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Simple Harmonic OscillationsDue: 12:00pm on Thursday, September 8, 2011
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Good Vibes: Introduction to Oscillations
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 , it will
ANSWER:
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 and then begin to move to the right.
Correct
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.
Correct
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: =
10Correct
Part D
If the frequency is 40 , what is the period ?Express your answer in seconds.
ANSWER: =
0.025Correct
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
both R and Q
Correct
Part F
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
Correct
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? Call that fraction . Then you can set . Dividing by the fraction
will give the period .
Express your answer in seconds.
ANSWER: =
0.02Correct
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: =
0.01Correct
Part I
What distance does the object cover during one period of oscillation?Express your answer in meters.
ANSWER:
=
0.48Answer Requested
Part J
What distance does the object cover between the moments labeled K and N on the graph?Express your answer in meters.
ANSWER: =
0.36Correct
Harmonic Oscillator Acceleration
Learning Goal: To understand the application of the general harmonic equation to finding the acceleration of a spring oscillator as a function of time.
One end of a spring with spring constant is attached to the wall. The other end is attached to a block of mass . The block rests on a frictionless horizontal surface. The equilibrium
position of the left side of the block is defined to be . The length of the relaxed spring is
.
The block is slowly pulled from its equilibrium position to some position along the x
axis. At time , the block is released with zero initial velocity.
The goal of this problem is to determine the acceleration of the block as a function of
time in terms of , , and .
It is known that a general solution for the position of a harmonic oscillator is
,
where , , and are constants.
Your task, therefore, is to determine the values of , , and in terms of , ,and and
then use the connection between and to find the acceleration.
Part A
Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the
block as a function of time.
Hint A.1
Physical laws
Hint not displayed
Express your answer in terms of , , and the coordinate of the block .
ANSWER:
=Correct
The negative sign in the answer is important: It indicates that the restoring force (the tension of the spring) is always directed opposite to the block's displacement. When the block is pulled to the right from the equilibrium position, the restoring force is pulling back, that is, to the left--and vice versa.
Part B
Using the fact that acceleration is the second derivative of position, find the acceleration of
the block as a function of time.
Express your answer in terms of , , and .
ANSWER:
= Correct
Part C
Find the angular frequency .
Hint C.1
Using the previous results
Hint not displayed
Express your answer in terms of and .
ANSWER: =
Correct
Note that the angular frequency and, therefore, the period of oscillations depend only on
the intrinsic physical characteristics of the system ( and ). Frequency and period do not depend on the initial conditions or the amplitude of the motion.
Harmonic Oscillator Kinematics
Learning Goal: To understand the application of the general harmonic equation to the kinematics of a spring oscillator.
One end of a spring with spring constant is attached to the wall. The other end is attached to a block of mass . The block rests on a frictionless horizontal surface. The equilibrium
position of the left side of the block is defined to be . The length of the relaxed spring is
.
The block is slowly pulled from its equilibrium position to some position along the x
axis. At time , the block is released with zero initial velocity.
The goal is to determine the position of the block as a function of time in terms of and .
It is known that a general solution for the displacement from equilibrium of a harmonic oscillator is
,
where , , and are constants.
Your task, therefore, is to determine the values of and in terms of and .
Part A
Using the general equation for given in the problem introduction, express the initial
position of the block in terms of , , and (Greek letter omega).
Hint A.1 Consider
Evaluate the general expression for when .
Hint A.2
Some useful trigonometry
Recall that and .
ANSWER:
=Correc
t
This result is a good first step. The constant in this case is simply , the initial position
of the block. What about ? To find the relationship between and other variables, let us
consider another initial condition that we know: At , the velocity of the block is zero.
Part B
Find the value of using the given condition that the initial velocity of the block is zero:
.
Hint B.1
How to approach the problem
Hint not displayed
Hint B.2
Differentiating harmonic functions
Hint not displayed
ANSWER:
0
Correct
Part C
What is the equation for the block?
