chapter 14 and 15 homework

5/1/2014 Chapter 14 and 15 Homework

http://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=2862739 1/29

Chapter 14 and 15 Homework

Due: 10:00pm on Friday, May 2, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

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 earthrevolving around the sun, an elastic ball bouncing up and down, or a block attached to a spring oscillating back andforth.The last example differs from the first two, in that it represents a special kind of periodic motion called simple harmonicmotion. 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 alwayspoint 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 amongthem.

Consider a block of mass attached to a spring with force constant , as shown in the figure. The spring can be eitherstretched or compressed. The block slides on a frictionlesshorizontal surface, as shown. When the spring is relaxed, theblock is located at . If the block is pulled to the right adistance and then released, will be the amplitude of theresulting oscillations.

Assume that the mechanical energy of the block-springsystem remains unchanged in the subsequent motion of theblock.

Part A

After the block is released from , it will

ANSWER:

'

4

4

'

) '

4

4



Correct

As the block begins its motion to the left, it accelerates. Although the restoring force decreases as the blockapproaches equilibrium, it still pulls the block to the left, so by the time the equilibrium position is reached, theblock has gained some speed. It will, therefore, pass the equilibrium position and keep moving, compressingthe spring. The spring will now be pushing the block to the right, and the block will slow down, temporarilycoming 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 ismeasured 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:

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:

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.

4

4

4

4

4

" " "

)[

unchanged.

doubled.

halved.

T T "

= 10 " )[



Correct

Part D

If the frequency is 40 , what is the period ?

Express your answer in seconds.

ANSWER:

Correct

The following questions refer to the figure that graphicallydepicts the oscillations of the block on the spring.

Note that the vertical axis represents the x coordinate of theoscillating object, and the horizontal axis represents time.

Part E

Which points on the x axis are located a distance from the equilibrium position?

ANSWER:

Correct

Part F

Suppose that the period is . Which of the following points on the t axis are separated by the time interval ?

)[

= 0.025 T

R only

Q only

both R and Q



ANSWER:

Correct

Now assume for the remaining Parts G - J, that the x coordinate of point R is 0.12 and the t coordinate of point K is0.0050 .

Part G

What is the period ?


Hint 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 .

ANSWER:

Correct

Part H

How much time does the block take to travel from the point of maximum displacement to the opposite point of

maximum displacement?


ANSWER:

Correct

K and L

K and M

K and P

L and N

M and P

N

T

0

T

= 0.02 T

0

= 0.01 0 T



Part I

What distance does the object cover during one period of oscillation?

Express your answer in meters.

ANSWER:

Correct

Part J

What distance does the object cover between the moments labeled K and N on the graph?

Express your answer in meters.

ANSWER:

Correct

Matching Initial Position and Velocity of Oscillator

Learning Goal:

Understand how to determine the constants in the general equation for simple harmonic motion, in terms of given initialconditions.

A common problem in physics is to match the particular initial conditions - generally given as an initial position andvelocity at - once you have obtained the general solution. You have dealt with this problem in kinematics,

where the formula

1.

has two arbitrary constants (technically constants of integration that arise when finding the position given that theacceleration is a constant). The constants in this case are the initial position and velocity, so "fitting" the generalsolution to the initial conditions is very simple.

For simple harmonic motion, it is more difficult to fit the initial conditions, which we take to be

, the position of the oscillator at , and, the velocity of the oscillator at .

There are two common forms for the general solution for the position of a harmonic oscillator as a function of time :

2. and

3. ,

where , , , and are constants, is the oscillation frequency, and is time.

= 0.48 N

= 0.36 N

4

2

0

40 04

2

0

4

0

2

0

0

40 DPT[0]

40 DPT[0 TJO [0

] [ 0



Although both expressions have two arbitrary constants--parameters that can be adjusted to fit the solution to the initialconditions--Equation 3 is much easier to use to accommodate and . (Equation 2 would be appropriate if the initialconditions were specified as the total energy and the time of the first zero crossing, for example.)

