PH 221-1D Spring 2013

Waves - I

Lectures 38-40

Chapter 16 (Halliday/Resnick/Walker, Fundamentals of Physics 9th edition)

1

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 wave Mechanical waves propagating along a stretched string Wave equation Principle of superposition of waves Wave interference Standing waves, resonance

2

Waves

A wave is a vibrational, trembling motion in an elastic, deformable body.

•The wave is initiated by some external force that acts on some parts of the body and deforms it.• The elastic restoring force communicates this initial disturbance from one part of the body to the next, adjacent part.• The disturbance gradually spreads outward through the entire elastic body.

The elastic body in which the wave propagate is called the medium.

When a wave propagates through a medium, the particles in the medium vibrate back and forth, but the medium as a whole does not perform translational motion.

3

Transverse Wave Motion

If we shake one end of the tightly stretched elastic string up and down with a flick of the wrist a disturbance travels along the string.

String is regarded as a row of particles connected by small, massless springs.

1 particle up => it will pull the 2nd

particle up.

Transverse wave motion – the direction of the disturbance propagation is perpendicular to the back and forth vibration of the particles.

1 up

2nd up

1 down

2 down

4

Longitudinal Wave Motion

Disturbance generated by pushing the 1st particle towards the 2nd

Longitudinal wave - direction of disturbance motion is parallel to the vibration of the particles.

The wave is a propagation of the disturbance through the medium without any net displacement of the medium.Consider a horizontal string connected to the mass executing simple harmonic motion in the vertical direction.

Each particle transmits the motion to the next particle along the entire length of the string. The resulting wave propagates in the horizontal direction with a velocity “v”, while any one particle of the string executes simple harmonic motion in the vertical direction.

The particle of the string is moving perpendicular to the direction of wave propagation and is not moving in the direction of the wave.

A transverse wave is a wave in which the particles of the medium execute simple harmonic motion in a direction perpendicular to its direction of propagation.

6

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. 7

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

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

longitudinal wavepiston 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 .v8

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 t

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 wave are identified in the lower figureFunction (y

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

x t x t

t tt t y y x t

x x

In this case , .oy y x t 9

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

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

period

) the 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 TT

y t T T

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

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

x

negative

ral wave that propagates along the

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

vk

11

A particular wave is given by y = (0.200m)sin[(0.500m-1)x – (8.20rad/s)t]Find:a) A e) f h) y at x = 10.0m and t = 0.500sb) K f) tc) λ g) v

12

Problem 5. If ( , ) (6.0 )sin( (600 / ) ) describes a wave travelingalong a string, how much time does any given point on the string take to move between displacements y=+2.0 mm and y=-2.00mm?

y x t mm kx rad s t

Let y1 = 2.0 mm (corresponding to time t1) and y2 = –2.0 mm (corresponding to time t2). Then we find

kx + 600t1 + = sin1(2.0/6.0) and

kx + 600t2 + = sin1(–2.0/6.0) .

Subtracting equations gives

600(t1 – t2) = sin1(2.0/6.0) – sin1(–2.0/6.0). Thus we find t1 – t2 = 0.011 s (or 1.1 ms).

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Problem 9. A transverse sinusoidal wave is moving along a string in the positive directionof an x axis with a speed of 80 m/s. At t=0, the string particle at x=0 has a tranverse displacement of 4.0 cm from its equilibrium position and is not moving. The maximum transverse speed of the string particle at x=0 is 16 m/s. (a) What is the frequency of the wave?(b) What is the wavelength of the wave? If

m

the wave equation is of the form ( , ) sin( )what are (c) y ,( ) , ( ) , (f) , and (g) the correct choice of sign in front of ?

my x t y kx td k e

(a) Recalling from Ch. 12 the simple harmonic motion relation um = ym, we have

1 6 4 0 0 r a d / s .0 .0 4 0

Since = 2f, we obtain f = 64 Hz. (b) Using v = f, we find = 80/64 = 1.26 m 1 .3 m . (c) The amplitude of the transverse displacement is 24 .0 c m 4 .0 1 0 m .my (d) The wave number is k = 2/ = 5.0 rad/m. (e) The angular frequency, as obtained in part (a), is 21 6 / 0 .0 4 0 4 .0 1 0 r a d / s . (f) The function describing the wave can be written as

0 .0 4 0 s i n 5 4 0 0y x t where distances are in meters and time is in seconds. We adjust the phase constant to satisfy the condition y = 0.040 at x = t = 0. Therefore, sin = 1, for which the “simplest”root is = /2. Consequently, the answer is

0 .0 4 0 s i n 5 4 0 0 .2

y x t

(g) The sign in front of is minus.

