Download PH202 Chapter 14 solutions

Chapter 14 – Solutions Good Vibes: Introduction to Oscillations
Description: Several conceptual and qualitative questions related to main characteristics of simple
harmonic motion: amplitude, displacement, period, frequency, angular frequency, etc. Both graphs and
equations are used.
Learning Goal: To learn the basic terminology and relationships among the main characteristics of
simple harmonic motion.
Motion that repeats itself over and over is called periodic motion. There are many examples of periodic
motion: the earth revolving around the sun, an elastic ball bouncing up and down, or a block attached to
a spring oscillating back and forth.
The last example differs from the first two, in that it represents a special kind of periodic motion called
simple harmonic motion. The conditions that lead to simple harmonic motion are as follows:
• There must be a position of stable equilibrium.
• There must be a restoring force acting on the oscillating object. The direction of this force must
always point toward the equilibrium, and its magnitude must be directly proportional to the
magnitude of the object's displacement from its equilibrium position. Mathematically, the
restoring force
is given by
, where
is the displacement from equilibrium
and is a constant that depends on the properties of the oscillating system.
• The resistive forces in the system must be reasonably small.
In this problem, we will introduce some of the basic quantities that describe oscillations and the
relationships among them.
Consider a block of mass
attached to a spring with force constant
, as shown in the figure
. The spring can be
either stretched or compressed. The block slides on a frictionless horizontal surface, as shown. When
the spring is relaxed, the block is located at
. If the block is pulled to the right a distance
and then released,
will be the amplitude of the resulting oscillations.
Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent
motion of the block.
Part A
After the block is released from
ANSWER:
, it will
remain at rest.
move to the left until it reaches equilibrium and stop there.
move to the left until it reaches
and stop there.
move to the left until it reaches
right.
and then begin to move to the
As the block begins its motion to the left, it accelerates. Although the restoring force decreases as the
block approaches equilibrium, it still pulls the block to the left, so by the time the equilibrium position
is reached, the block has gained some speed. It will, therefore, pass the equilibrium position and keep
moving, compressing the spring. The spring will now be pushing the block to the right, and the block
will slow down, temporarily coming to rest at
After
.
is reached, the block will begin its motion to the right, pushed by the spring. The
block will pass the equilibrium position and continue until it reaches
, completing one cycle
of motion. The motion will then repeat; if, as we've assumed, there is no friction, the motion will
repeat indefinitely.
The time it takes the block to complete one cycle is called the period. Usually, the period is denoted
and is measured in seconds.
The frequency, denoted
In SI units,
, is the number of cycles that are completed per unit of time:
is measured in inverse seconds, or hertz (
).
Part B
If the period is doubled, the frequency is
ANSWER:
unchanged.
doubled.
halved.
Part C
An oscillating object takes 0.10
frequency ?
Express your answer in hertz.
ANSWER:
=
to complete one cycle; that is, its period is 0.10
. What is its
.
frequency ?
Express your answer in hertz.
ANSWER:
=
Part D
If the frequency is 40
, what is the period
Express your answer in seconds.
?
ANSWER:
=
The following questions refer to the figure
that graphically depicts
the oscillations of the block on the spring.
Note that the vertical axis represents the x coordinate of the oscillating object, and the horizontal axis
represents time.
Part E
Which points on the x axis are located a distance
ANSWER:
R only
Q only
from the equilibrium position?
both R and Q
Part F
Suppose that the period is
interval
. Which of the following points on the t axis are separated by the time
?
ANSWER:
K and L
K and M
K and P
L and N
M and P
Now assume that the x coordinate of point R is 0.12
and the t coordinate of point K is 0.0050
.
Part G
What is the period
Hint
G.1
?
How to approach the problem
In moving from the point
Call that fraction
to the point K, what fraction of a full wavelength is covered?
. Then you can set
. Dividing by the fraction
will give the period
Express your answer in seconds.
ANSWER:
=
Part H
How much time does the block take to travel from the point of maximum displacement to the
opposite point of maximum displacement?
