not the case because the SHM
8
2
graph is not linear between x = A and x = 0. We need to use
x ( t ) = A cos (2pt/T ). First, we write the equation with x = 1 A:
2
SOLvE
x = A cos
An object oscillating on a spring is in SHM.
a. The oscillation frequency is
12
1
2
c. The object starts at x = + A at t = 0 s. This is exactly the
oscillation described by Equations 14.2 and 14.6. The position
at t = 0.800 s is
2p (0.800 s)
2pt
= (0.200 m) cos
T
0.667 s
12
1
2
= (0.200 m)cos (7.54 rad) = 0.0625 m = 6.25 cm
The velocity at this instant of time is
vx =  vmax sin
2p(0.800 s)
2pt
=  (1.88 m/s) sin
T
0.667 s
=  (1.88 m/s) sin (7.54 rad) =  1.79 m/s =  179 cm/s
At t = 0.800 s, which is slightly more than one period, the object is 6.25 cm to the right of equilibrium and moving to the left
at 179 cm/s. Notice the use of radians in the calculations.
12
12
A
2pt
= A cos
2
T
Then we solve for the time at which this position is reached:
T
1
Tp 1
t=
cos 1
=
=T
2p
2
2p 3
6
x=
The motion is slow at the beginning and then speeds up,
so it takes longer to move from x = A to x = 1 A than it does to
2
move from x = 1 A to x = 0. Notice that the answer is indepen2
dent of the amplitude A.
ASSESS
14.2 . Simple Harmonic Motion and Circular Motion
381
An object moves with simple harmonic motion. If the amplitude
and the period are both doubled, the object’s maximum speed is
Stop to think 14.1
a. Quadrupled.
d. Halved.
b. Doubled.
e. Quartered.
c. Unchanged.
14.2 Simple Harmonic Motion
and Circular Motion
The graphs of Figure 14.5 and the position function x ( t ) = A cos vt are for an oscillation in which the object just happened to be at x0 = A at t = 0. But you will recall that
t = 0 is an arbitrary choice, the instant of time when you or someone else starts a stopwatch. What if you had started the stopwatch when the object was at x0 =  A, or when
the object was somewhere in the middle of an oscillation? In other words, what if the
oscillator had different initial conditions. The position graph would still show an oscillation, but neither Figure 14.5 nor x ( t ) = A cos vt would describe the motion correctly.
To learn how to describe the oscillation for other initial conditions it will help to
turn to a topic you studied in Chapter 4—circular motion. There’s a very close connection between simple harmonic motion and circular motion.
Imagine you have a turntable with a small ball glued to the edge. FiguRE 14.6a shows
how to make a “shadow movie” of the ball by projecting a light past the ball and onto
a screen. The ball’s shadow oscillates back and forth as the turntable rotates. This is
certainly periodic motion, with the same period as the turntable, but is it simple harmonic motion?
To find out, you could place a real object on a real spring directly below the shadow, as shown in FiguRE 14.6b. If you did so, and if you adjusted the turntable to have the
same period as the spring, you would find that the shadow’s motion exactly matches
the simple harmonic motion of the object on the spring. Uniform circular motion
projected onto one dimension is simple harmonic motion.
To understand this, consider the particle in FiguRE 14.7. It is in uniform circular
motion, moving counterclockwise in a circle with radius A. As in Chapter 4, we can
locate the particle by the angle f measured ccw from the xaxis. Projecting the ball’s
shadow onto a screen in Figure 14.6 is equivalent to observing just the xcomponent
of the particle’s motion. Figure 14.7 shows that the xcomponent, when the particle is
at angle f, is
x = A cos f
(14.9)
Recall that the particle’s angular velocity, in rad/s, is
v=
df
dt
A projection of the circular
motion of a rotating ball matches the
simple harmonic motion of an object on
a spring.
FiguRE 14.6
(a)
Light from projector
Ball
Shadow
(b)
Simple harmonic motion of block
A particle in uniform
circular motion with radius A and
angular velocity v.
FiguRE 14.7
y
(14.10)
Particle in uniform
circular motion
(14.11)
v
A
f
A
0
A cos f
As f increases, the particle’s xcomponent is
x ( t ) = A cos vt
Screen
Oscillation of ball’s shadow
This is the rate at which the angle f is increasing. If the particle starts from f0 = 0 at
t = 0, its angle at a later time t is simply
f = vt
Turntable
Circular
motion
of ball
x
v
(14.12)
This is identical to Equation 14.5 for the position of a mass on a spring! Thus the
xcomponent of a particle in uniform circular motion is simple harmonic motion.
When used to describe oscillatory motion, v is called the angular fre
quency rather than the angular velocity. The angular frequency of an oscillator has
the same numerical value, in rad/s, as the angular velocity of the corresponding
particle in circular motion.
A
The xcomponent of
the particle’s position
describes the position
of the ball’s shadow.
NOTE
A
0
x
A cos f
A
New concepts are introduced
through observations about
the real world and theories,
grounded by making sense of
observations. This inductive
approach illustrates how
science operates, and has
been shown to improve
student learning by reconciling
new ideas with what they
already know.
382 c h a p t e r 14 . Oscillations
Analogy is used throughout the
text and figures to consolidate
student understanding by
comparing with a more familiar
concept or situation.
A cup on the turntable in a microwave
oven moves in a circle. But from the
outside, you see the cup sliding back and
forth—in simple harmonic motion!
The names and units can be a bit confusing until you get used to them. It may help
to notice that cycle and oscillation are not true units. Unlike the “standard meter” or the
“standard kilogram,” to which you could compare a length or a mass, there is no “standard cycle” to which you can compare an oscillation. Cycles and oscillations are simply
counted events. Thus the frequency f has units of hertz, where 1 Hz = 1 s 1. We may
say “cycles per second” just to be clear, but the actual units are only “per second.”
The radian is the SI unit of angle. However, the radian is a defined unit. Further, its
definition as a ratio of two lengths (u = s/r) makes it a pure number without dimensions. As we noted in Chapter 4, the unit of angle, be it radians or degrees, is really just
a name to remind us that we’re dealing with an angle. The 2p in the equation v = 2pf
(and in similar situations), which is stated without units, means 2p rad/cycle. When
multiplied by the frequency f in cycles/s, it gives the frequency in rad/s. That is why,
in this context, v is called the angular frequency.
NOTE Hertz is specifically “cycles per second” or “oscillations per second.” It is
used for f but not for v. We’ll always be careful to use rad/s for v, but you should
be aware that many books give the units of v as simply s 1.
The Phase Constant
A particle in uniform
circular motion with initial angle f0.
FiguRE 14.8
Angle at time t is
f vt f0.
y
Initial position of
particle at t 0
v
A
f0
0 A cos f
f = vt + f0
(14.13)
In this case, the particle’s projection onto the xaxis at time t is
A
f
Now we’re ready to consider the issue of other initial conditions. The particle in
Figure 14.7 started at f0 = 0. This was equivalent to an oscillator starting at the far
right edge, x0 = A. FiguRE 14.8 shows a more general situation in which the initial angle
f0 can have any value. The angle at a later time t is then
x0
x
A
A cos f0
v
The initial xcomponent of the
particle’s position can be anywhere
between A and A, depending on f0.
x ( t ) = A cos ( vt + f0 )
(14.14)
If Equation 14.14 describes the particle’s projection, then it must also be the position of an oscillator in simple harmonic motion. The oscillator’s velocity vx is found
by taking the derivative dx/dt. The resulting equations,
x ( t ) = A cos ( vt + f0 )
vx ( t ) =  vA sin( vt + f0 ) =  vmax sin( vt + f0 )
(14.15)
are the two primary kinematic equations of simple harmonic motion.
The quantity f = vt + f0, which steadily increases with time, is called the phase
of the oscillation. The phase is simply the angle of the circularmotion particle whose
shadow matches the oscillator. The constant f0 is called the phase constant. It specifies the initial conditions of the oscillator.
To see what the phase constant means, set t = 0 in Equations 14.15:
x0 = A cos f0
v0x =  vA sin f0
(14.16)
The position x0 and velocity v0x at t = 0 are the initial conditions. Different values of
the phase constant correspond to different starting points on the circle and thus
to different initial conditions.
The perfect cosine function of Figure 14.5 and the equation x ( t ) = A cos vt are for
an oscillation with f0 = 0 rad. You can see from Equations 14.16 that f0 = 0 rad implies x0 = A and v0 = 0. That is, the particle starts from rest at the point of maximum
displacement.
FiguRE 14.9 illustrates these ideas by looking at three values of the phase constant:
f0 = p/3 rad (60 ),  p/3 rad (  60 ), and p rad (180 ). Notice that f0 = p/3 rad and
f0 =  p/3 rad have the same starting position, x0 = 1 A. This is a property of the co2
sine function in Equation 14.16. But these are not the same initial conditions. In one case
the oscillator starts at 1 A while moving to the right, in the other case it starts at 1 A while
2
2
moving to the left. You can distinguish between the two by visualizing the motion.
14.2 . Simple Harmonic Motion and Circular Motion
FiguRE 14.9
383
Oscillations described by the phase constants f0 = p/3 rad,  p/3 rad, and p rad.
f0
p/3 rad
y
f0
1
2A
p
3
A
1
2A
x
p
3
f0
t
The starting point of the oscillation is
shown on the circle and on the graph.
A
0
1
2A
x
A
A
0
1
2A
A
t
T
x
A
0
t
t
vmax
t
0
T
T
A
v
vmax
T
x
A
x
v
vmax
x
A
0
v
p
0
A
A
A
f0
Extensive use is made of
multiple representations—
placing different representations
side by side to help students
develop the key skill of
translating between words,
math, and figures. Essential to
good problemsolving, this skill
is overlooked in most physics
textbooks.
0
A
A
0
p rad
y
0
At
x
x
x
The graphs each have the same
amplitude and period. They are
shifted relative to the f0 0 rad
graphs of Figure 14.5 because they
have different initial conditions.
f0
0
t
f0
p/3 rad
y
0
vmax
vmax
t
t
0
T
T
vmax
All values of the phase constant f0 between 0 and p rad correspond to a particle in
the upper half of the circle and moving to the left. Thus v0x is negative. All values of the
phase constant f0 between p and 2p rad (or, as they are usually stated, between  p
and 0 rad) have the particle in the lower half of the circle and moving to the right. Thus
v0x is positive. If you’re told that the oscillator is at x = 1 A and moving to the right at
2
t = 0, then the phase constant must be f0 =  p/3 rad, not + p/3 rad.
ExAMPLE 14.4
using the initial conditions
An object on a spring oscillates with a period of 0.80 s and an
amplitude of 10 cm. At t = 0 s, it is 5.0 cm to the left of equilibrium and moving to the left. What are its position and direction of
motion at t = 2.0 s?
MODEL
An object oscillating on a spring is in simple harmonic
motion.
12
12
SOLvE We can find the phase constant f0 from the initial condition x0 =  5.0 cm = A cos f0. This condition gives
f0 = cos 1
x0
1
2
= cos 1 = { p rad = { 120
A
2
3
Because the oscillator is moving to the left at t = 0, it is in the
upper half of the circularmotion diagram and must have a phase
constant between 0 and p rad. Thus f0 is 2 p rad. The angular
3
frequency is
v=
2p
2p
=
= 7.85 rad/s
T
0.80 s
1
Thus the object’s position at time t = 2.0 s is
x ( t ) = A cos ( vt + f0 )
= (10 cm) cos (7.85 rad/s) (2.0 s) +
= (10 cm) cos (17.8 rad) = 5.0 cm
2
p
3
2
The object is now 5.0 cm to the right of equilibrium. But which
way is it moving? There are two ways to find out. The direct way
is to calculate the velocity at t = 2.0 s:
vx =  vA sin ( vt + f0 ) = + 68 cm/s
The velocity is positive, so the motion is to the right. Alternatively, we could note that the phase at t = 2.0 s is f = 17.8 rad.
