Review of Lecture 12

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Review of Lecture 12

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Title: Review of Lecture 12

1
Review of Lecture 12

Equilibrium
The Center of Gravity
Examples of Static Equilibrium
Indeterminate Structures
Elasticity / Stress Strain
Tension and Compression
Shearing
Hydraulic Stress

2
Oscillations

Oscillations (motions that repeat themselves) are
everywhere
A child on a swing
A violin string being played
The orbit of the moon around the earth
Our ability to hear depends on oscillations as
does our sense of sight temperature

3
Oscillations

We will study simple harmonic motion the
underpinnings of all kinds of oscillations and
then look at some examples that should be
familiar to us
Then we will look at a more realistic example of
what happens in the real world damped simple
harmonic motion

4
Oscillations

Finally, we will briefly examine the concept of
forced oscillations and resonance

5
Simple Harmonic Motion

Here we can see that a particle is moving back
and forth around a central point as time marches
on (down in this case)
The particle is moving continuously we are
seeing are snapshots of its position at various
times

6
Simple Harmonic Motion

The particle is oscillating and the frequency (f)
is the number of oscillations the particle makes
in one second
Note that the particle has returned to its
starting position at xm in time T

7
Simple Harmonic Motion

The SI unit for frequency is the hertz (Hz) and
is defined as one oscillation per second
Related to the frequency is the period (T) of the
oscillation this is how long it takes to
complete one oscillation
So the relationship between period and frequency
is

8
Simple Harmonic Motion

Since the motion is repeating itself
periodically, it is called periodic motion or
harmonic motion
In this chapter we will be studying motion where
the displacement of an object as a function of
time is denoted by

9
Simple Harmonic Motion

When the displacement of an object as a function
of time can be given by this formulait is
called simple harmonic motion

10
Simple Harmonic Motion

Simple harmonic motion (SHM) is when the motion
of an object is a sinusoidal function of time as
is shown below
In this case the sinusoid is described by the
cosine function

11
Simple Harmonic Motion

Lets look at our equation a bit more
The factors xm, ? and ? are all constants
xm is the amplitude of the oscillation
The time varying quantity (?t ?) is called the
phase of the motion and ? is called the phase
constant or phase angle

12
Simple Harmonic Motion

Finally, lets look at the last constant in our
equation
The constant ? is called the angular frequency of
the motion

13
Simple Harmonic Motion

Lets let ? 0 in the following discussion
We can then see from our equationthat x(0)
xm
And given our definition of the period T, it
should be obvious that x(tT) has the same result
as x(t) for all t

14
Simple Harmonic Motion

The cosine function repeats itself when its
argument (the phase) has increased by the value
2p - so we can say that

15
Simple Harmonic Motion

So lets see what happens when we hold two of the
parameters in the equationconstant and vary
the remaining one
We will begin by varying xm

16
Simple Harmonic Motion

We can see that the curves look identical except
that one is taller than the other
These two curves have a different maximum
displacement or amplitude

17
Simple Harmonic Motion

Now lets vary the period (or frequency) of the
motion we do this by adjusting the value ?
In this case we let T T/2

18
Simple Harmonic Motion

Finally, we hold both xm and ? constant and vary
the phase angle ?
Here we can see that the 2nd curve has slid
over by a constant amount relative to the 1st
curve

19
Checkpoint 1

A particle undergoing SHM of period T is at xm
at t 0 (similar to what you see at the right..)
Is it at xm, xm, at 0, between xm and 0, or
between 0 and xm when at
t 2.00T
t 3.5T
t 5.25T

20
The Velocity of SHM

We know that the displacement of a particle
exhibiting SHM is
To get the velocity of the particle we then
simply differentiate the displacement function

21
The Velocity of SHM

So we havewhich gives us

22
The Velocity of SHM

The displacement and velocity functions are
plotted here for comparison
Note the maximum magnitude of the velocity
function is ?xm

23
The Velocity of SHM

Note also that the velocity is now a sine
function which means that it is one-quarter
period out of phase with the displacement function