Express your answer in seconds.
ANSWER:
=
Part I
.
What distance does the object cover during one period of oscillation?
Express your answer in meters.
ANSWER:
=
Part J
What distance does the object cover between the moments labeled K and N on the graph?
Express your answer in meters.
ANSWER:
=
± Introduction to Simple Harmonic Motion
Description: ± Includes Math Remediation. Mostly conceptual questions to introduce basic relations
between the restoring force, acceleration, and displacement in SHM of a mass-spring system. No previous
knowledge of SHM is required.
Consider the system shown in the figure.
It consists of a block of
mass
attached to a spring of negligible mass and force constant . The block is free to move on a
frictionless horizontal surface, while the left end of the spring is held fixed. When the spring is neither
compressed nor stretched, the block is in equilibrium. If the spring is stretched, the block is displaced to the
right and when it is released, a force acts on it to pull it back toward equilibrium. By the time the block has
returned to the equilibrium position, it has picked up some kinetic energy, so it overshoots, stopping
somewhere on the other side, where it is again pulled back toward equilibrium. As a result, the block moves
back and forth from one side of the equilibrium position to the other, undergoing oscillations. Since we are
ignoring friction (a good approximation to many cases), the mechanical energy of the system is conserved
and the oscillations repeat themselves over and over.
The motion that we have just described is typical of most systems when they are displaced from
equilibrium and experience a restoring force that tends to bring them back to their equilibrium position.
The resulting oscillations take the name of periodic motion. An important example of periodic motion is
simple harmonic motion (SHM) and we will use the mass-spring system described here to introduce some
of its properties.
Part A
Which of the following statements best describes the characteristic of the restoring force in the springmass system described in the introduction?
Hint
A.1
Find which force is the restoring force
Which of the following forces plays the role of the restoring force?
ANSWER:
gravity
friction
the force exerted by the spring
the normal force
Hint
A.2
Hooke's law
The expression known as Hooke's law says that a spring stretched or compressed by a distance
exerts a force given by
, where is a constant characteristic of the spring called the
spring constant. The negative sign expresses the fact that the force exerted by the spring acts in the
direction opposite the direction in which the displacement has occurred. Also note that the spring
exerts a varying force that is proportional to displacement.
ANSWER:
The restoring force is constant.
The restoring force is directly proportional to the displacement of the block.
The restoring force is proportional to the mass of the block.
The restoring force is maximum when the block is in the equilibrium position.
Whenever the oscillations are caused by a restoring force that is directly proportional to displacement,
the resulting periodic motion is referred to as simple harmonic motion.
Part B
As shown in the figure
,
a coordinate system with the origin at the equilibrium position is chosen so that the x coordinate
represents the displacement from the equilibrium position. (The positive direction is to the right.) What is
the initial acceleration of the block,
position?
Hint
B.1
, when the block is released at a distance
from its equilibrium
Find the restoring force
Find
, the x component of the net force acting on the block, when the block is at a distance
from its equilibrium position. Note that if the block is displaced a certain distance from its equilibrium
position, the spring is stretched by the same distance.
Hint
B.1.1
Forces exerted on the block in the x direction
The x component of the net force acting on the block is due exclusively to the force exerted by the
spring, since all the other forces (gravity and the normal force) act in the vertical direction.
Express your answer in terms of some or all of the variables
,
, and
.
ANSWER:
=
Express your answer in terms of some or all of the variables
ANSWER:
=
,
, and
.
ANSWER:
=
Part C
What is the acceleration
Hint
C.1
of the block when it passes through its equilibrium position?
A characteristic of equilibrium
By definition, an object in equilibrium does not accelerate.
Express your answer in terms of some or all of the variables
,
, and
.
ANSWER:
=
Your results from Parts B and C show that the acceleration of the block is negative when the block has
undergone a positive displacement. Then, the acceleration's magnitude decreases to zero as the block
goes through its equilibrium position. What do you expect the block's acceleration will be when the
block is to the left of its equilibrium position and has undergone a negative displacement?