Dividing by p, you can see that
f = 17.8 rad = 5.67p rad = (4p + 1.67p ) rad
The 4p rad represents two complete revolutions. The “extra”
phase of 1.67p rad falls between p and 2p rad, so the particle in
the circularmotion diagram is in the lower half of the circle and
moving to the right.
384 c h a p t e r 14 . Oscillations
cos 1 is a twovalued function. Your calculator
returns a single value, an angle between 0 rad and p rad. But the negative of this
angle is also a solution. As Example 14.4 demonstrates, you must use additional
information to choose between them.
NOTE The inversecosine function
Stop to think 14.2
The figure shows
four oscillators at t = 0. Which one
has the phase constant f0 = p/4 rad?
Stop to Think questions at
the end of a section allow
students to quickly check their
understanding. Using powerful
rankingtask and graphical
techniques, they are designed
to efficiently probe key
misconceptions and encourage
active reading. (Answers are
provided at the end of the
chapter.)
A
A
(a)
(b)
(c)
(d)
100
71
0
71 100
x (mm)
14.3 Energy in Simple Harmonic Motion
The energy is transformed
between kinetic energy and potential
energy as the object oscillates, but the
mechanical energy E = K + U doesn’t
change.
FiguRE 14.10
Energy is transformed between
kinetic and potential, but the total
mechanical energy E doesn’t change.
(a)
k
A
1
2
2 mv
E
12
2 kx
m
r
v
x
0
x
A
(b)
Potentialenergy
curve
Energy
E
Total
energy
line
Turning
point
A
0
Turning
point
x
A
Energy here is
purely kinetic.
Energy here is purely potential.
We’ve begun to develop the mathematical language of simple harmonic motion, but
thus far we haven’t included any physics. We’ve made no mention of the mass of
the object or the spring constant of the spring. An energy analysis, using the tools of
Chapters 10 and 11, is a good starting place.
FiguRE 14.10a shows an object oscillating on a spring, our prototype of simple
harmonic motion. Now we’ll specify that the object has mass m, the spring has
spring constant k, and the motion takes place on a frictionless surface. You learned
in Chapter 10 that the elastic potential energy when the object is at position x is
Us = 1 k( x)2, where x = x  xe is the displacement from the equilibrium position
2
xe. In this chapter we’ll always use a coordinate system in which xe = 0, making
x = x. There’s no chance for confusion with gravitational potential energy, so we
can omit the subscript s and write the elastic potential energy as
1
U = kx 2
(14.17)
2
Thus the mechanical energy of an object oscillating on a spring is
1
1
E = K + U = mv 2 + kx 2
(14.18)
2
2
1
2
FiguRE 14.10b is an energy diagram, showing the potentialenergy curve U = 2 kx as
a parabola. Recall that a particle oscillates between the turning points where the total
energy line E crosses the potentialenergy curve. The left turning point is at x =  A,
and the right turning point is at x = + A. To go beyond these points would require a
negative kinetic energy, which is physically impossible.
You can see that the particle has purely potential energy at x
t A and purely kinetic energy as it passes through the equilibrium point at x
0. At maximum
displacement, with x = { A and v = 0, the energy is
1
E (at x = { A ) = U = kA2
(14.19)
2
At x = 0, where v = { vmax, the energy is
1
E (at x = 0) = K = m ( vmax)2
(14.20)
2
14.3 . Energy in Simple Harmonic Motion
385
The system’s mechanical energy is conserved because the surface is frictionless
and there are no external forces, so the energy at maximum displacement and the energy at maximum speed, Equations 14.19 and 14.20, must be equal. That is
1
1
(14.21)
m(vmax)2 = kA2
2
2
Thus the maximum speed is related to the amplitude by
k
vmax =
A
(14.22)
Am
This is a relationship based on the physics of the situation.
Earlier, using kinematics, we found that
2pA
vmax =
= 2pfA = vA
(14.23)
T
Comparing Equations 14.22 and 14.23, we see that frequency and period of an oscillating spring are determined by the spring constant k and the object’s mass m:
v=
k
m
A
f=
1
k
m
2p A
T = 2p
m
Ak
(14.24)
These three expressions are really only one equation. They say the same thing, but
each expresses it in slightly different terms.
Equations 14.24 are the answer to the second question we posed at the beginning of
the chapter, where we asked how the period and frequency are related to the object’s
mass m, the spring constant k, and the amplitude A. It is perhaps surprising, but the
period and frequency do not depend on the amplitude A. A small oscillation and a
large oscillation have the same period.
Because energy is conserved, we can combine Equations 14.18, 14.19, and 14.20
to write
1
1
1
1
E = mv 2 + kx 2 = kA2 = m(vmax)2 (conservation of energy)
2
2
2
2
Kinetic energy, potential
energy, and the total mechanical energy
for simple harmonic motion.
FiguRE 14.11
The total mechanical
energy E is constant.
Potential energy
Kinetic energy
Energy
T
(14.25)
k2
( A  x2 ) = v 2A2  x2
(14.26)
Bm
FiguRE 14.11 shows graphically how the kinetic and potential energy change with
time. They both oscillate but remain positive because x and v are squared. Energy is
continuously being transformed back and forth between the kinetic energy of the moving block and the stored potential energy of the spring, but their sum remains constant.
Notice that K and U both oscillate twice each period; make sure you understand why.
Any pair of these expressions may be useful, depending on the known information.
For example, you can use the amplitude A to find the speed at any point x by combining the first and second expressions for E. The speed v at position x is
t
0
Position
v=
ExAMPLE 14.5
using conservation of energy
A 500 g block on a spring is pulled a distance of 20 cm and released.
The subsequent oscillations are measured to have a period of 0.80 s.
a. At what position or positions is the block’s speed 1.0 m/s?
b. What is the spring constant?
MODEL
The motion is SHM. Energy is conserved.
a. The block starts from the point of maximum displacement, where E = U = 1 kA2. At a later time, when the position is
2
x and the speed is v, energy conservation requires
121212
mv + k x = k A
2
2
2
Solving for x, we find
SOLvE
x=
B
A2 
12
mv 2
v
=
A2 v
k
B
2
t
0
1
2
where we used k/m = v2 from Equation 14.24. The angular frequency is easily found from the period: v = 2p/T = 7.85 rad/s.
Thus
x=
B
(0.20 m)2 
1.0 m/s
7.85 rad/s
2
= { 0.15 m = { 15 cm
There are two positions because the block has this speed on
either side of equilibrium.
b. Although part a did not require that we know the spring constant, it is straightforward to find from Equation 14.24:
m
Ak
4p2 (0.50 kg)
4 p2 m
k=
=
= 31 N/m
2
T
(0.80 s)2
T = 2p
386 c h a p t e r 14 . Oscillations
Stop to think 14.3
The four springs shown here have been compressed from their equilibrium position at
x = 0 cm. When released, the attached mass
will start to oscillate. Rank in order, from
highest to lowest, the maximum speeds of
the masses.
k
(a)
(b)
1
2k
4m
k
(c)
4m
2m
k
(d)
20
m
15
10
5
0
x (cm)
14.4 The Dynamics of Simple
Harmonic Motion
Our analysis thus far has been based on the experimental observation that the oscillation of a spring “looks” sinusoidal. It’s time to show that Newton’s second law predicts
sinusoidal motion.
A motion diagram will help us visualize the object’s acceleration. FiguRE 14.12 shows
one cycle of the motion, separating motion to the left and motion to the right to make
the diagram clear. As you can see, the object’s velocity is large as it passes through the
u
equilibrium point at x = 0, but v is not changing at that point. Acceleration measures
u
u
the change of the velocity; hence a = 0 at x = 0.
Motion diagram of simple harmonic motion. The left and right motions are
separated vertically for clarity but really occur along the same line.
FiguRE 14.12
To the right
r
a
r
a
r
r
a
0
r
v
r
v
r
r
r
r
a
a
v
Position x
0
Equilibrium
To the left
0
2A
r
a
Same
point
x
A
T
A
0
2T
T
t
A
Acceleration ax
r
a
Same
point
Position and acceleration
graphs for an oscillating spring. We’ve
chosen f0 = 0.
FiguRE 14.13
v
r
r
a
v2A when x
amax
A
amax
0
2T
T
t
amax
amin
v2A when x
A
In contrast, the velocity is changing rapidly at the turning points. At the right turnu
ing point, v changes from a rightpointing vector to a leftpointing vector. Thus the
u
acceleration a at the right turning point is large and to the left. In onedimensional
motion, the acceleration component ax has a large negative value at the right turning
u
point. Similarly, the acceleration a at the left turning point is large and to the right.
Consequently, ax has a large positive value at the left turning point.
Our motiondiagram analysis suggests that the acceleration ax is most positive
when the displacement is most negative, most negative when the displacement is a
maximum, and zero when x = 0. This is confirmed by taking the derivative of the
velocity:
ax =
dvx
d
= (  vA sin vt ) =  v2 A cos vt
dt
dt
(14.27)
then graphing it.
FiguRE 14.13 shows the position graph that we started with in Figure 14.4 and the corresponding acceleration graph. Comparing the two, you can see that the acceleration
14.4 . The Dynamics of Simple Harmonic Motion
graph looks like an upsidedown position graph. In fact, because x = A cos vt, Equation 14.27 for the acceleration can be written
ax =  v2 x
(Fsp )x =  k x
(14.29)
The minus sign indicates that the spring force is a restoring force, a force that always points back toward the equilibrium position. If we place the origin of the coordinate system at the equilibrium position, as we’ve done throughout this chapter, then
x = x and Hooke’s law is simply ( Fsp )x =  kx.
The xcomponent of Newton’s second law for the object attached to the spring is
(Fnet )x = (Fsp )x =  kx = max
(14.30)
Equation 14.30 is easily rearranged to read
You can see that Equation 14.31 is identical to Equation 14.28 if the system oscillates
with angular frequency v = 1k/m . We previously found this expression for v from
an energy analysis. Our experimental observation that the acceleration is proportional
to the negative of the displacement is exactly what Hooke’s law would lead us to expect. That’s the good news.
The bad news is that ax is not a constant. As the object’s position changes, so does
the acceleration. Nearly all of our kinematic tools have been based on constant acceleration. We can’t use those tools to analyze oscillations, so we must go back to the
very definition of acceleration:
ax = 
ax =
k
x
m
(14.31)
dvx d 2 x
=2
dt
dt
Acceleration is the second derivative of position with respect to time. If we use this
definition in Equation 14.31, it becomes
k
d 2x
=  x (equation of motion for a mass on a spring)
m
dt 2
The prototype of simple
harmonic motion: a mass oscillating on
a horizontal spring without friction.
FiguRE 14.14
(14.28)
That is, the acceleration is proportional to the negative of the displacement. The
acceleration is, indeed, most positive when the displacement is most negative and is
most negative when the displacement is most positive.
Recall that the acceleration is related to the net force by Newton’s second law. Consider again our prototype mass on a spring, shown in FiguRE 14.14. This is the simplest
possible oscillation, with no distractions due to friction or gravitational forces. We will
assume the spring itself to be massless.
As you learned in Chapter 10, the spring force is given by Hooke’s law:
(14.32)
Equation 14.32, which is called the equation of motion, is a secondorder differential
equation. Unlike other equations we’ve dealt with, Equation 14.32 cannot be solved
by direct integration. We’ll need to take a different approach.
Solving the Equation of Motion
The solution to an algebraic equation such as x 2 = 4 is a number. The solution to a
differential equation is a function. The x in Equation 14.32 is really x ( t ), the position
as a function of time. The solution to this equation is a function x ( t ) whose second
derivative is the function itself multiplied by (  k/m ).
One important property of differential equations that you will learn about in math
is that the solutions are unique. That is, there is only one solution to Equation 14.32
that satisfies the initial conditions. If we were able to guess a solution, the uniqueness
property would tell us that we had found the only solution. That might seem a rather
387
r
Spring
constant k
Fsp
m
2A
0
x
Oscillation
x
A
388 c h a p t e r 14 . Oscillations
strange way to solve equations, but in fact differential equations are frequently solved
by using your knowledge of what the solution needs to look like to guess an appropriate function. Let us give it a try!
