16/Maio/2018 – Aula 19

21/Maio/2018 – Aula 20 20 Doppler e Interferência 20.1 Efeito Doppler 20.2 Batimentos (interferência no tempo) 20.3 Interferência (no espaço) 20.4 Difração

19 Ondas estacionárias (OE) 19.1 Sobreposição de ondas 19.2 OE em cordas 19.3 OE em tubos 19.4 Potência e energia

1

2

19.1 Sobreposição de ondas

Sobreposição de duas ondas

simulação

Quando uma onda encontra uma descontinuidade no meio onde se propaga, pode ser refletida. Se não puder oscilar nesse ponto (zero de amplitude), a onda é refletida como o “simétrico” da onda incidente.

A onda refletida tem a mesma velocidade e o mesmo comprimento de onda da onda incidente.

Aula anterior

3

19.1 Sobreposição de ondas

As ondas estacionárias resultam da soma da onda incidente com a onda refletida:

Como

ψ x,t( ) = Asen kx −ωt( )+ Asen kx +ωt( )

sena+ senb = 2sen a+b2

⎛

⎝⎜

⎞

⎠⎟cos

a−b2

⎛

⎝⎜

⎞

⎠⎟

ψ x,t( ) = 2Asen kx( )cos ωt( )

Aula anterior

4

Como

f = vλ

⇒

sen k L( ) = 0 k L =mπ

k= 2πλ

λm =2Lm, m =1,2,3,...

Nas extremidades de uma corda, de comprimento L:

Dado que a frequência está relacionada com o comprimento de onda, só ondas com certas frequências podem existir numa corda:

fm =mv2L, m =1, 2, 3, 4,!

Aula anterior 19.2 Ondas estacionárias em cordas

5

1.  O número de máximos de amplitude (antinodos) é igual a m.

2.  O número de zeros de amplitude (nodos) é igual a m + 1.

3.  A frequência fundamental (m = 1) tem comprimento de onda λ1 = 2L.

4.  Numa corda de densidade µ, esticada por uma força de tensão FT, a velocidade v é dada por

v =FTµ

19.2 Ondas estacionárias em cordas

Corda vib1 simulação

Aula anterior

16.4 Standing Sound Waves and Normal Modes 523

be used to create sound waves in the surrounding air. This is the operating princi-ple of the human voice as well as many musical instruments, including wood-winds, brasses, and pipe organs.

Transverse waves on a string, including standing waves, are usually describedonly in terms of the displacement of the string. But, as we have seen, sound wavesin a fluid may be described either in terms of the displacement of the fluid or interms of the pressure variation in the fluid. To avoid confusion, we’ll use the termsdisplacement node and displacement antinode to refer to points where particlesof the fluid have zero displacement and maximum displacement, respectively.

We can demonstrate standing sound waves in a column of gas using an apparatuscalled a Kundt’s tube (Fig. 16.12). A horizontal glass tube a meter or so long isclosed at one end and has a flexible diaphragm at the other end that can transmitvibrations. A nearby loudspeaker is driven by an audio oscillator and amplifier; thisproduces sound waves that force the diaphragm to vibrate sinusoidally with a fre-quency that we can vary. The sound waves within the tube are reflected at the other,closed end of the tube. We spread a small amount of light powder uniformly alongthe bottom of the tube. As we vary the frequency of the sound, we pass through fre-quencies at which the amplitude of the standing waves becomes large enough for thepowder to be swept along the tube at those points where the gas is in motion. Thepowder therefore collects at the displacement nodes (where the gas is not moving).Adjacent nodes are separated by a distance equal to and we can measure thisdistance. Given the wavelength, we can use this experiment to determine the wavespeed: We read the frequency from the oscillator dial, and we can then calculatethe speed of the waves from the relationship

Figure 16.13 shows the motions of nine different particles within a gas-filledtube in which there is a standing sound wave. A particle at a displacement node

does not move, while a particle at a displacement antinode oscillates withmaximum amplitude. Note that particles on opposite sides of a displacement nodevibrate in opposite phase. When these particles approach each other, the gasbetween them is compressed and the pressure rises; when they recede from eachother, there is an expansion and the pressure drops. Hence at a displacement nodethe gas undergoes the maximum amount of compression and expansion, and thevariations in pressure and density above and below the average have their maxi-mum value. By contrast, particles on opposite sides of a displacement antinodevibrate in phase; the distance between the particles is nearly constant, and there isno variation in pressure or density at a displacement antinode.

