Mastering HW 1

10/13/2017

Mastering HW 1

Mastering HW 1 Due: 11:59pm on Friday, September 29, 2017 To understand how points are awarded, read the Grading Policy for Policy for this assignment.

Cosine Wave The graph shows the position graph is

of an oscillating object as a function of time . The equation of the

where is the amplitude, is the angular frequency, and , and are measurements to be used in your answers.

, is a phase constant. The quantities

,

Part A What is

in the equation?

Hint 1. Maximum of What is the maximum value of on the graph and what is the maximum of multiplied by a cosine function. Cosine can only range from to .

as described by the equation? The equation is just a constant

ANSWER:

Correct

Part B What is

in the equation?

Hint 1. Period Think of the simpler equation

. The period

is the same as before. What does

equal when

? Use the result to solve for

.

ANSWER:

https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588

1/29

10/13/2017

Mastering HW 1

Correct

Part C What is

in the equation?

Hint 1. Using the graph and trigonometry What is

equal to when

? Use your result for

to solve for

in terms of

,

, and

.

Hint 2. Using the graph and Part B You might be able to find

in terms of

and then use your result from Part B.

ANSWER:

Correct

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 called periodic motion. 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. motion. The conditions that lead to simple harmonic motion are as follows: There must be a position of stable equilibrium. 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 system. The resistive forces in the system must be reasonably small.

is a constant that depends on the properties of the oscillating

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 amplitude of of the resulting oscillations. Assume that the mechanical energy of the block-spring system remains unchanged in the subsequent motion of the block.


2/29

10/13/2017

Mastering HW 1

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 The frequency , denoted , is the number of cycles that are completed per unit of time:

and is measured in seconds.

. 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:


3/29

10/13/2017

Mastering HW 1

= 10

Correct

Part D If the frequency is 40

, what is the period

?

Express your answer in seconds. ANSWER: = 0.025

Correct

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


4/29

10/13/2017

Mastering HW 1

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 is 0.0050 .

Part G What is the period

?

Express your answer in seconds.

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: = 0.02

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?


Correct

Part I What distance

does the object cover during one period of oscillation?

Express your answer in meters. ANSWER: = 0.48

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: = 0.36

Correct

Harmonic Oscillator Kinematics https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588

5/29

10/13/2017

Mastering HW 1

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 time , the block is released with zero initial velocity. The goal is to determine the position of the block

along the x axis. At

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 omega).

given in the problem introduction, express the initial position of the block

in terms of

,

, and

(Greek letter

Hint 1. Consider Evaluate the general expression for

when

.

Hint 2. Some useful trigonometry Recall that

and

.

ANSWER: =

Correct 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 1. How to approach the problem Using the general equation expression for when

, obtain the expression for the block's velocity .

in terms of

,

,

, and . Then evaluate the general

Hint 2. Differentiating harmonic functions Recall that


6/29

10/13/2017

Mastering HW 1

and . Note the negative sign in the first formula.

ANSWER:

0

Correct

Part C What is the equation

for the block?

Express your answer in terms of ,

, and

.

Hint 1. Start with the general solution Use the general solution and the values for

and

obtained in the previous parts.

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

Express your answer in terms of

,

in the new coordinate system.

,

(Greek letter omega), and .


7/29

10/13/2017

Mastering HW 1

Hint 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: =

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.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 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 determined based on the information provided.

cannot be determined, the ranking cannot be

Hint 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,


8/29

10/13/2017

Mastering HW 1 .

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 3. Determining As defined in the problem, is the uncompressed length of the spring and crate falls before beginning to move back upward is given by

is the maximum compression of the spring. The total distance the

.

ANSWER:

Reset

Help

The correct ranking cannot be determined.

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.


9/29

10/13/2017

Mastering HW 1

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:


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

?


10/29

10/13/2017

Mastering HW 1

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 Where on the graph is the velocity

?

ANSWER:


Correct

Part F Where on the graph is the velocity

?

Hint 1. How to tell if The velocity is zero when the slope of the x vs. t curve is zero:

.

ANSWER:


11/29

10/13/2017

Mastering HW 1

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 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. The acceleration of a curve is negative for downward curvature and positive for upward curvature. Where is the curvature greater than 0?