We know from experimental evidence that the oscillatory motion of a spring appears to be sinusoidal. Let us guess that the solution to Equation 14.32 should have
the functional form
x ( t ) = A cos ( vt + f0 )
(14.33)
where A, v, and f0 are unspecified constants that we can adjust to any values that
might be necessary to satisfy the differential equation.
If you were to guess that a solution to the algebraic equation x 2 = 4 is x = 2, you
would verify your guess by substituting it into the original equation to see if it works.
We need to do the same thing here: Substitute our guess for x ( t ) into Equation 14.32
to see if, for an appropriate choice of the three constants, it works. To do so, we need
the second derivative of x ( t ). That is straightforward:
An optical technique called
interferometry reveals the belllike
vibrations of a wine glass.
x ( t ) = A cos ( vt + f0 )
dx
=  vA sin( vt + f0 )
dt
(14.34)
d 2x
=  v2A cos ( vt + f0 )
dt 2
If we now substitute the first and third of Equations 14.34 into Equation 14.32, we find
k
 v2A cos ( vt + f0 ) =  A cos ( vt + f0 )
(14.35)
m
Equation 14.35 will be true at all instants of time if and only if v2 = k/m. There do not
seem to be any restrictions on the two constants A and f0—they are determined by the
initial conditions.
So we have found—by guessing!—that the solution to the equation of motion for a
mass oscillating on a spring is
k
Bm
x ( t ) = A cos ( vt + f0 )
where the angular frequency
v = 2pf =
(14.36)
(14.37)
is determined by the mass and the spring constant.
NOTE Once again we see that the oscillation frequency is independent of the
amplitude A.
Equations 14.36 and 14.37 seem somewhat anticlimactic because we’ve been
using these results for the last several pages. But keep in mind that we had been
assuming x = A cos vt simply because the experimental observations “looked” like a
cosine function. We’ve now justified that assumption by showing that Equation 14.36
really is the solution to Newton’s second law for a mass on a spring. The theory of
oscillation, based on Hooke’s law for a spring and Newton’s second law, is in good
agreement with the experimental observations. This conclusion gives an affirmative
answer to the last of the three questions that we asked early in the chapter, which was
whether the sinusoidal oscillation of SHM is a consequence of Newton’s laws.
ExAMPLE 14.6
Analyzing an oscillator
At t = 0 s, a 500 g block oscillating on a spring is observed moving to the right at x = 15 cm. It reaches a maximum displacement
of 25 cm at t = 0.30 s.
a. Draw a positionversustime graph for one cycle of the motion.
b. At what times during the first cycle does the mass pass through
x = 20 cm?
14.5 . Vertical Oscillations
MODEL
The motion is simple harmonic motion.
12
a. The position equation of the block is x ( t ) = A cos ( vt +
f0 ). We know that the amplitude is A = 0.25 m and that
x0 = 0.15 m. From these two pieces of information we obtain
the phase constant:
SOLvE
f0 = cos 1
Positionversustime graph
for the oscillator of Example 14.6.
FiguRE 14.15
x (cm)
25
20
15
10
5
0
5
10
15
20
25
x0
= cos 1 (0.60) = { 0.927 rad
A
The object is initially moving to the right, which tells us that
the phase constant must be between  p and 0 rad. Thus f0 =
 0.927 rad. The block reaches its maximum displacement
xmax = A at time t = 0.30 s. At that instant of time
xmax = A = A cos ( vt + f0 )
This can be true only if cos ( vt + f0 ) = 1, which requires
vt + f0 = 0. Thus
 f0
 (  0.927 rad)
=
= 3.09 rad/s
v=
t
0.30 s
Now that we know v, it is straightforward to compute the
period:
2p
T=
= 2.0 s
v
x ( t ) = (25 cm) cos (3.09t  0.927), where
t is in s, from t = 0 s to t = 2.0 s.
b. From x = A cos ( vt + f0 ), the time at which the mass reaches
position x = 20 cm is
FiguRE 14.15 graphs
t=
=
2.0 s
T
0.3 0.5
1.0
1.5
2.0
1 12 2
112
t (s)
2
x
1
cos 1
 f0
v
A
1
20 cm
cos 1
+ 0.927 rad = 0.51 s
3.09 rad/s
25 cm
A calculator returns only one value of cos 1, in the range 0 to
p rad, but we noted earlier that cos 1 actually has two values.
Indeed, you can see in Figure 14.15 that there are two times at
which the mass passes x = 20 cm. Because they are symmetrical on either side of t = 0.30 s, when x = A, the first point is
(0.51 s  0.30 s) = 0.21 s before the maximum. Thus the mass
passes through x = 20 cm at t = 0.09 s and again at t = 0.51 s.
Stop to think 14.4
This is the position graph of a mass on a spring. What can you
say about the velocity and the force at the instant indicated by the dashed line?
a.
b.
c.
d.
e.
f.
g.
389
Velocity is positive; force is to the right.
Velocity is negative; force is to the right.
Velocity is zero; force is to the right.
Velocity is positive; force is to the left.
Velocity is negative; force is to the left.
Velocity is zero; force is to the left.
Velocity and force are both zero.
x
A
t
0
Extensive use is made of
multiple representations—
placing different representations
side by side to help students
develop the key skill of
translating between words,
math, and figures. Essential to
good problemsolving, this skill
is overlooked in most physics
textbooks.
A
14.5 vertical Oscillations
We have focused our analysis on a horizontally oscillating spring. But the typical
demonstration you’ll see in class is a mass bobbing up and down on a spring hung
vertically from a support. Is it safe to assume that a vertical oscillation has the same
mathematical description as a horizontal oscillation? Or does the additional force of
gravity change the motion? Let us look at this more carefully.
FiguRE 14.16 shows a block of mass m hanging from a spring of spring constant k. An
important fact to notice is that the equilibrium position of the block is not where the
spring is at its unstretched length. At the equilibrium position of the block, where it
hangs motionless, the spring has stretched by L.
Finding L is a staticequilibrium problem in which the upward spring force balances the downward gravitational force on the block. The ycomponent of the spring
force is given by Hooke’s law:
( Fsp )y =  k y = + k L
(14.38)
FiguRE 14.16
spring.
Gravity stretches the
k
Unstretched
spring
L
r
Fsp
m
The block hanging
at rest has stretched
the spring by L.
r
FG
390 c h a p t e r 14 . Oscillations
Equation 14.38 makes a distinction between L, which is simply a distance and is
a positive number, and the displacement y. The block is displaced downward, so
y =  L. Newton’s first law for the block in equilibrium is
(Fnet ) y = (Fsp ) y + (FG ) y = k L  mg = 0
(14.39)
from which we can find
mg
(14.40)
k
This is the distance the spring stretches when the block is attached to it.
Let the block oscillate around this equilibrium position, as shown in FiguRE 14.17.
We’ve now placed the origin of the yaxis at the block’s equilibrium position in order to
be consistent with our analyses of oscillations throughout this chapter. If the block moves
upward, as the figure shows, the spring gets shorter compared to its equilibrium length,
but the spring is still stretched compared to its unstretched length in Figure 14.16. When
the block is at position y, the spring is stretched by an amount L  y and hence exerts
an upward spring force Fsp = k ( L  y ). The net force on the block at this point is
L=
The block oscillates around
the equilibrium position.
FiguRE 14.17
Spring
stretched
by L
Spring
stretched
by L y
r
Fsp
A
y
m
Block’s
equilibrium
position
A
Fnet
r
0
FG
Oscillation around the
equilibrium position
is symmetrical.
(14.41)
But k L  mg is zero, from Equation 14.40, so the net force on the block is simply
(Fnet ) y =  ky
(14.42)
Equation 14.42 for vertical oscillations is exactly the same as Equation 14.30 for
horizontal oscillations, where we found (Fnet ) x =  kx. That is, the restoring force for
vertical oscillations is identical to the restoring force for horizontal oscillations. The
role of gravity is to determine where the equilibrium position is, but it doesn’t affect
the oscillatory motion around the equilibrium position.
Because the net force is the same, Newton’s second law has exactly the same oscillatory solution:
with, again, v = 2k/m. The vertical oscillations of a mass on a spring are the
same simple harmonic motion as those of a block on a horizontal spring. This is
an important finding because it was not obvious that the motion would still be simple
harmonic motion when gravity was included.
Handdrawn sketches are
incorporated into select
worked examples to provide a
clear model of what students
should draw during their own
problem solving.
ExAMPLE 14.7
(Fnet ) y = (Fsp ) y + (FG ) y = k( L  y )  mg = ( k L  mg )  ky
r
m
y ( t ) = A cos ( vt + f0 )
(14.43)
Bungee oscillations
An 83 kg student hangs from a bungee cord with spring constant
270 N/m. The student is pulled down to a point where the cord is
5.0 m longer than its unstretched length, then released. Where is
the student, and what is his velocity 2.0 s later?
A student on a bungee cord oscillates
about the equilibrium position.
FiguRE 14.18
The bungee cord is
modeled as a spring.
A bungee cord can be modeled as a spring. Vertical oscillations on the bungee cord are SHM.
MODEL
viSuALizE FiguRE 14.18
shows the situation.
Although the cord is stretched by 5.0 m when the student
is released, this is not the amplitude of the oscillation. Oscillations occur around the equilibrium position, so we have to begin
by finding the equilibrium point where the student hangs motionless. The cord stretch at equilibrium is given by Equation 14.40:
mg
L=
= 3.0 m
k
Stretching the cord 5.0 m pulls the student 2.0 m below the equilibrium point, so A = 2.0 m. That is, the student oscillates with
amplitude A = 2.0 m about a point 3.0 m beneath the bungee
SOLvE
cord’s original end point. The student’s position as a function of
time, as measured from the equilibrium position, is
y ( t ) = (2.0 m) cos ( vt + f0 )
where v = 2k/m = 1.80 rad/s. The initial condition
y0 = A cos f0 =  A
requires the phase constant to be f0 = p rad. At t = 2.0 s the student’s position and velocity are
14.6 . The Pendulum
391
y = (2.0 m) cos 1 (1.80 rad/s) (2.0 s) + p rad 2 = 1.8 m
vy =  vA sin ( vt + f0 ) =  1.6 m/s
The student is 1.8 m above the equilibrium position, or 1.2m be
low the original end of the cord. Because his velocity is negative,
he’s passed through the highest point and is heading down.
14.6 The Pendulum
Now let’s look at another very common oscillator: a pendulum. FiguRE 14.19a shows a
mass m attached to a string of length L and free to swing back and forth. The pendulum’s position can be described by the arc of length s, which is zero when the pendulum hangs straight down. Because angles are measured ccw, s and u are positive when
the pendulum is to the right of center, negative when it is to the left.
u
u
Two forces are acting on the mass: the string tension T and gravity FG. It will be
convenient to repeat what we did in our study of circular motion: Divide the forces
into tangential components, parallel to the motion, and radial components parallel to
the string. These are shown on the freebody diagram of FiguRE 14.19b.
Newton’s second law for the tangential component, parallel to the motion, is
(Fnet ) t = a Ft = (FG ) t =  mg sin u = mat
pendulum.
The motion of a
(a)
u and s are
negative on
the left.
u and s are
positive on
the right.
u
L
m
(14.44)
Using at = d 2 s/dt 2 for acceleration “around” the circle, and noting that the mass cancels, we can write Equation 14.44 as
d 2s
=  g sin u
dt 2
FiguRE 14.19
s
Arc length
0
(b)
Center
of circle
(14.45)
The tension has
only a radial
component.
u
r
This is the equation of motion for an oscillating pendulum. The sine function makes
this equation more complicated than the equation of motion for an oscillating
spring.
T
The SmallAngle Approximation
Suppose we restrict the pendulum’s oscillations to small angles of less than about
10 . This restriction allows us to make use of an interesting and important piece of
geometry.
FiguRE 14.20 shows an angle u and a circular arc of length s = r u. A right triangle
has been constructed by dropping a perpendicular from the top of the arc to the axis.