24
The Acceleration of SHM

Now lets differentiate once again to get the
acceleration functionwhich gives us

Now we see all three functions
The acceleration function is one-half period (or
p radians) out of phase with the displacement and
the maximum magnitude of the acceleration is ?2xm

26
The Acceleration of SHM

We can combine our original equation for the
displacement function and our equation for the
acceleration function to get

27
The Acceleration of SHM

This relationship of the acceleration being
proportional but opposite in sign to the
displacement is the hallmark of SHM
Specifically, the constant of proportionality is
the square of the angular frequency

So we can see that when the displacement is at a
maximum, the acceleration is also at a maximum
(but opposite in sign)
And when the displacement and acceleration are at
a maximum, the velocity is zero

Similarly, when the displacement is zero, the
velocity is at a maximum
Does this remind you of anything you have seen
previously?

30
The Force Law for SHM

Since we now have an equation for the
acceleration a particle feels when in SHM, lets
now apply that to Newtons 2nd law

31
The Force Law for SHM

Since we now have an equation for the
acceleration a particle feels when in SHM, lets
now apply that to Newtons 2nd law
If we call the term in parens k, we then get the
familiar equation for Hookes law

32
The Force Law for SHM

So we can also say that SHM is defined as when a
particle of mass m experiences a force that is
proportional to the displacement of the particle,
but opposite in sign

33
The Force Law for SHM

The block-spring system shown here is a classic
linear simple harmonic oscillator where linear
means that the force F is proportional to the
displacement x rather than to some other power of
x

34
The Force Law for SHM

The angular frequency ? and period T are
therefore related to the spring constant k by the
formulas

35
Checkpoint 2

Which of these relationships implies SHM?
F -5x
F -400x2
F 10x
F 3x2

36
Sample Problem 15-2

At t 0 the displacement x(0) of the block is
-8.50 cm, the velocity is v(0) -0.920 m/s and
its acceleration is a(0) 47.0 m/s
What are the values for ?, xm and ??

37
Sample Problem 15-2

We know that, at t 0, we have
Unfortunately, we dont know either xm or ?

38
Sample Problem 15-2

But we do have a relationship between the
acceleration and displacement functions that we
can exploitWe can rearrange this to get

39
Sample Problem 15-2

Plugging in our known values of x(0) and a(0) we
get

40
Sample Problem 15-2

Now that we know ?, we can solve for ? in a
similar way
In this case we use the displacement and velocity
functions

41
Sample Problem 15-2

Plugging in the values we get

42
Sample Problem 15-2

This will in fact give us two solutions for ? (a
calculator will typically only show the first
one)
We determine which is the correct solution by
using the value of ? to determine the value of xm

43
Sample Problem 15-2

Taking our equation for the displacement and
rearranging it a bit we get

44
Sample Problem 15-2

We know from our force equation F -kx that k
must be a positive constant
Therefore the correct solution for the phase
angle ? must be 155º and xm must therefore be
0.094 m or 9.4 cm

45
Energy in SHM

We already know from our work in Chapter 8 on
Potential Energy and Conservation of Energy that
kinetic and potential energy get transferred back
and forth as a linear oscillator moves
Remember also that the total mechanical energy E
remains constant

46
Energy in SHM

Well now look at this again in light of our new
knowledge of SHM
We know from Chapter 8 that the formula for
elastic potential energy isand we now know
the form that the function x(t) takes

47
Energy in SHM

So combining the two equations we getwhere
we note that cos2 (A) is the same as (cos A)2 and
not cos (A2)

48
Energy in SHM

Now lets look at the potential energy which we
know is
Using the velocity function and also making a
substitution of ?2 k/m we get

49
Energy in SHM

Finally, the mechanical energy iswhich gives
us

50
Energy in SHM

For any angle a, it is true that
So we finally getwhich tells us that the
total energy of a linear oscillator is constant
and independent of time

51
Energy in SHM

So any oscillating system contains an element of
springiness (which stores the potential energy)
and an element of inertia (which stores the
kinetic energy) even if the system is not
mechanical in nature
The element of springiness in an electrical
system is the capacitor and the element of
inertia is a choke (a.k.a. a coil)

52
Checkpoint 3

In the figure the block has a kinetic energy of 3
J and the spring has an elastic potential energy
of 2 J when the block is at x 2.0 cm
(a) What is the KE when the block is at x 0?
What are the elastic potential energies when the
block is at (b) x -2.0 cm and (c) x -xm?