Hint C.1
Start with the general solution
Use the general solution and the values for and obtained in the previous parts.
Express your answer in terms of , , and .
ANSWER: = Correct
In this problem the initial velocity is zero, so the quantity is the maximum displacement of the block from the equilibrium position. The magnitude of the maximum
displacement is called the amplitude, often denoted . Using this notation, the formula for
can be rewritten as
.
Now, imagine that we have exactly the same physical situation but that the x axis is
translated, so that the position of the wall is now defined to be .
The initial position of the block is the same as before, but in the new coordinate system, the
block's starting position is given by .
Part D
Find the equation for the block's position in the new coordinate system.
Hint D.1
Equilibrium position
Changing the origin of the coordinate system has no effect on the physical parameters of the problem (e.g., the frequency or the amplitude of the block's oscillations). The initial
velocity is still zero. The only difference is that now the block is oscillating around
whereas before it was oscillating around . What is the difference, at any moment,
between in the new coordinate system and in the old coordinate system?
ANSWER: = Correct
Use this relationship and the expression for , the block's position in the old coordinate
system, to derive .
Express your answer in terms of , , (Greek letter omega), and .
ANSWER: = Correct
Position, Velocity, and Acceleration of an Oscillator
Learning Goal: To learn to find kinematic variables from a graph of position vs. time.The graph of the position of an oscillating object as a function of time is shown.
Some of the questions ask you to determine ranges on the graph over which a statement is true. When answering these questions, choose the most complete answer. For example, if the answer "B to D" were correct, then "B to C" would technically also be correct--but you will only recieve credit for choosing the most complete answer.
Part A
Where on the graph is ?
ANSWER:
A to B
A to C
C to D
C to E
B to D
A to B and D to E
Correct
Part B
Where on the graph is ?
ANSWER:
A to B
A to C
C to D
C to E
B to D
A to B and D to E
Correct
Part C
Where on the graph is ?
ANSWER:
A only
C only
E only
A and C
A and C and E
B and D
Correct
Part D
Where on the graph is the velocity ?
Hint D.1 Finding instantaneous velocity
Hint not displayed
ANSWER:
A to B
A to C
C to D
C to E
B to D
A to B and D to E
Correct
Part E
Where on the graph is the velocity ?
ANSWER:
A to B
A to C
C to D
C to E
B to D
A to B and D to E
Correct
Part F
Where on the graph is the velocity ?
Hint F.1 How to tell if
Hint not displayed
ANSWER:
A only
B only
C only
D only
E only
A and C
A and C and E
B and D
Correct
Part G
Where on the graph is the acceleration ?
Hint G.1 Finding acceleration
Hint not displayed
ANSWER:
A to B
A to C
C to D
C to E
B to D
A to B and D to E
Correct
Part H
Where on the graph is the acceleration ?
ANSWER:
A to B
A to C
C to D
C to E
B to D
A to B and D to E
Correct
Part I
Where on the graph is the acceleration ?
Hint I.1 How to tell if
Hint not displayed
ANSWER:
A only
B only
C only
D only
E only
A and C
A and C and E
B and D
Correct
Relating Two General Simple Harmonic Motion Solutions
Learning Goal: To understand how the two standard ways to write the general solution to a harmonic oscillator are related.There are two common forms for the general solution for the position of a harmonic oscillator as a function of time t:
1. and
2. .
Either of these equations is a general solution of a second-order differential equation (
); hence both must have at least two--arbitrary constants--parameters that can be adjusted to fit the solution to the particular motion at hand. (Some texts refer to these arbitrary constants as boundary values.)
Part A
What are the arbitrary constants in Equation 1?
Hint A.1
What is considered a constant?
A constant is something that is defined by the physical situation under consideration (e.g., a spring constant, acceleration due to gravity, frequency of oscillation, or mass) and it does not change even if the motion is different owing to different initial conditions.
Hint A.2
What is considered arbitrary?