Part A

Find and in terms of the initial position and velocity of the oscillator.

Give your answers in terms of , , and . Separate your answers with a comma.

Hint 1. The only good way to start

Which of the following procedures would solve this problem in the most straightforward manner?

ANSWER:

Hint 2. Using kinematic relationships

Find , the velocity as a function of time from Equation 3.

Hint 1. Derivative of a trig function

From the chain rule of calculus, find the derivative of with respect to time.

ANSWER:

ANSWER:

Hint 3. Initial position

Now you have general expressions for and . Find the position at .

ANSWER:

4

2

4

2

[

Differentiate twice to find . Then integrate it twice. Plug in and as the constants of

integration.

Differentiate once to find . Evaluate and and then solve for the

desired quantities.

Dimensional analysis suffices since and have different dimensions.

Use Equation 1. Plug in where .

40 0 2

4

40 20 40 20

4

2

'40) ') [

20

DPT)0

= E DPT NU

EU

)TJO)0

= 20 [TJO[0[DPT[0

40 20 0



Hint 4. Initial velocity

Find the velocity at time .

ANSWER:

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 thegraph over which a statement is true. When answering thesequestions, choose the most complete answer. For example, ifthe answer "B to D" were correct, then "B to C" wouldtechnically also be correct--but you will only recieve credit forchoosing the most complete answer.

Part A

Where on the graph is ?

ANSWER:

= 40

0

= 20

[

, =

4

2

[

4



Correct

Part B


ANSWER:

Correct

Part C


ANSWER:

Correct

Part D

A to B

A to C

C to D

C to E

B to D

A to B and D to E

4

A to B

A to C

C to D

C to E

B to D

A to B and D to E

4

A only

C only

E only

A and C

A and C and E

B and D

2



Where on the graph is the velocity ?

Hint 1. Finding instantaneous velocity

Instantaneous velocity is the derivative of the position function with respect to time,

.

Thus, you can find the velocity at any time by calculating the slope of the vs. graph. When is the slope

greater than 0 on this graph?

ANSWER:

Correct

Part E


ANSWER:

Correct

Part F


Hint 1. How to tell if

2

20

40

0

4 0

A to B

A to C

C to D

C to E

B to D

A to B and D to E

2

A to B

A to C

C to D

C to E

B to D

A to B and D to E

2

2

40



The velocity is zero when the slope of the x vs. t curve is zero: .

ANSWER:

Correct

Part G

Where on the graph is the acceleration ?

Hint 1. Finding acceleration

Acceleration is the second derivative of the position function with respect to time:

.

This means that the sign of the acceleration is the same as the sign of the curvature of the x vs. t graph. Theacceleration of a curve is negative for downward curvature and positive for upward curvature. Where is thecurvature greater than 0?

ANSWER:

Correct

Part H

40

0

A only

B only

C only

D only

E only

A and C

A and C and E

B and D

40

0

A to B

A to C

C to D

C to E

B to D

A to B and D to E




ANSWER:

Correct

Part I


Hint 1. How to tell if

The acceleration is zero at the inflection points of the x vs. t graph. Inflection points are where the curvatureof the graph changes sign.

ANSWER:

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 itsequilibrium position and released from rest. Set the origin of the coordinate system at the equilibrium position of theobject 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.

A to B

A to C

C to D

C to E

B to D

A to B and D to E

A only

B only

C only

D only

E only

A and C

A and C and E

B and D

)



Part A

Beginning the instant the object is released, select the graph that best matches the position vs. time graph for theobject.


To find the graph that best matches the object's position vs. time, first determine the initial value of theposition. This will narrow down your choices of possible graphs. Then, interpret what the remaining graphssay about the subsequent motion of the object. You should find that only one graph describes the position ofthe object correctly.

Hint 2. Find the initial position

The origin of the coordinate system is set at the equilibrium position of the object, with the positive directionupward. The object is pulled below equilibrium and released. Therefore, is the initial position positive,negative, or zero?