We use the term pressure node to describe a point in a standing sound wave atwhich the pressure and density do not vary and the term pressure antinode todescribe a point at which the variations in pressure and density are greatest.Using these terms, we can summarize our observations about standing soundwaves as follows:

A pressure node is always a displacement antinode, and a pressure antinode isalways a displacement node.

(A)(N)

v = lƒ.vƒ

l>2,

Gas inlettube

Diaphragm vibratesin response to soundfrom speaker.

Speaker

NA

AN

A

AN

N

N Sound of an appropriate frequency producesstanding waves with displacement nodes (N)and antinodes (A). The powder collects at thenodes.

16.12 Demonstrating standing soundwaves using a Kundt’s tube. The blueshading represents the density of the gas at an instant when the gas pressure at thedisplacement nodes is a maximum or aminimum.

0

T18

T28

T38

T48

T58

T68

T

T78

18

A standing wave shown at intervalsof T for one period T

N 5 a displacement node 5 a pressure antinodeA 5 a displacement antinode 5 a pressure node

AN AN N

l

16.13 In a standing sound wave, a dis-placement node N is a pressure antinode (a point where the pressure fluctuates themost) and a displacement antinode A is apressure node (a point where the pressuredoes not fluctuate at all).

6

Tal como numa corda, também num tubo onde se propaguem ondas (sonoras) há sobreposição entre ondas incidentes e ondas refletidas, que conduzem ao aparecimento de ondas estacionárias.

16.4 Standing Sound Waves and Normal Modes 525

half-wavelength, and in this case that is equal to the length L of the pipe; The corresponding frequency, obtained from the relationship is

(16.16)

Figures 16.17b and 16.17c show the second and third harmonics (first and sec-ond overtones); their vibration patterns have two and three displacement nodes,respectively. For these, a half-wavelength is equal to and respectively,and the frequencies are twice and three times the fundamental, respectively. Thatis, and For every normal mode of an open pipe the length Lmust be an integer number of half-wavelengths, and the possible wavelengths are given by

(16.17)

The corresponding frequencies are given by so all the normal-mode frequencies for a pipe that is open at both ends are given by

(16.18)

The value corresponds to the fundamental frequency, to the secondharmonic (or first overtone), and so on. Alternatively, we can say

(16.19)

with given by Eq. (16.16).Figure 16.18 shows a pipe that is open at the left end but closed at the right

end. This is called a stopped pipe. The left (open) end is a displacement antinode(pressure node), but the right (closed) end is a displacement node (pressure antin-ode). The distance between a node and the adjacent antinode is always onequarter-wavelength. Figure 16.18a shows the lowest-frequency mode; the length

ƒ1

ƒn = nƒ1 1n = 1, 2, 3, Á2 1open pipe2n = 2n = 1

ƒn = nv2L

1n = 1, 2, 3, Á2 1open pipe2ƒn = v>ln,ƒn

L = nln

2 or ln = 2L

n 1n = 1, 2, 3, Á2 (open pipe)

ln

ƒ3 = 3ƒ1.ƒ2 = 2ƒ1

L>3,L>2ƒ1 = v

2L (open pipe)

ƒ = v>l, l>2 = L.

Vibrations from turbulent airflowset up standing waves in the pipe.

