HW 11 - Fa14 - Energy Transport via Waves

8/10/2019 HW 11 - Fa14 - Energy Transport via Waves

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HW 11 - fa14 - energy transport via waves: Chapter 16: 30, 32, 38, 60; Chapter 17: 7, 13, 22, 59, 60, 69.

Pool of problems: 16.32, 16.38, 16.60, 17.13, 17.22, 17.59, 17.60

Chapter 16, problem 30: Use the wave equation to find the speed of a wave ( , ) y y x t given in terms of the

general function h ; the wave is given as, 3( , ) ( , ); 4.00 10 ; ( , ) ( ); 30 ; 6.0 ;rad rad

m s y y x t Af x t A m f x t h kx t k (1.1)

Trick: There is something special about functions of the argument1 arg [argument of function ]kx t h :

their outer-function-chain-rule-derivatives are equal . Proof: let z kx t . Then,

( ) ( ) ( ) ( )( ) ; ;

( )

dh d h kx t kx t h z z dh h z z h kx t

dt dt kx t t z t dx z x

(1.2)

Taking one more derivative (using the product-rule),22 2 2

2 2 2

2 2

2

( ) ( ) ( ) ( ) ( ) ( )0 ;

( ) ( ) ( ) (

d h d h z z d h z z h z d z h z z z h z h z z

dt dt z t dt z t z dt t z t t z z t

d h d h z z d h z z h z d z h z dx dx z t dx z x z dx x

22

2 2

) ( ) ( )0 ; z z h z h z z z x x z z x

(1.3)

With these machinations (1.2) and (1.3) in mind, the wave equation is2 2

2 2

2d y d y

dt dxc ; hence, the speed c is,

22 2 2

22 2 2

2 2 2

2 22

( )( , ) ( )

( , ) ( )

h z d y d d z d d d

d y d d

dx dx

t

dx

t t Af x A h kx

c Af x A h k

t

t t x

t

2

2

( ) 2

( )

( )kx

h z

t

z

t

( ) 2

( )

( )( )

6.000 10.2 ;

1 0 30

kx x

x

rad

r

t

s

d

t

am

kx

kx

m

k k

t

s

t

k

(1.4)

This means that any function of the argument kx t has velocity c given by the wave equation! Another

important note is that the velocity we calculated in (1.4) would be negative for arg kx t and positive for

arg kx t ; however, the problem asks for the speed , so we take the positive root: 0.2 m s

c .

• chapter 16, problem 32: What phase difference between two identical traveling waves, moving in the same

direction along a stretched string, results in the combined wave having an amplitude 1.50 times that of the

common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, and (c)

wavelengths.

(a) Let the phase difference be , and abbreviate kx t X . Then, the two waves (to be combined) are,

1 By “argument”, it is meant “whatever is plugged into a function”: for instance, ( ) sin f x x , ( ) sin f y y ,

( ) sin f apple apple (although “apple” has physical dimensions, and you can’t take the sine of something bearing physical

dimensions (e.g., “meters”, “kilograms”, etc.; though, this meansapplesapples

appl

6

es

6

2( ) ( f f

2 apples

62

) ( ) (3) sin 3 0.1411 f f , since

the units of “apple” cancel in the ratio). In this case, we have: ( ) sin( ) f kx t kx t .



1 2 1 2

1 2 1 21 22 2

3

2

lim lim 2 cos sin sin( )

0 sin( ) [amplitude] 3.00 10

y A kx t kx t

kx t m

(1.11)

(c) What phase difference 2 results in the largest amplitude of the resultant wave? We this time apply the

maximizing operation to both sides of (1.9) and use max(cos ) 1 x , yielding a repeat of the steps in (1.10),

1 2 1 2 1 2 1 2

1 1 1 1 1 22 2 2 2max max 2 cos 2 max cos 2 1 2 cos 1 0 0 ; A A A A A (1.12

(d) What is that largest amplitude? Plugging the phase-difference (1.12) into (1.9) and effecting the “max”

operation to both sides to see the maximum value the transverse-displacement (as defined in (1.8)) takes on,