Part D
Select the correct expression that gives the block's acceleration at a distance
from the equilibrium
position. Note that can be either positive or negative; that is, the block can be either to the right or left
of its equilibrium position.
Hint D.1
How to approach the problem
Hooke's law gives you an expression for the force
exerted on the mass at a given displacement.
Newton's 2nd law tells you that
, where is the acceleration and
is the mass. Using
this equation, you can find a formula for the acceleration of the mass attached to the spring.
ANSWER:
Whether the block undergoes a positive or negative displacement, its acceleration is always opposite in
sign with respect to displacement. Moreover, the block's acceleration is not constant; instead, it is
directly proportional to displacement. This is a fundamental property of simple harmonic motion.
Whether the block undergoes a positive or negative displacement, its acceleration is always opposite in
sign with respect to displacement. Moreover, the block's acceleration is not constant; instead, it is
directly proportional to displacement. This is a fundamental property of simple harmonic motion.
Using the information found so far, select the correct phrases to complete the following statements.
Part E
Hint E.1
How to approach the problem
In Part D, you found that
. Since the acceleration is directly proportional to
displacement, it must reach its maximum value when displacement is maximum.
ANSWE
R:
The
magnitude
of the
block's
accelerati
on
reaches its
maximum
value
when the
block is
Part F
Hint F.1
How to approach the problem
When the block is in motion, its speed can be zero only when its velocity changes sign, that is, when
the direction of motion changes.
ANSWER
:
The
spee
d of
the
bloc
k is
zero
whe
n it
is
Part G
Hint G.1
How to approach the problem
As the block moves from its rightmost position to its leftmost position, its speed increases from zero to
a certain value and then decreases back to zero. This means that as the block moves away from its
rightmost position toward its leftmost position, its acceleration decreases from positive values to
negative values. In particular, the location where the block's acceleration changes sign must also be the
location where its speed reaches its maximum value, where it stops increasing and starts to decrease.
ANSWER:
The speed of the block
reaches its maximum
value when the block is
negative values. In particular, the location where the block's acceleration changes sign must also be the
location where its speed reaches its maximum value, where it stops increasing and starts to decrease.
ANSWER:
The speed of the block
reaches its maximum
value when the block is
Part H
Because of the periodic properties of SHM, the mathematical equations that describe this motion involve
sine and cosine functions. For example, if the block is released at a distance
position, its displacement
varies with time
from its equilibrium
according to the equation
,
where
is a constant characteristic of the system. If time is measured is seconds,
must be expressed
in radians per second so that the quantity
is expressed in radians.
Use this equation and the information you now have on the acceleration and speed of the block as it
moves back and forth from one side of its equilibrium position to the other to determine the correct set of
equations for the block's x components of velocity and acceleration,
expressions below,
Hint
H.1
and
and
, respectively. In the
are nonzero positive constants.
How to find the equation for acceleration
To determine the correct equation for the acceleration, simply substitute the equation
into the expression for
found in Part D and group all positive constants together.
You can verify then whether your result is correct by calculating the acceleration at
comparing it with your result in Part B.
Hint
H.2
and
How to find the equation for velocity
To determine the correct equation for the velocity, recall that when
the speed of the block is
zero. Mathematically, you can calculate when
from the given equation for displacement.
When you do that, you will not find a unique value for
; rather, you will have a set of values of
which
and
at
. At this point you simply need to determine which function among
is zero at those calculated values of
.
ANSWER:
,
Further calculations would show that the constants
and
can be expressed in terms of
and
.
,
,
,
Further calculations would show that the constants
and
can be expressed in terms of
and
Mass Hitting a Spring
Description: A mass sliding on a frictionless floor hits a spring. Identify (unlabelled) graphs of position
vs. time, velocity vs. time, and force vs. position by the shape of the plot.
