# Harmomic Oscillator

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```					Harmonic Oscillator

Nature’s Clocks
Mechanical Applications
Mechanical
Application – Bridge Failure
So how often do we deal with masses on springs?

Tacoma Narrows Bridge, 11/7/1940
Electronic Applications
Passive Electronics
Active Electronic Application
Active Electronics
Audio Applications
Audio

Sometimes with painful results
Climate Application
Future –
Global Warming                     Runaway
150 Years ago                      Hot?
Now –
warming up         Or oscillate
back to ice
age?
Or will Earth go back
nice and easy to what
it was before?
Predicted by Harmonic Oscillator Equation
Forcing
Function

The Physics
k

Mechanical
Homogeneous Equation            b       m

m  - bx - kx
x       
Non-homogeneous Equation
(AKA inhomogeneous)

m  - bx  kx  F(x, t)
x       
Difference is forcing function on RHS
b = resistance,
imagine
The Math    spring and
mass are in
k
water, for
Homogeneous Equation first   example.        m

Rearrange:       m  - bx - kx
x       
Becomes: d x  b  dx  k 
2
        x  0
 m  dt  m 
2
dt
b k
Characteristic       r  r 0
2
Equation:               m m
Math Continued
b k
Roots of               r2    r     0            are R1 & R2
m m
Or:                      r - R1 r - R 2   0
d       d     
Does this make sense?                       - R 1  - R 2  x  0
 dt     dt    
Yes!
b k
r - R1 r - R 2   r - R1  R 2 r  R1R 2  r    r     0
2                               2

m m
d          d         d2x                                 b     k
- R 1  - R 2  x  2 - R 1  R 2   R 1R 2 x      x    x  0
dx
                                                       x        
 dt        dt        dt               dt                 m     m
Why Do This?
It makes a 2nd Order
Differential Equation into 2
First Order Differential
Equations!
Now lets work with:
d       d     
 - R 1  - R 2  x  0
 dt     dt    
First Order Equations
d          d     
Recall:         - R 1  - R 2  x  0
 dt        dt                              2
-b           k  b 
What are   R1 and R2: R1,2   i                  -   
2m           m  2m 
Case 1: R1  R2 or b2  4mk
d                     d         
Means:            - R1 x  0   or       - R 2 x  0
 dt                    dt       
We know the solutions to these 1st order
equations:
x 1  x 1o e R t
1
or x 2  x 2o e R t      2
We have two solutions, x1 and x2. Let’s take a
linear combination:
x = a1x1 + a2x2
d 2 x  b  dx  k 
Does x solve?                                       x  0
 m  dt  m 
2
dt
Let’s try:
d 2 a1x1  a 2 x 2   b  da 1x1  a 2 x 2   k 
                           a x
1 1    a 2x 2   0
m                    m
2
dt                               dt
 d 2 x1  b  dx1  k            d 2 x 2  b  dx 2  k  
a1  2   
 dt                 x1   a 2  2   
        dt                  x 2   0

         m  dt  m                      m  dt  m  
a1  0  a 2  0  0   TRUE!
Linear combinations of solutions to the
homogeneous equation are ALSO a
solution!
General Solution to homogeneous Equation:
x  x 1o e R1t  x 2o e R 2 t
Before talking about what this solution looks like,
let’s go back and think about the case where the
roots are equal, R1 = R2 = R
Multiple Roots
Case 2: R1 = R2 = R or b2 = 4mk
d       d     
Recall:    - R 1  - R 2  x  0
dt       
dt  
2
d   
For R1 = R2 = R:       - R x  0
 dt 
One solution is unchanged:      x 1  x 1o e Rt
But second solution is equivalent to the
first:            x 2  x 2o e Rt