53
An AngularSimple Harmonic Oscillator

The figure shows an angular version of a simple
harmonic oscillator
In this case the mass rotates around its center
point and twists the suspending wire
This is called a torsion pendulum with torsion
referring to the twisting motion

54
An AngularSimple Harmonic Oscillator

If the disk is rotated through an angle (in
either direction) of ?, the restoring force is
given by the equation

55
An AngularSimple Harmonic Oscillator

Comparing this equationto the standard
Hookes law equationleads one to suspect that
a torsion pendulum follows the same rules where t
has replaced F, ? has replaced k, and ? has
replaced x

56
An AngularSimple Harmonic Oscillator

And similarly, we can surmise that the period of
the torsion pendulum will bewhere ? has
replaced k, and I (the moment of inertia of the
rotating bob) has replaced its counterpart the
mass m

57
Pendulums

When we were discussing the energy in a simple
harmonic system, we talked about the
springiness of the system as storing the
potential energy
But when we talk about a regular pendulum there
is nothing springy so where is the potential
energy stored?

58
The Simple Pendulum

As we have already seen, the potential energy in
a simple pendulum is stored in raising the bob up
against the gravitational force
The pendulum bob is clearly oscillating as it
moves back and forth but is it exhibiting SHM?

59
The Simple Pendulum

Lets look at the free-body diagram of the
pendulum
The tension in the string (of length L) must be
counterbalanced by the force of gravity acting on
the bob (of mass m)

60
The Simple Pendulum

The gravitational force is resolved into two
components one along the string (Fg cos ?) and
the other tangential to the path of the bob (Fg
sin ?)
We can see that the tangential component is
acting as a restoring force it works against
any displacement of the bob

61
The Simple Pendulum

Going back to our definition of torque, we can
see that the restoring force is producing a
torque around the pivot point ofwhere L is
the moment arm of the applied force

62
The Simple Pendulum

If we substitute t Ia, we get
This doesnt appear too promising until we make
the following assumption that ? is small
If ? is small we can use the approximation
thatsin ? ? ? (as long as we remember to express
? in radians)

63
The Simple Pendulum

Making the substitution we then getwhich we
can then rearrange to getwhich is the
angular equivalent to a -?2x

64
The Simple Pendulum

So, we can reasonably say that the motion of a
pendulum is approximately SHM if the maximum
angular amplitude is small
The period of a pendulum is given bywhere I
is the moment of inertia of the pendulum

65
The Simple Pendulum

If all of the mass of the pendulum is
concentrated in the bob, then I mL2 and we get

66
The Physical Pendulum

Now suppose that the mass is not all concentrated
in the bob?
In this case the equations are exactly the same,
but the restoring force acts through the center
of mass of the body (C in the diagram) which is a
distance h from the pivot point

67
The Physical Pendulum

So we go back to our previous equation for the
period and replace L with h to get

68
The Physical Pendulum

The other difference in this case is that the
rotational inertia will not be a simple I mL2
but rather something more complicated which will
depend on the shape of the body

69
The Physical Pendulum

For any physical pendulum that oscillates around
a point O with period T, there is a simple
pendulum of length L0 which oscillates with the
same period
The point on the physical pendulum a distance L0
from O is called the center of oscillation

70
Measuring g

The equation for the period of a physical
pendulum gives us a very nice and neat
relationship between T, I and g
We can exploit that relationship to allow us to
precisely measure g

71
Measuring g

Suppose we have a uniform rod of length L which
we allow to rotate from one end as shown
Using the parallel axis theorem, we can calculate
the moment of inertia to be I mL2/3
We also know that h (the distance to the center
of mass) will be L/2

72
Measuring g

Making those substitutions, we get

73
Measuring g

Solving for g we getwhich allows us to find
g if we can precisely measure L and T

74
Sample Problem 15-5

A 1 meter stick swings about a pivot point at one
end at a distance h from its center of mass
What is the period of oscillation?