Air fromblower

Body

Mouth

A

N

N

A

A

A

N

N

A

A

16.16 Cross sections of an organ pipe attwo instants one half-period apart. The N’sand A’s are displacement nodes and anti-nodes; as the blue shading shows, these arepoints of maximum pressure variation andzero pressure variation, respectively.

(b) Second harmonic: f2 5 2 5 2f1v

2L

L 5 2 l2

l2

l2

A AN

AN

(c) Third harmonic: f3 5 3 5 3f1v

2L

L 5 3 l2

l2

l2

l2

A AN

AN N

A

(a) Fundamental: f1 5v

2L

L 5 l2

Open end is always a displacement antinode.

A AN

16.17 A cross section of an open pipe showing the first three normal modes. The shading indicates the pressure variations. The redcurves are graphs of the displacement along the pipe axis at two instants separated in time by one half-period. The N’s and A’s are thedisplacement nodes and antinodes; interchange these to show the pressure nodes and antinodes.

16.18 A cross section of a stopped pipe showing the first three normal modes as well as the displacement nodes and antinodes. Onlyodd harmonics are possible.

(b) Third harmonic: f3 5 3 5 3f1v

4L

L 5 3 l4

l4

l4

l4

AN

AN

(c) Fifth harmonic: f5 5 5 5 5f1v

4L

L 5 5 l4

l4

l4

l4

l4

l4

AN

AN N

A

(a) Fundamental: f1 5v

4L

L 5l4

AN

Closed end is always a displacement node.

fm =mv2L, m =1, 2, 3,!

Tubo aberto:

Aula anterior 19.3 Ondas estacionárias em tubos

16.4 Standing Sound Waves and Normal Modes 523

be used to create sound waves in the surrounding air. This is the operating princi-ple of the human voice as well as many musical instruments, including wood-winds, brasses, and pipe organs.

Transverse waves on a string, including standing waves, are usually describedonly in terms of the displacement of the string. But, as we have seen, sound wavesin a fluid may be described either in terms of the displacement of the fluid or interms of the pressure variation in the fluid. To avoid confusion, we’ll use the termsdisplacement node and displacement antinode to refer to points where particlesof the fluid have zero displacement and maximum displacement, respectively.

We can demonstrate standing sound waves in a column of gas using an apparatuscalled a Kundt’s tube (Fig. 16.12). A horizontal glass tube a meter or so long isclosed at one end and has a flexible diaphragm at the other end that can transmitvibrations. A nearby loudspeaker is driven by an audio oscillator and amplifier; thisproduces sound waves that force the diaphragm to vibrate sinusoidally with a fre-quency that we can vary. The sound waves within the tube are reflected at the other,closed end of the tube. We spread a small amount of light powder uniformly alongthe bottom of the tube. As we vary the frequency of the sound, we pass through fre-quencies at which the amplitude of the standing waves becomes large enough for thepowder to be swept along the tube at those points where the gas is in motion. Thepowder therefore collects at the displacement nodes (where the gas is not moving).Adjacent nodes are separated by a distance equal to and we can measure thisdistance. Given the wavelength, we can use this experiment to determine the wavespeed: We read the frequency from the oscillator dial, and we can then calculatethe speed of the waves from the relationship

Figure 16.13 shows the motions of nine different particles within a gas-filledtube in which there is a standing sound wave. A particle at a displacement node

does not move, while a particle at a displacement antinode oscillates withmaximum amplitude. Note that particles on opposite sides of a displacement nodevibrate in opposite phase. When these particles approach each other, the gasbetween them is compressed and the pressure rises; when they recede from eachother, there is an expansion and the pressure drops. Hence at a displacement nodethe gas undergoes the maximum amount of compression and expansion, and thevariations in pressure and density above and below the average have their maxi-mum value. By contrast, particles on opposite sides of a displacement antinodevibrate in phase; the distance between the particles is nearly constant, and there isno variation in pressure or density at a displacement antinode.