The height of the triangle is h = r sin u. Suppose that the angle u is “small.” In that
case there is very little difference between h and s. If h s, then r sin u r u. It follows that
sin u
u
NOTE
u
The gravitational force
has a tangential
component mg sin u.
The smallangle approximation is valid only if angle u is in radians!
How small does u have to be to justify using the smallangle approximation? It’s
easy to use your calculator to find that the smallangle approximation is good to three
r
FG
(FG)r
The geometrical basis of
the smallangle approximation.
FiguRE 14.20
r
( u in radians)
The result that sin u u for small angles is called the smallangle approximation.
We can similarly note that l r for small angles. Because l = r cos u, it follows that
cos u 1. Finally, we can take the ratio of sine and cosine to find tan u
sin u u.
Table 14.3 summarizes the smallangle approximation. We will have other occasions
to use the smallangle approximation throughout the remainder of this text.
t
(FG)t
Tangential
axis
h
r sin u
s
ru
u
l
r cos u
Smallangle approximations.
u must be in radians.
TABLE 14.3
sin u
u
cos u
1
tan u
sin u
u
392 c h a p t e r 14 . Oscillations
significant figures, an error of … 0.1%, up to angles of 0.10 rad ( 5 ) . In practice,
we will use the approximation up to about 10 , but for angles any larger it rapidly loses
validity and produces unacceptable results.
If we restrict the pendulum to u 6 10 , we can use sin u u. In that case, Equation 14.44 for the net force on the mass is
mg
(Fnet ) t =  mg sin u
 mgu = s
L
where, in the last step, we used the fact that angle u is related to the arc length by
u = s/L. Then the equation of motion becomes
g
d 2s
= s
2
L
dt
The pendulum clock has been used
for hundreds of years.
(14.46)
This is exactly the same as Equation 14.32 for a mass oscillating on a spring. The
names are different, with x replaced by s and k/m by g/L, but that does not make it a
different equation.
Because we know the solution to the spring problem, we can immediately write the
solution to the pendulum problem just by changing variables and constants:
s ( t ) = A cos ( vt + f0 )
u ( t ) = umax cos ( vt + f0 )
or
The angular frequency
v = 2pf =
g
AL
(14.47)
(14.48)
is determined by the length of the string. The pendulum is interesting in that the frequency, and hence the period, is independent of the mass. It depends only on the
length of the pendulum. The amplitude A and the phase constant f0 are determined by
the initial conditions, just as they were for an oscillating spring.
ExAMPLE 14.8
The maximum angle of a pendulum
A 300 g mass on a 30cmlong string oscillates as a pendulum. It
has a speed of 0.25 m/s as it passes through the lowest point. What
maximum angle does the pendulum reach?
Databased Examples help
students with the skill of
drawing conclusions from
laboratory data. Designed
to supplement labbased
instruction, these examples
also help students in general
with mathematical reasoning,
graphical interpretation, and
assessment of results.
Assume that the angle remains small, in which case the
motion is simple harmonic motion.
MODEL
SOLvE
g
9.8 m/s 2
=
= 5.72 rad/s
BL
B 0.30 m
The speed at the lowest point is vmax = vA, so the amplitude is
A = smax =
The maximum angle, at the maximum arc length smax, is
The angular frequency of the pendulum is
v=
ExAMPLE 14.9
vmax
0.25 m/s
=
= 0.0437 m
v
5.72 rad/s
umax =
smax
0.0437 m
=
= 0.146 rad = 8.3
L
0.30 m
Because the maximum angle is less than 10 , our analysis
based on the smallangle approximation is reasonable.
A SSESS
The gravimeter
Deposits of minerals and ore can alter the local value of the freefall acceleration because they tend to be denser than surrounding
rocks. Geologists use a gravimeter—an instrument that accurately
measures the local freefall acceleration—to search for ore deposits. One of the simplest gravimeters is a pendulum. To achieve
the highest accuracy, a stopwatch is used to time 100 oscillations
of a pendulum of different lengths. At one location in the field, a
geologist makes the following measurements:
Length (m)
Time (s)
0.500
1.000
1.500
2.000
141.7
200.6
245.8
283.5
What is the local value of g?
14.6 . The Pendulum
Assume the oscillation angle is small, in which case the
motion is simple harmonic motion with a period independent of
the mass of the pendulum. Because the data are known to four significant figures ( { 1 mm on the length and { 0.1 s on the timing,
both of which are easily achievable), we expect to determine g to
four significant figures.
M ODEL
1
2
L 2 4p2
T = 2p
=
L
g
Ag
That is, the square of a pendulum’s period is proportional to its
length. Consequently, a graph of T 2 versus L should be a straight line
passing through the origin with slope 4p2 /g. We can use the experimentally measured slope to determine g. FiguRE 14.21 is a graph of
the data, with the period found by dividing the measured time by 100.
As expected, the graph is a straight line passing through the
origin. The slope of the bestfit line is 4.021 s 2/m. Consequently,
SOLvE
From Equation 14.48, using f = 1/T, we find
2
g=
4p2
4p2
=
= 9.818 m/s 2
slope 4.021 s 2/m
The fact that the graph is linear and passes through the
origin confirms our model of the situation. Had this not been the
ASSESS
393
Graph of the square of the pendulum’s
period versus its length.
FiguRE 14.21
T 2 (s2)
8
y
4.021x
0.001
6
4
Bestfit line
2
0
L (m)
0.0
0.5
1.0
1.5
2.0
case, we would have had to conclude either that our model of the
pendulum as a simple, smallangle pendulum was not valid or that
our measurements were bad. This is an important reason for having multiple data points rather than using only one length.
The Conditions for Simple Harmonic Motion
You can begin to see how, in a sense, we have solved all simpleharmonicmotion
problems once we have solved the problem of the horizontal spring. The restoring
force of a spring, Fsp =  kx, is directly proportional to the displacement x from equilibrium. The pendulum’s restoring force, in the smallangle approximation, is directly
proportional to the displacement s. A restoring force that is directly proportional to the
displacement from equilibrium is called a linear restoring force. For any linear restoring force, the equation of motion is identical to the spring equation (other than perhaps
using different symbols). Consequently, any system with a linear restoring force will
undergo simple harmonic motion around the equilibrium position.
This is why an oscillating spring is the prototype of SHM. Everything that we learn
about an oscillating spring can be applied to the oscillations of any other linear restoring force, ranging from the vibration of airplane wings to the motion of electrons in
electric circuits. Let’s summarize this information with a Tactics Box.
TACTiCS
BOx 14.1
identifying and analyzing simple harmonic motion
1
● If the net force acting on a particle is a linear restoring force, the motion will
be simple harmonic motion around the equilibrium position.
2
● The position as a function of time is x ( t ) = A cos ( vt + f0 ). The velocity
as a function of time is vx ( t ) =  vA sin( vt + f0 ). The maximum speed is
vmax = vA. The equations are given here in terms of x, but they can be written
in terms of y, u, or some other parameter if the situation calls for it.
3
● The amplitude A and the phase constant f0 are determined by the initial conditions through x0 = A cos f0 and v0x =  vA sin f0.
4
● The angular frequency v (and hence the period T = 2p/v ) depends on the
physics of the particular situation. But v does not depend on A or f0.
5
● Mechanical energy is conserved. Thus 1 mvx2 + 1 kx 2 = 1 kA2 = 1 m(vmax )2.
2
2
2
2
Energy conservation provides a relationship between position and velocity
that is independent of time.
Exercises 7–12, 15–19
Tactics Boxes give stepbystep procedures for developing
specific skills (drawing freebody
diagrams, using ray tracing, etc.).
394 c h a p t e r 14 . Oscillations
The Physical Pendulum
FiguRE 14.22
A physical pendulum.
l
u
d
Moment arm of
gravitational torque
Mg
Distance from
pivot to center of
mass
A mass on a string is often called a simple pendulum. But you can also make a pendulum from any solid object that swings back and forth on a pivot under the influence of
gravity. This is called a physical pendulum.
FiguRE 14.22 shows a physical pendulum of mass M for which the distance between
the pivot and the center of mass is l. The moment arm of the gravitational force acting
at the center of mass is d = l sin u, so the gravitational torque is
t =  Mgd =  Mgl sin u
The torque is negative because, for positive u, it’s causing a clockwise rotation. If we
restrict the angle to being small ( u 6 10 ) , as we did for the simple pendulum, we can
use the smallangle approximation to write
t =  Mglu
(14.49)
Gravity causes a linear restoring torque on the pendulum—that is, the torque is directly proportional to the angular displacement u—so we expect the physical pendulum
to undergo SHM.
From Chapter 12, Newton’s second law for rotational motion is
d 2u t
a= 2 =
I
dt
where I is the object’s moment of inertia about the pivot point. Using Equation 14.49
for the torque, we find
 Mgl
d 2u
=
u
2
I
dt
Mgl
BI
(14.50)
Comparison with Equation 14.32 shows that this is again the SHM equation of motion,
this time with angular frequency
v = 2pf =
(14.51)
It appears that the frequency depends on the mass of the pendulum, but recall that
the moment of inertia is directly proportional to M. Thus M cancels and the frequency
of a physical pendulum, like that of a simple pendulum, is independent of mass.
ExAMPLE 14.10
A swinging leg as a pendulum
A student in a biomechanics lab measures the length of his leg,
from hip to heel, to be 0.90 m. What is the frequency of the pendulum motion of the student’s leg? What is the period?
We can model a human leg reasonably well as a rod of
uniform cross section, pivoted at one end (the hip) to form a physical pendulum. The center of mass of a uniform leg is at the midpoint, so l = L/2.
MODEL
Mgl
Mg(L /2)
3g
1
1
1
=
=
= 0.64 Hz
2p B I
2p B ML2 /3
2p B 2L
The moment of inertia of a rod pivoted about one end is
I = 1 ML2, so the pendulum frequency is
3
SOLvE
f=
Lifescience and bioengineering
worked examples and
applications focus on the
physics of lifescience situations
in order to serve the needs of
lifescience students taking a
calculusbased physics class.
The corresponding period is T = 1/f = 1.6 s. Notice that we didn’t
need to know the mass.
As you walk, your legs do swing as physical pendulums
as you bring them forward. The frequency is fixed by the length
of your legs and their distribution of mass; it doesn’t depend on
amplitude. Consequently, you don’t increase your walking speed
by taking more rapid steps—changing the frequency is difficult.
You simply take longer strides, changing the amplitude but not
the frequency.
ASSESS
One person swings on a swing and finds that the period is 3.0 s. A
second person of equal mass joins him. With two people swinging, the period is
Stop to think 14.5
a. 6.0 s
c. 3.0 s
e. 1.5 s
b. 7 3.0 s but not necessarily 6.0 s
d. 6 3.0 s but not necessarily 1.5 s
f. Can’t tell without knowing the length
14.7 . Damped Oscillations
395
14.7 Damped Oscillations
A pendulum left to itself gradually slows down and stops. The sound of a ringing bell
gradually dies away. All real oscillators do run down—some very slowly but others
quite quickly—as friction or other dissipative forces transform their mechanical energy into the thermal energy of the oscillator and its environment. An oscillation that
runs down and stops is called a damped oscillation.
There are many possible reasons for the dissipation of energy, such as air resistance, friction, and internal forces within a metal spring as it flexes. The forces involved in dissipation are complex, but a simple linear drag model gives a quite accurate description of most damped oscillations. That is, we’ll assume a drag force that
depends linearly on the velocity as
u
u
D =  bv
(model of the drag force)
(14.52)
where the minus sign is the mathematical statement that the force is always opposite
in direction to the velocity in order to slow the object.
The damping constant b depends in a complicated way on the shape of the object
and on the viscosity of the air or other medium in which the particle moves. The damping constant plays the same role in our model of air resistance that the coefficient of
friction does in our model of friction.
The units of b need to be such that they will give units of force when multiplied by
units of velocity. As you can confirm, these units are kg/s. A value b = 0 kg/s corresponds to the limiting case of no resistance, in which case the mechanical energy
is conserved. A typical value of b for a spring or a pendulum in air is … 0.10 kg/s.
Objects moving in a liquid can have significantly larger values of b.
