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Chapters 16 18

Waves

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Types of waves

Mechanical governed by Newtons laws and existin a material medium (water, air, rock, ect.)

Electromagnetic governed by electricity and

magnetism equations, may exist without any medium

Matter governed by quantum mechanical

equations

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Types of waves

Depending on the direction of the displacementrelative to the direction of propagation, we can define

wave motion as:

Transverse if the direction of displacement is

perpendicularto the direction of propagation

Longitudinal if the direction of displacement is

parallel to the direction of propagation

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Types of waves

Depending on the direction of the displacementrelative to the direction of propagation, we can define

wave motion as:

Transverse if the direction of displacement is

perpendicularto the direction of propagation

Longitudinal if the direction of displacement is

parallel to the direction of propagation

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The linear wave equation

Let us consider transverse waves propagatingwithout change in shape and with a constant wave

velocityv

We will describe waves via vertical displacement

y(x,t)

For an observermoving with the wave

the wave shape doesnt depend on timey(x) = f(x)

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For an observerat rest:

the wave shape depends on timey(x,t)

the reference frame linked to the wave is moving

with the velocity of the wave v

vtxx += ' vtxx ='

)()'( vtxfxf = )(),( vtxftxy =

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We considered a wave propagating with velocity v

For a medium with isotropic (symmetric) properties,

the wave equation should have a symmetric solution

for a wave propagating with velocityv

)(),(1 vtxftxy =

))((),(2 tvxftxy =

)( vtxf +=

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Therefore, solutions of the wave equation shouldhave a form

Considering partial derivatives

)(),( vtxftxy =

x

vtxf

x

txy

=

)(),(

x

vtx

vtx

vtxf

=

)(

)(

)()(' vtxf =

t

vtxf

t

txy

=

)(),(

t

vtx

vtx

vtxf

= )(

)(

)()()(' vvtxf =

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Therefore, solutions of the wave equation shouldhave a form


)(),( vtxftxy =

=

x

vtxf

xx

txy )(),(2

2

( ))(' vtxfx

= )('' vtxf =

= t

vtxftt

txy )(),(2

2

( ))()(' vvtxft

=

2)('' vvtxf =

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Therefore, solutions of the wave equation should

have a form


)(),( vtxftxy =

)(''),(

2

2

vtxfx

txy=

2

2

2

)(''),(

vvtxft

txy=

2

22 ),(

x

txyv

=

2

22

2

2 ),(),(

x

txyv

t

txy

=

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The linear wave equation (not the only one having

solutions of the formy(x,t) = f(x vt)):

It works forlongitudinal waves as well

v is a constant and is determined by the propertiesof the medium. E.g., for a stretched string with linear

density = m/lunder tension T

Tv =

2

22

2

2 ),(),(

x

txyv

t

txy

=

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Superposition of waves

Let us consider two different solutions of the linear

wave equation

Superposition principle a sum of two solutions of

the linear wave equation is a solution of the linear

wave equation

2

1

22

2

1

2

x

yv

t

y

=

2

2

22

2

2

2

x

yv

t

y

=

2

2

22

2

1

22

2

2

2

2

1

2

x

yv

x

yv

t

y

t

y

+

=

+

2

21

22

2

21

2 )()(

x

yyv

t

yy

+

=++

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Superposition of waves

Overlapping solutions of the linear wave equation

algebraically add to produce a resultant (net) wave

Overlapping solutions of the linear wave equation

do not in any way alter the travel of each other

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Chapter 16

Problem 44(a) Show that the function y(x, t) = x2 + v2t2 is a solution to the wave equation.

(b) Show that the function in part (a) can be written as f(x + vt) + g(x vt) anddetermine the functional forms for f and g. (c) Repeat parts (a) and (b) for the

function y(x, t) = sin (x) cos (vt).

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Reflection of waves at boundaries

Within media with boundaries, solutions to the wave

equation should satisfy boundary conditions. As a

results, waves may be reflected from boundaries

Hard reflection a fixed zero value of deformation at

the boundary a reflected wave is inverted

Soft reflection a free value of deformation at the

boundary a reflected wave is not inverted

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Sinusoidal waves

One of the most characteristic solutions of the

linear wave equation is a sinusoidal wave:

A amplitude, phase constant

)2/)(cos(

))(sin()(

+=

+=

vtxkA

vtxkAvtxy

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Wavelength

Freezing the solution at t = 0 we obtain a

sinusoidal function ofx:

Wavelength smallest distance (parallel to thedirection of waves travel) between repetitions of the

wave shape

))(cos(),( += vtxkAtxy

)cos()0,( += kxAxy

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Wave number

On the other hand:

Angular wave number: k = 2 /

)cos()0,( +=

kxAxy ))(cos( ++=

xkA

)cos( ++= kkxA

)2cos()cos( ++=+ kxkx /2=k

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Angular frequency

Considering motion of the point atx = 0we observe a simple harmonic motion (oscillation) :

For simple harmonic motion (Chapter 15):