We use the term pressure node to describe a point in a standing sound wave atwhich the pressure and density do not vary and the term pressure antinode todescribe a point at which the variations in pressure and density are greatest.Using these terms, we can summarize our observations about standing soundwaves as follows:

A pressure node is always a displacement antinode, and a pressure antinode isalways a displacement node.

(A)(N)

v = lƒ.vƒ

l>2,

Gas inlettube

Diaphragm vibratesin response to soundfrom speaker.

Speaker

NA

AN

A

AN

N

N Sound of an appropriate frequency producesstanding waves with displacement nodes (N)and antinodes (A). The powder collects at thenodes.

16.12 Demonstrating standing soundwaves using a Kundt’s tube. The blueshading represents the density of the gas at an instant when the gas pressure at thedisplacement nodes is a maximum or aminimum.

0

T18

T28

T38

T48

T58

T68

T

T78

18

A standing wave shown at intervalsof T for one period T

N 5 a displacement node 5 a pressure antinodeA 5 a displacement antinode 5 a pressure node

AN AN N

l

16.13 In a standing sound wave, a dis-placement node N is a pressure antinode (a point where the pressure fluctuates themost) and a displacement antinode A is apressure node (a point where the pressuredoes not fluctuate at all).

7

Tal como numa corda, também num tubo onde se propaguem ondas (sonoras) há sobreposição entre ondas incidentes e ondas refletidas, que conduzem ao aparecimento de ondas estacionárias.

16.4 Standing Sound Waves and Normal Modes 525

half-wavelength, and in this case that is equal to the length L of the pipe; The corresponding frequency, obtained from the relationship is

(16.16)

Figures 16.17b and 16.17c show the second and third harmonics (first and sec-ond overtones); their vibration patterns have two and three displacement nodes,respectively. For these, a half-wavelength is equal to and respectively,and the frequencies are twice and three times the fundamental, respectively. Thatis, and For every normal mode of an open pipe the length Lmust be an integer number of half-wavelengths, and the possible wavelengths are given by

(16.17)

The corresponding frequencies are given by so all the normal-mode frequencies for a pipe that is open at both ends are given by

(16.18)

The value corresponds to the fundamental frequency, to the secondharmonic (or first overtone), and so on. Alternatively, we can say

(16.19)

with given by Eq. (16.16).Figure 16.18 shows a pipe that is open at the left end but closed at the right

end. This is called a stopped pipe. The left (open) end is a displacement antinode(pressure node), but the right (closed) end is a displacement node (pressure antin-ode). The distance between a node and the adjacent antinode is always onequarter-wavelength. Figure 16.18a shows the lowest-frequency mode; the length

ƒ1

ƒn = nƒ1 1n = 1, 2, 3, Á2 1open pipe2n = 2n = 1

ƒn = nv2L

1n = 1, 2, 3, Á2 1open pipe2ƒn = v>ln,ƒn

L = nln

2 or ln = 2L

n 1n = 1, 2, 3, Á2 (open pipe)

ln

ƒ3 = 3ƒ1.ƒ2 = 2ƒ1

L>3,L>2ƒ1 = v

2L (open pipe)

ƒ = v>l, l>2 = L.

Vibrations from turbulent airflowset up standing waves in the pipe.

Air fromblower

Body

Mouth

A

N

N

A

A

A

N

N

A

A

16.16 Cross sections of an organ pipe attwo instants one half-period apart. The N’sand A’s are displacement nodes and anti-nodes; as the blue shading shows, these arepoints of maximum pressure variation andzero pressure variation, respectively.

(b) Second harmonic: f2 5 2 5 2f1v

2L

L 5 2 l2

l2

l2

A AN

AN

(c) Third harmonic: f3 5 3 5 3f1v

2L

L 5 3 l2

l2

l2

l2

A AN

AN N

A

(a) Fundamental: f1 5v

2L

L 5 l2

Open end is always a displacement antinode.