FiguRE 14.23 shows a mass oscillating on a spring in the presence of a drag force.
With the drag included, Newton’s second law is
The shock absorbers in cars and trucks
are heavily damped springs. The vehicle’s
vertical motion, after hitting a rock or a
pothole, is a damped oscillation.
An oscillating mass in the
presence of a drag force.
FiguRE 14.23
r
(Fnet ) x = (Fsp ) x + Dx =  kx  bvx = max
Fsp
(14.53)
Using vx = dx/dt and ax = d 2 x /dt 2, we can write Equation 14.53 as
d 2 x b dx k
+
+ x=0
dt 2 m dt m
Spring
constant k
(14.54)
r
D
m
r
v
Equation 14.54 is the equation of motion of a damped oscillator. If you compare it to
Equation 14.32, the equation of motion for a block on a frictionless surface, you’ll see
that it differs by the inclusion of the term involving dx/dt.
Equation 14.54 is another secondorder differential equation. We will simply assert
(and, as a homework problem, you can confirm) that the solution is
x ( t ) = Ae bt/2m cos ( vt + f0 )
( damped oscillator)
b2
k
b2
= v02 2
B m 4m
B
4m2
(14.55)
where the angular frequency is given by
Here v0 = 2k/m is the angular frequency of an undamped oscillator (b = 0). The
constant e is the base of natural logarithms, so e bt/2m is an exponential function.
Because e 0 = 1, Equation 14.55 reduces to our previous solution, x ( t ) = A cos ( vt +
f0 ), when b = 0. This makes sense and gives us confidence in Equation 14.55. A
lightly damped system, which oscillates many times before stopping, is one for which
b/2m V v0. In that case, v v0 is a good approximation. That is, light damping
does not affect the oscillation frequency.
FiguRE 14.24 is a graph of the position x ( t ) for a lightly damped oscillator, as given
by Equation 14.55. Notice that the term Ae bt/2m, which is shown by the dashed line,
v=
Positionversustime graph
for a damped oscillator.
FiguRE 14.24
x
(14.56)
A
0
A
A is the initial amplitude.
The envelope of the
amplitude decays
exponentially:
xmax Ae bt/2m
t
396 c h a p t e r 14 . Oscillations
Several oscillation
envelopes, corresponding to different
values of the damping constant b.
acts as a slowly varying amplitude:
FiguRE 14.25
xmax ( t ) = Ae bt/2m
Energy is conserved
if there is no damping.
For mass
m 1.0 kg
A smaller b causes
less damping.
Amplitude
b
b
0
20
0.3 kg/s
0.03 kg/s
b
0
0 kg/s
b
A
0.1 kg/s
40
60
t (s)
Envelope from Figure 14.24
(14.57)
where A is the initial amplitude, at t = 0. The oscillation keeps bumping up against
this line, slowly dying out with time.
A slowly changing line that provides a border to a rapid oscillation is called the
envelope of the oscillations. In this case, the oscillations have an exponentially
decaying envelope. Make sure you study Figure 14.24 long enough to see how both
the oscillations and the decaying amplitude are related to Equation 14.55.
Changing the amount of damping, by changing the value of b, affects how quickly
the oscillations decay. FiguRE 14.25 shows just the envelope xmax ( t ) for several oscillators that are identical except for the value of the damping constant b. (You need
to imagine a rapid oscillation within each envelope, as in Figure 14.24.) Increasing
b causes the oscillations to damp more quickly, while decreasing b makes them last
longer.
A larger b causes the oscillations
to damp more quickly.
mathematical aside
Exponential decay
Exponential decay occurs in a vast number of physical systems of
importance in science and engineering. Mechanical vibrations, electric circuits, and nuclear radioactivity all exhibit exponential decay.
The number e = 2.71828 p is the base of natural logarithms
in the same way that 10 is the base of ordinary logarithms. It arises
naturally in calculus from the integral
du
3 u = ln u
This integral—which shows up in the analysis of many physical
systems—frequently leads to solutions of the form
u = Ae v/v0 = A exp (  v/v0 )
where exp is the exponential function.
u
u starts at A.
A
u decays to 37% of
its initial value at v
v0.
u decays to 13% of its
initial value at v 2v0.
e 1A
A graph of u illustrates what we mean by exponential decay.
It starts with u = A at v = 0 (because e 0 = 1 ) and then steadily
decays, asymptotically approaching zero. The quantity v0 is called
the decay constant. When v = v0, u = e 1A = 0.37A. When
v = 2v0, u = e 2A = 0.13A.
Arguments of functions must be pure numbers, without units.
That is, we can evaluate e 2, but e 2 kg makes no sense. If v/v0
is a pure number, which it must be, then the decay constant v0
must have the same units as v. If v represents position, then v0 is a
length; if v represents time, then v0 is a time interval. In a specific
situation, v0 is often called the decay length or the decay time. It
is the length or time in which the quantity decays to 37% of its
initial value.
No matter what the process is or what u represents, a quantity that decays exponentially decays to 37% of its initial value
when one decay constant has passed. Thus exponential decay
is a universal behavior. Every time you meet a new system that
exhibits exponential decay, its behavior will be exactly the same
as every other exponential decay. The decay curve always looks
exactly like the figure shown here. Once you’ve learned the properties of exponential decay, you’ll immediately know how to apply this knowledge to a new situation.
e 2A
0
0
v0
2v0
v
Energy in Damped Systems
When considering the oscillator’s mechanical energy, it is useful to define the time
constant t (also called the decay time) to be
m
t=
(14.58)
b
Because b has units of kg/s, t has units of seconds. With this definition, we can write
the oscillation amplitude as xmax ( t ) = Ae t/2t.
14.7 . Damped Oscillations
12
397
Because of the drag force, the mechanical energy is no longer conserved. At any
particular time we can compute the mechanical energy from
1
1
1 2 t/t
k(xmax )2 = k(Ae t/2t )2 =
kA e
= E0 e t/t
2
2
2
E (t) =
1
2
2 kA
m2
(14.59)
where E0 =
is the initial energy at t = 0 and where we used (z ) = z . In other
words, the oscillator’s mechanical energy decays exponentially with time constant T.
As FiguRE 14.26 shows, the time constant is the amount of time needed for the energy
to decay to e 1, or 37%, of its initial value. We say that the time constant t measures the “characteristic time” during which the energy of the oscillation is dissipated.
Roughly twothirds of the initial energy is gone after one time constant has elapsed,
and nearly 90% has dissipated after two time constants have gone by.
For practical purposes, we can speak of the time constant as the lifetime of the
oscillation—about how long it lasts. Mathematically, there is never a time when the
oscillation is “over.” The decay approaches zero asymptotically, but it never gets there
in any finite time. The best we can do is define a characteristic time when the motion
is “almost over,” and that is what the time constant t does.
ExAMPLE 14.11
The energy has decreased to
37% of its initial value at t t.
The energy has
decreased to 13%
of its initial value
at t 2t.
0.37E0
0.13E0
0
0
t
t
t
2t
E ( t ) = E0 e t/t
The time at which an exponential decay is reduced to 1 E0, half
2
its initial value, has a special name. It is called the halflife
and given the symbol t1/2. The concept of the halflife is widely
used in applications such as radioactive decay. To relate t1/2 to
t, we first write
1
E (at t = t1/2 ) = E0 = E0 e t1/2 /t
2
The E0 cancels, giving
1
= e t1/2 /t
2
Again, we take the natural logarithm of both sides:
The motion is a damped oscillation.
a. The initial amplitude at t = 0 is xmax = A. At t = 35.0 s
the amplitude is xmax = 1 A. The amplitude of oscillation at
2
time t is given by Equation 14.57:
SOLvE
xmax ( t ) = Ae bt/2m = Ae t/2t
In this case,
1
A = Ae  (35.0 s)/2t
2
Notice that we do not need to know A itself because it cancels
out. To solve for t, we take the natural logarithm of both sides
of the equation:
12
ln
12
1
=  ln 2 = ln e t1/2 /t =  t1/2 /t
2
Finally, we solve for t1/2:
1
35.0 s
=  ln 2 = ln e  (35.0 s)/2t = 2
2t
This is easily rearranged to give
35.0 s
t=
= 25.2 s
2 ln 2
If desired, we could now determine the damping constant to be
b = m/t = 0.020 kg/s.
t1/2 = t ln 2 = 0.693t
This result that t1/2 is 69% of t is valid for any exponential
decay. In this particular problem, half the energy is gone at
t1/2 = (0.693) (25.2 s) = 17.5 s
The oscillator loses energy faster than it loses amplitude.
This is what we should expect because the energy depends on the
square of the amplitude.
ASSESS
Rank in order, from largest to smallest, the time constants ta to td
of the decays shown in the figure. All the graphs have the same scale.
Stop to think 14.6
E
E
E
t
(a)
E0
b. The energy at time t is given by
a. What is the time constant for this oscillator?
b. At what time will the energy have decayed to half its initial
value?
ln
The oscillator starts
with energy E0.
Energy
A damped pendulum
A 500 g mass swings on a 60cmstring as a pendulum. The amplitude is observed to decay to half its initial value after 35.0 s.
MODEL
Exponential decay of the
mechanical energy of an oscillator.
FiguRE 14.26
2m
E
t
(b)
t
(c)
t
(d)
t
398 c h a p t e r 14 . Oscillations
14.8 Driven Oscillations and Resonance
The response curve shows
the amplitude of a driven oscillator at
frequencies near its natural frequency
of 2.0 Hz.
FiguRE 14.27
Amplitude
The oscillation has
maximum amplitude
when fext f0. This
is resonance.
The oscillation has
only a small amplitude
when fext differs
substantially from f0.
1
2
fext (Hz)
3
This is the natural
frequency.
The resonance amplitude
becomes higher and narrower as the
damping constant decreases.
FiguRE 14.28
Amplitude
f0
b
0.08 kg/s
b
0.20 kg/s
b
1
2.0 Hz
A lightly damped system
has a very tall and very
narrow response curve.
0.80 kg/s
A heavily damped
system has little
response.
2
3
fext (Hz)
Thus far we have focused on the free oscillations of an isolated system. Some initial
disturbance displaces the system from equilibrium, and it then oscillates freely until its
energy is dissipated. These are very important situations, but they do not exhaust the
possibilities. Another important situation is an oscillator that is subjected to a periodic
external force. Its motion is called a driven oscillation.
A simple example of a driven oscillation is pushing a child on a swing, where your
push is a periodic external force applied to the swing. A more complex example is a
car driving over a series of equally spaced bumps. Each bump causes a periodic upward force on the car’s shock absorbers, which are big, heavily damped springs. The
electromagnetic coil on the back of a loudspeaker cone provides a periodic magnetic
force to drive the cone back and forth, causing it to send out sound waves. Air turbulence moving across the wings of an aircraft can exert periodic forces on the wings and
other aerodynamic surfaces, causing them to vibrate if they are not properly designed.
As these examples suggest, driven oscillations have many important applications.
However, driven oscillations are a mathematically complex subject. We will simply
hint at some of the results, saving the details for more advanced classes.
Consider an oscillating system that, when left to itself, oscillates at a frequency f0.
We will call this the natural frequency of the oscillator. The natural frequency for
a mass on a spring is 1k/m /2p, but it might be given by some other expression for
another type of oscillator. Regardless of the expression, f0 is simply the frequency of
the system if it is displaced from equilibrium and released.
Suppose that this system is subjected to a periodic external force of frequency fext.
This frequency, which is called the driving frequency, is completely independent
of the oscillator’s natural frequency f0. Somebody or something in the environment
selects the frequency fext of the external force, causing the force to push on the system
fext times every second.
Although it is possible to solve Newton’s second law with an external driving
force, we will be content to look at a graphical representation of the solution. The
most important result is that the oscillation amplitude depends very sensitively on the
frequency fext of the driving force. The response to the driving frequency is shown
in FiguRE 14.27 for a system with m = 1.0 kg, a natural frequency f0 = 2.0 Hz, and a
damping constant b = 0.20 kg/s. This graph of amplitude versus driving frequency,
called the response curve, occurs in many different applications.
