**Mecflu - din?mica de um sistema de part?culas**

respostas do Marion

Continuous Systems; Waves

13-1. The initial velocities are zero and so all of the rν vanish [see Eq. (13.8b)]. The rµ are given by [see Eq. (13.8a)] sin sinL r

Ax r x dx

so that 3

The characteristic frequency 3ω is [see Eq. (13.1)]

and therefore,

,c os sin x qx t A t

For the particular set of initial conditions used, only one normal mode is excited. Why?

13-2.

436 CHAPTER 13 The initial conditions are hL x hL Lx x L

63 sin sin r L hr x h r x xd x L x

hr r

= (3) We see that 0rµ= for r = 3, 6, 9, etc. The displacement function is

93 12 1 4 , sin cos sin cos sin cos h x x qx t t t t

where

The frequencies 3ω, 6ω, 9ω, etc. are absent because the initial displacement at 3L prevents that point from being a node. Thus, none of the harmonics with a node at 3L are excited.

13-3. The displacement function is

, 13 1 5 sin cos sin cos sin cos qx t x x t

where

1r L r

For t = 0, x x

The figure below shows the first term, the first two terms, and the first three terms of this function. It is evident that the triangular shape is well represented by the first three terms.

CONTINUOUS SYSTEMS; WAVES 437 1 term

0 2 terms

0 3 terms

The time development of q(x,t) is shown below at intervals of 1 of the fundamental period. 8 t = 0, T tT=

tT=

tT=

13-4. The coefficients rν are all zero and the rµ are given by Eq. (13.8a):

L r rx xL x dx

0, even

32 , odd rµ π

Since

() ,s in cosr r the amplitude of the n-th mode is just nµ. The characteristic frequencies are given by Eq. (13.1):

13-5. The initial conditions are ( )

0, otherwise vx s qx t

The rµ are all zero and the rν are given by [see Eq. (13.8b)]

4 sin sin r r Ls rx vd v r s rL from which

0, even

4 1sin, odrr v rs r rL ππω

(Notice that the even modes are all missing, as expected from the symmetrical nature of the initial conditions.)

Now, from Eq. (13.1),

and 1rrωω=. Therefore,

4 1sin, odrr v rs rrL dπν πω

According to Eq. (13.5),

CONTINUOUS SYSTEMS; WAVES 439

() ,s in sin sinit r r

r r rx qx t e r x t

4 13 3 , sin sin sin sin sin sin v sx s x t t t

13-6. The initial conditions are ( ) v L x v L qx x x

L xL

L v

L4 L2 34L The rµ are identically zero and the rν are given by:

Lr rx qx dx rL L

v r r

Observe that for r = 4n, rν is zero. This happens because at t = 0 the string was struck at 4L, and none of the harmonics with modes at that point can be excited.

Evaluation of the first few rν gives

v v and so, v x qx t t t

x t ππωω − (4)

From these amplitudes we can find how many db down the fundamental are the various harmonics:

Second harmonic:

Third harmonic:

These values are much smaller than those found for the case of example (13.1). Why is this so? (Compare the degree of symmetry of the initial conditions in each problem.)

13-7.

h LO

Since (),00qx= , we know that all of the rν are zero and the rµ are given by Eq. (13.8a):

,0 sinLr rx qx dx

CONTINUOUS SYSTEMS; WAVES 441 h x h x L L x L

h Lx L x L

Evaluating the rµ we find

sin sin37r hr r r 7 π πµ π

Obviously, 0rµ= when 47r and 37r simultaneously are integers. This will occur when r is any multiple of 7 and so we conclude that the modes with frequencies that are multiples of 17ω will be absent.

13-8. For the loaded string, we have [see Eq. (12.152)]

21 sin r r dn

n r Ln

The function r r n

is plotted in the figure for n = 3, 5, and 10. For comparison, the characteristic frequency for a continuous string is also plotted:

Continuous string n = 10 n = 5 n = 3

Of course, the curves have meaning only at the points for which r is an integer.