Arbitrary constants are used to "fit" the general solution to a particular set of initial conditions, such as how far you pull the oscillator from its equilibrium position, and how fast it is moving when you let it go.
ANSWER:
only
only
and
and
and
and and
Correct
Part B
What are the arbitrary constants in Equation 2?
ANSWER:
only
only
only
and
and
and
Correct
Because both Equation 1 and Equation 2 are general solutions, they can both represent any set of initial conditions (i.e., initial position and velocity). Therefore, one equation could be expressed in terms of the other.
Part C
Find analytic expressions for the arbitrary constants and in Equation 2 (found in Part B)
in terms of the constants and in Equation 1 (found in Part A), which are now considered as given parameters.
Hint C.1
A useful trig identity
What is the angle-sum trigonometric identity for ? Hint: If you can remember the general form but are unsure of the sign or whether a particular term is cos or sin, try
your expression for simple values like 0 and/or .
Give your answer in terms of , , , and .
ANSWER: = Correct
Hint C.2
How to use the trig indentity
The left side of the equation, , is in the form of Equation 1, whereas the right
side is in the form of Equation 2. If you make a proper substitution for and first on the left and then correspondingly on the right side you can solve for the arbitrary constants in Equation 2.
Give your answers for the coefficients of and , separated by a comma.
Express your answers in terms of and .
ANSWER: , = ,
Answer Requested
Part D
Find analytic expressions for the arbitrary constants and in Equation 1 (found in Part A)
in terms of the constants and in Equation 2 (found in Part B), which are now considered as given parameters.
Hint D.1 Find a relationship between , and
Examine the sum where you substitute the preceding answers for and .
Hint D.1.1
Useful trig identity
Recall that .
Give your answer in terms of and other given quantities.
ANSWER:
= Correct
Hint D.2 Useful trig indentityfor finding
Try using
.
Of course you don't want functions of on the right; however, you do want and on the right.
Express the amplitude and phase (separated by a comma) in terms of and .
ANSWER:
, = ,Answer Requested
This problem was very mathematical. To understand its utility, realize that for a typical mechanical oscillator, the variables have the following meaning:
is the amplitude of oscillation,
is the initial phase angle,
is the initial position at , and
is the initial velocity at divided by .
Therefore, if you are given the initial amplitude and phase, you can find the initial position and velocity. Similarly, if you are given the initial position and velocity you can find the initial amplitude (whose square is related to the total energy) and the phase angle, which permits you to answer questions like "When does the particle reach its maximum
displacement?" or "When does the particle first return to ?"
Analyzing Simple Harmonic Motion
This applet shows two masses on springs, each accompanied by a graph of its position versus time.
Part A
What is an expression for , the position of mass I as a function of time? Assume that position is measured in meters and time is measured in seconds.
Hint A.1
How to approach the problem
Hint not displayed
Hint A.2
Find the amplitude
Hint not displayed
Hint A.3
Find the angular frequency
Hint not displayed
Express your answer as a function of . Express numerical constants to three significant figures.
ANSWER: = Correct
Part B
What is , the position of mass II as a function of time? Assume that position is measured in meters and time is measured in seconds.
Hint B.1
How to approach the problem
Hint not displayed
Hint B.2
Find the amplitude
Hint not displayed
Hint B.3
Find the angular frequency
Hint not displayed
Express your answer as a function of . Express numerical constants to three significant figures.
ANSWER:
= Correct
Period of a Mass-Spring System Ranking Task
Different mass crates are placed on top of springs of uncompressed length and stiffness .
The crates are released and the springs
compress to a length before bringing the crates back up to their original positions.
Part A
Rank the time required for the crates to return to their initial positions from largest to smallest.
Hint A.1
Formula for the period
The period is defined as the time it takes for an oscillator to go through one complete cycle of its motion. Therefore, the time for each crate to return to its initial position is one period. The period of a mass-spring system is given by
.