When the driving frequency is substantially different from the oscillator’s natural
frequency, at the right and left edges of Figure 14.27, the system oscillates but the amplitude is very small. The system simply does not respond well to a driving frequency
that differs much from f0. As the driving frequency gets closer and closer to the natural
frequency, the amplitude of the oscillation rises dramatically. After all, f0 is the frequency at which the system “wants” to oscillate, so it is quite happy to respond to a
driving frequency near f0. Hence the amplitude reaches a maximum when the driving
frequency exactly matches the system’s natural frequency: fext = f0.
The amplitude can become exceedingly large when the frequencies match, especially if the damping constant is very small. FiguRE 14.28 shows the same oscillator
with three different values of the damping constant. There’s very little response if the
damping constant is increased to 0.80 kg/s, but the amplitude for fext = f0 becomes
very large when the damping constant is reduced to 0.08 kg/s. This largeamplitude
response to a driving force whose frequency matches the natural frequency of the system is a phenomenon called resonance. The condition for resonance is
fext = f0 (resonance condition)
(14.60)
Within the context of driven oscillations, the natural frequency f0 is often called the
resonance frequency.
An important feature of Figure 14.28 is how the amplitude and width of the resonance depend on the damping constant. A heavily damped system responds fairly
Challenge Example
little, even at resonance, but it responds to a wide range of driving frequencies. Very
lightly damped systems can reach exceptionally high amplitudes, but notice that the
range of frequencies to which the system responds becomes narrower and narrower
as b decreases.
This allows us to understand why a few singers can break crystal goblets but not
inexpensive, everyday glasses. An inexpensive glass gives a “thud” when tapped, but a
fine crystal goblet “rings” for several seconds. In physics terms, the goblet has a much
longer time constant than the glass. That, in turn, implies that the goblet is very lightly
damped while the ordinary glass is heavily damped (because the internal forces within
the glass are not those of a highquality crystal structure).
The singer causes a sound wave to impinge on the goblet, exerting a small driving
force at the frequency of the note she is singing. If the singer’s frequency matches the
natural frequency of the goblet—resonance! Only the lightly damped goblet, like the
top curve in Figure 14.28, can reach amplitudes large enough to shatter. The restriction, though, is that its natural frequency has to be matched very precisely. The sound
also has to be very loud.
CHALLENgE E xAMPLE 14.12
A swinging pendulum
A pendulum consists of a massless, rigid rod with a mass at one
end. The other end is pivoted on a frictionless pivot so that the rod
can rotate in a complete circle. The pendulum is inverted, with
the mass directly above the pivot point, then released. The speed
of the mass as it passes through the lowest point is 5.0 m/s. If the
pendulum later undergoes smallamplitude oscillations at the bottom of the arc, what will its frequency be?
This is a simple pendulum because the rod is massless.
However, our analysis of a pendulum used the smallangle approximation. It applies only to the smallamplitude oscillations at
the end, not to the pendulum swinging down from the inverted
position. Fortunately, energy is conserved throughout, so we can
analyze the big swing using conservation of mechanical energy.
A singer or musical instrument can shatter
a crystal goblet by matching the goblet’s
natural oscillation frequency.
The frequency of a simple pendulum is f = 1g/L /2p.
We’re not given L, but we can find it by analyzing the pendulum’s
swing down from an inverted position. Mechanical energy is conserved, and the only potential energy is gravitational potential
energy. Conservation of mechanical energy Kf + Ugf = Ki + Ugi,
with Ug = mgy, is
SOLvE
12
1
mv + mgyf = mvi2 + mgyi
2f
2
MODEL
is a pictorial representation of the pendulum swinging down from the inverted position. The pendulum
length is L, so the initial height is 2L.
399
The mass cancels, which is good since we don’t know it, and two
terms are zero. Thus
12
v = g(2L) = 2gL
2f
viSuALizE FiguRE 14.29
Solving for L, we find
L=
Beforeandafter pictorial representation of the
pendulum swinging down from an inverted position.
FiguRE 14.29
vf2
(5.0 m/s)2
=
= 0.638 m
4g 4(9.80 m/s 2)
g
1
9.80 m/s 2
1
=
= 0.62 Hz
2p B L 2p B 0.638 m
Now we can calculate the frequency:
f=
The frequency corresponds to a period of about 1.5 s,
which seems reasonable.
ASSESS
A consistent 4step approach
provides a problemsolving
framework throughout the
book (and all supplements):
students learn the importance
of making assumptions (in
the MODEL step), gathering
information, and making
sketches (in the VISUALIZE
step) before treating the
problem mathematically
(SOLVE) and then analyzing
their result (ASSESS).
Challenge Examples illustrate
how to integrate multiple
concepts and use more
sophisticated reasoning in
problemsolving, ensuring
an optimal range of worked
examples for students to study
in preparation for homework
problems.
400 c h a p t e r 14 . Oscillations
SuMMARy
The goal of Chapter 14 has been to understand systems that oscillate with simple harmonic motion.
general Principles
Dynamics
Energy
SHM occurs when a linear restoring force acts to return a
system to an equilibrium position.
If there is no friction
or dissipation, kinetic
and potential energy are
alternately transformed
into each other, but the
total mechanical energy
E = K + U is conserved.
1
1
E = mv 2 + k x 2
2
2
1
= m(vmax) 2
2
1
= k A2
2
k
Horizontal spring
m
(Fnet )x =  kx
0
x
Vertical spring
The origin is at the equilibrium
position L = mg/k.
Unique and critically acclaimed
visual chapter summaries
consolidate understanding
by providing each concept in
words, math, and figures and
organizing these into a vertical
hierarchy—from General
Principles (top) to Applications
(bottom).
k
Both: v =
Bm
m
T = 2p
Bk
(Fnet )y =  ky
Pendulum
g
v=
BL
(Fnet )t = 
12
k
0
m
y
L
T = 2p
Bg
mg
s
L
0
A
A
x
All potential
E
E0
In a damped system, the
energy decays exponentially
L
0
All kinetic
0.37E0
E = E0 e t/t
0
t
0
t
where t is the time constant.
s
important Concepts
Simple harmonic motion (SHM) is a sinusoidal oscillation with
period T and amplitude A.
Frequency f =
1
T
Angular frequency
2p
v = 2pf =
T
x
1
2pt
+ f0
T
2
y
f = vt + f0 is the phase
A
Position x ( t ) = A cos ( vt + f0 )
= A cos
T
SHM is the projection
onto the xaxis of
uniform circular motion.
t
0
A
A
f
The position at time t is
0x
x ( t ) = A cos f
= A cos ( vt + f0 )
f0
x
A cos f
x0 A cos f0
The phase constant f0
determines the initial conditions:
x0 = A cos f0
Velocity vx ( t ) =  vmax sin ( vt + f0 ) with maximum speed
vmax = vA
v0x =  vA sin f0
Acceleration ax ( t ) =  v2 x ( t ) =  v2Acos ( vt + f0 )
Applications
Resonance
When a system is driven by
a periodic external force, it
responds with a largeamplitude
oscillation if fext f0, where
f0 is the system’s natural
oscillation frequency, or
resonant frequency.
Damping
Amplitude
u
u
If there is a drag force D =  bv ,
where b is the damping constant,
then (for lightly damped systems)
x ( t ) = Ae bt/2m cos ( vt + f0 )
f0
fext
A
0
The time constant for energy loss
is t = m/b.
A
x
t
Conceptual Questions
401
Terms and Notation
oscillatory motion
oscillator
period, T
frequency, f
hertz, Hz
simple harmonic motion,
SHM
linear restoring force
damped oscillation
damping constant, b
envelope
time constant, t
halflife, t1/2
driven oscillation
amplitude, A
angular frequency, v
phase, f
phase constant, f0
restoring force
equation of motion
smallangle approximation
natural frequency, f0
driving frequency, fext
response curve
resonance
resonance frequency, f0
CONCEPTuAL QuESTiONS
1. A block oscillating on a spring has period T = 2 s. What is the
period if:
a. The block’s mass is doubled? Explain. Note that you do not
know the value of either m or k, so do not assume any particular values for them. The required analysis involves thinking
about ratios.
b. The value of the spring constant is quadrupled?
c. The oscillation amplitude is doubled while m and k are
unchanged?
2. A pendulum on Planet X, where the value of g is unknown, oscillates with a period T = 2 s. What is the period of this pendulum if:
a. Its mass is doubled? Explain. Note that you do not know the
value of m, L, or g, so do not assume any specific values. The
required analysis involves thinking about ratios.
b. Its length is doubled?
c. Its oscillation amplitude is doubled?
x (cm)
3. FiguRE Q14.3 shows a positionversustime graph for a particle in
10
SHM. What are (a) the amplitude
A, (b) the angular frequency v, and
t (s)
0
(c) the phase constant f0? Explain.
2
4
10
FiguRE Q14.3
2
t
0
A
FiguRE Q14.7
v
vmax
2
t
0
1
vmax
3
FiguRE Q14.8
9.
shows the potentialenergy diagram and the total
energy line of a particle oscillating on a spring.
a. What is the spring’s equilibrium length?
b. Where are the turning points of the motion? Explain.
c. What is the particle’s maximum kinetic energy?
d. What will be the turning points if the particle’s total energy is
doubled?
FiguRE Q14.9
Energy (J)
x
1
FiguRE Q14.8 shows a velocityversustime graph for a particle in
SHM.
a. What is the phase constant f0? Explain.
b. What is the phase of the particle at each of the three numbered points on the graph?
20
4. Equation 14.25 states that 1 kA2 = 1 m ( vmax)2. What does this
2
2
mean? Write a couple of sentences explaining how to interpret
this equation.
5. A block oscillating on a spring has an amplitude of 20 cm. What
will the amplitude be if the total energy is doubled? Explain.
6. A block oscillating on a spring has a maximum speed of 20 cm/s.
What will the block’s maximum speed be if the total energy is
doubled? Explain.
7. FiguRE Q14.7 shows a positionversustime graph for a particle in
SHM.
a. What is the phase constant f0? Explain.
b. What is the phase of the particle at each of the three numbered points on the graph?
A
8.
3
PE
15
10
TE
5
0
12
16
20
24
28
x (cm)
FiguRE Q14.9
10. Suppose the damping constant b of an oscillator increases.
a. Is the medium more resistive or less resistive?
b. Do the oscillations damp out more quickly or less quickly?
c. Is the time constant t increased or decreased?
11. a. Describe the difference between t and T. Don’t just name
them; say what is different about the physical concepts they
represent.
b. Describe the difference between t and t1/2.
12. What is the difference between the driving frequency and the
natural frequency of an oscillator?
Conceptual Questions require
careful reasoning and can be
used for group discussions or
individual work.
402 c h a p t e r 14 . Oscillations
The endofchapter problems
are rated by students to
show difficulty level with the
variety expanded to include
more realworld, challenging,
and explicitly calculusbased
problems.
Exercises (for each section)
allow students to build
their skills and confidence
with straightforward, onestep
questions.
ExErcisEs and ProblEms
Problems labeled
integrate material from earlier chapters.
Section 14.4 The Dynamics of Simple Harmonic Motion
Exercises
11.
Section 14.1 Simple Harmonic Motion
When a guitar string plays the note “A,” the string vibrates at
440 Hz. What is the period of the vibration?
2.  An airtrack glider attached to a spring oscillates between the
10 cm mark and the 60 cm mark on the track. The glider completes 10 oscillations in 33 s. What are the (a) period, (b) frequency, (c) angular frequency, (d) amplitude, and (e) maximum
speed of the glider?
3.  An airtrack glider is attached to a spring. The glider is pulled
to the right and released from rest at t = 0 s. It then oscillates
with a period of 2.0 s and a maximum speed of 40 cm/s.
a. What is the amplitude of the oscillation?
b. What is the glider’s position at t = 0.25 s?
1.
Section 14.3 Energy in Simple Harmonic Motion

12.
13.
Section 14.2 Simple Harmonic Motion and Circular Motion
4.
What are the (a) amplitude, (b) frequency, and (c) phase constant of the oscillation shown in FigurE Ex14.4?