Angular frequency

))(cos(),( += vtxkAtxy

)cos(),0( += kvtAty )cos( += kvtA

)cos()( += tAty

/2 vkv ==

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Frequency, period

Definitions offrequency and period are the same as

for the case of rotational motion or simple harmonic

motion:

Therefore, for the wave velocity

2//1 == Tf /2=T

fTkv === //

)cos(),( += tkxAtxy

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Chapter 16

Problem 18A transverse sinusoidal wave on a string has a period T = 25.0 ms and travels

in the negative x direction with a speed of 30.0 m/s. At t = 0, an element of thestring at x = 0 has a transverse position of 2.00 cm and is traveling downward

with a speed of 2.00 m/s. (a) What is the amplitude of the wave? (b) What is the

initial phase angle? (c) What is the maximum transverse speed of an element of

the string? (d) Write the wave function for the wave.

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Interference of waves

Interference a phenomenon of combining waves,

which follows from the superposition principle

Considering two sinusoidal waves of the same

amplitude, wavelength, and direction of propagation

The resultant wave:

)cos(),(2 += tkxAtxy)cos(),(1 tkxAtxy =

),(),(),( 21 txytxytxy +=

)cos()cos( ++= tkxAtkxA

+

=+ 2cos

2cos2coscos

)2/cos()2/cos(2 += tkxA

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If = 0 (Fully constructive)

If = (Fully destructive)

If = 2/3 (Intermediate)

)2/cos()2/cos(2),( += tkxAtxy

)cos(2),( tkxAtxy =

0),( =txy

)3/cos(

)3/cos(2),(

+=

tkx

Atxy

)3/cos( += tkxA

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Considering two sinusoidal waves of the same

amplitude, wavelength, but running in opposite

directions

The resultant wave:

)cos(),(2 ++= tkxAtxy)cos(),(1 tkxAtxy =

),(),(),( 21 txytxytxy +=

)cos()cos( +++= tkxAtkxA

+

=+ 2cos

2cos2coscos

)2/cos()2/cos(2 += tkxA

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If two sinusoidal waves of the same amplitude and

wavelength travel in opposite directions, their

interference with each other produces a standing

wave

)sin()sin(2),( kxtAtxy ==

...2,1,0

)21(

=+=

n

nkx

22

1

+= nx

Antinodes

1|sin| =kx

tAy sin2=

...2,1,0==

n

nkx

0sin

=kx

0=y

2

nx =

Nodes

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Chapter 18

Problem 25A standing wave pattern is observed in a thin wire with a length of 3.00 m. The

wave function is y = (0.002 m) sin (x) cos (100t), where x is in meters and t isin seconds. (a) How many loops does this pattern exhibit? (b) What is the

fundamental frequency of vibration of the wire? (c) If the original frequency is

held constant and the tension in the wire is increased by a factor of 9, how

many loops are present in the new pattern?

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Standing waves and resonance

For a medium with fixed boundaries (hard reflection)

standing waves can be generated because of the

reflection from both boundaries: resonance

Depending on the number of antinodes, different

resonances can occur

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Standing waves and resonance

Resonance wavelengths

Resonance frequencies

L2

=

2

2L=

3

2L=

...3,2,1,

2==

nn

L

vf= ...3,2,1,

2

== n

L

nv

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Harmonic series

Harmonic series collection of all possible modes -

resonant oscillations (n harmonic number)

First harmonic (fundamental mode):

...3,2,1,2

== nL

vnfn

L

v

f 21 =

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More about standing waves

Longitudinal standing waves can also be produced

Standing waves can be produced in 2 and 3

dimensions as well

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More about standing waves

Longitudinal standing waves can also be produced

Standing waves can be produced in 2 and 3

dimensions as well

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Rate of energy transmission

As the wave travels it transports energy, even

though the particles of the medium dont propagate

with the wave

The average powerof energy transmission for the

sinusoidal solution of the wave equation

Exact expression depends on the medium or the

system through which the wave is propagating

vAPavg22

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Sound waves

Sound longitudinal waves in a substance (air,

water, metal, etc.) with frequencies detectable by

human ears (between ~ 20 Hz and ~ 20 KHz)

Ultrasound longitudinal waves in a substance (air,

water, metal, etc.) with frequencies higher thandetectable by human ears (> 20 KHz)

Infrasound longitudinal waves in a substance (air,

water, metal, etc.) with frequencies lower thandetectable by human ears (< 20 Hz)

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Speed of sound

Speed of sound:

density of a medium,B bulk modulus of a

medium

Traveling sound waves

Bv =

V

VBP

=

)cos(

))(cos(),(

tkxs

vtxkstxs

m

m

=

==

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Chapter 17

Problem 12As a certain sound wave travels through the air, it produces pressure

variations (above and below atmospheric pressure) given by P = 1.27 sin (x 340t) in SI units. Find (a) the amplitude of the pressure variations, (b) the

frequency, (c) the wavelength in air, and (d) the speed of the sound wave.

