Impulse superposition Green’s function for underdamped oscillator Exponential driving force Green’s function for an undamped oscillator Solution for constant force Step function method

1. Impulsive Methods 1
Impulsive Methods
• The big picture
– Principle of Superposition
– Overview of two methods.
• Impulse superposition
– Green’s function for underdamped oscillator
• Exponential driving force
– Green’s function for an undamped oscillator
• Solution for constant force
• Step function method
– Why it works?
– Undamped example
• Purely time dependent force as Green’s integral
– Equivalence to double integral solution
• Non-Quiescent initial conditions

2. Impulsive Methods 2
What will we do in this chapter?
We develop the impulse (Green’s
function) method for getting solutions for
the harmonic oscillator with an arbitrary
time dependent driving force. We do this
using two techniques. In the first method
we write the solution as a superposition
of solutions with zero initial
displacement but velocities given by the
impulses acting on the oscillator due to
the external force. An arbitrary driving
force is written as a sum of impulses, The
single impulse x(t) responses are added
together in the form of a continuous
integral.
Another way to get this integral is based
on adding the responses of the oscillator
to a rectangular driving force by
considering this as the superposition of a
positive and negative step function. We
next take the limit where the time base of
the rectangle becomes infinitesimal and
compute the response of the driving force
to this “impulse” .
Our solutions will be for quiescent initial
conditions (zero initial displacement and
velocity) We conclude by extending the
treatment to different initial conditions.
This basic Green’s function method is
used in nearly all branches of physics.

3. Impulsive Methods 3
The big picture
We are going to work out a general
expression for the response of a damped
mass-spring system to an arbitrary force as
a function of time making some very
clever uses of Superposition. We will view
the force as a sum of rectangular
infinitesimal impulses and add the x(t)
solutions for each impulsive force. For an
initially quiescent oscillator each impulse
produces a solution equivalent to a free
oscillator with initial velocity of v0 =
FDt/m. The solution becomes a sum
(integral) over such impulse responses.
An alternative method of solution is to
solve for the x(t) response of each impulse
by viewing an impulse as the sum of a step
function and an inverted step function. The
difference of these step function responses
is related to the derivative of the step
function response.
( )
F t
t
0
t 

0
t
0
t 

0
t
( )
f t
0
t
0
( )
t t
 
Step function
impulse

4. Impulsive Methods 4
Impulse Superposition Method
( ')
F t
t
D
Consider an underdamped harmonic oscillator at 0
and at rest at 0 subjected to an sharp impulse
delivered at '. Immediately after the impulse the
oscillator aquires a velocity of
lim ( )
t
x
t
t t
v a t dt
D




' '
' '
o 0
( ) ( ')
But as we still have x(t'+ ) 0 . At a time
' we just have the familiar solution for an
( ')
underdamped oscillator with x 0 and v :
1 ( ')
( ) sin
t t
t
F t F t
dt
m m
t t
F t
m
F t
x t
m


D D

D
 
D   D 

D
 
D

 
 
 
 
 
i
i
1
( ') exp ')
If we had a series of impulses of duration at t at a
time t than max t , superposition gives us:
( )
1
( ) sin ( ) exp )
N i
i i
i
t t t t
F t
x t t t t t
m

 





    
D
D
     

( )
F t ( )
i
F t
t
i
t N
t
 
0
In the limit the solution becomes the integral:
sin ( ') exp ')
1
( ) ' ( ')
This result is called a for the undamped
oscillator. This is a powerful re
Green's Integ
sult since it
r
r
al
t t t t t
x t dt F t
m
 



D  
    
 
epresents the
solution to the differential equation for an arbitrary force
2 ( ).
The only catch is that solution is for an initially
quiescent oscillator (0) (0) 0. We sometimes
write th
mx m x kx F t
x x

  
 
 
 
1
0
'
1
is as an integral over a "Green's" function
1
x(t)= ' ( ') ( , ') where
1
( , ') sin '
m
( ')
t t
dt F t
G t t e
G
t t t t
m
t t






   


5. Impulsive Methods 5
Example: exponential driving force
 
 
       