1 2 1 2

1 2 1 2

1 2

1 22 2

1 2 2 1 2

3 3

1 1 1

max lim max lim 2 cos sin sin( )

max 2 sin sin( ) max 2 sin( )

max lim 2 1 2 2(5.00 10 ) 3.00 10 13 [amplitude for

y A kx t kx t

A kx t kx t A kx t

y A A m m mm

2]

(1.13

(e) What is the resultant amplitude if the phase angle is 1 2( ) / 2 ? This seems to be a typo on the part of the

textbook —the author probably intended for “ 1 2( ) / 2 ” to be equal to something or another

4. The solutions

manual hints that 11 2 2

; plugging this into (1.9),

1 2 24 4( , ) 2 cos sin sin( ) y y x t A kx t kx t (1.14)

Applying the “max” operation to both sides of (1.14) to find the amplitude, we note the condition for the 1st

sinusoid is,

2

1 1 14 4 4 2 4 2

3 32

2

max max 2 cos sin sin( ) 2 cos sin sin 2

2 5.00 10 3.00 10 9.1924 ;

y A kx t kx t A A

m m mm

(1.15)

••• Chapter 16, problem 60: In the Figure, a string, tied to a sinusoidal oscillator at P and running over a

support at Q, is stretched by a block of mass m. The separation L between P and Q is 1.20 L m , and the

frequency f of the oscillator is fixed at 120 f Hz . The amplitude of the motion at P is small enough for that

point to be considered a node. A node also exists at Q. A standing wave appears when the mass of the hanging

block is either 286.1nm g or 1 447.0nm g , but not for any intermediate mass. What is the linear density of

the string?

The nth standing-wave-frequency n f of a string of linear-mass-density / M L (e.g., a string of length L and mass M)

subject to tension mg is given by 0 / (2 ) f v L , where /v ,

4 I hope you guys now understand my frustration when you don’t use “=” signs‼ It honestly drives me bonkers.



0 ; 1,2,3, ;

2 2 2 2

v n n n mg f nf n v n

L L L L

(1.16)

Using (1.16) for nm and 1nm , and dividing the resulting expressions, we can work to calculate n ,

1

1

2

11

12

11 ( 1)

( 1) 1

n

n n

n

m g n L n

n nm g mn

n L m

n m f n m n m n

f n m

(1.17)

Going back to (1.16) and solving for , we immediately have 2 2

solve for 2

2 4

n nm g m g n L L f

f n

, into which

we immediately plug in numbers,

2

1

223

2 4

2 2 2 2 2 21286.1

447.0

(286.1 10 )(9.81 )1 18.462 10 ;

4 4 4(1.20 ) (120 )1 1n

n

m

n n s

m g sm g

kg m g m g kg n

L f L f m m

(1.18)

Chapter 17, problem 7: A stone is dropped into a depth- d well. The splash is heard 3.00 f st t s t

later.What is the depth d of the well? As indicated, f t is the time for the stone to fall to the water and t s is the time

for the sound (let the speed of sound be sv ) of the splash to travel from the water to the top of the well; the total

of these times must be t . Then, we have the following two expressions for d on a planet with gravity- g , each

following from21

0 0, 2 y y y v t gt , which can be easily combined with f st t t ,

210 2

; 3.00 2 / / ; f s s f s s y y d gt v t t s t t d g d v (1.19)

There are now two ways to solve this problem: (1) a graphical way, and (2) a formal and algebraic way (the

latter of these is messier). The graphical solution to (1.19) is to plot the equation 2 / / 0 sd g d v t as a

function of d and pick the d where it passes through zero5,

(1.20)

5 This is far easier, and would be the method of solution I recommend if this problem is put on the final exam‼

40.5 41.0 41.5 42.0d

0.02

0.01

0.01

0.02

0.03

0.04

0.05

2 d g d v s t



We see this immediately procures the approximate answer 40.7d m shown in the solutions manual. This

solution-method is eminently sensible if this question (1) appears on an exam (it is my understanding that Prof.