ANSWER:


Correct

Part H Where on the graph is the acceleration

?

ANSWER:



12/29

10/13/2017

Mastering HW 1

Correct

Part I Where on the graph is the acceleration

?

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 curvature of the graph changes sign.

ANSWER:

A only B only C only D only E only A and C A and C and E B and D

Correct

Problem 14.5

Part A What is the amplitude of the oscillation shown in the figure?

Express your answer to two significant figures and include the appropriate units. ANSWER: = 10

Correct


13/29

10/13/2017

Mastering HW 1

Part B What is the frequency of the oscillation shown in the figure?

Express your answer to two significant figures and include the appropriate units. ANSWER: = 0.50

Correct

Part C What is the phase constant of the oscillation shown in the figure?

Express your answer to two significant figures and include the appropriate units. ANSWER: = 120

Correct

Energy of Harmonic Oscillators 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.


14/29

10/13/2017

Mastering HW 1

Part A Which moment corresponds to the maximum potential energy of the system?

Hint 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

Correct

Part B Which moment corresponds to the minimum kinetic energy of the system?

Hint 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


15/29

10/13/2017

Mastering HW 1

Correct When the block is displaced a distance

from equilibrium, the spring is stretched (or compressed) the most, and the block is momentarily at rest.

Therefore, the maximum potential energy is

. At that moment, of course,

. Recall that

. Therefore,

. In general, the mechanical energy of a harmonic oscillator equals its potential energy at the maximum or minimum displacement.

Part C Consider the block in the process of oscillating. ANSWER:

at the equilibrium position. at the amplitude displacement. If the kinetic energy of the block is increasing, the block must be

moving to the right. moving to the left. moving away from equilibrium. moving toward equilibrium.

Correct

Part D Which moment corresponds to the maximum kinetic energy of the system?

Hint 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

Correct

Part E Which moment corresponds to the minimum potential energy of the system?

Hint 1. Consider the distance from equilibrium The smallest potential energy corresponds to the smallest distance from equilibrium.

ANSWER:


16/29

10/13/2017

Mastering HW 1

A B C D

Correct 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 ( and that

). The maximum kinetic energy can then be written as

. . Recall that

at the equilibrium position. Therefore, .

Recalling what we found out before, , we can now conclude that , or .

Part F At which moment is

?

Hint 1. Consider the potential energy At this moment,

. Use the formula for

to obtain the corresponding distance from equilibrium.

ANSWER:

A B C D

Correct

Part G Find the kinetic energy

of the block at the moment labeled B.


and

.

Hint 1. How to approach the problem Find the potential energy first; then use conservation of energy.

Hint 2. Find the potential energy Find the potential energy

of the block at the moment labeled B.


17/29

10/13/2017

Mastering HW 1

Express your answer in terms of and

.

ANSWER:

=

ANSWER:

=

Correct

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 time , the block is released with zero initial velocity. The goal of this problem is to determine the acceleration of the block terms of , , and .

along the x axis. At

as a function of time in

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

Express your answer in terms of ,

, and the coordinate of the block

as a function of time.

.

Hint 1. Physical laws Combine the expressions for Hooke's law and Newton's 2nd law of the spring at any time equals the coordinate of the block .

. Note that, since the initial coordinate is zero, the deformation

ANSWER:

=


18/29

10/13/2017

Mastering HW 1

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


, , and

as a function of time.

.

ANSWER: =

Correct

Part C Find the angular frequency

.


and

.

Hint 1. Using the previous results In the previous parts, you obtained two expressions for

:

and . Compare these expressions to determine

.

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.

± Introduction to Simple Harmonic Motion 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 th e 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 https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588

19/29

10/13/2017

Mastering HW 1

Which of the following statements best describes the characteristic of the restoring force in the spring-mass system described in the introduction?

Hint 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 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.

Correct 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?

Express your answer in terms of some or all of the variables

,

, and .

Hint 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.


,

, and .

Hint 1. Forces exerted on the block in the x direction https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=5701588

20/29

10/13/2017

Mastering HW 1

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.

