Chapter  14  –  Solutions
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 , is the number of cycles that are completed per unit of time:
). In SI units, 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 to complete one cycle; that is, its period is 0.10 . What is its frequency ?
Express your answer in hertz.
ANSWER:
=

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 from the equilibrium position?
ANSWER:
R only
Q only

Part F
both R and Q
Suppose that the period is . Which of the following points on the t axis are separated by the time interval ?
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 to the point K, what fraction of a full wavelength is covered?
. Dividing by the fraction will give the period . Call that fraction . Then you can set
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:
=
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, , when the block is released at a distance from its equilibrium position?
Hint
B.1
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 , , and .
ANSWER:
=

ANSWER:
Part C
=
What is the acceleration of the block when it passes through its equilibrium position?
Hint
C.1
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 from its equilibrium position, its displacement varies with time 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, and , respectively. In the expressions below, and are nonzero positive constants.
Hint
H.1
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. and 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
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 at which and
ANSWER:
. At this point you simply need to determine which function among is zero at those calculated values of .
,
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 .
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.
; that is, the x axis represents and the y axis represents For all graphs, treat the origin as
.

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

Part B force vs. time
force vs. position
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

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:

ANSWER:
Part B
negative
zero
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:

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 is the distance from the equilibrium position.
The kinetic energy of the system is, as always,
, 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, , and the amplitude of vibrations, , are given. Answer the following questions.
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
When the block is displaced a distance from equilibrium, the spring is stretched (or compressed)

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,
. Therefore,
. Recall that
.
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:
If the kinetic energy of the block is increasing, the block must be at the equilibrium position. at the amplitude displacement. 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
D

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 at the equilibrium position. Therefore,
. Recall that
.
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:
A
B
C
D
Part G
. Use the formula for to obtain the corresponding distance
Find the kinetic energy of the block at the moment labeled B.
Hint
G.1
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:
= and that Using the facts that the total energy for the kinetic energy at moment B.
Express your answer in terms of and .
ANSWER:
=
, you can now solve

ANSWER:
=
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 , , or some other variable if the situation calls for it.
3.
The amplitude 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 ) depends on the physical properties of the particular oscillator, but does not depend on . For a mass on a spring, the frequency is given by
.
5.
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
, where is The position of a 45 oscillating mass is given by 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:
=

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 of this spring be?
Hint
B.1
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 of the mass described in the previous parts?
Hint
C.1
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 from its equilibrium position is given by
,
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 v x
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.
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:
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 We will use Equation 14.1 to find frequency.
Solve: (a) The amplitude A = 20 cm.
(b) The period T = 4.0 s, thus
Assess: 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/s 2 .

Solve: The period is and is given by the formula
Assess: 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
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 m
A
= m
B
, then
Assess:
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 m b
and m
B
, which have initial velocities v b
and v
B
. After m b
collides with and sticks to m
B
, the two move together with velocity v f
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 is
(b) No. The oscillation frequency depends on the masses but not on the speeds.
Assess: 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, And
Solve: 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;
A
1
= 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.