A block sliding with velocity
along a frictionless floor hits a spring at time
(configuration 1). The spring compresses until the block comes to a momentary stop (configuration 2).
Finally, the spring expands, pushing the block back in the direction from which it came.
In this problem you will be shown a series of plots related to the motion of the block and spring, and
you will be asked to identify what the plots represent. In each plot, the point labeled "1" refers to
configuration 1 (when the block first comes in contact with the spring). The point labeled "2" refers to
configuration 2 (when the block comes to rest with the spring compressed).
In the questions that follow, "force" refers to the x component of the force that the spring exerts on the
block and "position" and "velocity" refer to the x components of the position and velocity of the block.
For all graphs, treat the origin as
.
; that is, the x axis represents
and the y axis represents
.
Part A
Consider graph A.
What might this graph represent?
Hint
A.1
Specify the initial condition
Besides time, what other quantities are equal to zero when the block hits the spring (configuration
1)?
ANSWER:
position only
velocity only
position and velocity
position and force
velocity and force
Hint
A.2
Determine what the slope means
Consider which of the quantities (position, velocity, force, and/or time) is monotonically increasing
between configuration 1 and configuration 2.
ANSWER:
position vs. time
velocity vs. time
force vs. time
force vs. position
Part B
Consider graph B.
What might this graph represent?
Hint B.1
Starting point and slope
You need to find a pair of quantities that have the following properties:
• Both are zero in configuration 1.
• The quantities are proportional to each other.
• One quantity becomes more and more negative while the other becomes more and more
positive.
ANSWER:
position vs. time
velocity vs. time
force vs. time
force vs. position
Part C
Consider graph C.
What might this graph represent?
Hint C.1
Specify the initial and final conditions
Specify which quantity or quantities are nonzero when the mass hits the spring and zero when the
spring is fully compressed.
ANSWER:
position only
velocity only
force only
force and position
force and velocity
position and velocity
ANSWER:
position vs. time
velocity vs. time
force vs. time
force vs. position
Simple Harmonic Motion Conceptual Question
Description: Identify graphs of position, velocity and acceleration vs. time describing an object
undergoing simple harmonic motion.
An object of mass
is attached to a vertically oriented spring. The object is pulled a short distance
below its equilibrium position and released from rest. Set the origin of the coordinate system at the
equilibrium position of the object and choose upward as the positive direction. Assume air resistance is
so small that it can be ignored.
Refer to these graphs
when answering the following questions.
Part A
Beginning the instant the object is released, select the graph that best matches the position vs. time
graph for the object.
Hint
A.1
How to approach the problem
To find the graph that best matches the object's position vs. time, first determine the initial value of
the position. This will narrow down your choices of possible graphs. Then, interpret what the
remaining graphs say about the subsequent motion of the object. You should find that only one
graph describes the position of the object correctly.
Hint
A.2
Find the initial position
The origin of the coordinate system is set at the equilibrium position of the object, with the positive
direction upward. The object is pulled below equilibrium and released. Therefore, is the initial
position positive, negative, or zero?
ANSWER:
positive
negative
zero
ANSWER:
negative
zero
ANSWER:
Part B
Beginning the instant the object is released, select the graph that best matches the velocity vs. time
graph for the object.
Hint
B.1
Find the initial velocity
The object is released from rest. Is the initial velocity positive, negative, or zero?
ANSWER:
positive
negative
zero
Hint
B.2
Find the velocity a short time later
After the object is released from rest, in which direction will it initially move?
ANSWER:
upward (positive)
downward (negative)
ANSWER:
Part C
Beginning the instant the object is released, select the graph that best matches the acceleration vs.
time graph for the object.
Hint C.1
Find the initial acceleration
The object is released from rest, and a short time later it is moving upward. Based on this
observation, what is the direction of the initial acceleration?
ANSWER:
positive
negative
neither positive nor negative (i.e., there is no acceleration)
ANSWER:
Energy of Harmonic Oscillators
Description: Several questions, both qualitative and computational, related to kinetic and potential
energy of a mass-spring system vibrating horizontally.