ANSWER: =

ANSWER: =

Correct

Part C What is the acceleration

of the block when it passes through its equilibrium position?


,

, and .

Hint 1. A characteristic of equilibrium By definition, an object in equilibrium does not accelerate.

ANSWER: = 0

Correct 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 a nd has undergone a negative displacement?

Part D Select the correct expression that gives the block's acceleration at a distance that is, the block can be either to the right or left of its equilibrium position.

from the equilibrium position. Note that

can be either positive or negative;

Hint 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 formu la for the acceleration of the mass attached to the spring.

ANSWER:

Correct 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.


21/29

10/13/2017

Mastering HW 1

Using the information found so far, select the correct phrases to complete the following statements.

Part E

Hint 1. How to approach the problem In Part D, you found that displacement is maximum.

. Since the acceleration is directly proportional to displacement, it must reach its maximum value when

ANSWER: The magnitude of the block's acceleration reaches its maximum value when the block is in the equilibrium position. at either its rightmost or leftmost position. between its rightmost position and the equilibrium position. between its leftmost position and the equilibrium position.

Correct

Part F

Hint 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:

in the equilibrium position. at either its rightmost or leftmost position. The speed of the block is zero when it is between its rightmost position and the equilibrium position. between its leftmost position and the equilibrium position.

Correct

Part G

Hint 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:


22/29

10/13/2017

Mastering HW 1 in the equilibrium position. at either its rightmost or leftmost position.

The speed of the block reaches its maximum value when the block is between the rightmost position and the equilibrium position. between the leftmost position and the equilibrium position.

Correct

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 measure d is seconds, expressed in radians.

must be expressed in radians per second so that the quantity

is

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 1. How to find the equation for acceleration To determine the correct equation for the acceleration, simply substitute the equation into the expression for group all positive constants together. You can verify then whether your result is correct by calculating the acceleration at with your result in Part B.

found in Part D and and comparing it

Hint 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 . At this point you simply need to determine which functio n among and is zero at those calculated values of .

ANSWER:

, , , ,

Correct Further calculations would show that the constants

and

can be expressed in terms of

and

.

Vertical Mass-and-Spring Oscillator A block of mass is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance from its equilibrium length. The spring has an unknown spring constant .


23/29

10/13/2017

Mastering HW 1

Part A What is the spring constant ?

Express the spring constant in terms of given quantities and , the magnitude of the acceleration due to gravity.

Hint 1. Sum of forces acting on the block Since the block is not accelerating, the net force acting on the block must be zero. Taking the positive y direction to be upward, write an expression for the net vertical force acting on the block.

Express the sum of the vertical forces in terms of

,

, , and , the magnitude of the acceleration due to gravity.

Hint 1. Force due to spring What is

, the force that the spring exerts on the block?

ANSWER: =

Hint 2. Force due to gravity What is

, the gravitational force on the block?

ANSWER: =

ANSWER: =

ANSWER:

=

Correct

Part B Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency equilibrium position.


of the block's oscillation about its

24/29

10/13/2017

Mastering HW 1

Express the frequency in terms of given quantities and , the magnitude of the acceleration due to gravity.

Hint 1. Formula for angular frequency The angular frequency of simple harmonic motion for a body of mass

acted on by a restoring force with force constant

is given by

.

ANSWER:

=

Correct It may seem that this result for the frequency does not depend on either the mass of the block or the spring constant, which might make little sense. However, these parameters are what would determine the extension of the spring when the block is hanging: . One way of thinking about this problem is to consider both and as unknowns. By measuring and (both fairly simple measurements), and knowing the mass, you can determine the value of the spring constant and the acceleration due to gravity experimentally.

± The Fish Scale A vertical scale on a spring balance reads from 0 to 180 the spring oscillates vertically at a frequency of 2.00 .

. The scale has a length of 15.0

from the 0 to 180

reading. A fish hanging from the bottom of

Part A Ignoring the mass of the spring, what is the mass

of the fish?

Express your answer in kilograms.

Hint 1. How to approach the problem Calculate the spring constant for the fish scale, then use this with the angular frequency of the bouncing fish to calculate its mass.