14.
x (cm)
20
10
0
2
10
FigurE Ex14.4
5.
4
6
t (s)
8
15.
20
What are the (a) amplitude, (b) frequency, and (c) phase constant of the oscillation shown in FigurE Ex14.5?

x (cm)
A block attached to a spring with unknown spring constant
oscillates with a period of 2.0 s. What is the period if
a. The mass is doubled?
b. The mass is halved?
c. The amplitude is doubled?
d. The spring constant is doubled?
Parts a to d are independent questions, each referring to the initial situation.
 A 200 g airtrack glider is attached to a spring. The glider is
pushed in 10 cm and released. A student with a stopwatch finds
that 10 oscillations take 12.0 s. What is the spring constant?
 A 200 g mass attached to a horizontal spring oscillates at a
frequency of 2.0 Hz. At t = 0 s, the mass is at x = 5.0 cm and
has vx =  30 cm/s. Determine:
a. The period.
b. The angular frequency.
c. The amplitude.
d. The phase constant.
e. The maximum speed.
f. The maximum acceleration.
g. The total energy.
h. The position at t = 0.40 s.
 The position of a 50 g oscillating mass is given by x ( t ) =
( 2.0 cm) cos (10t  p/4), where t is in s. Determine:
a. The amplitude.
b. The period.
c. The spring constant.
d. The phase constant.
e. The initial conditions.
f. The maximum speed.
g. The total energy.
h. The velocity at t = 0.40 s.
 A 1.0 kg block is attached to a spring with spring constant
16 N/m. While the block is sitting at rest, a student hits it with a
hammer and almost instantaneously gives it a speed of 40 cm/s.
What are
a. The amplitude of the subsequent oscillations?
b. The block’s speed at the point where x = 1 A?
2

10
5
0
5
FigurE Ex14.5
6.
7.
8.
9.
10.
1
2
3
4
t (s)
10
 An object in simple harmonic motion has an amplitude of
4.0 cm, a frequency of 2.0 Hz, and a phase constant of 2p/3 rad.
Draw a position graph showing two cycles of the motion.
 An object in simple harmonic motion has an amplitude of
8.0 cm, a frequency of 0.25 Hz, and a phase constant of  p/2 rad.
Draw a position graph showing two cycles of the motion.
 An object in simple harmonic motion has amplitude 4.0 cm
and frequency 4.0 Hz, and at t = 0 s it passes through the equilibrium point moving to the right. Write the function x ( t ) that
describes the object’s position.
 An object in simple harmonic motion has amplitude 8.0 cm
and frequency 0.50 Hz. At t = 0 s it has its most negative position. Write the function x ( t ) that describes the object’s position.
 An airtrack glider attached to a spring oscillates with a period
of 1.5 s. At t = 0 s the glider is 5.00 cm left of the equilibrium
position and moving to the right at 36.3 cm/s.
a. What is the phase constant?
b. What is the phase at t = 0 s, 0.5 s, 1.0 s, and 1.5 s?
Section 14.5 Vertical Oscillations
16.
 A spring is hanging from the ceiling. Attaching a 500 g
physics book to the spring causes it to stretch 20 cm in order to
come to equilibrium.
a. What is the spring constant?
b. From equilibrium, the book is pulled down 10 cm and released. What is the period of oscillation?
c. What is the book’s maximum speed?
17.  A spring with spring constant 15 N/m hangs from the ceiling.
A ball is attached to the spring and allowed to come to rest. It is
then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?
18.  A spring is hung from the ceiling. When a block is attached
to its end, it stretches 2.0 cm before reaching its new equilibrium length. The block is then pulled down slightly and released.
What is the frequency of oscillation?
Section 14.6 The Pendulum
19.
 A mass on a string of unknown length oscillates as a pendulum with a period of 4.0 s. What is the period if
a. The mass is doubled?
Exercises and Problems
20.
21.
22.
23.
24.
b. The string length is doubled?
c. The string length is halved?
d. The amplitude is doubled?
Parts a to d are independent questions, each referring to the initial situation.
 A 200 g ball is tied to a string. It is pulled to an angle of 8.0 and
released to swing as a pendulum. A student with a stopwatch
finds that 10 oscillations take 12 s. How long is the string?
 What is the period of a 1.0mlong pendulum on (a) the earth
and (b) Venus?
 What is the length of a pendulum whose period on the moon
matches the period of a 2.0mlong pendulum on the earth?
 Astronauts on the first trip to Mars take along a pendulum that
has a period on earth of 1.50 s. The period on Mars turns out to
be 2.45 s. What is the freefall acceleration on Mars?
 A uniform steel bar swings from a pivot at one end with a
period of 1.2 s. How long is the bar?
Section 14.7 Damped Oscillations
Section 14.8 Driven Oscillations and Resonance
25.
26.
27.
28.
29.
BIO
 A 2.0 g spider is dangling at the end of a silk thread. You can
make the spider bounce up and down on the thread by tapping
lightly on his feet with a pencil. You soon discover that you can
give the spider the largest amplitude on his little bungee cord if
you tap exactly once every second. What is the spring constant
of the silk thread?
 The amplitude of an oscillator decreases to 36.8% of its initial
value in 10.0 s. What is the value of the time constant?
 Sketch a position graph from t = 0 s to t = 10 s of a damped
oscillator having a frequency of 1.0 Hz and a time constant
of 4.0 s.
 In a science museum, a 110 kg brass pendulum bob swings
at the end of a 15.0mlong wire. The pendulum is started at exactly 8:00 a.m. every morning by pulling it 1.5 m to the side and
releasing it. Because of its compact shape and smooth surface,
the pendulum’s damping constant is only 0.010 kg/s. At exactly
12:00 noon, how many oscillations will the pendulum have completed and what is its amplitude?
 Vision is blurred if the head is vibrated at 29 Hz because the
vibrations are resonant with the natural frequency of the eyeball
in its socket. If the mass of the eyeball is 7.5 g, a typical value,
what is the effective spring constant of the musculature that
holds the eyeball in the socket?
Problems
30.
FiguRE P14.30 is the velocityversustime graph of a particle in
simple harmonic motion.
a. What is the amplitude of the oscillation?
b. What is the phase constant?
c. What is the position at t = 0 s?

vx (cm/s)
60
30
0
30
FiguRE P14.30
60
3
6
9
12
t (s)
31.
403
 FiguRE P14.31 is the positionversustime graph of a particle in
simple harmonic motion.
a. What is the phase constant?
b. What is the velocity at t = 0 s?
c. What is vmax?
y
x (cm)
A
10
B
5
0
5
1
2
3
4
t (s)
0
6
12
t (s)
10
FiguRE P14.31
32.
33.
34.
35.
BIO
FiguRE P14.32
The two graphs in FiguRE P14.32 are for two different vertical
massspring systems. If both systems have the same mass, what
is the ratio kA /kB of their spring constants?
 An object in SHM oscillates with a period of 4.0 s and an amplitude of 10 cm. How long does the object take to move from
x = 0.0 cm to x = 6.0 cm?
 A 1.0 kg block oscillates on a spring with spring constant
20 N/m. At t = 0 s the block is 20 cm to the right of the equilibrium position and moving to the left at a speed of 100 cm/s.
Determine (a) the period and (b) the amplitude.
 Astronauts in space cannot weigh themselves by standing on a
bathroom scale. Instead, they determine their mass by oscillating
on a large spring. Suppose an astronaut attaches one end of a
large spring to her belt and the other end to a hook on the wall of
the space capsule. A fellow astronaut then pulls her away from
the wall and releases her. The spring’s length as a function of
time is shown in FiguRE P14.35.
a. What is her mass if the spring constant is 240 N/m?
b. What is her speed when the spring’s length is 1.2 m?

L (m)
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
FiguRE P14.35
0
3
6
t (s)
 The motion of a particle is given by x ( t ) = (25 cm)cos (10t ),
where t is in s. At what time is the kinetic energy twice the potential energy?
37.  a. When the displacement of a mass on a spring is 1 A, what
2
fraction of the energy is kinetic energy and what fraction is
potential energy?
b. At what displacement, as a fraction of A, is the energy half
kinetic and half potential?
38.  For a particle in simple harmonic motion, show that vmax =
( p/2) vavg where vavg is the average speed during one cycle of the
motion.
39.  A 100 g block attached to a spring with spring constant
2.5 N/m oscillates horizontally on a frictionless table. Its velocity is 20 cm/s when x =  5.0 cm.
a. What is the amplitude of oscillation?
b. What is the block’s maximum acceleration?
c. What is the block’s position when the acceleration is maximum?
d. What is the speed of the block when x = 3.0 cm?
36.
Problems (spanning concepts
from the whole chapter),
require indepth reasoning
and planning, and allow
students to practice their
problemsolving strategies.
Contextrich problems require
students to simplify and model
more complex realworld
situations. Specifically labeled
problems integr ate concepts
from multiple previous
chapters.
404 c h a p t e r 14 . Oscillations
40.
41.
42.
BIO
Bio problems are set in lifescience, bioengineering, or
biomedical contexts.
43.
44.
A block on a spring is pulled to the right and released at
t = 0 s. It passes x = 3.00 cm at t = 0.685 s, and it passes
x =  3.00 cm at t = 0.886 s.
a. What is the angular frequency?
b. What is the amplitude?
Hint: cos ( p  u ) =  cos u.
 A 300 g oscillator has a speed of 95.4 cm/s when its displacement is 3.0 cm and 71.4 cm/s when its displacement is 6.0 cm.
What is the oscillator’s maximum speed?
 An ultrasonic transducer, of the type used in medical ultrasound imaging, is a very thin disk (m = 0.10 g) driven back and
forth in SHM at 1.0 MHz by an electromagnetic coil.
a. The maximum restoring force that can be applied to the disk
without breaking it is 40,000 N. What is the maximum oscillation amplitude that won’t rupture the disk?
b. What is the disk’s maximum speed at this amplitude?
 A 5.0 kg block hangs from a spring with spring constant
2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back toward
equilibrium. What are the (a) frequency, (b) amplitude, and
(c) total mechanical energy of the motion?
 Your lab instructor has asked you to measure a spring constant
using a dynamic method—letting it oscillate—rather than a static method of stretching it. You and your lab partner suspend the
spring from a hook, hang different masses on the lower end, and
start them oscillating. One of you uses a meter stick to measure
the amplitude, the other uses a stopwatch to time 10 oscillations.
Your data are as follows:

a. What is the spring constant of each spring if the empty car
bounces up and down 2.0 times each second?
b. What will be the car’s oscillation frequency while carrying
four 70 kg passengers?
49.  The two blocks in FiguRE P14.49 oscillate on a frictionless surface with a period of 1.5 s. The upper block just begins to slip
when the amplitude is increased to 40 cm. What is the coefficient
of static friction between the two blocks?
FiguRE P14.49
50.
BIO
 It has recently become possible to “weigh” DNA molecules
by measuring the influence of their mass on a nanooscillator.
FiguRE P14.50 shows a thin rectangular cantilever etched out of
silicon (density 2300 kg/m3 ) with a small gold dot at the end. If
pulled down and released, the end of the cantilever vibrates with
simple harmonic motion, moving up and down like a diving
board after a jump. When bathed with DNA molecules whose
ends have been modified to bind with gold, one or more molecules may attach to the gold dot. The addition of their mass
causes a very slight—but measurable—decrease in the oscillation frequency.
4000 nm
Mass (g)
Time (s)
100
6.5
7.8
150
Databased problems allow
students to practice drawing
conclusions from data (as
demonstrated in the new databased examples in the text).
Amplitude (cm)
5.5
9.8
400 nm
Thickness = 100 nm
200
46.
47.
48.
10.9
250
45.
6.0
FiguRE P14.50
3.5
12.4
A vibrating cantilever of mass M can be modeled as a block of
mass 1 M attached to a spring. (The factor of 1 arises from the mo3
3
ment of inertia of a bar pivoted at one end.) Neither the mass nor
the spring constant can be determined very accurately— perhaps
to only two significant figures—but the oscillation frequency can
be measured with very high precision simply by counting the oscillations. In one experiment, the cantilever was initially vibrating
at exactly 12 MHz. Attachment of a DNA molecule caused the
frequency to decrease by 50 Hz. What was the mass of the DNA?
