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Figure 14.1: The girl on a swing.

cos (!0 t) and sin (!0 t). For 0 t < the solution can be found by making the substitution

x = x0 + F0=m!o . 2

2

When t < 0 and t the function x(t) satis es the di erential equation x+!0 x = 0.

This di erential equation has general solutions of the form x(t) = A cos (!0 t)+B sin (!0 t).

For t < 0 the displacement vanishes, hence the constants A and B vanish. This determines

the solution for t < 0 and t . For 0 t < the displacement satis es the di erential

2 2

equation x + !0 x = F0 =m. The solution can be found by writing x(t) = F0 =m!0 + y(t).

2

The function y(t) then satis es the equation y + !0 y = 0, which has the general solution

C cos (!0t) + D sin (!0t). This means that the general solution of the oscillator is given by

80

> for t < 0

< F0

x(t) = > m!0 + C cos (!0t) + D sin (!0 t) for 0 t < (14.3)

: A cos (!0t) + B sin (!0t)

2

for t

where A, B, C and D are integration constants that are not yet known.

These integration constants follow from the requirement that the motion x(t) of the

oscillator is at all time continuous and that the velocity x(t) of the oscillator is at all time

_

continuous. The last condition follows from the consideration that when the force is nite,

the acceleration is nite and the velocity is therefore continuous. The requirement that the

14.1. THE GIRL ON A SWING 175

both x(t) and x(t) are continuous at t = 0 and at t =

_ lead to the following equations:

F02

m!0 + C = 0

!0 D = 0 (14.4)

F02 + C cos (!0 ) + D sin (!0 ) = A cos (!0 ) + B sin (!0 )

m!0

) + D cos (!0 ) = ;A sin (!0 ) + B cos (!0 )

;C sin (!0

These equations are four linear equations for the four unknown integration constants

A, B, C and D. The two ; upper equations can be solved directly for the constants C and D

2

to give the values C = F0 =m!0 and D = 0. These values for C and D can be inserted

;

in the lower two equations. Solving these equations then for the constant A and B gives

; ;

2 2

the values A = F0 =m!0 (1 cos (!0 )) and B = F0 =m!0 sin (!0 ). Inserting these

; ;

values of the constants in (14.3) shows that the motion of the oscillator is given by:

8

>0 for t < 0

< F0 cos (!0 t)g for 0 t <

x(t) = > m!0 (14.5)

f1 ;

2

: m!02

F (!0 (t )) cos (!0 t)g for t

fcos ; ;

0

This is the solution for a push with duration delivered at time t = 0. Suppose now

that the push is very short. When the duration of the push is much shorter than the

period of the oscillator !0 1. In that case one can use a Taylor expansion in !0 for

the term cos (!0 (t )) in (14.5). This can be achieved by using that cos (!0 (t )) =

; ;

cos (!0 t) cos (!0 ) sin (!0 t) sin (!0 ) and by using the Taylor expansions sin(x) = x

; ;

x3 =6 + O(x5 ) and cos(x) = 1 x2 =2 + O(x4 ) for sin (!0 ) and cos (!0 ). Retaining term

;

of order (!0 ) and ignoring terms of higher order in (!0 ) shows that for an impulsive

push (!0 1) the solution is given by:

(0 for t < 0

x(t) = (14.6)

F0 2 ) sin (!0 t) for t >

m!0 (!0

We will not bother anymore with the solution between 0 t < because in the limit

0 this interval is of vanishing duration.

!

At this point we have all the ingredients needed to determine the response of the

oscillator for a general driving force F(t). Suppose we divide the time-axis in intervals of

duration . In the i-th interval, the force is given by Fi = F(ti ) where ti is the time of

the i-th interval. We know from expression (14.6) the response to a force of duration

at time t = 0. The response to a force Fi at time ti follows by replacing F0 by Fi and by

replacing t by t ti. Making these replacements it this follows that the respond to a force

;

Fi delivered over a time interval at time ti is given by:

(

for t < ti

x(t) = 0 1 sin (! (t t )) F(t ) (14.7)

for t > ti

0 i i

m!0 ;

This is the response due to the force acting at time ti only. To obtain the response

to the full force F(t) one should sum over the forces delivered at all the times ti . In the

language of the girl on the swing one would say that equation (14.6) gives the motion

of the swing for a single impulsive push, and that expression (14.7) gives the response of

CHAPTER 14. GREEN'S FUNCTIONS, PRINCIPLES

176

the swing to a sequence of pushes given by the mother. Since the di erential equation

(14.1) is linear we can use the superposition principle that states that the response to

the superposition of two pushes is the sum of the response to the individual pushes. (In

the language of section 11.7 we would say that the swing is a linear system.) This means

that when the swing receives a number of pushes at di erent times ti the response can be

written as the sum of the response to every individual push. With (14.7) this gives:

X1

x(t) = m! sin (!0 (t ti)) F(ti ) : (14.8)

;

0

ti <t

Note that in (14.7) the response to a push at time t before the push at time ti vanishes.

For this reason one only needs to sum in (14.8) over the pushes at earlier times because

the pushes at later times give a vanishing contribution. For this reason the summation

extended to times t ti .

Suppose now that the swing is not given a nite number of impulse pushes but that the

driving force is a continuous function. This case can be handled by taking the limit 0.

!

The summation in (14.8) then needs to be replaced by an integration. This can naturally

be achieved because the duration is equal to the in nitesimal interval dt used in the

integration. What we really are doing here is replacing the continuous function F(t) by a

function that is constant within every interval at times ti , see gure (14.2), and then

taking the limit where the width of the intervals goes to zero 0. A similar treatment

!

may be familiar to you from the theory of integration. When the limit 0 is taken

!

F(t) F(t)

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