The Physical PendulumDue: 8:00pm on Friday, September 9, 2011

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

[Switch to Standard Assignment View] 

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.

Part A

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

Hint A.1 Consider the position of the block

Hint not displayed

ANSWER:

A

B

C

D

Correct

Part B

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

Hint B.1 How does the velocity change?

Hint not displayed

ANSWER:

A

B

C

D

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:

If the kinetic energy of the block is increasing, the block must be

at the equilibrium position.

at the amplitude displacement.

moving to the right.

moving to the left.

moving away from equilibrium.

moving toward equilibrium.

Correct

Part D

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

Hint D.1 Consider the velocity of the block

Hint not displayed

ANSWER:

A

B

C

D

Correct

Part E

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

Hint E.1 Consider the distance from equilibrium

Hint not displayed

ANSWER:

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

( ). The maximum kinetic energy can then be written as . Recall that

and 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 F.1 Consider the potential energy

Hint not displayed

ANSWER:

A

B

C

D

Correct

Part G

Find the kinetic energy of the block at the moment labeled B.

Hint G.1

How to approach the problem

Hint not displayed

Hint G.2

Find the potential energy

Hint not displayed

Express your answer in terms of and .

ANSWER:    =

Correct

 

± PSS: Simple Harmonic Motion II: Energy

Learning Goal: To practice Problem-Solving Strategy: Simple Harmonic Motion II: Energy.A child's toy consists of a spherical object of mass 50  attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the

oscillation is 6  and the maximum velocity achieved by the toy is 3.2  . What is the

kinetic energy of the toy when the spring is compressed 4.6  from its equilibrium

position?Problem-Solving Strategy: Simple Harmonic Motion II: Energy

The energy equation, , is a useful alternative relationship between velocity and position, especially when energy quantities are also required. If the problem involves a relationship among position, velocity, and acceleration without reference to time,

it is usually easier to use the equation for simple harmonic motion, (from Newton’s second law) or the energy equation above (from energy conservation) than to use the general expressions for , , and as functions of time. Because the energy equation

involves and , it cannot tell you the sign of or of ; you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position toward the point of greatest positive displacement, then is positive and is positive.

IDENTIFY the relevant concepts

Energy quantities are required in this problem, therefore it is appropriate to use the energy equation for simple harmonic motion.SET UP the problem using the following steps

Part A

The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.Select all that apply.

ANSWER:

amplitude

total energy

force constant maximum velocity

mass

potential energy at

kinetic energy at position from equilibrium

Correct

Your target variable is the kinetic energy of the toy at a distance 4.6  from its equilibrium position.

EXECUTE the solution as follows

Part B

What is the kinetic energy of the object on the spring when the spring is compressed 4.6  from its equilibrium position?

Hint B.1

How to approach the problem

Hint not displayed

Hint B.2 Determine an expression for the kinetic energy

Hint not displayed

Hint B.3 Calculate the force constant

Hint not displayed

Express your answer in joules using three significant figures.

ANSWER:    =

0.106Correct

 

EVALUATE your answer

Part C

What is the potential energy of the toy when the spring is compressed 4.6  from its equilibrium position?Express your answer in joules using three significant figures.

ANSWER:    =

0.150Correct  

The total energy in the system, , remains constant as the toy oscillates on the spring, and is equal to the sum of the kinetic energy and the potential energy in the system at any given position . To check your results for consistency, you should find that the sum of the

kinetic and potential energy calculated in Parts B and C equal 0.256  .

 

Energy of a Spring

An object of mass attached to a spring of force constant oscillates with simple harmonic

motion. The maximum displacement from equilibrium is and the total mechanical energy

of the system is .

Part A

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

Hint A.1

How to approach the problem

Hint not displayed

Hint A.2

Find the fraction of total energy that is potential energy

Hint not displayed

Hint A.3 Find the total energy of the system

Hint not displayed

ANSWER:

Correct

Part B

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

Hint B.1

How to approach the problem

Hint not displayed

Hint B.2

Find the kinetic energy

Hint not displayed

Hint B.3

Formula for the velocity in terms of position

Hint not displayed

Hint B.4

Find the object's position

Hint not displayed

ANSWER:

Correct

 

Torsional Pendulum

A thin metal disk of mass and radius is attached at its center

to a long fiber. When the disk is turned

from the relaxed state through a small angle , the torque exerted by the fiber on the disk is

proportional to :

.