A AN

16.17 A cross section of an open pipe showing the first three normal modes. The shading indicates the pressure variations. The redcurves are graphs of the displacement along the pipe axis at two instants separated in time by one half-period. The N’s and A’s are thedisplacement nodes and antinodes; interchange these to show the pressure nodes and antinodes.

16.18 A cross section of a stopped pipe showing the first three normal modes as well as the displacement nodes and antinodes. Onlyodd harmonics are possible.

(b) Third harmonic: f3 5 3 5 3f1v

4L

L 5 3 l4

l4

l4

l4

AN

AN

(c) Fifth harmonic: f5 5 5 5 5f1v

4L

L 5 5 l4

l4

l4

l4

l4

l4

AN

AN N

A

(a) Fundamental: f1 5v

4L

L 5l4

AN

Closed end is always a displacement node.

fm =mv4L, m =1, 3, 5,!

Tubo fechado:

19.3 Ondas estacionárias em tubos Aula anterior

16.4 Standing Sound Waves and Normal Modes 523

be used to create sound waves in the surrounding air. This is the operating princi-ple of the human voice as well as many musical instruments, including wood-winds, brasses, and pipe organs.

Transverse waves on a string, including standing waves, are usually describedonly in terms of the displacement of the string. But, as we have seen, sound wavesin a fluid may be described either in terms of the displacement of the fluid or interms of the pressure variation in the fluid. To avoid confusion, we’ll use the termsdisplacement node and displacement antinode to refer to points where particlesof the fluid have zero displacement and maximum displacement, respectively.

We can demonstrate standing sound waves in a column of gas using an apparatuscalled a Kundt’s tube (Fig. 16.12). A horizontal glass tube a meter or so long isclosed at one end and has a flexible diaphragm at the other end that can transmitvibrations. A nearby loudspeaker is driven by an audio oscillator and amplifier; thisproduces sound waves that force the diaphragm to vibrate sinusoidally with a fre-quency that we can vary. The sound waves within the tube are reflected at the other,closed end of the tube. We spread a small amount of light powder uniformly alongthe bottom of the tube. As we vary the frequency of the sound, we pass through fre-quencies at which the amplitude of the standing waves becomes large enough for thepowder to be swept along the tube at those points where the gas is in motion. Thepowder therefore collects at the displacement nodes (where the gas is not moving).Adjacent nodes are separated by a distance equal to and we can measure thisdistance. Given the wavelength, we can use this experiment to determine the wavespeed: We read the frequency from the oscillator dial, and we can then calculatethe speed of the waves from the relationship

Figure 16.13 shows the motions of nine different particles within a gas-filledtube in which there is a standing sound wave. A particle at a displacement node

does not move, while a particle at a displacement antinode oscillates withmaximum amplitude. Note that particles on opposite sides of a displacement nodevibrate in opposite phase. When these particles approach each other, the gasbetween them is compressed and the pressure rises; when they recede from eachother, there is an expansion and the pressure drops. Hence at a displacement nodethe gas undergoes the maximum amount of compression and expansion, and thevariations in pressure and density above and below the average have their maxi-mum value. By contrast, particles on opposite sides of a displacement antinodevibrate in phase; the distance between the particles is nearly constant, and there isno variation in pressure or density at a displacement antinode.

We use the term pressure node to describe a point in a standing sound wave atwhich the pressure and density do not vary and the term pressure antinode todescribe a point at which the variations in pressure and density are greatest.Using these terms, we can summarize our observations about standing soundwaves as follows:

A pressure node is always a displacement antinode, and a pressure antinode isalways a displacement node.

(A)(N)

v = lƒ.vƒ

l>2,

Gas inlettube

Diaphragm vibratesin response to soundfrom speaker.

Speaker

NA

AN

A

AN

N

N Sound of an appropriate frequency producesstanding waves with displacement nodes (N)and antinodes (A). The powder collects at thenodes.

