# Modeling by gjjur4356

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```									                Week 1 Lecture 2
Problems 2, 5

 What if something oscillates with no obvious
spring? What is ? (problem set problem)

 Try and get to SHM form

Ex. Full beer can in lake, oscillating

get                            (where x = distance from
equilibrium point and A=cross
k            sectional area)
This is now in SHM form

rearrange:

so, effective “k”

1
Note that „g‟ in this SHM equation comes from
the Archimedes contribution
Rotational Oscillations (review)
 These are systems which oscillate rotationally
about an axis, rather than linearly
Eg. Torsional Pendulum
This stuff is all
Trick here: note that linear                                   review from first
variables simply change to angular                            year – look at your
ones:                                                             old notes
- all Fs become ts
- all ms become Is
- all xs become qs     and so on….

Linear:           F = -kx                    F = ma

Rotational:

restoring          angular          torque              angular
torque             displacement              rotational acceleration
torsional
spring                              inertia
constant

Combining as before:

or
Defining
angular                                      equation for
frequency,                                   angular SHM
2
Like the linear form                    2
except that x s are now qs
Going back to linear SHM again…
How do we figure out where the block is at
any given time, i.e. what is x(t)?

(defining equation
for linear SHM)

is a 2nd order differential equation.
To find x as a function of time (solving the
differential equations)
rewrite as:                        where

The general solution* to this diff eq‟n is given by
(1)

To evaluate constants c 1 and c2, we need
Boundary Conditions (BCs).

So, look again at our oscillating block:
 Pull it out to A and let go at t = 0                          +x
 So at t = 0, x = A (BC #1)                                      A     t= 0
 A = max displacement
 Also at t = 0, v = dx/dt = 0                                          m

(BC #2)
equilibrium
3
* This is something that you learn in a differential equations course
– if you haven‟t seen it yet you should soon
Sub BC #1 into (1)
gives
or

Differentiate equation (1), then sub in BC #2

At t = 0, dx/dt = 0

So, subbing c1 and c2 back into (1) gives

This gives us x as a
or                                  function of t for this case

Note that this equation only holds for the
boundary conditions we assumed:
at t = 0, x = A and v = 0.

Really important to keep in mind!                             4
The Phase Constant 
Usually we write x = Acos(t + ) (rather than just
x=Acos t)
The value of  depends on where we choose t = 0
 If we choose t = 0 when x = A, then:
 x = Acost
x
(these were the boundary
conditions we considered      A
t
on the previous page)

 However, if we choose t = 0 when x = 0, then
our plot of x vs. t looks like:
 The curve is shifted         x

right by /2 (but of course            A
t
in the general case this
shift can be any value)       shifted
right by /2 = -/2

Mathematically, this can be written as
( = - /2)
Or

 Therefore, depending on where you define
t = 0, SHM can be represented by either a sin
or cos function                              5
Therefore, there are three equivalent ways
to express the solution of

1)
here we use BCs to set c 1 and c2 and have no
phase constant.
Use BCs to get C
2)                                       and  (note: since C
is the amplitude it will
be the same for both
3)
of these but  will be
900 different)

Note: We will use the solution a lot, so we have     Hey! Is there a relationship
to pick which one of the above three we will use.    Between c 1 , c2 and C?
We choose this one – mainly because it is the        You bet!
form used in the textbook French. Other books
(like Pain) use the sin form more frequently.
You can see the proof of this
on page 5 of the courseware
or prove it with phasors
We will mainly use:                           (cos and sin are 90o out of phase)

C
c2

c1

Although we usually use the above cos form - you should be                           6
comfortable with finding each of the 3 forms of solution for any
oscillating system!!! - see the example on the next page
Ex. A spring/mass system is pulled 0.5mm from
equilibrium and then given a push to give it an
initial velocity of   mm/s as shown. Its
resulting angular frequency is  = 2rad/s.
Develop an expression for x(t).

0.5mm

equilibrium

___________________________________________________

BCs @ t= 0 x = +0.5 mm (BC#1)
and v = + 2√2 mm/s (BC#2)

Assume SHM applies

Examine the three equivalent methods:

Method 1 uses

BC#1 gives 0.5 = C1sin 0 + C2cos 0             C2 = 0.5mm

BC#2 need to differentiate, giving v = C1cos t – C2sin t
Subbing in BC#2 and w=2 rad/s gives 2√2 = C1 - 0               C1 = √2

7
Result is    x = √2 sin 2t + 0.5 cos 2t
Method 2 uses

BC 1 gives

BC 2 need to diff, gives

Sub in BC 2 and                  gives

get

Solving the two eq ns simultaneously gives

so

Out by /2
Method 3 uses
(1.57) as
BC 1 gives                                   expected

BC 2 – diff gives

or

Solving the two eq ns simultaneously gives

so
8
Maple Plots for 3 Equivalent Methods

These are all identical!

9
Plotting Displacement, Velocity and
Acceleration [back to using x(t)=Acos(t+)]

 Plots of the displacement x, the velocity v, and
the acceleration a, as functions of time ( = 0).
Displacement (x)
x

A

t

Velocity (v)
Note that v      v
(left) by /2
compared to
x(t)
A                                                       t

vmax
at x=0
Acceleration (a)
Note that a        a                       a=max
by                                        at xmax
compared to
x(t)
2A                                                        t
a=0
at vmax
at x=0                                           10
Hey – check out IP demo#1 for this
and follow along with the bouncing spring…
Representing SHM as a Complex Exponential

It is useful to be able to define SHM in terms of a
complex exponential. Why? Because when we try
and solve more complicated versions of our differential
equation (say, ones which include damping terms or
forcing terms) the math becomes much easier if we
can turn our equation into a complex exponential.

The complex exponential can also be represented as a
vector (phasor). This is handy when we start to add
waves together – adding vectors is much easier than

To go from the SHM equations to a complex
exponential, we need to go through a couple of steps,
shown below. The next few pages will illustrate these
steps.
SHM       (x = C cost)

rotating vector

complex number

complex exponential and complex vector

makes math                  easier to add        11
easier                      waves as
vectors
Simple Harmonic Motion and Circular Motion:
The Rotating Vector Representation

 Imagine particle P (or vector OP) moving with
constant speed vo counter-clockwise around
the circle as shown (radius A).
 Point Q is the projection onto the x axis, and
as the vector OP rotates, the point Q
oscillates in SHM from –xo to +xo.
 The y position also exhibits SHM
 At any position:
x = Acost
y = Asint
y

P

A
y = Asint
t
x
O            Q
xo

x = Acost
Check out the Virtual Physic Laboratory website on my webpage
links. Look under Mechanics, then Simple Harmonic motion. This will
help you visualize the motion. Note that the only difference between 12
this demo and the figure above is that this demo has the particle
oscillating along the y axis rather than the x axis.

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