75
Sample Problem 15-5

The stick is clearly not a simple pendulum so we
use the formula for the period of a physical
pendulum
From the previous discussion, we know that the
moment of inertia for a stick rotated around one
end is I mL2/3

76
Sample Problem 15-5

We also know that h L/2, so plugging those
values in we get

77
Sample Problem 15-5

What is the distance L0 between the pivot point
of the stick and the center of oscillation of the
stick?

78
Sample Problem 15-5

Here we use the relationship
By inspection we can see that L0 2L/3 66.7 cm
which is marked as P

79
Checkpoint 4

Three physical pendulums, of masses m0, 2m0 and
3m0 have the same shape and size and suspended
from the same point
Rank the masses according to the periods of the
pendulums, greatest period first

80
Simple Harmonic Motion Uniform Circular Motion

At the right we have a plot of data recorded by
Galileo of an object (the moon Callisto) that
moved back and forth relative to the disk of
Jupiter

81
Simple Harmonic Motion Uniform Circular Motion

The circles are Galileos data points and the
curve is a best fit to that data
This would strongly suggest that Callisto
exhibits SHM

82
Simple Harmonic Motion Uniform Circular Motion

But in fact Callisto is moving with pretty much a
constant speed in a nearly circular orbit about
Jupiter
So what is it that we are seeing in the data?

83
Simple Harmonic Motion Uniform Circular Motion

What Gallileo saw and what the data tells us
is that SHM is the projection of uniform circular
motion in the plane of the motion
In other words, SHM is uniform circular motion
viewed edge-on
More formally SHM is the projection of uniform
circular motion on a diameter of the circle in
which the latter motion takes place

84
Simple Harmonic Motion Uniform Circular Motion

We have at the right a reference circle the
particle at point P is moving on that circle at
a constant angular speed ?
The radius of our reference circle is xm
Finally, the projection of the position of P
onto the x axis is the point P

85
Simple Harmonic Motion Uniform Circular Motion

We can easily see that the position of the
projection point P is given by the
formulawhich is the formula for the
displacement of an object exhibiting SHM

86
Simple Harmonic Motion Uniform Circular Motion

Similarly, if we look at the velocity of our
particle (and use the relationshipv ?r) we can
see that it obeys

87
Simple Harmonic Motion Uniform Circular Motion

And finally, if we look at the acceleration of
our particle (and use the relationship ar ?2r)
we can see that it obeys

So regardless of whether we look at displacement,
velocity or acceleration, we see that the
projection of uniform circular motion does indeed
obey the rules of SHM

88
Damped SHM

So far we have been examining SHM with the
implicit and rather unrealistic assumption
that there is no friction involved
However friction is inherent in pretty much any
real-world system and so we should examine the
effect that it has on SHM

89
Damped SHM

Lets look at an idealized model of our (now more
realistic) view of the world
Here we have a mass suspended from a spring which
is in turn suspended from a rigid support
We would expect this system to exhibit SHM in the
vertical direction

90
Damped SHM

Now to add realism (friction) we introduce a vane
which is submerged in a liquid
As the vane moves up and down the liquid exerts a
damping force that is proportional to the
velocity of the vane (and block)

91
Damped SHM

The resistive force can therefore by expressed in
the form Fd -bv where b is a damping constant
that depends on the characteristics of both the
vane and the liquid

92
Damped SHM

We know from Hookes law that the relationship
between the force F applied to an object and the
objects displacement x is given by F -kx
Furthermore, we know from Newtons 2nd law that F
ma