 It is said that Galileo discovered a basic principle of the
pendulum—that the period is independent of the amplitude—by
using his pulse to time the period of swinging lamps in the cathedral as they swayed in the breeze. Suppose that one oscillation of
a swinging lamp takes 5.5 s.
a. How long is the lamp chain?
b. What maximum speed does the lamp have if its maximum
angle from vertical is 3.0 ?
 A 100 g mass on a 1.0mlong string is pulled 8.0 to one side
and released. How long does it take for the pendulum to reach
4.0 on the opposite side?
 Orangutans can move by brachiation, swinging like a pendulum beneath successive handholds. If an orangutan has arms that
are 0.90 m long and repeatedly swings to a 20 angle, taking one
swing after another, estimate its speed of forward motion in m/s.
While this is somewhat beyond the range of validity of the smallangle approximation, the standard results for a pendulum are
adequate for making an estimate.
Use the bestfit line of an appropriate graph to determine the
spring constant.
 A 200 g block hangs from a spring with spring constant
10 N/m. At t = 0 s the block is 20 cm below the equilibrium
point and moving upward with a speed of 100 cm/s. What are
the block’s
a. Oscillation frequency?
b. Distance from equilibrium when the speed is 50 cm/s?
c. Distance from equilibrium at t = 1.0 s?
 A spring with spring constant k is suspended vertically from
a support and a mass m is attached. The mass is held at the point
where the spring is not stretched. Then the mass is released and
begins to oscillate. The lowest point in the oscillation is 20 cm
below the point where the mass was released. What is the oscillation frequency?
 While grocery shopping, you put several apples in the spring
scale in the produce department. The scale reads 20 N, and you
use your ruler (which you always carry with you) to discover
that the pan goes down 9.0 cm when the apples are added. If you
tap the bottom of the applefilled pan to make it bounce up and
down a little, what is its oscillation frequency? Ignore the mass
of the pan.
 A compact car has a mass of 1200 kg. Assume that the car has
one spring on each wheel, that the springs are identical, and that
the mass is equally distributed over the four springs.
51.
52.
53.
BIO
Exercises and Problems
54.
 Show that Equation 14.51 for the angular frequency of a physical pendulum gives Equation 14.48 when applied to a simple
pendulum of a mass on a string.
55.  A 15@cm@long, 200 g rod is pivoted at one end. A 20 g ball of
clay is stuck on the other end. What is the period if the rod and
clay swing as a pendulum?
56.  A uniform rod of mass M and length L swings as a pendulum
on a pivot at distance L/4 from one end of the rod. Find an expression for the frequency f of smallangle oscillations.
57.  A solid sphere of mass M and radius R is suspended from a
thin rod, as shown in FiguRE P14.57. The sphere can swing back
and forth at the bottom of the rod. Find an expression for the
frequency f of smallangle oscillations.
Potential energy (J)
4
Rubber bands
FiguRE P14.62
63.
R
FiguRE P14.57
 A geologist needs to determine the local value of g . Unfortunately, his only tools are a meter stick, a saw, and a stopwatch.
He starts by hanging the meter stick from one end and measuring
its frequency as it swings. He then saws off 20 cm—using the
centimeter markings—and measures the frequency again. After
two more cuts, these are his data:
Length (cm)
100
60.
61.
62.
0.79
40
59.
0.67
60
64.
0.61
80
BIO
Frequency (Hz)
0.96
Use the bestfit line of an appropriate graph to determine the
local value of g.
 Interestingly, there have been several studies using cadavers
to determine the moments of inertia of human body parts, information that is important in biomechanics. In one study, the center of mass of a 5.0 kg lower leg was found to be 18 cm from the
knee. When the leg was allowed to pivot at the knee and swing
freely as a pendulum, the oscillation frequency was 1.6 Hz. What
was the moment of inertia of the lower leg about the knee joint?
 A 500 g airtrack glider attached to a spring with spring constant 10 N/m is sitting at rest on a frictionless air track. A 250 g
glider is pushed toward it from the far end of the track at a speed
of 120 cm/s. It collides with and sticks to the 500 g glider. What
are the amplitude and period of the subsequent oscillations?
 A 200 g block attached to a horizontal spring is oscillating
with an amplitude of 2.0 cm and a frequency of 2.0 Hz. Just as it
passes through the equilibrium point, moving to the right, a sharp
blow directed to the left exerts a 20 N force for 1.0 ms. What are
the new (a) frequency and (b) amplitude?
 FiguRE P14.62 is a top view of an object of mass m connected
between two stretched rubber bands of length L. The object rests
on a frictionless surface. At equilibrium, the tension in each rubber band is T. Find an expression for the frequency of oscillations perpendicular to the rubber bands. Assume the amplitude
is sufficiently small that the magnitude of the tension in the rubber bands is essentially unchanged as the mass oscillates.
L
65.
66.
67.
68.
69.
19
10
19
2
10
19
1
L
10
3
Pivot
58.
405
10
19
0.08 0.10 0.12 0.14 0.16
Bond length (nm)
FiguRE P14.63
 A molecular bond can be modeled as a spring between two
atoms that vibrate with simple harmonic motion. FiguRE P14.63
shows an SHM approximation for the potential energy of an
HCl molecule. For E 6 4 * 1019 J it is a good approximation to
the more accurate HCl potentialenergy curve that was shown in
Figure 10.31. Because the chlorine atom is so much more massive than the hydrogen atom, it is reasonable to assume that the
hydrogen atom (m = 1.67 * 1027 kg) vibrates back and forth
while the chlorine atom remains at rest. Use the graph to estimate the vibrational frequency of the HCl molecule.
 An ice cube can slide around the inside of a vertical circular hoop of radius R. It undergoes smallamplitude oscillations
if displaced slightly from the equilibrium position at the lowest
point. Find an expression for the period of these smallamplitude
oscillations.
 A penny rides on top of a piston as it undergoes vertical simple
harmonic motion with an amplitude of 4.0 cm. If the frequency
is low, the penny rides up and down without difficulty. If the
frequency is steadily increased, there comes a point at which the
penny leaves the surface.
a. At what point in the cycle does the penny first lose contact
with the piston?
b. What is the maximum frequency for which the penny just
barely remains in place for the full cycle?
 On your first trip to Planet X you happen to take along a
200 g mass, a 40cmlong spring, a meter stick, and a stopwatch.
You’re curious about the freefall acceleration on Planet X,
where ordinary tasks seem easier than on earth, but you can’t
find this information in your Visitor’s Guide. One night you suspend the spring from the ceiling in your room and hang the mass
from it. You find that the mass stretches the spring by 31.2 cm.
You then pull the mass down 10.0 cm and release it. With the
stopwatch you find that 10 oscillations take 14.5 s. Based on this
information, what is g?
 The 15 g head of a bobblehead doll oscillates in SHM at a
frequency of 4.0 Hz.
a. What is the spring constant of the spring on which the head is
mounted?
b. The amplitude of the head’s oscillations decreases to 0.5 cm
in 4.0 s. What is the head’s damping constant?
 An oscillator with a mass of 500 g and a period of 0.50 s has
an amplitude that decreases by 2.0% during each complete oscillation. If the initial amplitude is 10 cm, what will be the amplitude after 25 oscillations?
 A spring with spring constant 15.0 N/m hangs from the ceiling.
A 500 g ball is attached to the spring and allowed to come to rest. It
is then pulled down 6.0 cm and released. What is the time constant
if the ball’s amplitude has decreased to 3.0 cm after 30 oscillations?
An Increased emphasis on
symbolic answers encourages
students to work algebraically.
The Student Workbook also
contains new exercises to
help students work through
symbolic solutions.
406 c h a p t e r 14 . Oscillations
 A 250 g airtrack glider is attached to a spring with spring
constant 4.0 N/m. The damping constant due to air resistance is
0.015 kg/s. The glider is pulled out 20 cm from equilibrium and
released. How many oscillations will it make during the time in
which the amplitude decays to e 1 of its initial value?
71.  A 200 g oscillator in a vacuum chamber has a frequency of
2.0 Hz. When air is admitted, the oscillation decreases to 60%
of its initial amplitude in 50 s. How many oscillations will have
been completed when the amplitude is 30% of its initial value?
72.  Prove that the expression for x ( t ) in Equation 14.55 is a solution to the equation of motion for a damped oscillator, Equation 14.54, if and only if the angular frequency v is given by the
expression in Equation 14.56.
73.  A block on a frictionless table is connected as shown in FiguRE P14.73 to two springs having spring constants k1 and k2 .
Show that the block’s oscillation frequency is given by
70.
f = 2f12 + f22
where f1 and f2 are the frequencies at which it would oscillate if
attached to spring 1 or spring 2 alone.
k1
k2
k1
m
FiguRE P14.73
74.
k2
m
FiguRE P14.74
A block on a frictionless table is connected as shown in
to two springs having spring constants k1 and k2.
Find an expression for the block’s oscillation frequency f in
terms of the frequencies f1 and f2 at which it would oscillate if
attached to spring 1 or spring 2 alone.

FiguRE P14.74
76. A 1.00 kg block is attached to a horizontal spring with spring
constant 2500 N/m. The block is at rest on a frictionless surface.
A 10 g bullet is fired into the block, in the face opposite the
spring, and sticks. What was the bullet’s speed if the subsequent
oscillations have an amplitude of 10.0 cm?
77. A spring is standing upright on a table with its bottom end fastened to the table. A block is dropped from a height 3.0 cm above
the top of the spring. The block sticks to the top end of the spring
and then oscillates with an amplitude of 10 cm. What is the oscillation frequency?
78. The analysis of a simple pendulum assumed that the mass was
a particle, with no size. A realistic pendulum is a small, uniform
sphere of mass M and radius R at the end of a massless string, with
L being the distance from the pivot to the center of the sphere.
a. Find an expression for the period of this pendulum.
b. Suppose M = 25 g, R = 1.0 cm, and L = 1.0 m, typical values for a real pendulum. What is the ratio Treal /Tsimple, where
Treal is your expression from part a and Tsimple is the expression derived in this chapter?
79. a. A mass m oscillating on a spring has period T. Suppose
the mass changes very slightly from m to m + m, where
m V m. Find an expression for T, the small change in the
period. Your expression should involve T, m, and m but not
the spring constant.
b. Suppose the period is 2.000 s and the mass increases by 0.1%.
What is the new period?
80. FiguRE CP14.80 shows a 200 g uniform rod pivoted at one end.
The other end is attached to a horizontal spring. The spring is
neither stretched nor compressed when the rod hangs straight
down. What is the rod’s oscillation period? You can assume that
the rod’s angle from vertical is always small.
Axle
20 cm
Challenge Problems
75. A block hangs in equilibrium from a vertical spring. When a second identical block is added, the original block sags by 5.0 cm.
What is the oscillation frequency of the twoblock system?
k
3.0 N/m
FiguRE CP14.80
STOP TO THiNK ANSwERS
Stop to Think 14.1: c. vmax = 2pA/T. Doubling A and T leaves vmax
unchanged.
Stop to Think 14.3: c + b + a
d. Energy conservation
=
1
2
2 m(vmax ) gives vmax = 1k/m A. k or m has to be increased or decreased by a factor of 4 to have the same effect as increasing or decreasing A by a factor of 2.
Stop to Think 14.2: d. Think of circular motion. At 45 , the particle
is in the first quadrant (positive x) and moving to the left (negative vx ) .
1
2
2 kA
Stop to Think 14.4: c. vx = 0 because the slope of the position graph
is zero. The negative value of x shows that the particle is left of the
equilibrium position, so the restoring force is to the right.
Stop to Think 14.5: c. The period of a pendulum does not depend
on its mass.
Stop to Think 14.6: Td + Tb
Tc + Ta. The time constant is the
time to decay to 37% of the initial height. The time constant is independent of the initial height.
...
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