13-9. From Eq. (13.49), we have:

From section 3.5, we know that underdamped motion requires:

24Ds b π τ ρρ<

4 or underdamped

4 ise critically damped

4 overdamped s D s D s D

The complementary solution to Eq. (13.48) for underdamped motion can be written down using Eq. (3.40). The result is:

CONTINUOUS SYSTEMS; WAVES 443 where 22102ωωβ=−, 0ω and β are as defined in (1), and C and sφ are arbitrary constants depending on the initial conditions. The complete solution to Eq. (13.48) is the sum of the particular and complementary solutions (analogous to Eq. (13.50)):

2s in cos

2cos s t s s s Ft tC e t sD b

tans D s b ωδ πτρ ωρ

From Eq. (13.40):

r x qx t t

Thus

2s in cos

2,e xp cos sin 24

(underdamped) r r

Dt s D r x qx t C t b rD b

13-10. From Eq. (13.4) the equation for the driving Fourier coefficient is:

,s inbs sx ft F x t dx

If the point x is a node for normal coordinate s, then where is an integerxn nsbs =≤

(This comes from the fact that normal mode s has s-half wavelengths in length b.)

For xnbs =,

()sinsin0; hence 0ssx nfb

Thus, if the string is driven at an arbitrary point, al modes with nodes at the driving point will be excited. none of the norm

,s inbs sx ft F x t dx

where (),Fxt is the driving force, and ()sft is the Fourier coefficient of the Fourier expansion of

0for sft s nsn

≠ = From the form of (1), we are led to try a solution of the form

() ( ),s in nx Fx t g t b where g(t) is a function of t only. Thus nx sx ft g t dxbb= ∫ ππ

For n ≠ s, the integral is proportional to () x ns xb π =

For n = s, we have nx b ftgtdxgtb==∫π≠

Only the n normal coordinate will be driven. th thThus, to drive the harmonic only, ,s in n nx

Fx t g t b =π

CONTINUOUS SYSTEMS; WAVES 445

Compare this equation to Eq. (3.35):

Thus, by analogy, the solution to (1) is

12 22exp exp44Dts

Ds Ds

13-13. Assuming k is real, while ω and v are complex, the wave function becomes

and the wave is damped in time, with damping coefficient β.

From the relation

we obtain

By equating the real and imaginary part of this equation we can solve for α and β in terms of u and w:

Since we have assumed β to be real, we choose the solution

Substituting this into (5), we have as expected. Then, the phase velocity is obtained from the oscillatory factor in (2) by its definition:

Re V

That is, Vu=

Consider the above circuit. The circuit in the inductor is I, and the voltage above ground at the point between the elements is V. Thus we have thn n thn n n Q VC=′ and n n dI LV V

We may also write

1n n ndQ

Differentiating (1) with respect to time and using (2) gives

or

CONTINUOUS SYSTEMS; WAVES 447

Let us define a parameter x which increases by x∆ in going from one loop to the next (this will become the coordinate x in the continuous case), and let us also define

which will become the inductance and the capacitance, respectively, per unit length in the limit . 0x∆→

From the above definitions and

or,

But by virtue of the above definitions, we can now pass to the continuous limit expressed by

Ix t Ix t LC and for , we obtain 0x∆→

where

Ai t kx

Bi t k k xψω ψω ω where ;kωω∆∆ . A and B are complex constants:

A i

The superposition of 1ψ and 2ψ is given by

tk x t kx iabk it k x A i e B i e ω i ω

(3) which can be rewritten as iiabk iw t k x A e B e ωφ φ ω φ φφφωψ

Define tx kω δ

1222 cos cos

1222 sin sin iabk and then,

Re cos cos 2 2

sin sin 2 2 abab k tk x k tk x

(1)

CONTINUOUS SYSTEMS; WAVES 449

From this expression we see that the wave function is modulated and that the phenomenon of beats occurs, but for A ≠ B, the waves never beat to zero amplitude; the minimum amplitude is, from Eq. (1), AB−, and the maximum amplitude is AB+. The wave function has the form shown in the figure.