Learning Goal: To learn to apply the law of conservation of energy to the analysis of harmonic
oscillators.
Systems in simple harmonic motion, or harmonic oscillators, obey the law of conservation of energy
just like all other systems do. Using energy considerations, one can analyze many aspects of motion of
the oscillator. Such an analysis can be simplified if one assumes that mechanical energy is not
dissipated. In other words,
,
where
is the total mechanical energy of the system,
is the kinetic energy, and
is the potential
energy.
As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this
problem, we will consider a horizontally moving block attached to a spring. Note that, since the
gravitational potential energy is not changing in this case, it can be excluded from the calculations.
For such a system, the potential energy is stored in the spring and is given by
,
where is the force constant of the spring and
The kinetic energy of the system is, as always,
is the distance from the equilibrium position.
,
where
is the mass of the block and
is the speed of the block.
We will also assume that there are no resistive forces; that is,
.
Consider a harmonic oscillator at four different moments, labeled A, B, C, and D, as shown in the figure
. Assume that the force
constant , the mass of the block,
following questions.
, and the amplitude of vibrations,
, are given. Answer the
Part A
Which moment corresponds to the maximum potential energy of the system?
Hint A.1
Consider the position of the block
Recall that
, where is the distance from equilibrium. Thus, the farther the block is
from equilibrium, the greater the potential energy. When is the block farthest from equilibrium?
ANSWER:
A
B
C
D
Part B
Which moment corresponds to the minimum kinetic energy of the system?
Hint B.1
How does the velocity change?
Recall that
, where is the speed of the block. When is the speed at a minimum?
Keep in mind that speed is the magnitude of the velocity, so the lowest value that it can take is
zero.
ANSWER:
A
B
C
D
zero.
ANSWER:
A
B
C
D
When the block is displaced a distance
from equilibrium, the spring is stretched (or compressed)
the most, and the block is momentarily at rest. Therefore, the maximum potential energy is
. At that moment, of course,
. Recall that
. Therefore,
.
In general, the mechanical energy of a harmonic oscillator equals its potential energy at the maximum
or minimum displacement.
Part C
Consider the block in the process of oscillating.
ANSWER:
at the equilibrium
position.
at the amplitude
displacement.
If the kinetic energy of the block is increasing,
the block must be
moving to the right.
moving to the left.
moving away from
equilibrium.
moving toward
equilibrium.
Part D
Which moment corresponds to the maximum kinetic energy of the system?
Hint D.1
Consider the velocity of the block
As the block begins to move away from the amplitude position, it gains speed. As the block
approaches equilibrium, the force applied by the spring—and, therefore, the acceleration of the
block—decrease. The speed of the block is at a maximum when the acceleration becomes zero. At
what position does the object begin to slow down?
ANSWER:
A
B
C
block—decrease. The speed of the block is at a maximum when the acceleration becomes zero. At
what position does the object begin to slow down?
ANSWER:
A
B
C
D
Part E
Which moment corresponds to the minimum potential energy of the system?
Hint E.1
Consider the distance from equilibrium
The smallest potential energy corresponds to the smallest distance from equilibrium.
ANSWER:
A
B
C
D
When the block is at the equilibrium position, the spring is not stretched (or compressed) at all. At
that moment, of course,
. Meanwhile, the block is at its maximum speed (
). The maximum kinetic energy can then be written as
and that
. Recall that
at the equilibrium position. Therefore,
.
Recalling what we found out before,
,
we can now conclude that
,
or
.
Part F
At which moment is
Hint F.1
?
Consider the potential energy
At this moment,
from equilibrium.
ANSWER:
. Use the formula for
to obtain the corresponding distance
A
B
C
D
Part G
Find the kinetic energy
Hint
G.1
of the block at the moment labeled B.
How to approach the problem
Find the potential energy first; then use conservation of energy.
Hint
G.2
Find the potential energy
Find the potential energy
of the block at the moment labeled B.