 
 
 
 
1 1
1
1
1
1
0
' '
o
1
1
' ' ' '
o
0
1
'
o
1 1 0
'
o
1 1 0
o
( ) ( )
x(t)=
F
' ( ') sin '
F
'
2
F
2
F
2
F
t
t t t t
t
t t i t t t i t t
t
i t
i t
t
t
i t
i t
t
t
F t F e t
dt t e t t
m
e
dt e e
m i
e e e
m i i
e e e
m i i
e

 

     
  


  






   
   

  


      
 


 



 
  
 
 
 
 
 
 
 
  
 
 



 
 
 
 
 
 
1
1
1
1
'
1 1
'
o
1 1
1
2
1
F
2 ' '
i t
i t
i t
i t
t
e e
m i i
e e
e
m i t i t
  

  


   
   
 

 


 

 
 
 
 
 

 

 
 
 
 
 
1
1
o
2 2
1 1
1
1
One then combines factors such as exp(-i )
and exp(+i ) into sines and cosine.
Reduction is tedious but straightforward x(t)=
F
cos sin
t
t
t
t
e
e t t
m




 
 
  



 

 
 

 

   
 
 
 
0
0.2
0.4
0.6
0.8
1
1.2
1.4
0 10 20 30 40 50
0
0.2
0.4
0.6
0.8
1
1.2
0 10 20 30 40 50
We show the response
to an exponential with
moderate and very
high damping. At high
damping the x(t)
response is nearly the
same as the driving
force except for the
clear x0 = 0 initial
condition.
( )
x t
exp( )
t


1
0.2
 

1
 

Sometimes difficult to work problems analytically but
often easy to integrate Green’s functions on the computer!

6. Impulsive Methods 6
A simpler example
 
0
0
'
0
2 2 ' 0
0
0 2
For the case of a constant force F ( ')
sin (t-t')
( ) '
sin (t-t')
' cos ( ')
1 cos
Thus ( ) F .
Check by solution of the inhomogeneous DE
with a constant pa
t
o
t t t
o
t
F t
F
x t dt
m
F F
d t t t
m m
t
x t
m



 
 




 

  




 
2
0
0 0
2 2
0
2
0
2
rticular sol.
First solve: x /
sin cos
/
sin cos
Solve for A and B using the initial conditions
(0) (0) 0 (0) 0
(0) 0 Thus 1 cos
This chec
H
P
x F m
x A t B t
F m F
x x A t B t
m
F
x x x B
m
F
x A x t
m

 
 
 

 

 
 
    
     
    
ks our Green's solution!
2
Consider solving the solving the undamped
oscillator subject to a constant force
- / and a quiescent initial state
using Green's function techniques. We can get
build the Green's integral from
o
x x F m
 
0
0 0
0
0
0
the homogenous
solution: ( ) cos sin for x =0
and / . For (0) (0) 0, the
response to an impulse delivered at t' is:
( ')
( ) sin ( ') ; . Hence by
1
superposition ( )
v
x t x t t
v F m x x
v F t
x t t t v
m
x t
 



 
 D  
D
  

 
 
 
'
0
1
0
0
1
1
' ( ')sin ( ')
We can check this by turning the damping off
of our damped general solution:
If =0, then and
1
( ) ' ( ') sin '
1
( ) ' ( ')
as be
sin
for
'
e
t t
t
t
t
d
x
t F t t t
m
x t dt F t
t dt F t e t t
m
t t
m



  




 

 







7. Impulsive Methods 7
A simple non-oscillator example
Drag
The Green's function approach works for other
inhomogenous linear differential equations.
An example is ( ) . This describes
a mass subjected to a viscous drag force of
F and a time depe
v v f t
m v


 
  ndent force of
( ) ( ). We wish to find a Green's
integral describing ( ) for an abtrary ( ).
As before our technique extends only to
the quiescent initial condition v(0) = 0 or
release from res
F t m f t
v t f t

o
t. To obtain the Green's integral
using the impulse method we need to find the
the homogeneous solution for v(t) for an
initial velocity v that we will ultimately set
to the impulse where is the
f D D
-
0
time
duration of the impulse.
For 0, the auxillary equation for
exp( ) is just 0 -
Hence ; (0) ,thus
t t
v v
v r r r
v Ae v A v v e
 