AB+ AB−Reψ wt – kx

13-16. As explained at the end of section 13.6, the wave will be reflected at and will then propagate in the –x direction. 0x x= mj n

where n is an integer. Following the procedure in Section 12.9, we write

Assume solutions of the form

Substituting (3a,b) into (2a,b), we obtain ikd ikd

ikd ikd

AB e A Be md

BA e B Ae md from which we can write

AB kd md md

Ak d B md md

The solution to this set of coupled equations is obtained by setting the determinant of the coefficients equal to zero. We then obtain the secular equation

Solving for ω, we find

(7) from which we find the two solutions kd dm m m m m m

kd dm m m m m m

If m′ < m″, and if we define

Then the ω vs. k curve has the form shown below in which two branches appear, the lower branch being similar to that for m′ = m″ (see Fig. 13-5).

0 kω π/2d

Using (9) we can write (6) as

From this expression and the figure above we see that for cωω> and for abωωω<<, the wave number k is complex. If we let , we then obtain from (10) kκ=+iβ

( ) ( )2 2 2 2sin sin cosh cos sinh 2 sin cos sinh coshid d d d d i d d d d Wκβ κ β κ β κ κ β β ω+= − + = (1) Equating the real and imaginary parts, we find

CONTINUOUS SYSTEMS; WAVES 451 sin cos sinh cosh 0 sin cosh cos sinh d d d d d d W κκ β β

(12)

We have the following possibilities that will satisfy the first of these equations:

a) sin κd = 0, which gives κ = 0. This condition also means that cos κd = 1; then β is determined from the second equation in (12):

Thus, cωω>, and κ is purely imaginary in this region.

b) cos κd = 0, which gives κ = π/2d. Then, sin κd = 1, and ()2coshdWβω=. Thus, abωωω<<, and κ is constant at the value π/2d in this region.

c) sinh βd = 0, which gives β = 0. Then, ()2ndWsiκω=. Thus, aωω< or bcωωω<<, and κ is real in this region.

Altogether we have the situation illustrated in the diagram. k

13-18. The phase and group velocities for the propagation of waves along a loaded string are c kdd Vk

cdd Uk kd

where

When kdπ=, U = 0 but cdVωπ=. In this situation, the group (i.e., the wave envelope) is stationary, but the wavelets (i.e., the wave structure inside the envelope) move forward with the velocity V.

452 CHAPTER 13 13-19. The linear mass density of the string is described by if 0;

Consider the string to be divided in three different parts: I for x < 0, I for 0 < x < L, and II for x > L.

Let be a wave train, oscillating with frequency ω, incident from the left on I. We can write for the different zones the corresponding wave functions as follows: (it k xAe ωφ −= ) it k x it k x it ik x ik x I it k x it k x it ik x ik x

it k x I

Ae Be e Ae Be Ce De e Ce De Ee

(1)

Where k ω ωτ τ

and where τ is the tension in the string (constant throughout). To solve the problem we need to state first the boundary conditions; these will be given by the continuity of the wave function and its derivative at the boundaries x = 0 and x = L. For x = 0, we have

ψψ ψψ (3) and for x = L, the conditions are ( ) ( )I II

xL x Lxx x == ψψ ψψ (4)

Substituting ψ as given by (1) into (3) and (4), we have

k A BC D and

CONTINUOUS SYSTEMS; WAVES 453 ik L ik L ik L ik L ik L ik L

Ce D e Ee k Ce D e Ee

From (6) we obtain

ik k L ik k L k CE e k DE e

Hence, ik Lkk

From (5) we have

Using (7) and rearranging the above equation ik Lkk A ek k

k k D

In the same way

112 ik LBk k e k k

From (10) and (1) we obtain

ik L k eB

On the other hand, from (6) and (8) we have

2 ik k LkD

which, together with (10) gives ()

4 ik k L ik L k eE

A k e k k

Since the incident intensity is proportional to 0I2A, the reflected intensity is 2rIB=, and the total transmitted intensity is 2tEI=, we can write

