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.