 

 
    
  
- ( ')
- ( ')
0
Hence the response of a quiescent mass to a
single impulse of duration delivered at t'
would be ( ) ( ') or for an
arbitrary ( ') : ( ) ' ( ')
As an example, we apply this s
t t
t
t t
v t f t e
f t v t dt f t e




D
 D
 
 
 
- ( ') ( ')
0 0
0
olution to a mass
released form rest in gravity. Here ( ') .
( ) ' ( ') '
exp( ')
( ) 1 exp
This is indeed the solution we obtained for this
problem i
t t
t t t t
t
t
f t g
v t dt f t e ge dt e
t g
v t ge t
  
 

 
 


 
 
   
 
 
 
term
n Physics 225. Recall the terminal
velocity /
v g 
 

8. Impulsive Methods 8
An alternative (step function) method
We start with a linear differential equation like
(1) We start by solving the differential equation X 1 subjected
to the initial condition X(0) 0 (0) 0 . (Thi
( )
s s l
( )
o u
A BX
F t
A x Bx Cx f t
m
CX
X
  
 
   
tion will be a sum of
a homogeneous (transient) and particular solution.
dX
(2) Let X(t) be the solution to part (1) and let X(t) =
dt
(3) The Green's function is just G(t,t')=X(t-t') (t-t')
And the G


-
reen's integral is simply ( ) ' ( ') ( - ')
t
x t dt f t X t t

 
It is also possible to obtain a Green’s integral through the response of the
system to a unit step function. We will state the technique, give an
example, and argue why it works.

9. Impulsive Methods 9
An example of the step function method
2
2
2
2
Consider the special case of an undamped
oscillator
We first solve: X 1
sin cos
1 1
sin cos
Solve for A and B using the initial conditions
1
(0) (0) 0
(
0
)
( )
H
P
X
X A t B t
X
F t
x x
X A t B t
X X X B
m

 
 
 

 
 
    
  

 

 
 
2
2
-
2
-
0
1
(0) 0 Thus X= 1 cos
1 sin
and 1 cos Thu
1 sin (t-t')
( ) ' ( ')
( ) ' ( ') ( - '
s:
as before!
)
t
t
x t dt f t X t
x t dt F t
X
t
A t
d t
X t
dt
m

 



 





   
 
  
 







( )
F t
t
0
t 

0
t
0
t 

0
t
( )
f t
0
t
0
( )
t t
 
unit step
function
Why does this work? We again write the force
as a sum of square impulses.
Each square impulse can be written as a
superposition of an upright and inverted step
function: (t-t0)

10. Impulsive Methods 10
Why did our simple prescription work?
 
 
0
0
0
( ) ( ) ( ) ( )
Let X(t) be the solution to
(or the solution to X 1 for 0)
By superposition:
x(t)
X ( )
( ) ( ) ( )
( ) (
i i i
i
i i i
i i
i
t t
f t Lim f t t t t t
A BX CX t
Lim f t X t t X t t
Lim X t t X
A BX C t
t
X
t
D
D
D

       D
   

   
    D
  


 
-
'
' (
)
( )
Thus ( ) ( ) ( ) ') ( - ')
t t
i
i i
t
i
t t
dX
dt
X t t
x t f t X t t dt f t X t t



 
 D  D
 
 
  D
  D  

t
1
t 2
t 3
t i
t
( )
i
f t
D
We write the force as a sum of upright
and inverted unit step functions
multiplied by the force in the center
We write the response as a sum of
responses for each rectangular impulse.
These response differences are just the
time derivatives in the infinitesimal limit.
The sum in the limit of small D becomes
an integral but only forces ahead of time
considered are included.