ANSWER:

ANSWER:

positive

negative

zero



Correct

Part B

Beginning the instant the object is released, select the graph that best matches the velocity vs. time graph for theobject.

Hint 1. Find the initial velocity

The object is released from rest. Is the initial velocity positive, negative, or zero?

ANSWER:

Hint 2. Find the velocity a short time later

After the object is released from rest, in which direction will it initially move?

ANSWER:

ANSWER:

A

B

C

D

E

F

G

H

positive

negative

zero

upward (positive)

downward (negative)



Correct

Part C

Beginning the instant the object is released, select the graph that best matches the acceleration vs. time graph forthe object.

Hint 1. Find the initial acceleration

The object is released from rest, and a short time later it is moving upward. Based on this observation, whatis the direction of the initial acceleration?

ANSWER:

ANSWER:

Correct

A

B

C

D

E

F

G

H

positive

negative

neither positive nor negative (i.e., there is no acceleration)

A

B

C

D

E

F

G

H



Exercise 14.6

In a physics lab, you attach a 0.200- air-track glider to the end of an ideal spring of negligible mass and start it

oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it movesthrough that point is 2.60 .

Part A

Find the spring's force constant.

ANSWER:

Correct

Exercise 14.9

An object is undergoing SHM with period 0.880 and amplitude 0.320 . At = 0, the object is at = 0.320 and isinstantaneously at rest.

Part A

Calculate the time it takes the object to go from = 0.320 , to = 0.160 .

Express your answer with the appropriate units.

ANSWER:

Correct

Part B

Calculate the time it takes the object to go from = 0.160 , to = 0.


ANSWER:

Correct

LH

T

= 0.292 ' /N

T N 0 4 N

4 N 4 N

= 0.147 0 T

4 N 4

= 7.33102 0 T



Exercise 14.11

A frictionless block of mass 2.50 is attached to an ideal spring with force constant 320 . At the spring is

neither stretched nor compressed and the block is moving in the negative direction at a speed of 13.0 .

Part A

Find the amplitude.

ANSWER:

Correct

Part B

Find the phase angle.

ANSWER:

Correct

Part C

Write an equation for the position as a function of time.

ANSWER:

Correct

Energy of a Spring

An object of mass attached to a spring of force constant oscillates with simple harmonic motion. The maximumdisplacement from equilibrium is and the total mechanical energy of the system is .

LH /N 0

NT

= 1.15 N

= 1.57 ] SBE

1.15 11.3

1.15 11.3

11.3 1.15

11.3 1.15

4 N TJO SBET 0

4 N DPT SBET 0

4 NTJO SBET0

4 NDPT SBET0

) '



Part A

What is the system's potential energy when its kinetic energy is equal to ?


Since the sum of kinetic and potential energies of the system is equal to the system's total energy, if youknow the fraction of total energy corresponding to kinetic energy you can calculate how much energy ispotential energy. Moreover, using conservation of energy you can calculate the system's total energy interms of the given quantities and . At this point you simply need to combine those results to find the

potential energy of the system in terms of and .

Hint 2. Find the fraction of total energy that is potential energy

When the kinetic energy of the system is equal to , what fraction of the total energy is potential

energy?

Express your answer numerically.

Hint 1. Conservation of mechanical energy

In a system where no forces other than gravitational and elastic forces do work, the sum of kineticenergy and potential energy is conserved. That is, the total energy of the system, given by

, is constant.

ANSWER:

Hint 3. Find the total energy of the system

What is the total mechanical energy of the system, ?

Express your answer in terms of some or all of the variables , , and .


If you apply conservation of energy to the system when the object reaches its maximumdisplacement, you can calculate the system's total energy in terms of the given quantities and

. In fact, when the object is at its maximum displacement from equilibrium, its speed is momentarily

zero and so is its kinetic energy. It follows that the system's energy at this point is entirely potential,that is, , where is the spring's elastic potential energy.

Hint 2. Elastic potential energy

The elastic potential energy of a spring that has been compressed or stretched by a distance is

given by

,

'

'

0.250

) '

'

4

'

4

'



where is the force constant of the spring.

