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

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.

Impulsive Methods 3

The big pictureWe 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 = Ftm. 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

0t 0t

0t 0t

( )f t

0t

0( )t t

Step function

impulse

Impulsive Methods 4

Impulse Superposition Method( ')F t

t

Consider an underdamped harmonic oscillator at 0and at rest at 0 subjected to an sharp impulse delivered at '. Immediately after the impulse the oscillator aquires a velocity of

lim ( )t

xtt t

v a t dt

' '

' '

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 tdt

m m

t tF t

mF t

x tm

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 ii

t t t t

F tx t t t t t

m

( )F t ( )iF t

tit Nt

0

In the limit the solution becomes the integral:

sin ( ') exp ')1( ) ' ( ')

This result is called a for the undampedoscillator. This is a powerful re

Green's Integsult since it

rr

al

t t t t tx t dt F t

m

epresents thesolution 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

1x(t)= ' ( ') ( , ') where

1( , ') sin '

m

( ')t t

dt F t

G t t e

G

t t t tm

t t

Impulsive Methods 5

Example: exponential driving force

1 1

11

11

0

' 'o1

1

' ' ' 'o

01

'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

tt t i t t t i t t

ti ti tt

ti ti tt

t

F t F e t

dt t e t tm

edt e e

m i

e e e

m i i

e e e

m i i

e

11

11

'

1 1

'

o

1 1

1

2

1F

2 ' '

i ti t

i ti tt

e e

m i i

e ee

m i t i t

1

1

o2 2

1 111

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 tm

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 texp( )t

10.2

1

Sometimes difficult to work problems analytically but often easy to integrate Green’s functions on the computer!

Impulsive Methods 6

A simpler example

0

0

'02 2 ' 00

0 2

For the case of a constant force F ( ')

sin (t-t')( ) '

sin (t-t')' cos ( ')

1 cosThus ( ) F .

Check by solution of the inhomogeneous DE

with a constant pa

to

t t tot

F t

Fx t dt

mF F

d t t tm m

tx t

m

20

0 02 2

02

02

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 mx A t B t

F m Fx x A t B t

m

Fx x x B

mF

x A x tm

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 getbuild the Green's integral from

ox x F m

00 0

0

00

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

1superposition ( )

vx t x t t

v F m x x

v F tx t t t v

m

x t

'

0

1

0

0

11

' ( ') sin ( ')

We can check this by turning the damping offof 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 tm

x t dt F t

t dt F t e t tm

t tm

Impulsive Methods 7

A simple non-oscillator example

Drag

The Green's function approach works for otherinhomogenous linear differential equations.An example is ( ) . This describesa 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 tv 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 setto the impulse where is the f

-0

timeduration of the impulse.For 0, the auxillary equation for

exp( ) is just 0 - Hence ; (0) , thus t t

v vv 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

- ( ') ( ')

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 tt t t t

t

t

f t g

v t dt f t e ge dt e

t gv t ge t

term

n Physics 225. Recall the terminal

velocity /v g

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 tA 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 ( ) ' ( ') ( - ')

tx 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.

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 1sin cos

1 1sin cos

Solve for A and B using the initial conditions1

(0) (0) 0

(

0

)

( )

H

P

XX A t B t

X

F tx x

X A t B t

X X X B

m

2

2

-

2

-

0

1(0) 0 Thus X= 1 cos

1 sinand 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 tX t

dt

m

( )F t

t

0t 0t0t

0t

( )f t

0t

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)

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 ii

i i i

i i

it 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

-

'

' (

)

( )

Thus ( ) ( ) ( ) ') ( - ')

t t

i

i i

t

it t

dX

dt

X t t

x t f t X t t dt f t X t t

t

1t 2t 3t it

( )if t

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 becomes an integral but only forces ahead of time considered are included.

Impulsive Methods 11

Another Non-Oscillator Example0t

t

( )( )

F tf 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

tt t t

vdu

x t t t f t dt dt f t dt

rce.

0

We could have also obtained the Green's integralfrom impulse method. In absence of external force

( ')( ) ( - ') ( - ') ( ') ( - ')

By superposition : ( ) '( - ') ( ')t

F tx t v t t t t f t t t

m

x t dt t t f t

12

Non-quiescent initial conditions2

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

vreality 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

vx 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 ( ')

tvx 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

0f

t

( )f t

0t

0

0

0

0 00 0

00

0

00 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

vx t t dt t t

t t t tv

x t t x t

v fx t t t t

v

t

0

00

0 0 0

0

s

After the impulse is applied ( ) sin

and ( ) cos since

t

vx t t

x t f v tF t

f vm

Impulsive Methods 13

Checking our solution in impulsive limit

0

0 0 0 00

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 tv f v t

x t t t t

0 00 0

0 00 0

0 00

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 fx t t t t t

v fx t t t t t t

v fx t t t

h equals our Green's solution.