Express your answer in terms of
and
.
ANSWER:
=
Using the facts that the total energy
for the kinetic energy
and that
at moment B.
Express your answer in terms of
ANSWER:
=
and
.
, you can now solve
ANSWER:
=
Tactics Box 14.1 Identifying and Analyzing Simple Harmonic Motion
Description: Knight/Jones/Field Tactics Box 14.1 Identifying and analyzing simple harmonic motion is
illustrated.
Learning Goal: To practice Tactics Box 14.1 Identifying and analyzing simple harmonic motion.
A complete description of simple harmonic motion must take into account several physical quantities
and various mathematical relations among them. This Tactics Box summarizes the essential information
needed to solve oscillation problems of this type.
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, velocity, and acceleration as a function of time are given by
,
The equations are given here in terms of x, but they can be written in terms of y,
other variable if the situation calls for it.
3.
The amplitude
, or some
is the maximum value of the displacement from equilibrium. The maximum
speed and the maximum magnitude of the acceleration are
and
, respectively.
4.
The frequency
(and hence the period
the particular oscillator, but
is given by
5.
) depends on the physical properties of
does not depend on
. For a mass on a spring, the frequency
.
The sum of potential energy plus kinetic energy is constant. As the oscillation proceeds, energy
is transformed from kinetic into potential energy and then back again.
Part A
The position of a 45
oscillating mass is given by
in seconds. Determine the velocity at
Hint
A.1
, where
.
Find the amplitude
What is the amplitude
of the oscillations performed by the mass?
Express your answer in meters.
is
in seconds. Determine the velocity at
Hint
A.1
.
Find the amplitude
What is the amplitude
of the oscillations performed by the mass?
Express your answer in meters.
ANSWER:
=
Hint
A.2
Find the frequency
What is the frequency of the oscillations performed by the mass?
Express your answer in hertz to two significant figures.
ANSWER:
=
Express your answer in meters per second to two significant figures.
ANSWER:
=
Part B
Assume that the oscillating mass described in Part A is attached to a spring. What would the spring
constant
Hint
B.1
of this spring be?
Find the frequency
What is the frequency of the oscillations performed by the mass?
Express your answer in hertz to three significant figures.
ANSWER:
=
Now use the equation for the frequency of a mass on a spring given in the Tactic Box above, and
solve for the spring constant.
Express your answer in newtons per meter to two significant figures.
ANSWER:
=
ANSWER:
=
Part C
What is the total energy
Hint
C.1
of the mass described in the previous parts?
How to approach the problem
As outlined in the Tactic Box above, in simple harmonic motion mechanical energy is conserved.
Therefore, the total energy of the mass is the same at any instant during an oscillation. Thus,
choose an instant in the oscillation and calculate the total energy of the mass at that time. For
example, you could find the energy of the mass at the equilibrium position, where energy is purely
kinetic, or at the position of maximum displacement, where energy is purely potential.
Hint
C.2
Kinetic energy
The kinetic energy
of a mass
with speed
is given by
.
Hint
C.3
Elastic potential energy
The elastic potential energy in a spring stretched by a distance
given by
from its equilibrium position is
,
where
is the spring constant.
Express your answer in joules to two significant figures.
ANSWER:
=
Q14.23. Reason: This is simple harmonic motion where the equilibrium position is at
(a) One can read the period off the graph by seeing how much time elapses from peak to peak or trough to
trough. Reading peak to peak gives about
So the correct choice is B.
(b) The amplitude is the maximum displacement from the equilibrium position. This is easily read at a number of
places on the graph, but the first positive peak occurs at about
the displacement there is
The
correct choice is C.
(c) At the time
the graph crosses a grid line and allows us to read the answer,
The correct
choice is B.
(d) The velocity of the object is given by the slope of the versus graph. We are looking for the first time that
slope is zero (the tangent line is horizontal); this occurs at
The correct choice is B.