ANSWER:

ANSWER:

Correct

Part B

What is the object's velocity when its potential energy is ?


You can calculate the object's velocity using energy considerations. Calculate the fraction of the system'stotal energy that is kinetic energy and then find the object's velocity from the definition of kinetic energy. Tosimplify your expression write the total energy in terms of and . Alternatively, you could directly use the

formula for the object's velocity in terms of the variables , , , and displacement derived from energy

considerations. The only unknown quantity in such a formula would be the object's displacement , which

can be calculated from the system's potential energy.

Hint 2. Find the kinetic energy

If the system's potential energy is , what is the system's kinetic energy?

Hint 1. Conservation of mechanical energy

In a system where no forces other than gravitational and elastic forces do work, the sum of kineticenergy and potential energy is conserved. That is, the total energy of the system, given by

, is constant.

Hint 2. Total energy of the system

The total energy of a system consisting of an object attached to a horizontal spring of force constant

'

= '

'

'

'

'

'

' ) 4

4

'



is given by

,

where is the maximum displacement of the object from its equilibrium position.

ANSWER:

Hint 3. Formula for the velocity in terms of position

The velocity of an object of mass undergoing simple harmonic motion is given by

,

where is the force constant of the system, is the object's position, and is maximum displacement.

Hint 4. Find the object's position

When the system's potential energy is , what is the displacement of the object from its equilibrium

position?

Hint 1. Elastic potential energy

The elastic potential energy of a spring that has been compressed or stretched by a distance is

given by

,

where is the force constant of the spring.

Hint 2. Total energy of the system

The total energy of a system consisting of an object attached to a horizontal spring of force constant is given by

,

where is the maximum displacement of the object from its equilibrium position.

ANSWER:

'

'

'

'

'

'

)

2e

'

)

4

' 4

4

4

'

4

'

'

'



ANSWER:

Correct

Exercise 14.23

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude ofthe motion is 0.140 . The maximum speed of the block is 3.81 .

Part A

What is the maximum magnitude of the acceleration of the block?


ANSWER:

Correct

Exercise 14.31

You are watching an object that is moving in SHM. When the object is displaced 0.600 to the right of its equilibriumposition, it has a velocity of to the right and an acceleration of to the left.

e

e

'

e

e

'

)

e

'

)

e

'

)

e

'

)

N NT

= 104 N

T

N

NT NT



Part A

How much farther from this point will the object move before it stops momentarily and then starts to move back tothe left?

ANSWER:

Correct

Gravity on Another Planet

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 . The explorerfinds that the pendulum completes 101 full swing cycles in a time of 145 .

Part A

What is the magnitude of the gravitational acceleration on this planet?

Express your answer in meters per second per second.


Calculate the period of the pendulum, and use this to calculate the magnitude of the gravitationalacceleration on the planet.

Hint 2. Calculate the period

Calculate the period of the pendulum.


ANSWER:

Hint 3. Equation for the period

The period of a simple pendulum is given by the equation , where is the length of the

pendulum and is the magnitude of the gravitational acceleration on the planet.

ANSWER:

0.240 N

DN

T

= 1.44 T

R #

QMBOFU

#

QMBOFU

= 9.58 #QMBOFU

NT



Correct

Properties of Ocean Waves

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves onthe surface of the water. It takes a time of 2.00 for the boat to travel from its highest point to its lowest, a total distanceof 0.670 . The fisherman sees that the wave crests are spaced a horizontal distance of 6.00 apart.

Part A

How fast are the waves traveling?

Express the speed in meters per second using three significant figures.


Calculate the period of the ocean waves, using the fisherman's observations. Then, use the period andwavelength to calculate the speed of the waves.

Hint 2. Calculate the period of the waves

Calculate the period of the ocean waves.

Express your answer in seconds using three significant figures.

Hint 1. Definition of period

The period of a wave is the time it takes for one full wavelength to pass a particular point. This is alsothe time it takes to go from one crest to the next, or from one trough to the next.