16.12 Demonstrating standing soundwaves using a Kundt’s tube. The blueshading represents the density of the gas at an instant when the gas pressure at thedisplacement nodes is a maximum or aminimum.

0

T18

T28

T38

T48

T58

T68

T

T78

18

A standing wave shown at intervalsof T for one period T

N 5 a displacement node 5 a pressure antinodeA 5 a displacement antinode 5 a pressure node

AN AN N

l

16.13 In a standing sound wave, a dis-placement node N is a pressure antinode (a point where the pressure fluctuates themost) and a displacement antinode A is apressure node (a point where the pressuredoes not fluctuate at all).

8

19.3 Ondas estacionárias em tubos Aula anterior

Se se somar todas as harmónicas, o que é que se obtém?

Sintetizador de Fourier

simulação

We have discussed the analysis of a wave pattern using Fourier’s theorem. Theanalysis involves determining the coefficients of the harmonics in Equation 18.16 froma knowledge of the wave pattern. The reverse process, called Fourier synthesis, can alsobe performed. In this process, the various harmonics are added together to form a re-sultant wave pattern. As an example of Fourier synthesis, consider the building of asquare wave, as shown in Figure 18.25. The symmetry of the square wave results in onlyodd multiples of the fundamental frequency combining in its synthesis. In Figure18.25a, the orange curve shows the combination of f and 3f. In Figure 18.25b, we haveadded 5f to the combination and obtained the green curve. Notice how the generalshape of the square wave is approximated, even though the upper and lower portionsare not flat as they should be.

Figure 18.25c shows the result of adding odd frequencies up to 9f. This ap-proximation (purple curve) to the square wave is better than the approximationsin parts a and b. To approximate the square wave as closely as possible, we would need to add all odd multiples of the fundamental frequency, up to infinite frequency.

Using modern technology, we can generate musical sounds electronically by mixing different amplitudes of any number of harmonics. These widely used elec-tronic music synthesizers are capable of producing an infinite variety of musical tones.

568 C H A P T E R 18 • Superposition and Standing Waves

(c)

f + 3f + 5f + 7f + 9f

Square wavef + 3f + 5f + 7f + 9f + ...

(b)

f + 3f + 5f

5f

f

3f

(a)

ff + 3f

3f

Active Figure 18.25 Fourier synthesis of a square wave, which is represented by the sum of oddmultiples of the first harmonic, which has frequency f. (a) Waves of frequency f and 3f are added.(b) One more odd harmonic of frequency 5f is added. (c) The synthesis curve approaches closerto the square wave when odd frequencies up to 9f are added.

At the Active Figures linkat http://www.pse6.com, youcan add in harmonics withfrequencies higher than 9f totry to synthesize a square wave.

“Soma” de ondas

simulação

9

19.4 Potência e energia

Numa corda de densidade µ, esticada por uma força de tensão FT, a onda que aí se propaga tem velocidade v. A potência transferida pela força de tensão é

P =!FT.!v (aula 6)

Pmed =12µ vω2A2

A potência média em qualquer ponto x ao longo da corda é

v =FTµ

A velocidade v é

10

19.4 Potência e energia

A energia média, num dado intervalo de tempo Δt, é

Emed = PmedΔt

= 12µvω2A2Δt

= 12µω2A2Δx

11

Periodic Waves S E C T I O N 1 5 - 2 | 505

Example 15-5 A Harmonic Wave on a String

The wave function is for a harmonic wave ona string. (a) In what direction does this wave travel and what is its speed? (b) Find the wave-length, frequency, and period of this wave. (c) What is the maximum displacement of anypoint on the string? (d) What is the maximum speed of any point on the string?