Substituting (12) and (14) into (15), we have, for the reflected intensity,

1ik L r ik L k e I

From which k k L

and for the transmitted intensity, we have ()

4 ik k L t ik L k e I k e k k

We observe that I, as it must. 0rtI+=

For maximum transmission we need minimum reflection; that is, the case of best possible transmission is that in which

In order that , (17) shows that L must satisfy the requirement 0rI=

The optical analog to the reflection and transmission of waves on a string is the behavior of light waves which are incident on a medium that consists of two parts of different optical densities (i.e., different indices of refraction). If a lens is given a coating of precisely the correct thickness of a material with the proper index of refraction, there will be almost no reflected wave.

CONTINUOUS SYSTEMS; WAVES 455

We divide the string into two zones: I: 0

it kx it kx it kx

The boundary condition is

That is, the string is continuous at x = 0. But because the mass M is attached at x = 0, the derivative of the wave function will not be continuous at this point. The condition on the derivative is obtained by integrating the wave equation from x = –ε to x = +ε and then taking the limit ε → 0.

Thus,

Substituting the wave functions from (1), we find which can be rewritten as ik M

From (4) and (6) we obtain

from which we write

M ikB M A ik M M ik ω τω τωω==−−τ (8)

Define

tan 2

Then, we can rewrite (8) as

Bi P

And if we substitute this result in (4), we obtain a relation between 1A and 2A:

The reflection coefficient, 2 1

R A =, will be, from (10),

or,

T A =, will be from (10)

or,

The phase changes for the reflected and transmitted waves can be calculated directly from (10) and (1) if we substitute i i i

B e B A e A A Ae A

Then,

and

CONTINUOUS SYSTEMS; WAVES 457

Hence, the phase changes are ()

1 tan tan cot

tan tan tanBAAA P P

13-21. The wave function can be written as {see Eq. (13.111a)]

Since A(k) has a non-vanishing value only in the vicinity of 0kk=, (1) becomes it kxkk xt e dkωψ +∆

According to Eq. (13.113), k iw k t i t x k

ik k t x i k k t x ik t i k t i tx k i tx k xt e e dk e e it x e e tx i ω (4) and writing the term in the brackets as a sine, we have

2s in, it k xtx k

The real part of the wave function at t = 0 is

2s in Re ,0 cosxk

If ∆, the cosine term will undergo many oscillations in one period of the sine term. That is, the sine term plays the role of a slowly varying amplitude and we have the situation in the figure below.

13-2. a) Using Eq. (13.111a), we can write (for t = 0)

,0 ikx k ikx

k i k k xik x ik x ui ux xA k e dk

Be e dk

Be e e dk Be e e du

This integral can be evaluated by completing the square in the exponent:

b ax x ax bx a b bax x a a a

b ax a e dx e dx e e dx and letting 2yxba=−, we have

4b ayax bx ae e dx e e dy

Using Eq. (E.18c) in Appendix E, we have

4b ax bx aee dx e

Therefore,

CONTINUOUS SYSTEMS; WAVES 459

The form of (),0xψ (the wave packet) is Gaussian with a 1 width of e4σ, as indicated in the diagram below.

b) The frequency can be expressed as in Eq. (13.113a):

, it kx it k k t kx

ik k t k k xit k x k k it k x it x u xt A k e dk

Ak e dk

Be e e dk Be e e du

Using the same integral as before, we find

Then,

, ik k t k k t k k xit k x

t iuit k x i w t x u xt e A k e dk

Be e e du ωσωψ (10)

We notice that if we make the change 02itσωσ−→′′, then (10) becomes identical to (7). Therefore,

, 2 it k x xtxt B i e where tx i t

xt t

The 1 width of the wave packet will now be e

2e twt σω σ or,

In first order, 1eW, shown in the figure above, does not depend upon the time, but in second order, 1eW depends upon t through the expression (14). But, as can be seen from (8) and (1), the group velocity is 0ω′, and is the same in both cases. Thus, the wave packet propagates with velocity 0ω′ but it spreads out as a function of time, as illustrated below.

O x t = t t = 0