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Intensity of sound

Intensity of sound average rate of sound energy

transmission per unit area

For a sinusoidal traveling wave:

Decibel scale

sound level;I0 = 10-12 W/m2 lower limit of human

hearing

AI

P=

22

2

1 mvsI=

0

log)10(

I

IdB=

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Sources of musical sound

Music produced by musical instruments is a

combination of sound waves with frequencies

corresponding to a superposition of harmonics

(resonances) of those musical instruments

In a musical instrument, energy of resonant

oscillations is transferred to a resonator of a fixed or

adjustable geometry

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Open pipe resonance

In an open pipe soft reflection of the waves at the

ends of the pipe (less effective than form the closed

ends) produces standing waves

Fundamental mode (first harmonic): n = 1

Higher harmonics:

...3,2,1

2,

2

=

==

n

L

vnf

n

L

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Organ pipes

Organ pipes are open on one end and closed on the

other

For such pipes the resonance condition is modified:

L

vnf

n

L

nnL

4,

4

...5,3,1;

4==

==

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Musical instruments

The size of the musical instrument reflects the range

of frequencies over which the instrument is designed

to function

Smallersize implies higherfrequencies, largersize

implies lowerfrequencies

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Musical instruments

Resonances in musical instruments are not

necessarily 1D, and often involve different parts of

the instrument

Guitarresonances (exaggerated) at low frequencies:

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Musical instruments



the instrument

Guitarresonances at medium frequencies:

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Musical instruments



the instrument

Guitarresonances at high frequencies:

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Beats

Beats interference of two waves with close

frequencies

tss m 11 cos=

tss m 22 cos=

+ tstssss mm 2121 coscos +=+=

+

= ttsm

2cos

2cos2 2121

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Sound from a point source

Point source source with size negligible compared

to the wavelength

Point sources produce spherical waves

Wavefronts surfaces over which oscillations havethe same value

Rays lines perpendicular to wavefronts indicating

direction of travel of wavefronts

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Interference of sound waves

Far from the point source wavefronts can be

approximated as planes planar waves

Phase difference and path length difference are

related:

Fully constructive interference

Fully destructive interference

2212

LLL =

=

,...2,1,0=

L

,...

2

5,

2

3,

2

1=

L

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Variation of intensity with distance

A single point emits sound isotropically with equal

intensity in all directions (mechanical energy of the

sound wave is conserved)

All the energy emitted by the source must pass

through the surface of imaginary sphere of radius r

Sound intensity

(inverse square law)

API= 24 r

Ps

=

C 1

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Chapter 17

Problem 26Two small speakers emit sound waves of different frequencies equally in all

directions. Speaker A has an output of 1.00 mW, and speaker B has an outputof 1.50 mW. Determine the sound level (in decibels) at point C assuming (a)

only speaker A emits sound, (b) only speaker B emits sound, and (c) both

speakers emit sound.

D l ff t

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Doppler effect

Doppler effect change in the frequency due to

relative motion of a source and an observer (detector)

Andreas Christian

Johann Doppler(1803 -1853)

D l ff t

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Doppler effect

For a moving detector(ear) and a stationary source

In the source (stationary) reference frame:

Speed of detector isvD

Speed of sound waves is v

In the detector(moving) reference frame:

Speed of detector is 0

Speed of sound waves is v + vD

fv =

vf=

''

vf=

Dvv +=

f

v=

v

vvf D

+=

D l ff t

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Doppler effect

For a moving detector(ear) and a stationary source

If the detector is moving away from the source:

For both cases:

v

vvff D

+='

v

vvff D

='

v

vvff D

='

D l ff t

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Doppler effect

For a stationarydetector(ear) and a moving source

In the detector(stationary) reference frame:

In the moving (source) frame:

*

'

vf=

*Svvf =

fvv S=*

Svv

vf

=

D l ff t

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Doppler effect

For a stationarydetectorand a moving source

If the source is moving away from the detector:

For both cases:

Svv

vff

='

Svv

vff

+='

Svv

vff

='

D l ff t

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Doppler effect

For a movingdetectorand a moving source

Doppler radar:

S

D

vv

vvff

='

Ch t 17

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Chapter 17

Problem 37A tuning fork vibrating at 512 Hz falls from rest and accelerates at 9.80 m/s2.

How far below the point of release is the tuning fork when waves of frequency485 Hz reach the release point? Take the speed of sound in air to be 340 m/s.

S i d

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Supersonic speeds

For a source moving faster than the speed of sound

the wavefronts form the Mach cone

Mach number

Ernst Mach

(1838-1916)v

vs

vt

tvs=

sin

1=

S personic speeds

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Supersonic speeds

The Mach cone produces a sonic boom

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Questions?

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Answers to the even-numbered problems

Chapter 16

Problem 80.800 m/s

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Chapter 17

Problem 16

(a) 5.00 10-17 W

(b) 5.00 10-5 W

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Chapter 17

Problem 40

46.4

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Answers to the even-numbered

problems

Chapter 17

Problem 54

The gap between bat and insect is

closing at 1.69 m/s.

chapter 16-18.ppt

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