The constant of proportionality is called the "torsional constant" and is a property of the fiber.

Part A

Find an expression for the torsional constant in terms of the moment of inertia of the disk and the angular frequency of small, free oscillations.

Hint A.1

Equation of motion

Hint not displayed

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

ANSWER:

   = Correct

Part B

The disk, when twisted and released, oscillates with a period of 1.00 s. Find the torsional

constant of the fiber.

Hint B.1

Moment of inertia of a disk

Hint not displayed

Hint B.2

Angular frequency in terms of the period

Hint not displayed

Give your numerical answer for the torsional constant to an accuracy of three significant figures.

ANSWER:    =

1.910×10−5

Correct  

 

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 .

Part A

What is the spring constant ?

Hint Sum of forces acting on the block

A.1

Hint not displayed

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

ANSWER:    

= Correct

Part B

Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency of the block's oscillation about its equilibrium position.

Hint B.1

Formula for angular frequency

Hint not displayed

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

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.

 

Weighing Lunch

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made

fresh for you with 0.300  of Italian ham. The slices of ham are weighed on a plate of mass

0.400  placed atop a vertical spring of negligible mass and force constant of 200  . The slices of ham are dropped on the plate all at the same time from a height of 0.250  . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.

Part A

What is the amplitude of oscillation of the scale after the slices of ham land on the plate?

Hint A.1

How to approach the problem

Hint not displayed

Hint A.2

Find the position of the plate and the ham immediately after the collision

Hint not displayed

Hint A.3

Find the speed of the plate and the ham immediately after the collision

Hint not displayed

Hint A.4 How to find by matching initial conditions

Hint not displayed

Hint A.5 Find using energy conservation

Hint not displayed

Express your answer numerically in meters and take free-fall acceleration to be  

= 9.80  .

ANSWER:    =

5.80×10−2

Correct 

Part B

What is the period of oscillation of the scale?

Hint B.1

Period of oscillation in SHM

Hint not displayed

Express your answer numerically in seconds.

ANSWER:    =

0.372Correct

 

 

± Gravity on Another Planet

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 52.0  . The explorer finds that the pendulum completes 91.0 full swing cycles in a time of 130  .

Part A

What is the value of the acceleration of gravity on this planet?

Hint How to approach the problem

A.1

Hint not displayed

Hint A.2

Calculate the period

Hint not displayed

Hint A.3

Equation for the period

Hint not displayed

Express your answer in meters per second per second.

ANSWER:

   =

10.1Correct  

 

A Pivoting Rod on a Spring

A slender, uniform metal rod of mass and length is pivoted without friction about an axis through its midpoint and perpendicular to the rod. A horizontal spring, assumed massless

and with force constant , is attached to the lower end of the rod, with the other end of the

spring attached to a rigid support.

Part A

We start by analyzing the torques acting on the rod when it is deflected by a small angle from the vertical. Consider first the torque due to gravity. Which of the following statements most accurately describes the effect of gravity on the rod?Choose the best answer.

ANSWER:

Under the action of gravity alone the rod would move to a horizontal position. But for small deflections from the vertical the torque due to gravity is sufficiently small to be ignored.

Under the action of gravity alone the rod would move to a vertical position. But for small deflections from the vertical the restoring force due to gravity is sufficiently small to be ignored.

There is no torque due to gravity on the rod.

Correct

Assume that the spring is relaxed (exerts no torque on the rod) when the rod is vertical. The

rod is displaced by a small angle from the vertical.

Part B

Find the torque due to the spring. Assume that is small enough that the spring remains

effectively horizontal and you can approximate (and ).

Hint B.1

Find the change in spring length

Hint not displayed

Hint B.2

Find the moment arm

Hint not displayed

Express the torque as a function of and other parameters of the problem.

ANSWER:    =

Correct

Since the torque is opposed to the deflection and increases linearly with it, the system will undergo angular simple harmonic motion.

Part C

What is the angular frequency of oscillations of the rod?

Hint C.1

How to find the oscillation frequency

Hint not displayed

Hint C.2

Solve the angular equation of motion

Hint not displayed

Hint C.3

Determine the moment of inertia of the rod

Hint not displayed

Express the angular frequency in terms of parameters given in the introduction.

