solutions to final

8/4/2019 Solutions to Final

1/8

Solutions to Final Exam

Physics 604Dec. 15, 2003

This CLOSED BOOK exam is to be completed in 2 hours. Choose THREE of the four problems if you attempt all four,

be sure to indicate clearly which three should be graded. You must explain your reasoning to receive full credit.

Initialization

ClearAll@"Global`"D;Off@General::spell, General::spell1D

$DefaultFont=

8"Times", 12


2/8

or

-w2c2

Cos@k x - w tD +Ha - bLHb Cos@k x - w tD + kSin@k x - w tDL + H-k b Sin@k x - w tD + k2Cos@k x - w tDL 0

Collecting the coefficients of the sine and cosine functions provides the two equations

-w2c2

+ Ha - bLb + k2 0Ha - bLk - k b 0

such that

b =a2

-w2c2

+a24

+ k2 0 k = $%%%%%%%%%%%%%%%%%%%%%w2c2

-a24

This solution is an attenuated travelling wave where the attenuation length, b-1 , is independent of frequency. Hence, the

balance between high and moderately low frequencies is preserved as the sound propagates.

b)

However, there is a low-frequency cut-off, wc = a c 2 = b c , below which waves do not propagate because kbecomescomplex. For very low frequencies, we substitute a trial solution of the form

y@x, tD = -b xCos@w tD - w2c2

+ bHa - bL 0to obtain

b =

a

2 $%%%%%%%%%%%%%%%%%%%%%a2

4 -

w2

c2

for standing waves. Both damping coefficients are positive and must be retained, but after some distance the smaller one

will dominate and determines the maximum penetration of low-frequencies vibrations into the horn. Notice that at w 0

there is a static solution with b 0 that is not attenuated.

c)

The phase and group velocities are given by

vp =wk

= c$%%%%%%%%%%%%%%%%%%%1 + a24k2

w w c2kk vg = wk

= c2

vp

such that

vpvg = c2

These velocities are sketched below as functions ofk a . Notice that vp diverges and vg vanishes at k= 0.

2 SolutionsToFinal.nb


3/8

PlotAEvaluateA9ikjjj1 + 1

4k2y{zzz

12,ikjjj1 + 1

4k2y{zzz12

=E,8k, 0.6, 5


4/8

Problem 2 (10 points)

Use the Laplace transform to solve the initial-value problem

y

@tD + 0t

y@tD t -t

, y@0D 1solution

The Laplace transform of this integro-differential equation reads

s y

- 1 + s-1y

Hs + 1L-1 y = sHs + 2LHs + 1LHs2 + 1LThe solution transform is resolved into partial fractions according to

As + 1

+B

s +

+C

s -

sHs + 2L

Hs + 1

L

Hs2 + 1

L AHs2 + 1L +BHs + 1LHs - L + CHs + 1LHs + L s2 + 2s

which yields the system of equations

A +B + C 1

BH1 - L + CH1 + L 2A - B + C 0

The second gives

B =2 - H1 + LC

1 -

and substitution into the third gives

H1 - LA - 2 + H1 + LC+ H1 - LC 0 A =2

1 - H1 - CLFinally, substitution into the first equation gives

2 H1 - CL + 2 - H1 - LC+ H1 - LC 0 C = 1 + 3 4

=3 -

4

such that

A =2

1 -

J 1 + 4

N = 4

H1 + L2 = - 12

B =8 - H1 + LH3 - L

4H1 - L = 8- H4 + 2 L

4H1 - L =

H4 - 2 LH1 + L

8=

3 +

4

Thus,

y =

sHs + 2LHs + 1LHs2 + 1L = - 12 1s + 1 + 3 + 4 1s + + 3 - 4 1s -

can be inverted using

1s + a

-a t y@tD = - 12

-t +3 +

4- t +

3 -

4 t



5/8

Therefore, we obtain

y@tD = 12

H3Cos@tD + Sin@tD- -tL

Problem 3 (10 points)

Consider an integral of the form

0

g@xD xwhereg@xD is well-behaved on the positive real axis but is not symmetric with respect to the sign ofx , such that theintegration interval cannot be extended for use with a great semicircle. If the correspondingg@zD is a meromorphicfunction that decreases sufficiently rapidly forz and is sufficiently small at the origin, one can often use

f@zD = g@zDLog@zD with a branch cut on the positive real axis and aPacMan contour of the form shown below.Apply this method for the following integral.

