Damped SHM

 F   bx  kx  mx

k o  k m  b m
“Natural Frequency” “Damping Parameter”
(rad/s) (s-1)
m

2
x   x  o x  0
“Damping Constant”
(kg/s) b
EOM: damped oscillator
Guess a complex solution: z (t )  Ae j  pt  

j 2 p 2 Ae j  pt    jpAe j  pt    o Ae j  pt    0
2


Ae j  pt    p 2  jp  o  0
2

A0
2
“trivial solution”  p 2  jp  o  0
Actually 2 equations:

Real = 0 Imaginary = 0

 p 2  o  0
2
jp  0
p  o  0

… also trivial !
Try a complex frequency: z (t )  Ae j n  js t  

 n  js    j n  js   o  0
2 2

2
 n 2  2 jns  s 2  jn  s  o  0

Real Imaginary
2
 n 2  s 2  s  o  0  2 jns  jn  0

 2
 2 
2
 n 2    o  0 s
4 2 2

2  2
n 2  o  A,  are free constants.
4
  2 
2
  
j    o   j t  
 4 2  
z (t )  Ae  

 2 
2 

 t j  o  t  

 4 
z (t )  Ae 2
e  


 t  2 2 
x(t )  Ae 2
cos o  t   
 4 
0.8

0.6

0.4

0.2
* amplitude decays due to damping

0
* frequency reduced due to damping
-0.2

-0.4

-0.6

-0.8
0 2 4 6 8 10 12
How damped?

Quality factor: unitless ratio of natural frequency to damping parameter

0
Q


  1 
 ot j  o 1 2 t  
2Q 4Q
z (t )  Ae e  

Often use it in the EOM:

o 2
x  x  o x  0
Q
1. “Under Damped” or “Lightly Damped”: Q  1

Oscillates at ~o (slightly less)

Looks like SHM (constant A) over a few cycles:

o = 1,  = .01, Q = 100, xo = 1
1

0.5
Position

-0.5

-1
0 5 10 15 20 25 30 35
Time

Amplitude drops by 1/e in Q/ cycles.

2. “Over Damped”: Q  1 o  

 2 
2 

 t j  o  t  

 4 
z (t )  Ae 2
e  

imaginary!

 2 

 t j j   o 2 t  
 4 
z (t )  Ae 2
e  


 t 
2
 o 2 t
part of A
z (t )  Ae 2
e 4
e j

Still need two constants for the 2nd order EOM:

 2   2 

   o 2  t 
   o 2 t
2 4  2 4 
z (t )  A1e  
 A2 e  

No oscillations!
 Over Damped
o = 1,  = 10, Q = .1, xo = 1
1

0.8
Position 0.6

0.4

0.2

0
0 5 10 15 20 25 30 35
Time
3 “Critically Damped”: Q  0.5   2o
  2    2 
   o 2  t    o 2 t
2 4  2 4 
z (t )  A1e  
 A2 e  

0 0

 
  t
z (t )   A1  A2 e 2

…really just one constant, and we need two. Real solution:

 
  t
z (t )   A  Bt e 2
 Critically Damped
o = 1,  = 2, Q = .5, xo = 1
1

0.8

0.6
Position
0.4

0.2

-0.2
0 5 10 15 20 25 30 35
Series

Fastest approach to zero with no overshoot.

Real oscillators lose energy due to damping. This can be
represented by a damping force in the equation of
motion, which leads to a decaying oscillation solution.
The relative size of the resonant frequency and damping
parameter define different behaviors: lightly damped,
critically damped, or over damped.