ANSWER:    =

Correct

Note that if the spring were simply attached to a mass , or if the mass of the rod were

concentrated at its ends, would be . The frequency is greater in this case because mass near the pivot point doesn't move as much as the end of the spring. What do you suppose the frequency of oscillation would be if the spring were attached near the pivot point?

 

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 A.1 Period of a simple pendulum

Hint not displayed

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 B.1 How to approach the problem

Hint not displayed

ANSWER:

Correct

Part C

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

Hint C.1 How to approach the problem

Hint not displayed

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.

 

Extreme Period for a Physical Pendulum

A solid, uniform disk of mass and radius may be rotated about any axis parallel to the disk axis, at variable distances from the center of the disk.

Part A

What is , the moment of inertia of the disk around its center of mass? You should know this formula well.Express your answer in terms of given variables.

ANSWER:

   =

Correc

t

Part B

If you use this disk as a pendulum bob, what is , the period of the pendulum, if the axis

is a distance from the center of mass of the disk?

Hint B.1 Formula for

Hint not displayed

Hint B.2

Moment of inertia

Hint not displayed

Express the period of the pendulum in terms of given variables.

ANSWER:

   =

Correct

Part C

The period of the pendulum has an extremum (a local maximum or a local minimum) for

some value of between zero and infinity. Is it a local maximum or a local minimum?

Hint C.1

Physical reasoning

Hint not displayed

Hint C.2 As approaches zero

Hint not displayed

Hint C.3 As approaches infinity

Hint not displayed

ANSWER:

maximum

minimum

Correct

Part D

What is , the minimum period of the pendulum?

Hint How to minimize the period

D.1

Hint not displayed

Your answer for the minimum period should include given variables.

ANSWER:

   =Correct

 

Oscillations of a Balanced Object

Two identical thin rods, each of mass and length , are joined at right angles to form an L-shaped object. This object is balanced on top of a sharp edge

. If the object is displaced slightly, it oscillates.

Assume that the magnitude of the acceleration due to gravity is .

Part A

Find , the angular frequency of oscillation of the object.

Hint A.1

Determine the angular frequency of a physical pendulum

This L-shaped object is an example of a physical pendulum. What is , the angular frequency of small-amplitude oscillations for a physical pendulum?

Answer in terms of , the distance between the center of mass and the pivot point, , the moment of inertia of the object, and other given variables.

ANSWER:

   =

Correct

Hint A.2 Calculate

What is the distance between the pivot point (the corner of the "L") and the center of mass of the object?

Hint A.2.1

How to look at the problem

Hint not displayed

Express your answer in terms of .

ANSWER:

   =Correct

Hint A.3

Find the moment of inertia

Hint not displayed

Your answer for the angular frequency may contain the given variables and as well as .

ANSWER:

   =

Correct

 

Test Your Understanding 13.4: Vertical Simple Harmonic Motion

An object of mass hangs from an ideal, vertical spring. The object oscillates up and down

in simple harmonic motion with amplitude .

Part A

For a fixed value of , which of the following aspects of the motion depends on the value of , the acceleration due to gravity?

ANSWER:

the oscillation period

the maximum acceleration of the object during a cycle of oscillation

the maximum speed of the object during a cycle of oscillation

the equilibrium position of the object

more than one of the above

Correct

The greater the value of , the lower the object hangs when at its equilibrium position. The other aspects of the motion are unaffected.

 

Test Your Understanding 13.6: The Physical Pendulum

A physical pendulum is undergoing angular oscillations around its equilibrium position, swinging from left to right and from right to left.

Part A

At the instant that the center of gravity of the pendulum is at the equilibrium position and moving from left to right, the acceleration of the center of gravity is

ANSWER:

zero

downward and to the right

downward and to the left

downward

upward

to the right

to the left

upward and to the left

upward and to the right

Correct

At the instant in question, the center of gravity is neither speeding up nor slowing down. Hence the acceleration has no component along the direction of motion (to the left or right). The center of gravity is following a circular path centered on the pivot, which is above the center of gravity. Hence the acceleration of the center of gravity is upward, toward the pivot.

Score Summary:Your score on this assignment is 94.8%.You received 123.21 out of a possible total of 130 points.

 [ Print ]