(e) Kinetic energy is maximum when the speed is greatest; this occurs as the object moves through the equilibrium
position, not at the end points of its motion where it is instantaneously at rest. Of the choices given, 8 s is the time
where the graph crosses the t-axis and where
and the object is in equilibrium (but moving quickly). The correct
choice is B.
Assess: It would be very instructive to construct a vx versus graph for this same situation (it would be the
slope of this versus graph) and think about the same questions in relation to the new graph.
Q14.24. Reason: Some of the question may be answered by comparing the expression given with the
general equation for simple harmonic motion.
The expression given for the displacement is
The genera expression for the displacement is
Comparing these two expressions for the displacement we see that the amplitude of oscillation is A = 0.350 m,
choice B.
The frequency of oscillation may be determined by
choice B.
The mass attached to the spring may be determined by
choice B.
The total mechanical energy of the oscillator is equal to its maximum potential energy, which may be determined
by
choice E.
The maximum speed may be determined by
choice E.
Assess: Working this problem brings our attention to two things. First, there is a lot of information tucked
away in the function given for the displacement of the oscillator. Second, there are a lot of details associated with
an oscillation. It is important to know these details and their interconnection.
Q14.25. Prepare: Your arms act like simple pendulums and carrying the weights in your hands merely
increases the mass of the bob, but that doesn’t affect the frequency of oscillation,
The answer is B.
Assess: The mass of the bob doesn’t appear in the equation for the frequency of a simple pendulum.
Problems P14.3. Prepare: Your pulse or heartbeat is 75 beats per minute or 75 beats/60 s = 1.25 beats/s. The period is
the inverse of the frequency, so we will use Equation 14.1.
Solve: The frequency of your heart’s oscillations is
The period is the inverse of the frequency, hence
Assess: A heartbeat of 1.3 beats per second means that one beat takes a little less than 1 second, which is what
we obtained above.
P14.6. Prepare: The air-track glider oscillating on a spring is in simple harmonic motion. The glider
completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark. We will use
Equations 14.1 and 14.13.
Solve: (a)
(b)
(c) The oscillation from one side to the other is equal to 60 cm – 10 cm = 50 cm = 0.50 m. Thus, the amplitude
is
(d) The maximum speed is
Assess:
m/s, or
The glider takes 3.3 seconds to complete on oscillation. Its maximum speed of 0.48
is reasonable.
P14.9. Prepare: Please refer to Figure P14.9. The oscillation is the result of simple harmonic motion. As the
graph shows, the time to complete one cycle (or the period) is
frequency.
Solve: (a) The amplitude A = 20 cm.
(b) The period T = 4.0 s, thus
Assess:
We will use Equation 14.1 to find
It is important to know how to find information from a graph.
P14.19. Prepare: The oscillating mass is in simple harmonic motion. The position of the oscillating mass is
given by
where t is in seconds. We will compare this with Equation 14.10.
Solve: (a) The amplitude A = 2.0 cm.
(b) The period is calculated as follows:
(c) The spring constant is calculated from Equation 14.27 as follows:
(d) The maximum speed from Equation 14.26 is
(e) The total energy from Equation 14.22 is
(f ) At t = 0.40 s, the velocity from Equation 14.12 is
Assess:
Velocity at t = 0.40 s is less than the maximum velocity, as would be expected.
P14.24. Prepare: Because the angle of displacement is less than 10°, the small-angle approximation holds
and the pendulum exhibits simple harmonic motion. We will use Equation 14.31 and g = 9.80 m/s2.
Solve:
Assess:
The period is
and is given by the formula
A length of 35.7 cm for the simple pendulum is reasonable.
P14.37. Prepare: The initial amplitude is
Solve:
The equation is
with
(a)
(b) Use
Assess: We expect the amplitude at 4.0 s
P14.39. Prepare: We will model the child on the swing as a simple small-angle pendulum. To make the
amplitude grow large quickly we want to drive (push) the oscillator (child) at the natural resonance frequency. In
other words, we want to wait the natural period between pushes.
