Chapter 16 Waves I

In this chapter we will start the discussion on wave phenomena. We will study the following topics:

Types of wavesAmplitude, phase, frequency, period, propagation speed of a waveMechanical waves propagating along a stretched string Wave equation Principle of superposition of waves Wave interference Standing waves, resonance

(16 – 1)

A wave is defined as a disturbance that is self-sustained and propagates in space with a constant speed

Waves can be classified in the following three categories:

1. Mechanical waves. These involve motions that are governed by Newton’s laws and can exist only within a material medium such as air, water, rock, etc. Common examples are: sound waves, seismic waves, etc.2. Electromagnetic waves. These waves involve propagating disturbances in the electric and magnetic field governed by Maxwell’s equations. They do not require a material medium in which to propagate but they travel through vacuum. Common examples are: radio waves of all types, visible, infra-red, and ultra-violet light, x-rays, gamma rays. All electromagnetic waves propagate in vacuum with the same speed c = 300,000 km/s3. Matter waves. All microscopic particles such as electrons, protons, neutrons, atoms etc have a wave associated with them governed by Schroedinger’s equation. (16 – 2)

Waves can be divided into the following two categoriesdepending on the orientation of the disturbance with respect to the wave propagation velocity .

If the disturba

v

Transverse and Longitudinal waves

r

nce associated with a particular wave isperpendicular to the wave propagation velocity, this wave is called " ". An example is given in the upper figure which depicts a mechanical wave that

transverse

propagates along a string. The movement of each particle on the string is along the -axis; the wave itselfpropagates along the -axis.

yx

A wave in which the associated disturbance is parallel to the wave propagationvelocity is known as a " ". An example of such a wave is given in the lower figure. It is produced by a p

lonitudinal waveiston oscillating in a tube filled with

air. The resulting wave involves movement of the air molecules along the axis of the tube which is also the direction of the wave propagation velocity .vr

(16 – 3)

Consider the transverse wave propagating along the string as shown in the figure. The position ofany point on the string can be described by a function ( , ). Further along the chapterwe shall s

y h x t=

( )

ee that function has to have a specificform to describe a wave. Once such suitable function is: ( , ) sin - Such

m

h

y x t y kx tω=

a wave which is described by a sine (or a cosine) function is known as" ".The various terms that appear in the expression for a harmonic waveare identified in the lower figureFunction (y x

harmonic wave

( )

, ) depends on and . There are two ways to visualize it. The first is to "freeze" time (i.e. set ). This is like taking a snapshot of thewave at . , The second is to set

.

o

o o

o

t x t

t tt t y y x t

x x

=

= =

= ( )In this case , .oy y x t=

(16 – 4)

( )

( )

The is the absolute value of themaximum displacement from the equilibriumposition. The is defined as the argument of the sine function.The

( , ) sin

is the

m

my

kx

y x t y kx t

t

ω

ω

λ

−

−

=

amplitude

phase

wavelength shortest distance between two repetitions of the wave at a fixed time.

( ) ( ) ( )1 1

1 1 1

We fix at 0. We have the condition: ( ,0) ( ,0)

sin sin sin

Since the sine function is periodic with period 2 2

A is the time it takes (with ed

2

fix

m m m

t t y x y x

y kx y k x y kx k

k

T

k

λ

λ λ

π λ πλ

π

= = + →

= + = +

→ = → =

period

( ) ( ) ( )

) tthe sine function to complete one oscillation. We take 0 (0, ) (0, )

sin sin sin 2 2m m m

xx y t y t T

y t y t T y t T TTπωω ω ω ω ω π

= → = + →

− = − + = − + → = → =

(16 – 5)

In the figure we show two snapshots of a harmonic wavetaken at times and . During the time interval the wave has traveled a distance . The wave speed

. On

t t t tx

xvt

+ ∆ ∆∆

∆=∆

The speed of a traveling wave

e method of finding is to imagine that

we move with the same speed along the -axis. In thiscase the wave will seem to us that it does not change.

v

x

( )Since ( , ) sin this means that the argument of the sine functionis constant. constant. We take the derivative with respect to .

0 The speed

A harmonic

my x t y kx tkx t t

dx dx dxk vdt dt k dt k

ωω

ω ωω

= −

− =

− = → = = =

( ) ( ) wave that propagates along the -axis is described by the equation:

( , ) sin . The function ( , ) describes a general wave that propagates along the positive -axis. A gene

m

xy x t y kx t y x t h kx t

xω ω= + = −

negative

( )ral wave that propagates along the

-axis is described by the equatio ( , )n: y x t hx kx tω= +negative

vkω

=(16 – 6)

Below we will determine the speed of a wave thatpropagates along a string whose linear mass densityis . The tension on the string is equal to . Consider a small secµ τ

Wave speed on a stretched string

tion of the string of length . ∆l

( )The shape of the element can be approximated to be an arc of a circle of radius Rwhose center is at O. The net force in the direction of O is 2 sin .