ANSWER:

Hint 3. Equation for the speed of a wave

The speed of a wave is given by , where is the frequency of the waves and = 6.00 is the

wavelength. The frequency is simply the reciprocal of the period, or .

ANSWER:

Correct

T

N N

2

= 4.00 T

2 "M " M N

"

= 1.50 2 NT



Part B

What is the amplitude of each wave?

Express your answer in meters using three significant figures.

Hint 1. Definition of amplitude

The amplitude of a wave is the vertical distance from the top of the crest to the neutral position, halfwaybetween the crest and trough. Equivalently, the amplitude is the vertical distance from the bottom of thetrough to the neutral position.

ANSWER:

Correct

The fisherman does not simply move up and down as the waves pass by. In fact, the motion of the fishermanwill be roughly circular with both upward and forward components (with respect to the direction of the wave) asthe wave rises and downward and backward components as the wave falls. The water that comprises theocean wave itself moves in this same way. Thus, an ocean wave is not a purely transverse wave; it also has alongitudinal component.

Exercise 15.4

Ultrasound is the name given to frequencies above the human range of hearing, which is about 20000 . Waves abovethis frequency can be used to penetrate the body and to produce images by reflecting from surfaces. In a typicalultrasound scan, the waves travel with a speed of 1500 . For a good detailed image, the wavelength should be no

more than 1.0 .

Part A

What frequency is required?

ANSWER:

Correct

Standard Expression for a Traveling Wave

Learning Goal:

To understand the standard formula for a sinusoidal traveling wave.

One formula for a wave with a y displacement (e.g., of a string) traveling in the x direction is

= 0.335 N

)[

NT

NN

= 1.50106 " )[

540 TJO'4[0



.

All the questions in this problem refer to this formula and to the wave it describes.

Part A

Which of the following are independent variables?

Hint 1. What are independent variables?

Independent variables are those that are freely varied to control the value of the function. The independentvariables typically appear on the horizontal axis of a plot of the function.

ANSWER:

Correct

Part B

Which of the following are parameters that determine the characteristics of the wave?

Hint 1. What are parameters?

Parameters are constants that determine the characteristics of a particular function. For a wave theseinclude the amplitude, frequency, wavelength, and period of the wave.

ANSWER:

540 TJO'4[0

only

only

only

only

only

and

and

and and

4

0

'

[

4 0

[ 0

' [



Correct

Part C

What is the phase of the wave?

Express the phase in terms of one or more given variables ( , , , , and ) and any needed constants

like .

Hint 1. Definition of phase

The phase is the argument of the trig function, which is expressed in radians.

ANSWER:

Correct

Part D

What is the wavelength of the wave?

Express the wavelength in terms of one or more given variables ( , , , , and ) and any needed

constants like .

Hint 1. Finding the wavelength

Consider the form of the wave at time . The wave crosses the y axis, sloping upward at . The

wavelength is the x position at which the wave next crosses the y axis, sloping upward (i.e., the length ofone complete cycle of oscillation).

ANSWER:

only

only

only

only

only

and

and

and and

4

0

'

[

4 0

[ 0

' [

]40

' 4 0 [

R

= ]40

'4[0

M

' 4 0 [

R

0 4



Correct

Part E

What is the period of this wave?

Express the period in terms of one or more given variables ( , , , , and ) and any needed constants

like .

ANSWER:

Correct

Part F

What is the speed of propagation of this wave?

Express the speed of propagation in terms of one or more given variables ( , , , , and ) and any

needed constants like .

Hint 1. How to find

If you've done the previous parts of this problem, you have found the wavelength and the period of this wave.The speed of propagation is a function of these two quantities: .

ANSWER:

Correct

Breaking Storm Waves

Large waves on the deep ocean propagate at the speed

,

where is the magnitude of the acceleration due to gravity and is the wavenumber.