Solve:
Assess: You could also increase the amplitude by pushing every other time (every
the amplitude grow as quickly as pushing every period.
The mass of the child was not needed; the answer is independent of the mass.
but that would not make
P14.49. Prepare: The vertical mass/spring systems are in simple harmonic motion. Please refer to Figure P14.49.
Solve: (a) For system A, y is positive for one second as the mass moves downward and reaches maximum
negative y after two seconds. It then moves upward and reaches the equilibrium position, y = 0, at t = 3 seconds.
The maximum speed while traveling in the upward direction thus occurs at t = 3.0 s. The frequency of oscillation
is 0.25 Hz.
(b) For system B, all the mechanical energy is potential energy when the position is at maximum amplitude, which for
the first time is at t = 1.5 s. The time period of system B is thus 6.0 s.
(c) Spring/mass A undergoes three oscillations in 12 s, giving it a period
Spring/mass B
undergoes two oscillations in 12 s, giving it a period
From Equation 14.27, we have
If mA = mB, then
Assess:
It is important to learn how to read a graph.
P14.50. Prepare: First we figure the mass of the chair alone, then the mass of the chair plus astronaut,
then subtract.
Solve:
which we report as 54.6
Assess: This is a reasonable weight for a light astronaut. P14.52. Prepare: The transducer undergoes simple harmonic motion. The maximum restoring force that the
disk can withstand is 40,000 N. By applying Newton’s second law to the disk of mass 0.10 g, we will first find
its maximum acceleration and then use Equation 14.18 to find the maximum oscillation amplitude. Once we find
the amplitude, we will use Equation 14.26 to find the disk’s maximum speed.
Solve: Newton’s second law for the transducer is
Because from Equation 14.18
(b) The maximum speed is
Assess: Apparently, the amplitude is very small and the maximum oscillation speed is large. However, to
oscillate at high frequencies such as 1.0 MHz, you would expect a small amplitude and a large speed.
P14.56. Prepare: A completely inelastic collision between the bullet and the block results in simple
harmonic motion. Let us denote the bullet’s and block’s masses as mb and mB, which have initial velocities vb and
vB. After mb collides with and sticks to mB, the two move together with velocity vf and exhibit simple harmonic
motion. Since the amplitude of the harmonic motion is given, we will first find the final velocity of the bullet +
block system using the mechanical energy conservation equation. The momentum conservation will then give us
the bullet’s speed.
Solve:
(a) The equation for conservation of energy after the collision is
The momentum conservation equation for the perfectly inelastic collision
(b) No. The oscillation frequency
Assess:
is
depends on the masses but not on the speeds.
The bullet’s speed is high but not unimaginable.
P14.73. Prepare: Since the web is horizontal, it will sag until the effective spring force of the web is equal to
the weight of the fly. That is
Solve: Solving the above expression for k, obtain
The correct choice is A.
Assess: Since the web is very delicate, we expect a small effective spring constant.
P14.74. Prepare: Example 14.7 shows how this is done,
Solve:
And
Combine the two equations above by inserting the first one for into the second one.
The correct answer is C.
Assess: This is not an extremely high frequency, but within the range of spider detection, and one you
could see.
P14.75. Prepare: Changing the orientation of the web (but otherwise leaving the web alone) changes nothing
in the equations.
Solve: The frequency remains the same. The answer is C.
Assess: We’ve seen that a mass on a vertical spring is analyzed the same way as a mass and spring on a horizontal frictionless surface. P14.76. Prepare: Equation 14.18 gives the maximum acceleration as
We could separately
compute
for each case, but since we don’t care what the actual values are it might be better to try the ratio
approach.
Let the subscript 1 refer to the low frequency case and 2 to the high frequency case;
A1 = 0.1 mm, and
Solve:
The
for the low-frequency oscillation is only one-thousandth as much as
for the high-frequency
oscillation.
The correct answer is B.
Assess: The answer makes sense: The spider can feel the high-frequency oscillations better because the maximum
acceleration is greater.