Here we assume that 1 sin 2

F

FR R

τ θ

θ θ θ τ

=

∆ ∆→ = → = (l l= ;

( )

( )

2 2

2

The force is also given by Newton's second law:

If we compare equations 1 and 2 we get:

The speed depends on the tension and the mass densit

v vF mR R

vvR R

v

τ

µ

µ τ

τµ

=

=

∆ ∆

∆∆ = →

eqs.1)

= (eqs.2)

Note :

l

ll

y but not on the wave frequency .f

µ

v τµ

=

(16 – 7)

( )

( )

Consider a transverse wave propagating along a string which is described by the equation: ( , ) sin - . The transverse velocity

- cos - At point a both

m

m

y x t y kx tyu y kx tt

ω

ω ω

=

∂= =∂

Rate of energy transmission

and are equal to zero. At point b both and have maxima. y u y u

( )

2

2

2

1In general the kinetic energy of an element of mass is given by: 2

1 1 - cos - The rate at which konetic energy propagates 2 2

1along the string is equal to 2

m

dm dK dmv

dK dx y kx t

dK vdt

ω ω

µ ω

=

=

= ( )

( )

2 2

2 2 2 2 2

cos - The average rate

1 1cos - As in the case of the oscillating 2 4

spring-mass system

m

m mavgavg

avgavg avg avg a

y kx t

dK v y kx t v ydt

dU dK dU dKPdt dt dt dt

ω

µ ω ω µ ω = =

= → = +

2 212g

mv

v yµ ω=

(16 – 8)

Consider a string of mass density and tension A transverse wave propagates along the string. The transverse motion is described by ( , ) Consider an element of length and mas

y x tdx

µ τThe wave equation

( )

1 2

2 2 1 1 2 1

1 2 1 11

2 2

s The forces the net force along the y-axis is given by the equation:

sin sin sin sin Here we

assume that 1 and 1 sin tan

and sin tan

ynet

dm dxF F

F F F

yx

yx

µτ

θ θ τ θ θ

θ θ θ θ

θ θ

== =

= − = −

∂ → = ∂

∂= ∂

= = ;

;2 2 1

ynety yFx x

τ ∂ ∂ → = − ∂ ∂

θ2

2

22 1

2 2 2 22 1

2 2 2 2 22 1

From Newton's second law we have:

1

ynet yy y yF dma dxt x x

y yy y y y y yx xdxx x t dx t t v t

µ τ

µτ µτ

∂ ∂ ∂ = = = − → ∂ ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ ∂ ∂∂ ∂ − = → = = = ∂ ∂ ∂ ∂ ∂ ∂

2 2

2 2 2

1y yt v t

∂ ∂=

∂ ∂

(16 – 9)

2 2 2

2 2 2 2

1The wave equation even though

it was derived for a transverse wave propagating alonga string under tension, is true for all types of wa

y y yt t v t

µτ

∂ ∂ ∂= =

∂ ∂ ∂

The principle of superposition for waves

1

2

1 1 2 2 1 2

ves.This equation is "linear" which means that if and are solutions of the wave equation, the function

is also a solution. Here and are constants.The principle of superposition

yyc y c y c c+

is a direct consequenceof the linearity of the wave equation. This principle can be expressed as follows:

1 2

Consider two waves of the same type that overlap at some point P in space.Assume that the functions ( , ) and ( , ) describe the displacementsif the wave arrived at P alone. The displacement at P

y x t y x t

1 2

when both wavesare present is given by:

Overlapping waves do not in any way alter the travel of each oth( , ) ( , ) ( , )

ery x t y x t y x t′ = +

Note : (16 – 10)

1 2( , ) ( , ) ( , ) y x t y x t y x t′ = +

Consider two harmonic waves of the same amplitude and frequency which propagate along the x-axis. The two waves have a phase difference . Wewill combine these waves using the

φ

Interference of waves

1

principle of superposition. The phenomenon of combing waves is knwon as and the two waves are said to . The displacement of the two waves are given by the functions: ( , ) my x t y=

interferenceinterfere

( )( )

( ) ( ) ( )

( )

2 1 2

sin

and ( , ) sin .

, sin sin

The resulting wave has the same frequency asthe original waves, and i

, 2 cos sin

ts amplitude

2 2

2 c

m

m m

m

m

m

y x t y kx t

kx t

y x t y kx t y y y

y x t y kx t y t

y y

kx

ω

ω φ

ω ω φ

φ φω

−

′= − + = +

′ = −

′ = − +

+

=

−

′

+

Its phase is equalos 2

to 2

φ φ

(16 – 11)

( ) [ ]

The amplitude of two interefering waves is given by:

2 cos It has its maximum value if 02

In this case The displacement of the resul

2ting wave is:

, 2 s

m m

m m

m

y y

y y

y x t y

φ φ

′ =

′ = =

′ =

Constructive interference

in2

This phenomenon is known as

kx t φω − +

fully constructive interference

(16 – 12)

( )

The amplitude of two interefering waves is given by:

2 cos It has its minimum value if 2

In this case The displacement of the resulting wave is:

, 0

0

This

m

m my y

y x t

y

φ φ π

=

′ = =

′ =

′

Destructive interference

phenomenon is known as fully destructive interference

(16 – 13)