Therefore, if can be determined from the provided information, a ranking can be determined. If cannot be determined, the ranking cannot be determined based on the information provided.
Hint A.2
Determining the mass
At equilibrium, the force of the spring upward is equal to the force of gravity downward:
.
Solving for the mass we get
.
Since the crate oscillates with equal amplitude above and below the equilibrium position, the compression of the spring at equilibrium is one-half the total distance the crate falls before beginning to move back upward; that is,
.
Combining these two ideas results in
.
Expressing in terms of known quantities, and substituting mass into the period formula, will allow you to determine the correct ranking.
Hint A.3
Determining
As defined in the problem, is the uncompressed length of the spring and is the maximum compression of the spring. The total distance the crate falls before beginning to move back upward is given by
.
Rank from largest to smallest. To rank items as equivalent, overlap them.
ANSWER:
View Correct
Simple Harmonic Motion Conceptual Question
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
Hint not displayed
Hint A.2
Find the initial position
Hint not displayed
ANSWER:
Correct
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
Hint not displayed
Hint B.2
Find the velocity a short time later
Hint not displayed
ANSWER:
Correct
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
Hint not displayed
ANSWER:
Correct
± The Fish Scale
A vertical scale on a spring balance reads from 0 to 215 . The scale has a length of 13.0
from the 0 to 215 reading. A fish hanging from the bottom of the spring oscillates
vertically at a frequency of 2.00 .
Part A
Ignoring the mass of the spring, what is the mass of the fish?
Hint A.1
How to approach the problem
Hint not displayed
Hint A.2
Calculate the spring constant
Hint not displayed
Hint A.3
Calculate the angular frequency
Hint not displayed
Hint A.4
Formula for the angular frequency of a mass on a spring
Hint not displayed
Express your answer in kilograms.
ANSWER: =
10.5Correct
Exercise 13.12
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 . At
the block has velocity -4.00 and displacement +0.200 .
Part A
Find (a) the amplitude and (b) the phase angle.
ANSWER: =
0.383Correct
Part B
ANSWER:
=
1.02Correct
Part C
Write an equation for the position as a function of time.
Assume in meters and in seconds.
ANSWER: = Correct
Exercise 13.27
You are watching an object that is moving in SHM. When the object is displaced 0.600 to
the right of its equilibrium position, it has a velocity of to the right and an
acceleration of to the left.
Part A
How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?
ANSWER:
0.240Correct
Test Your Understanding 13.1: Describing Oscillation
An object oscillates back and forth along the -axis. Its equilibrium position is at .
Part A
At an instant when the object is at , what are the signs of the object's -acceleration and -velocity?
ANSWER:
and
and
; not enough information is given to determine the sign of
and
and
and
; not enough information is given to determine the sign of
and not enough information is given to determine the sign of or
Correct
In oscillation, the sign of the x-acceleration is always the opposite of the sign of the
displacement. Since , it must be that . We are not told whether the object is moving in the positive -direction, moving in the negative -direction, or instantaneously at rest, so we can't say anything about the sign of the -velocity .
Test Your Understanding 13.2: Simple Harmonic Motion
An object oscillates back and forth along the -axis in simple harmonic motion. Its
displacement as a function of time is given by . At time the object has a positive -acceleration.
Part A
Which of the following is a possible value of the phase angle ?
ANSWER: radians
radians
radiansmore than one of the above
Correct
The x-acceleration is the second derivative of the displacement:
At t = 0, the x-acceleration is
The amplitude A is positive, so in order to have it must be true that . Note
that , , , and . Hence the only
possible answer among the choices provided is .
Exercise 13.20
An object is undergoing SHM with period 0.255 and amplitude 6.20 . At the object is instantaneously at rest at 6.20 .
Part A
Calculate the time it takes the object to go from 6.20 to -1.40 .
ANSWER: =
7.30×10−2
Correct
Score Summary:Your score on this assignment is 94.7%.You received 123.14 out of a possible total of 130 points.