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1 1 0.5

Problem 29. Use the wave equation to find the speed of a wave given by ( , ) (2.00 )[(20 ) (4.0 ) ] .y x t mm m x s t

The wave 1 1 1/2( , ) (2.00 mm)[(20 m ) (4.0 s ) ]y x t x t is of the form ( )h kx t with angular wave number 120 mk and angular frequency 4.0 rad/s . Thus, the speed of the wave is

1/ (4.0 rad/s)/(20 m ) 0.20 m/s.v k

15

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 sec

Wave speed on a stretched string

tion of the string of length .

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

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 densi

v

v vF mR R

vR R

v

(eqs.1)

= (eqs.2)

Note :

ty but not on the wave frequency .f

v

16

Problem : The drawing shows a 15.0 kg ball being whirled in a circular path on the end of a string. The motion occurs on a frictionless horizontal table. The angular speed of the ball is ω = 12.0 rad/s. The string has a mass of 0.0230 kg. How much time does it take for a wave on the string to travel from the center of the circle to the ball?

17

The Speed of a Wave on a StringProblem To measure the acceleration due to gravity on a distant planet, an astronaut hangs a 0.086 kg ball from the end of a wire. The wire has length of 1.5 m and a linear density of 3.1 x 10-4 kg/m. Using electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of 0.083 s. The mass of the wire is negligible compared to the mass of the ball. Determine the acceleration due to gravity.

18

The Speed of a Wave on a StringProblem A horizontal wire is under a tension of 315 N and has a mass per unit length of 6.50 x 10-3

kg/m. A transverse wave with an amplitude of 2.50 mm and a frequency of 585 Hz is traveling on this wire. As the wave passes, a particle of the wire moves up and down in simple harmonic motion. Obtain:a) The speed of the waveb) The maximum speed with which the particle moves up and down

19

Problem 23. A 100 g wire is held under a tension of 250 N with one end at x=0 and the other at x=10.0m. At time t=0, pulse 1 is sent along the wire from the end at x=10.0 m.At time t=30.0 ms, pulse 2 is sent along the wire from the end at x=0. At what position xdo the pulses begin to meet?

The pulses have the same speed v. Suppose one pulse starts from the left end of the wireat time t = 0. Its coordinate at time t is x1 = vt. The other pulse starts from the right end, atx = L, where L is the length of the wire, at time t = 30 ms. If this time is denoted by t0then the coordinate of this wave at time t is x2 = L – v(t – t0). They meet when x1 = x2, or, what is the same, when vt = L – v(t – t0). We solve for the time they meet: t = (L + vt0)/2vand the coordinate of the meeting point is x = vt = (L + vt0)/2. Now, we calculate the wavespeed:

(250 N) (10.0 m) 158m/s.0.100 kg

Lv

m

Here is the tension in the wire and m/L is the linear mass density of the wire. Thecoordinate of the meeting point is

310.0 m (158m/s) (30.0 10 s) 7.37 m.2

x

This is the distance from the left end of the wire. The distance from the right end is L – x= (10.0 m – 7.37 m ) = 2.63 m. 20

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

Rate of energy transmission

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

2

2

2

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

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

1along the string is equal to 2

m

dm dK dmv

dK dx y kx t

dK v ydt

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 av

kx t

dK v y kx t v ydt

dU dK dU dKPdt dt dt dt

2 212 m

g

v y

Problem 26. A string along which waves can travel is 2.70 m long and has a mass of 260 g. The tension in the string is 36.0 N. What must be the frequency of traveling waves of amplitude 7.70 mm for the average power to be 85.0 W?

avg3

21 1 2(85.0 W) 198 Hz.2 2 (7.70 10 m)/ (36.0 N) (0.260 kg / 2.70 m )m

Pf

y

Using Eq. 16–33 for the average power and Eq. 16–26 for the speed of the wave, we solve for f = /2:

2 212avg myP v

v

22

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

The 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

2 2 1

ynety yFx x

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 dx

t x x

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

2 2

2 2 2

1y yx v t

23

The Superposition of Waves

A superposition principle: when two or more waves arrive at any given point simultaneously, the resultant instantaneous deformation is the sum of the individual instantaneous deformations.