93
Damped SHM

And we now also have a resistive force of the
form Fd -bv involved
Putting all of this together (and ignoring the
effects of gravity) we get

94
Damped SHM

Remembering thatwe get

95
Damped SHM

This equationis a second-order, linear,
homogeneous differential equation

96
Damped SHM

The solution to the previous equation is

97
Damped SHM

Now lets look at the other factor in our
solution
At t 0, we can see this function has a value of
1
Furthermore, we can also see that as t ? ?, the
value of the function asymptotically approaches
zero
We can also see that the rate at which this
function approaches zero is determined by the
damping constant b and the mass of the object m

98
Damped SHM

Taken together, what we have is a SHM oscillator
whose amplitude gradually decreases over time

99
Damped SHM

Because friction is present, the energy of this
damped oscillator is not constant
If the damping constant b is small, we find that
the energy of the oscillator over time is given
by

100
Checkpoint 5

Below are 3 sets of values for the spring
constant, damping constant and mass for a damped
oscillator
Rank them according to the time required for the
mechanical energy to decrease to ¼ of its initial
value, greatest first

101
Forced Oscillations Resonance

An oscillator that has been set in motion and
then left alone is exhibiting what is known as
free oscillation
An example of this is a child in a swing who was
given an initial push but then no additional
pushes

102
Forced Oscillations Resonance

The converse situation when someone continues
to push the swing periodically is called a
forced, or driven oscillation
We wont go into the details of driven
oscillations other than to note that there are
two frequencies involved the natural angular
frequency (?) of the oscillator (when undriven)
and the angular frequency (?d) of the driving
force

103
Forced Oscillations Resonance

Going back to this diagram, imagine that the
structure marked rigid support was in fact
moving up and down with some constant angular
frequency ?d

104
Forced Oscillations Resonance

Ignoring the initial transients, the system will
eventually oscillate at the driven frequency ?d
as given by the equationwhere xm is the
amplitude of the oscillation

105
Forced Oscillations Resonance

However, the value xm is not the same as the
amplitude would be in an undriven system
Instead the value of the displacement amplitude
is a complicated function which involves both the
natural frequency (?) of the oscillator as well
as the driving frequency (?d)

106
Forced Oscillations Resonance

Lets go back to the example of the swing
If you gave it a push only very occasionally you
would not expect the average amplitude of the
swing to be very much

107
Forced Oscillations Resonance

Now imagine the other case suppose you gave it
a push very frequently
In this case you should be able to see that
sometimes your push would be in the direction of
the swings motion and so would help it along
On the other hand, about half the time the
converse would be true your pushes would be
working against the direction of travel of the
swing and would tend to slow it down

108
Forced Oscillations Resonance

Now imagine that you push the swing at a
frequency that closely matches the natural
frequency of the swing
In this case your pushes would reinforce the
motion of the swing and should increase the
amplitude of its oscillations
So you would expect that the greatest amplitude
from a driven oscillator would be at the natural
frequency of the oscillator

109
Forced Oscillations Resonance

And that would be true if there werent any
damping, but we are assuming there is so its a
little more complicated
But as it turns out, the maximum amplitude of a
driven, damped oscillator occurs when the driving
frequency is very close to the natural frequency
of the oscillator

110
Forced Oscillations Resonance

The figure at the right shows the relationship
called a resonance peak
The x axis is the ratio ?d/? and the y axis is
the amplitude of the oscillations

111
Forced Oscillations Resonance

You can see that the peak occurs at a little less
than ?d/? 1

112
Forced Oscillations Resonance

More importantly, you can see that the height and
breadth of the peak depends on the damping
constant b the higher the value of the
constant, the lower and broader is the peak

113
Forced Oscillations Resonance

All mechanical structures have resonance
frequencies and if the structure is driven at a
frequency near to a resonance frequency, the
structure may begin to oscillate uncontrollably
This was very ably demonstrated years ago by Ella
Fitzgerald in a series of TV commercials for
Memorex audio tape

114
Forced Oscillations Resonance