Field pulls no punches in burying correct answers in heavy algebra; I have seen exam-questions of his

demanding use of the quadratic formula!), or (2) is asked of you by WileyPLUS. However, if this question

arises in the context of a research problem, it’s better to indeed obtain a closed-form solution via the quadratic

formula. The total time is to be put into the standard form of a quadratic equation for d ,

square both sides 2 2 2 2 2 22 / / 2 / /

2 / 2( / ) (1 ) 2 ( ) 0; s s

s s s s s

t d g d v d g t d vd g t t v d v d gd v gt v d gv t

(1.21)

This (1.21) is a quadratic equation for d . Its solutions are,

2 2 2 2 2 2 2 22 ( ) ( 2 ( )) 4( )( ) 2 ( ) 4 ( ) 4;

2 2

s s s s s s s s s sv gt v v gt v g gv t v gt v v gt v g v t d

g g

(1.22)

The physical solution must yield d = 0 for t = 0, so we take the solution with the negative sign in front of the

square root. We can also simplify (1.22) drastically,

2 2

2 40.7( ) ( ) ( )

1 1 ( ) 203 ; s

gt s s s s s gt v

gt v gt v gt vd v gt v

g g m

(1.23)

Fascinatingly, a great deal more effort6 in (1.21) through (1.23) procured nothing more or less than merely a

few more significant figures…which may be beyond the resolving power of whatever experimental apparatus

you used to measure the 3.00t s , the 343 m s sv , and 29.81 m

s g (i.e., the entire cadre of problem-givens).

Hopefully, this drives home the message that: “physics is not math”!

•• Chapter 17, problem 13: A sound wave of the form ( , ) cos( )m

s s x t s kx t travels at 7

343 /m s s

v k through air in a long horizontal tube. At one instant t , air molecule A at 2.000 A x m is at its

maximum positive displacement of ( , ) 6.00 A m s x t nm s and air molecule B at 2.070 B

x m is at a positive

displacement of 13

( , ) 2.00 B m s x t nm s . All the molecules between A and B are at intermediate displacements.

What is the frequency of the wave? The frequency is given by2 2

skv f

; we find k using,

( , )1 ( , ) ( , )1 1

solve for

( , )1

cos ; cos cos;

cos ;

A B A

m m m

B

m

s x t s x t s x t A s s sk

s x t B A B s

kx t k

x xkx t

(1.24)

Combining these two statements,

/3( , ) ( , )1 1 1 1cos cos cos cos 343959.975 ;

2 2 2 2.07 2.00 2

m m B A

m m m m

s s s x t s x t m

s s s s s s s

B A

kv v f Hz

x x m m

(1.25)

6 The heartbreaking (yet interesting) part of physics is that “more effort” does not necessarily procure better results!

7 Recall that /c k follows from the wave-equation2 2

2 2

2d y d y

dt dxc where ( , ) ( ) y x t f kx t .



•• Chapter 17, problem 22: In the Figure, sound

with a 240.0 10 m wavelength travels

rightward from a source and through a tube that

consists of a straight portion and a half-circle. Part

of the sound wave travels through the half-circle

and then rejoins the rest of the wave, which goes

directly through the straight portion. This rejoining

results in interference. What is the smallest radius

r that results in an intensity minimum at the

detector?

Let the soundwave be of amplitude A , wavevector k , and frequency . The resultant wave that combines at

the circle-pipe interface just before the detector, using2 2

sin sin 2sin cos

(a familiar 8 mathematical

identity), will be the following,

11 2 2 2

cos( ) cos( ) 2cos sin( ); ( ) ( 2 ); pipe y y y A kx t A kx t kx t kd k C k r r

(

For an intensity maximum, we need2

max max(2cos sin( )) y kx t ; combining this with

2 /k , we have,2

solve for2 40.0 10( 2 ) ( 2) 0.175 ;2( 2) 2( 2)

r mk r r r r m

(1.27)

•• Chapter 17, problem 59: In Fig. 17-41, a French submarine and a U.S. submarine move toward each other

during maneuvers in motionless water in the North Atlantic. The French sub moves at speed 50.00 km F hr

v , and

the U.S. sub at 70.00 kmUSA hr

v . The French sub sends out a sonar signal 9 (sound wave in water) at

3

1 1.000 10 f Hz . Sonar waves travel at (phase velocity) of 5470 kmhr

v . (a) What is the signal’s frequency as

detected by the U.S. sub? (b) What frequency is detected by the French sub in the signal reflected back to it by

the U.S. sub?