• The waves do not interact, they have no effect on one another• Each wave propagates as though the others were not present• The net displacement of the medium is the vector sum of the individual displacements

The superposition principle is very well satisfied for waves of low amplitude.

For waves of very large amplitude the superposition principle fails, because the first wave alters the properties of the medium and therefore affects the behavior of a second wave propagating on the same string, air, etc. (medium)

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 yx t v t

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

yy

c y c y c c is a direct consequence

of 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 :

1 2( , ) ( , ) ( , ) y x t y x t y x t

25

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

Interference of waves

1

principle of superposition. The phenomenon of combing waves is known 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

Its phase is equalos 2

to 2

26

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

Constructive interference

in2

This phenomenon is known as

kx t

fully constructive interference

27

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

Destructive interference

phenomenon is known as fully destructive interference

28

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

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

Note :n wavelength

In this case re2

member that:radians 1

29

Pr oblem 32. What phase difference between two identical traveling waves moving in the samedirection along a stretched string, results in the combined wave having an amplitude 1.50 timesthat of the common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, (c) wavelengths?

(a) Let the phase difference be . Then from Eq. 16–52, 2ym cos(/2) = 1.50ym, which gives

1 1.502cos 82.8 .2

m

m

yy

(b) Converting to radians, we have = 1.45 rad. (c) In terms of wavelength (the length of each cycle, where each cycle corresponds to 2rad), this is equivalent to 1.45 rad/2 = 0.230 wavelength.

2 cos2m my y

30

1 1

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

Its magnitude is equal to the wave's amp

my x t y kx t

Phasors

1. 1litude 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. i

my

y

2.

1

2 2

ts 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

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

31

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

Wave addition using phasors

2 2

and

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

m

m m

t

y x t y kx t

y y kx t y

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

my

32

m1 2 1 2

Problem 35. Two sinusoidal waves of the same frequency travel in the same direction along a string. If y 3.0 , 4.0 cm, 0, and / 2 , what is the amlitude of the resultant wave?mcm y rad

33

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 y

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

s known as a standing wave.

, 2 sin cosmy x t y kx t

34

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 t

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

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

Note 1 : Note 2 :

Consider a string under tension on which is clamped at points A and B separated by a distance . We senda harmonic wave traveling to the right. The wave isreflected at poi

L

Standing waves and resonance

nt 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

36

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

2

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

In this case 2 2

L L

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

. the resonant frequencies 2n

n

v vf nL

Problem 43. A string fixed at both ends is 8.40 m long and has a mass of 0.120 kg. It is subjected to a tension of 96.0 N and set oscillating. (a) What is the speed of the waves on the string? (b) What is the longest possible wavelength for a standing wave? (c) Find the frequency of the wave.

(a) The wave speed is given by ,v where is the tension in the string and is the linear mass density of the string. Since the mass density is the mass per unit length, = M/L, where M is the mass of the string and L is its length. Thus

(96.0 N) (8.40 m) 82.0 m/s.0.120 kg

LvM

(b) The longest possible wavelength for a standing wave is related to the length of thestring by L = /2, so = 2L = 2(8.40 m) = 16.8 m. (c) The frequency is f = v/ = (82.0 m/s)/(16.8 m) = 4.88 Hz.

38

Problem 50. A rope, under a tension of 200N and fixed at both ends, oscillates in a second harmonic standing wave pattern. The displacement of the rope is given by (0.10 )(sin / 2)sin12 , where x=0

y m x t at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope,

(b) the speed of the wave on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third harmonic standing wave pattern, what will be the period of oscillation?

Since the rope is fixed at both ends, then the phrase “second-harmonic standing wave pattern” describes the oscillation shown in Figure 16–23(b), where (see Eq. 16–65)

a n d = vL fL

.

(a) Comparing the given function with Eq. 16-60, we obtain k = /2 and = 12 rad/s. Since k = 2/ then

2 4 .0 m 4 .0 m .2

L

(b) Since = 2f then 2 1 2 r a d / s ,f which yields

6 .0 H z 2 4 m / s .f v f (c) Using Eq. 16–26, we have

2 0 0 N 2 4 m /s/ ( 4 .0 m )

vm

which leads to m = 1.4 kg. (d) With

3 3 ( 2 4 m /s ) 9 .0 H z2 2 ( 4 .0 m )

vfL

The period is T = 1/f = 0.11 s.

, 2 sin cosmy x t y kx t

39