0 xHx + 1LHx2 + 2x + 2LFor this integral one could resolve the integrand into partial fractions instead, but to obtain full credit you must

employ the method described above, which can be useful for more complicated problems that are not suitable for

partial-fraction decomposition. Be sure to justify neglect of any portions of the contour that do not contribute.

x

y

solution

Let

f@zD = Log@zDHz+ 1LHz2 + 2z+ 2L = Log@zD

Hz+ 1LHz+ 1 + LHz+ 1 - Lwith a branch cut on the positive real axis, such that

SolutionsToFinal.nb 5


6/8

z= x + f@zD = Log@xDHx + 1LHx2 + 2x + 2Lz= x - f@zD = Log@xD + 2 p Hx + 1LHx2 + 2x + 2L

For largeR we use

z= R q f@zD z q

> Log@RD + qR2

0

while near the origin

z= q f@zD z q

HLog@D + qL 0Thus, we can neglect both circular portions and obtain

0

xHx + 1LHx2 + 2x + 2L = - 12 p C

Log@zDHz+ 1LHz+ 1 + LHz+ 1 - L z

The right-hand side can now be evaluated as the sum of the residues forz= -1, -1 - , -1 + , such that

0

xHx + 1LHx2 + 2x + 2L = -J Log@-1DHLH-L + Log@-1 - DH-LH-2 L + Log@-1 + DHLH2 L N

= -J p - 12

JLogA!!!2 E+ 5 p4

N - 12

JLogA!!!2 E + 3 p4

NNTherefore,

0

xHx + 1LHx2 + 2x + 2L = LogA!!!2 E

Problem 4 (5 points per part)

The curved surface of a cylinder of radius a is grounded while the endcaps atz= L 2 are maintained at oppositepotentials y@x, f, L 2D = V@x, fD .a) Develop an expansion for the electrostatic potential y@x, f, zD within the cylinder and express the coefficients interms of the appropriate integral overV@x, fD .

b) Determine the coefficients for the simple case V@x, fD = V0 where V0 is constant.Possibly useful information:

2 =1x

x

i

k

jjx x

y

{

zz + 1x2

2

f2

+2

z2

2R x2

+1x

R x

+ ikjjjk2 -m2x2

y{zzzR 0 R = A Jm@kxD + B Nm@kxDJm@km,naD = 0

0

a

Jm@km,n xD2 x x = a22

Jm @km,n aD2

m 1

x

HxmJm@xDL = xmJm-1@xD



7/8

a)

The Laplacian takes the form

2 =1

x

x

ikjjx

x

y{zz

+1

x2

2

f2

+2

z2

in cylindrical coordinates. Laplace's equation

2 y 0 y@x, f, zD = R@xDF@fDZ@zDwith boundary conditions

R@0D finite, R@aD 0F@f + 2 pD F@fD

ZB-L2F -ZB L

2F

then separates into

1F

2 F

f2 -m2 F = m f , m = 0, 1, 2,

1Z

2Zz2

k2 Z = Sinh@k zD2R x2

+1x

R x

+ikjjjk2 -

m2x2

y{zzzR 0 R = Jm@kxD

whereJm is the Bessel function of the first kind, the one that is finite at the origin. Note that although the Bessel equation

is insensitive to the sign ofm , the conventional definitionJ-m@xD = H-LmJm@xD simplifies some of the relationships amongBessel functions. The boundary condition at the curved surface requires k km,n where the index n enumerates the roots

of

Jm@km,n aD 0Thus, the general solution takes the form

y@x, f, zD = m=-

n=1

ym,nJm@km,n xDSinh@km,nzD m f

where the yn,m are numerical coefficients obtained by matching the boundary conditions atz= L 2 according toym,n = Ip a2Jm @km,naD2 Sinh@km,nL 2DM-1

0

2 p

f0

a

x x - m fJm@km,n xDV@x, fD

b)

The expansion coefficients forV@x, fD = V0 require the integrals

0

2 p

f - m f = 2 p dm,0

0

a

J0@k0,n xD x x = 1k0,n

2

0

k0,n a

J0@xDx x = 1k0,n

2

0

k0,na x

Hx J1@xDLx = ak0,n

J1@k0,naD

SolutionsToFinal.nb 7


8/8

Thus, the electrostatic potential becomes

y@x, f, zD = 2V0a

n=1

J0@k0,n xD

km,nJ1@k0,naD

Sinh@k0,nzDSinh@k0,nL 2D


solutions to final

Documents