What do we do?
Double Root Second Solution
d                           d       
Observe:         - R x 2  x 1   , and       - R  x1  0
 dt                          dt     
2
d    
Therefore:     - R  x2  0
 dt  
Solvable using integrating factors
learned when discussing 1st order
differential equations.
Therefore skipping proof and presenting
x 2  x 2o te Rt and: x 1  x 1o e
Rt
result:
What Roots Look Like
Note Real and Imaginary Parts of R1 & R2
2
-b      k  b 
R1,2     i     -     - γ  i
2m      m  2m 
2
b                    k  b 
Where:     γ          and       ω  -    
2m                    m  2m 
For b2  4mk,  is a Real number
Otherwise Real roots (b2  4mk):
2     0 if b, m, k  0
-b    b     k 
R 1,2           -    0 if m, k  0 & b  0
2m    2m  m 
may have roots  0 otherwise
Roots > 0 mean exponential growth!
Summary So Far
b k
Equation:                 x    x  0
x        
m m
Homogenous Solutions
Double Real
Unequal Roots
Root

                             
NA
Complex      e -γ t x1o e iω t  x 2o e -iω t
Roots

Real Roots      e R1t  x 2o e R 2 t x1o e -γ t  x2o te-γ t
x1o                                   
Graph It!
Two initial conditions for many values of b:
Constant initial velocity, zero initial displacement
– give system initial push
Constant initial displacement, zero initial velocity
– plucking
Other choices cause phase shifting – we’re
advanced students, we understand phase shifts
Initial conditions determine values of x1o and
x2o
Many potential forms of the forcing
function (F)
Advanced techniques to handle them –
LaPlace transformations
For this talk, consider two
Sinusoidal
Constant – parachutist
F Sinusoidal
Two different frequencies
Rename  = n (natural frequency)
Forcing frequency = f
Non-homogeneous Equation
b      k       F(x, t)
    x    x 
x        
m      m         m

 A sin ωf t 
F(x, t)
Let:
m
Sinusoidal F Continued
Simple situations, like this, solutions are
same form as forcing function, F
Let: x p  x sinωf t  x cosωf t
p                         p
1o                        2o

Called “Particular Solution”
sin ωf t  φ 
A
xp 
2               2
 bωf   k       
Tedious, but                      - ωf2 
  m  m         
straightforward
where tan φ  
- bωf m
solution:                               k     2
 - ωf 
m      
Solutions
Remember linearity where we found
linear combinations of different solutions
to the homogeneous equation also solved
the equation
Similarly, we can add solution to
homogeneous equation, xh, to the
particular solution, xp, and this solves
non-homogeneous equation
General Solution
b       k
xh solves:           h    x h    x h  0
x          
m       m
xp solves:                b       k       F(x, t)
 p    x p    x p 
x          
m       m         m
Does xg = xp+xh solve:
d2
x p  x h     x p  x h    x p  x h  
bd               k               F(x, t)
 m  dt           m
2
dt                                                         m
Rearranging:
0
       b      k              b       k   F(x, t)
 p   x p   x p    h   x h   x h  
x                          x           
        m      m              m      m      m
TRUE!                b       k       F(x, t)
 g    x g    x g 
x          
m       m         m
Particular Solution
sin ωf t  φ  where tan φ  
A                                                  - bωf m
xp 
2
 bωf   k    2
2                                      k     2
        - ωf                                            - ωf 
 m  m                                                   m      

2
k 1 b 
Amplitude maximized at:                       ωf,res     -  
m 2m
2
k 1 b 
Compare:       ωn   -  
m 4m
What does this look like?
Steady State x h t   0, x h t   0

System at rest, force turned on at t = 0
What about Amplitude vs. Phase angle?
Parachutist
Special Case – Jumping out of an
airplane
b
Initial x=0, v=0. Equation:     x  g
x        
m
First order in x , but for instructive

purposes, use formalism derived
x1o  x 2o e -bt/m
Solution: xp = mgt/b, xh = 

xg 
mgt
b
m 2g
 2 1- e
b
- bt/m

Plots
Conclusion
Developed and Observed Behavior of
Harmonic Oscillator
Large Number of Applications
Note use of differential operator.
Powerful for other problems
Phase Lag for forced oscillations
Further study: LaPlace Transforms

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 views: 4 posted: 6/13/2012 language: pages: 27