11. Impulsive Methods 11
Another Non-Oscillator Example
0
t
t
( )
( )
F t
f t
m

0
Consider a free particle subjected to an external
force satisfying the Diff Eq: ( ) / ( ).
Assume (0) (0) 0 and ( ) 0. We
already know the solution to this problem is a
double integral of t
x F t m f t
x x f t t
 
   
0 0 '
0
0
0
0 0
'
0
0
(
he form:
( ) ' '' ( '') . Since
( ) 0, we can write this solution as
Let us try to solve this using Green's functions.
W
) (
e
) '' ( ') ' '' (
solve 1 for (0)
'')
t t t
t t
x t t t dt f
x t dt dt f t
f t t
X
t dt dt f t
X

 
 
   
 
 
0
2
0
(0) 0. The
solution is ( ) / 2. The Green function
is then ( - ') ( - ') ( - ') ( ')
and ( ) ' ( - ') ( ')
t
x t dt
X
X t t
G t t X
t
t t t t
t f t
t t


   
 
0 0
0
0 0
'
0
'
0 0
These two forms do not look alike but we can
integrate the Green solution by parts:
( ) ' ( - ') ( ')
Let ( ') ' and -( '- )
( ) ( '') '' ; '
( - ') ( '') ''
t t
t
t t
x t dt t t f t udv
dv f t dt u t t
v t f t dt du dt
udv t t f t dt
 
 
  

 

 
0
0
0 0
'
0
' 0
'
0
0 0 0
( ) ( - ) ( '') '' ' ( '') ''
Hence we essentially can write the single
Green's integral as the familiar double integral
solution for a purely time dependent fo
t t t
t
t t t
vdu
x t t t f t dt dt f t dt


  
 
 
 

  
rce.
0
We could have also obtained the Green's integral
from impulse method. In absence of external force
( ')
( ) ( - ') ( - ') ( ') ( - ')
By superposition : ( ) '( - ') ( ')
t
F t
x t v t t t t f t t t
m
x t dt t t f t
D
    D 
 

12. 12
Non-quiescent initial conditions
2
0
t
0
Lets consider solving ( ) with
(0) 0 ; (0) and the driving acceleration
f(t) first becomes non-zero at t > 0. We cannot
write solution in the form x(t)= dt' G(t-t') f(t')
since if f(t)=0
x x f t
x x v

 
 

0
t
0
0
0
this form gives x(t)=0 when in
v
reality x(t) = sin . The solution is to
consider dt' G(t-t') f(t') as the "particular"
solution and the complete solution as
( ) sin ' ( - ') ( ')
whe
t
t
v
x t t dt G t t f t




 


0
0
re G(t-t') is the Green's function for
the initial condition x(0)=x(0)=0. Hence
1
( ) sin ' ( ')sin ( ')
t
v
x t t dt f t t t
 
 
  

This is clearly the correct solution when ( ') 0.
We can also check the solution for the below case:
f t 
0
f
t
D
( )
f t
0
t
0
D 
0
0
0 0
0 0
0
0
0
0
0 0
1
( ) sin ' f sin ( ')
We can now try to check this answer by starting
our clock at - . At
( ) sin s
0 the mass is at
( 0) sin and ( 0) c
i ( )
o
n
t
t
v
x t t dt t t
t t t t
v
x t t x t
v f
x t t t t
v
t
 

 
 



D
  
 
  
D
   


0
0
0
0 0 0
0
s
After the impulse is applied ( ) sin
and ( ) cos since
t
v
x t t
x t f v t
F t
f v
m






 D 
 D  D 

D  

13. Impulsive Methods 13
Checking our solution in impulsive limit
0
0 0 0 0
0
We can think of ( ) and ( ) as a new set of "initial conditions"
that starts the oscillator at 0 The oscillator moves according to:
cos
( ) sin cos sin
where
x t x t
t t or t
v f v t
x t t t t

  
 
 D  D
 
D 
 
 
 
 
0 0
0 0
0 0
0 0
0 0
0
you can easily verify the "initial" conditions.
We can re-arrange the formula as:
sin cos cos sin sin
sin sin . Using - we have
sin sin - whic
v f
x t t t t t
v f
x t t t t t t
v f
x t t t
    
 
 
 
 
 
D
  
D
   
D
  h equals our Green's solution.