Driven/Damped Harmonic Oscillator, by Dr. Colton
Physics 471 – Optics
“Lorentz model”: charge on a spring driven back & forth
…by oscillating electric field E = E0 cos(–t)
→ I’m using cos(–t) instead of cos(+t) so it matches time dependence
of standard traveling EM wave, cos(kx–t)
…with velocity-dependent damping,  (units of  chosen such that force = mv)
F  mx
Fdriving  Fspring  Fdamping  mx
qE0 cos(t )  kx  mx  mx
qE
k
x  0 cos( t )
m
m
qE
x  x   0 2 x  0 cos( t )
m
x  x 
→ Let 0 
k
m
This is the equation of motion we need to solve
Guess x  x0 cos(t   ) as solution
→
Derivatives:
x  x0ei ei (t )
~
x~
x0 e i ( t ) ( is lumped in with complex ~
x0 )
~
x  (i ) ~
x ei ( t )
0
~
x  (i ) 2 ~
x0 ei ( t )
Plug into boxed equation, also changing cosine into exponential:
qE
2
(i ) 2 ~
x0 e i ( t )   (i ) ~
x0 e i ( t )   0 ~
x0 e i ( t )  0 e i ( t )
m
x0 :
Divide by e i ( t ) , factor out ~


qE
2
~
x0   2   (i )   0  0
m
qE
1
~
x0  0 2
The answer in compact form!!
m 0   2  i
qE
1
Or more specifically, ~
x 0 2
e i ( wt )
2
m 0    i
What does it mean? To understand better, we need to convert to polar form. Need first to get
real & imaginary parts.
Work with the complex part: (a little like “rationalizing the denominator”)
 0 2   2  i
0 2   2  i
1

(now in real + imaginary form)

2
2
2 2
0 2   2  i 0 2   2  i
 0     


qE0
0   2
Real part of ~
x0 =
m  2   2 2   2
0
qE0

Imaginary part 
m 0 2   2 2   2
2




Re 2  Im2 
Need to convert to polar form:
Re 2  Im 2 
A little algebra:
qE 0
m
1

2
0
 2

2
  
2
 

 Im 
1

  tan  2
2 



 Re 
 0

(Note: this angle is not correct if in II or III quadrant.)
  tan 1 
qE0
So we have: x0  m
1

0
2
 2

2
  
2

 
exp i tan 1  2
   2

 0

 
 
x 0 is the amplitude of oscillation and the phase of ~
x 0 is the
Remembering that the magnitude of ~
phase of the oscillation, our final answer is:
x(t ) 
qE0
m

1
2
0

  2   
2
2

   

cos  t  tan 1  2
2 




 0


(assuming phase angle is in I or IV quadrant)
x0
Most of this was really just algebra… the physics was finished after the boxed equation for ~
labeled “the answer in compact form” on the previous page.