The amplitude of two interefering waves is given by:

2 cos When interference is neither fully2

constructive nor fully destructive it is called

An

m my y φ′ =

Intermediate interference

intermediate interference

( ) [ ]

2 example is given in the figure for 3

In this case The displacement of the resulting wave is:

, sin3

Sometimes the phase difference is expressed as a difference i

m

m m

y x t y kx t

y y

πφ

πω

=

′ = − +

′

=

Note :n wavelength

In this case re2

membre that:radians 1π λ

λ

↔

(16 – 14)

( ) ( )1 1

A phasor is a method for representing a wave whose diasplacement is: , sinThe phasor is defined as a vector with the followingproperties:

Its magnitude is equal to the wave's am

my x t y kx tω= −

Phasors

1. 1plitude The phasor has its tail at the origin O and rotates in

the clockwise direction about an axis through Owith angular speed .Thus defined, the projection of the phasor on the -axis (i.e.

my

yω

2.

( )

( ) ( )

1

2 2

its y-component) is equal to sinA phasor diagram can be used to represent more than one waves. (see fig.b). The displacement of the second wave is: , sin The phasor of the

m

m

y kx t

y x t y kx t

ω

ω φ

−

= − + secondwave forms an angle with the phasor of the first wave indicating that it lags behind wave 1 by a phase angle .

φφ

(16 – 15)

( )1 1

Consider two waves that have the same frequencybut different amplitudes. They also have a phase difference . The displacements of the two wavesare: , sinmy x t y kx

φω= −

Wave addition using phasors

( )( ) ( )

( )

2 2

and

, sin . The superposition of the twowaves yields a wave that has the sameangular frequency and is described by: sin Here is thewave amplitude and is the phas

m

m m

t

y x t y kx t

y y kx t y

ω φω

ω ββ

= − +

′ ′ ′= − +

e angle. To determine and we add the two phasors representing the waves as vectors (see fig.c). Note: The phasor mathod can be used to add vectors that have different amplitudes.

my β′

(16 – 16)

( ) ( ) ( )1 2

Consider the superposition of two waves that have the samefrequency and amplitude but travel in opposite directions. The displacements of two waves are: , sin , ,my x t y kx t y x t yω= − =

Standing Waves :

( )( ) ( ) ( )

( ) ( ) ( ) [ ]1 2

sin

The displacement of the resulting wave , , ,

, sin sin 2 sin cosThis is not a traveling wave but an oscillation that has a position dependent amplitude. It i

m

m m m

kx t

y x t y x t y x t

y x t y kx t y kx t y kx t

ω

ω ω ω

+

′ = +

′ = − + + =

s known as a standing wave.

( ) [ ], 2 sin cosmy x t y kx tω′ =

(16 – 17)

n n n

a

a

a

n

( ) [ ]The displacement of a standing wave is given by the equation

, 2 sin cosThe position dependant amplitude is equal to

These are defined as positions

2 si

where the stand

:

ing wa

n

ve a

m

my kxy x t y kx tω′ =

Nodes : mplitude vanishes. They occur when 0,1, 2,

2 0,1, 2,...

These are defined as positions where the standing wave amplitude is maximum.

1They occur when

2

2

n

kx n n

x n n nx

kx n

λπ

π πλ

= =

→ = → =

= +

=

Antinodes :

0,1,2,...

2 1 0,1, 2,...2

The distance between ajacent nodes and antinodes is /2 The

12 2

distance between a node and an ajacent antinode is /4

n

n

x n x nn

π

π π

λ

λλ

λ

=

→ = + ′ = +→ =

Note 1 : Note 2 :

(16 – 18)

Consider a string under tension on which is clamped at points A and B separated by a sistance . We senda harmonic wave traveling the thr right. the wave isreflected at po

L

Standing waves and resonance

int B and the reflected wave travels to the left. The left going wave reflects back at point Aand creates a thrird wave traveling to the right. Thus we have a large number of overlapping waves half of which travel to the right and the rest to the left.

A

A

A

B

B

B

For certain frequencies the interference produces a standing wave. Such astanding wave is said to be . The frequencis at which the standingwave occurs are known as the

at resonanceresonant frequencie of the system. s

(16 – 19)

A

A

A

B

B

B

Resonances occur when the resulting standing wavesatisfies the boundary condition of the problem. These are that the Amplitude must be zero at point Aand point B and arise from the fact that the strin

1

1

g isclamped at both points and therefore cannot move. The first resonance is shown in fig.a. The standing

wave has two nodes at points A and B. Thus 2

. The second standing wave is sh2 own

L

L

λ

λ =

=

→

2

in fig.b. It has three nodes (two of them at A and B)

In this case 2 2

L Lλ λλ = = →

=

3

The third standing wave is shown in fig.c. It has four nodes (two of them at A and B)

In this case 3 The general expression for the resonant2

2 wavelen

2

gths is: 1, 2,3,

3

..

n

LL

L nn

λ

λ

λλ = =

=

=→

= . the resonant frequencies 2n

n

v vf nLλ

= =

(16 – 20)