ñòð. 4 |

the large and di cult problem that you cannot solve in smaller and simpler problems

that you can solve. By assembling these smaller sub-problems one can then often solve

the large problem. This is exactly what we will do here. First we will solve how much

time it takes for the ball to bounce once given its velocity. Given a prescription of the

energy-loss in one bounce we will determine a relation between the velocity of subsequent

bounces. From these ingredients we can determine the relation between the times needed

for subsequent bounces. By summing this series over an in nite number of bounces we

can determine the total time that the ball has bounced. Keep this general strategy in mind

when solving complex problems. Almost all of us are better at solving a number of small

problems rather than a single large problem!

CHAPTER 2. SUMMATION OF SERIES

12

....

Figure 2.3: The motion of a bouncing ball that looses energy with every bounce.

Problem a: A ball moves upward from the level z = 0 with velocity v. Determine the

height the ball reaches and the time it takes for the ball to return to its starting

point.

At this point we have determined the relevant properties for a single bounce. During each

bounce the ball looses energy due to the fact that the ball is deformed anelastically during

the bounce. We will assume that during each bounce the ball looses a fraction of its

energy.

Problem b: Let the velocity at the beginning of the n;th bounce be vn. Show that

with assumed rule for energy loss this velocity is related to the velocity vn;1 of the

previous bounce by p

vn = 1 vn;1 : (2.22)

;

Hint: when the ball bounces upward from z = 0 all its energy is kinetic energy.

In problem a you determined the time it took the ball to bounce once, given the initial

velocity, while expression (2.22) gives a recursive relation for the velocity between subse-

quent bounces. By assembling these results we can nd a relation for the time tn for the

n;th bounce and the time tn;1 for the previous bounce.

Problem c: Determine this relation. In addition, let us assume that the ball is thrown

up the rst time from z = 0 to reach a height z = H. Compute the time t0 needed

for the ball to make the rst bounce and combine these results to show that

s

t = 8H (1 )n=2 (2.23)

n g ;

where g is the acceleration of gravity.

We can now use this expression to determine the total time TN it takes to carry out N

P

bounces. This time is given by TN = N tn. By setting N equal to in nity we can

n=0

compute the time T1 it takes to bounce in nitely often.

Problem d: Determine this time by carrying out the summation and show that this time

is given by: s

T1 = 8H 1 11 : (2.24)

g

p

; ;

2.3. REFLECTION AND TRANSMISSION BY A STACK OF LAYERS 13

;p n and treat 1

Hint: write (1 )n=2 as 1 as the parameter x in the

p

; ; ;

appropriate Taylor series of section (2.1).

This result seems to suggest that the time it takes to bounce in nitely often is indeed

nite.

Problem e: Show that this is indeed the case, except when the ball looses no energy

between subsequent bounces. Hint: translate the condition that the ball looses no

energy in one of the quantities in the equation (2.24).

Expression (2.24) looks messy. It happens often in mathematical physics that a nal

expression is complex very often nal results look so messy it is di cult to understand

them. However, often we know that certain terms in an expression can assumed to be

very small (or very large). This may allow us to obtain an approximate expression that

is of a simpler form. In this way we trade accuracy for simplicity and understanding. In

practice, this often turns out to be a good deal! In our example of the bouncing ball we

assume that the energy-loss at each bounce is small, i.e. that is small.

q 8H 2

Problem f: Show that in this case T1 by using the leading terms of the appro-

g

priate Taylor series of section (2.1).

This result is actually quite useful. It tells us how the total bounce time approaches in nity

when the energy loss goes to zero.

In this problem we have solved the problem in little steps. In general we will take

larger steps during this course, you will have to discover how to divide a large step in

smaller steps. The next problem is a \large" problem, solve it by dividing it in smaller

problems. First formulate the smaller problems as ingredients for the large problem before

you actually start working on the smaller problems. Make it a habit whenever you solve

problems to rst formulate a strategy how you are going to attack a problem before you

actually start working on the sub-problems. Make a list if this helps you and don't be

deterred if you cannot solve a particular sub-problem. Perhaps you can solve the other

sub-problems and somebody else can help you with the one you cannot solve. Keeping this

in mind solve the following \large" problem:

Problem g: Let the total distance travelled by the ball during in nitely many bounces

be denoted by S. Show that S = 2H= .

2.3 Re ection and transmission by a stack of layers

Lord Rayleigh 48] addressed in 1917 the question why some birds or insects have beautiful

iridescent colors. He explained this by studying the re ective properties of a stack of thin

re ective layers. This problem is also of interest in geophysics in exploration seismology

one is also interested in the re ection and transmission properties of stacks of re ective

layers in the earth. Lord Rayleigh solved this problem in the following way. Suppose

we have one stack of layers on the left with re ection coe cient RL and transmission

coe cient TL and another stack of layers on the right with re ection coe cient RR and

CHAPTER 2. SUMMATION OF SERIES

14

R B

T

1 A

L(eft) R(ight)

Figure 2.4: Geometry of the problem where stacks of n and m re ective layers are com-

bined. The notation of the strength of left- and rightgoing waves is indicated.

transmission coe cient TR . If we add these two stacks together to obtain a larger stack

of layers, what are the re ection coe cient R and transmission coe cient T of the total

stack of layers? See gure (2.4) for the scheme of this problem. Note that the re ection

coe cient is de ned as the ratio of the strength of the re ected wave and the incident

wave, similarly the transmission coe cient is de ned as the ratio of the strength of the

transmitted wave and the incident wave. For simplicity we will simplify the analysis and

ignore that the re ection coe cient for waves incident from the left and the right are in

general not the same. However, this simpli cation does not change the essence of the

coming arguments.

Before we start solving the problem, let us speculate what the transmission coe cient

of the combined stack is. Since the transmission coe cient TL of the left stack determines

the ratio of the transmitted wave to the incident wave, and since TR is the same quantity

of the right stack, it seems natural to assume that the transmission coe cient of the

combined stack is the product of the transmission coe cient of the individual stacks:

T = TL TR . However, this result is wrong and we will try to discover why this is so.

Consider gure (2.4) again. The unknown quantities are R, T and the coe cients A

and B for the right-going and left-going waves between the stacks. An incident wave with

strength 1 impinges on the stack from the left. Let us rst determine the coe cient A

of the right-going waves between the stacks. The right-going wave between the stacks

contains two contributions the wave transmitted from the left (this contribution has a

strength 1 TL ) and the wave re ected towards the right due the incident left-going wave

with strength B (this contribution has a strength B RL ). This implies that:

A = TL + BRL : (2.25)

Problem a: Using similar arguments show that:

B = ARR (2.26)

ñòð. 4 |