For an observer taken to be stationary, a source moving with velocity S v emitting a wave with phase velocity

v at frequency 0 f has its frequency observed to actually be 0S

v

v v f f

for a source velocity approaching, and

0S

v

v v f f

for a source velocity receding. This can be summarized in the following plot,

8 Proof:( )/2 ( )/2 ( )/2 ( )/2

12 2 2 2 2

2sin cos 2( )( ) ( ) sin sini i i i i i i ie e e e

i i e e e e

, where

throughout we have used Euler’s identity cos sinixe x i x . You are not expected to prove either of these equations on the quiz

since this is a mathematical identity and this is not a mathematics class; however, as an engineer, you will encounter such

trigonometric machinations via “complex exponential” again, and so, as I always say “you heard it here first!”. 9 In the form of an outr-r-r-r-rageous accent, of course.



(1.28)

(a) Since the subs are approaching one another, we use ,

, ,

0

eff

ef f S ef f

v

v v f f

, in which S USAv v , and the “effective”

phase velocity ,eff v is boosted by the velocity of the USA-sub, so ,eff USAv v v and ,S eff USA F v v v . The

frequency as detected by the U.S. sub then is,

1 , 1 1

1

, ,

( ) ( ) 1 (5470 70.00 )1.022

( ) 5470 50.00

km kmeff USA USA hr hr

km kmeff S eff USA USA F F hr hr

f v f v v f v v kHz f kHz

v v v v v v v v

(1.29)

(b) The frequency calculated in (1.29) is reflected back to the sub, meaning we have two Doppler shifts “in

series”.

, 1

1 1

, ,

( )

( )

5470 70.00 5470 50.001

5470 70.00 5470 501.04

.5 ;

00

eff USA F USA F

eff S eff F F USA F USA F

v f v v v v v v v v

f f f v v v v v v v v v v

k

v v

kHz Hz

(1.30)

The solution manual speaks of “if the French sub were still”. This piece of information is staggeringly

irrelevant, and directs the problem away from the underlying symmetry highlighted in (1.30).

x<0 (source approaches

observer)x>0 (source recedes from

observer)

sonic boom singularity

(division by 0).

no Doppler shift at x = 0 = vS v

1.0 0.5 0.0 0.5 1.0x v S v

1

2

3

4

5

6

f' f source v S v 1



•• chapter 17, problem 60: A stationary motion detector sends sound waves of frequency 0.150 f MHz

toward a truck approaching at a speed of 45.0 mtr s

v . What is the frequency of the waves reflected back to the

detector? This problem is a special case of the submarine-problem 17.59 (above), except with, effectively, a

zero-velocity USA-sub, and a finite-velocity French sub,

343 45.0(0.150 ) 0.195 ;

343 45.0

m m s tr s s

m m s tr s s

v v f f MHz MHz

v v

(1.31)

Chapter 17, problem 69: A jet plane passes over

you at a height of 5000h m and a speed of Mach

1.5. (a) Find the Mach cone angle (the sound speed

is 331 m s sv ). (b) How long after the jet passes

directly overhead does the shock wave reach you?

(a) The half angle of the Mach cone is given by sin = v/vS , where v is the speed of sound and vS is the speed

of the plane. Since vS = 1.5v, sin = v/1.5v = 1/1.5. This means = 42°.

(b) Let h be the altitude of the plane and suppose the Mach cone intersects Earth's surface a distance d behind

the plane. The situation is shown on the diagram, with P indicating the plane and O indicating the observer. The

cone angle is related to h and d by tan = h/d , so d = h/tan . The shock wave reaches O in the time the plane

takes to fly the distance d :

5000 m11s

tan 1.5(331 m/s)tan42

d ht

v v

.

HW 11 - Fa14 - Energy Transport via Waves

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