Seafaring mariners report that in great storms when the average peak-to-peak wave height becomes about 1/7 of the

= MR

'

' 4 0 [

R

= R[

2

' 4 0 [

R

2

2 M

= 2[

'

2

#

'

# '



wavelength, the tops of the largest ocean waves can become separated from the rest of the wave. They claim that thewind and the wave's forward velocity cause huge "hunks" of water to tumble down the face of the wave. Some arereportedly large enough to damage or capsize small vessels.

The reason these "rogue waves" appear is that the amplitude of the water waves becomes so large that the accelerationof the water in the top of the wave would have to be greater than for the wave to stay in one piece. Because gravity is

the only significant vertical force on the water, the acceleration cannot exceed , so instead the water at the top of the

wave breaks off and is blown down the side of the wave.

In this problem, you will compute the ratio of amplitude to wavelength of a rogue wave.

The analytic expression for the vertical displacement of the water surface when an ocean wave of amplitude ispropagating in the +x direction is

.

Part A

Find the angular frequency of water waves.

Express the angular frequency in terms of the wavenumber and constants such as and . All of these

may not be present in your answer.

Hint 1. Define

Express the wavenumber in terms of the angular frequency of the wave and the wave velocity .

ANSWER:

ANSWER:

Correct

Part B

Find the vertical acceleration of this wave at position and time .

Express the vertical acceleration in terms of (Greek letter 'omega'), , , and the independent

variables and .


The acceleration of a point on the wave is equal to the derivative of the velocity of that point (with respect totime).

#

#

640 DPT'4[0

[

' # R

'

' [ 2

= '[

2

= [ #'

40

6

4 0

[ '

4 0



Hint 2. Find an expression for the velocity

Find the vertical velocity of this wave at position and time . We'll call this velocity to distinguish it

from the wave speed .

Express the vertical velocity in terms of (Greek letter 'omega'), , , and the independent

variables and .

Hint 1. How to find the velocity

The velocity is the partial derivative of with respect to . This is simply the derivative of

with respect to , treating as a constant.

Hint 2. A helpful derivative

.

ANSWER:

Hint 3. A helpful derivative

.

ANSWER:

Correct

Part C

Now find the critical ratio of to such that this wave will have a maximum acceleration at the top of . Then

the water at the top would have to accelerate faster than to stay connected with the wave, which is clearly not

possible!

Express in terms of constants (such as , , etc.).

Hint 1. Find the maximum acceleration

From the expression for acceleration found in the last part, what is the maximum negative value of theacceleration?

Express your answer in terms of and .

4 0 402

6

2

[ '

4 0

640 0 640

0 4

DPT 0 TJO 0

0

= 4026

[TJO'4[0

TJO 0 DPT 0

0

= 406

DPT'4[0[

M #

#

M # R

[



Hint 1. Maximum value of

The maximum value of is 1.

ANSWER:

Hint 2. Relationship between and

Express in terms of and constants such as and .

ANSWER:

ANSWER:

Correct

This ratio is about 1/6, which implies a ratio of peak-to-peak wave height to wavelength of about 1/3 (sinceamplitude is 1/2 peak-to-peak height). The factor of 2 discrepancy with sailors' experience probably resultsfrom the fact that actually not all the waves are of the same height. In fact, a fraction of only about 1% of thewaves in a ocean storm are "rogue waves," and are about twice the average wave height. So when most of thewaves in a storm have a height to wavelength ratio of 1/6 or 1/7, the rogue waves will just start to have a heightto wavelength ratio of 1/3, where they start doing damage.

Note that and do not appear in your answer: You are not focusing on the acceleration of a particular point

on the wave, but rather on the maximum acceleration that any arbitrary point on the wave will experience.

Score Summary:

Your score on this assignment is 110%.You received 14.33 out of a possible total of 14 points, plus 1.05 points of extra credit.

DPT'4 [0

DPT'4[0

= NBY

[

' M

M ' # R

= MR

'

= 0.159

M

4 0

chapter 14 and 15 homework

Documents

block of mass

block slides

block willpass

cycle of motion

equilibrium position

subsequent motion of

examples of periodic

spring oscillating