Hint 2. Calculate the spring constant Calculate the spring constant

for the spring in the fish scale.

Express your answer in newtons per meter.

Hint 1. Using the reading At rest, the weight of the object will be counteracted by the restoring force in the spring, which can be seen by drawing a force diagram of the fish on the spring. Hence . Because we know both the maximum weight the scale can show and the length the spring is stretched at that weight, the spring constant can be calculated from this equation.

ANSWER: = 1200

Correct

Hint 3. Calculate the angular frequency Calculate the angular frequency

for the fish oscillating on the spring.

Express your answer numerically in radians per second.

Hint 1. Relating frequency and angular frequency Recall that, for a given oscillation,

, where

is the angular frequency and

= 2.00


is the frequency.

25/29

10/13/2017

Mastering HW 1

ANSWER: = 12.6

Correct

Hint 4. Formula for the angular frequency of a mass on a spring An object of mass

on the end of a spring with spring constant

will oscillate with frequency

.

ANSWER: = 7.60

Correct

Changing the Period of a Pendulum A simple pendulum consisting of a bob of mass

attached to a string of length

swings with a period

.

Part A If the bob's mass is doubled, approximately what will the pendulum's new period be?

Hint 1. Period of a simple pendulum The period

of a simple pendulum of length

is given by ,

where

is the acceleration due to gravity.

ANSWER:

Correct

Part B If the pendulum is brought on the moon where the gravitational acceleration is about

, approximately what will its period now be?

Hint 1. How to approach the problem Recall the formula of the period of a simple pendulum. Since the gravitational acceleration appears in the denominator, the period must increase when the gravitational acceleration decrease s.

ANSWER:


26/29

10/13/2017

Mastering HW 1

Correct

Part C If the pendulum is taken into the orbiting space station what will happen to the bob?

Hint 1. How to approach the problem Recall that the oscillations of a simple pendulum occur when a pendulum bob is raised above its equilibrium position and let go, causing the pendulum bob to fall. The gravitational force acts to bring the bob back to its equilibrium position. In the space station, the earth's gravity acts on both the station and everything inside it, giving them the same acceleration. These objects are said to be in free fall.

ANSWER:

It will continue to oscillate in a vertical plane with the same period. It will no longer oscillate because there is no gravity in space. It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall. It will oscillate much faster with a period that approaches zero.

Correct In the space station, where all objects undergo the same acceleration due to the earth's gravity, the tension in the string is zero and the bob does not fall relative to the point to which the string is attached.

Period of a Pendulum Ranking Task

Part A Six pendulums of mass and length as shown are released from rest at the same angle of complete cycles of motion each pendulum goes through per minute.

from vertical. Rank the pendulums according to the number

Rank from most to least complete cycles of motion per minute. To rank items as equivalent, overlap them.

Hint 1. Frequency For a simple pendulum, , where is the period (the time for one complet e cycle), is the length of the pendulum, and is the acceleration due to gravity. The frequency the inverse of the period. The frequency of an oscillator is the number of complete cycles it goes through in a given time interval.

is

Hint 2. The roles of mass and length The force of gravity is the only force with a component along the direction of motion of the pendulum bob. Because this force provides the same acceleration to objects regardless of mass, all of the masses will accelerate at the same rate when released. Pendulums with larger lengths require that the bob attached to the pendulum must travel a larger distance before completing a cycle.

ANSWER:


27/29

10/13/2017

Mastering HW 1 Reset

Help

The correct ranking cannot be determined.

Correct

± Gravity on Another Planet After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 52.0 full swing cycles in a time of 133 .

. The explorer finds that the pendulum completes 94.0

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

Express your answer in meters per second per second.

Hint 1. How to approach the problem Calculate the period of the pendulum, and use this to calculate the magnitude of the gravitational acceleration on the planet.

Hint 2. Calculate the period Calculate the period

of the pendulum.


Incorrect; Try Again; 4 attempts remaining

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:


28/29

10/13/2017

Mastering HW 1 = 10.3

Correct Score Summary: Your score on this assignment is 91.6%. You received 48.56 out of a possible total of 53 points.


29/29

Mastering HW 1

Recommend Documents