PICTURE (a) To find the direction of travel, express as either a function of oras a function of and use Equations 15-1 and 15-2. To find the wave speed, use (Equation 15-16). (b) The wavelength, frequency, and period can be found from the wavenumber and the angular frequency (c) The maximum displacement of a point on thestring is the amplitude (d) The velocity of a point on the string is

SOLVE

!y>!t.A.v.k

v " kv(x # vt)(x $ vt)y(x,t)

(3.5 s$1)t]$sin[(2.2 m$1)x%(0.030 m)"y(x,t)

(a) 1. The given wave function is of the form Using (Equation 15-16), write the wave function as afunction of Then, use Equations 15-1 and 15-2 to findthe direction of travel:

x $ vt.v " kv

y(x,t) " A sin(kx $ vt). andsoThe wave travels in the #x direction.y(x,t) " A sin(kx $ kvt) " A sin[k(x $ vt)]

v " kvy(x,t) " A sin(kx $ vt)

2. Because the form is we know as well asboth and Use these to calculate the speed:k.v

Ay " A sin(kx $ vt),

1.6 m>s"

v "l

T"l

2p2pT

"v

k"

3.5 s$1

2.2 m$1 " 1.59 m>s(b) The wavelength is related to the wave number and the period

and frequency are related to v:fTk,l

0.56 Hzf "1T

"1

1.80 s" 0.557 Hz "

1.8 sT "2pv

"2p

3.5 s$1 " 1.80 s "

2.9 ml "2pk

"2p

2.2 m$1 " 2.86 m "

(c) The maximum displacement of a string segment is theamplitude A:

0.030 mA "

(d) 1. Compute to find the velocity of a point on the string:!y>!t

" $(0.105 m>s) cos(2.2 m$1x $ 3.5 s$1t)" (0.030 m)($3.5 s$1) cos(2.2 m$1x $ 3.5 s$1t)

vy "!y!t

" (0.030 m)![sin(2.2 m$1x $ 3.5 s$1t)]

!t

2. The maximum transverse speed occurs when the cosinefunction has the value of &1:

0.11 m>svy,max " 0.105 m>s "

CHECK We have included the units explicitly to show how they work out. They serve as aplausibility check. Often we will omit the units for brevity.

Energy transfer via waves on a string Consider again a string attached to atuning fork. As the fork vibrates, it transfers energy to the segment of the string at-tached to it. For example, as the fork moves upward from its equilibrium positionit stretches the adjacent string segment slightly—increasing its elastic potential en-ergy. In addition, the fork slows as it moves upward from it equilibrium, so it slowsthe string segment closest to it. This decreases the kinetic energy of the segment.As a wave moves along the string, energy is transferred from one segment to thenext in a similar manner.

Power is the rate of energy transfer. We can calculate the power by consideringwork done by the force that one segment of the string exerts on a neighboring seg-ment. The rate of work done by this force is the power. Figure 15-9 shows a har-monic wave moving to the right along a string segment. That is, we assume a wavefunction of the form

15-19y(x,t) " A sin(kx $ vt)

y

v

xFT vtr

θ

F I G U R E 1 5 - 9 The tension force hasa component in the direction of the transversevelocity so at this instant the force is doingwork on the end of the string that has apositive value.

vStr ,

FS

T

Exemplo Uma onda harmónica tem comprimento de onda λ = 0,25 m e amplitude A = 0,012 m. Esta onda move-se ao longo de uma corda, com 60 m de comprimento e 0,32 kg de massa, que está sujeita a uma tensão de 12 N. Determine: a)  a velocidade e a frequência angular da onda; b)  a energia da onda, ao longo de um segmento da corda com 15 m de comprimento.

a) v =FTµ

v = f λ

v =FT Lm

= 47,4 m/s

ω = 2π vλ=1190 rad/s

b) Emed = PmedΔt =12µvω2A2Δt = 1

2µω2A2Δx

Emed =12mL2π v

λ

⎛

⎝⎜

⎞

⎠⎟2A2Δx = 8,19 J

12

20.1 Efeito Doppler

Quando um emissor (fonte) de ondas e um recetor estiverem parados, a frequência recebida é a mesma que a emitida.

Velocidade do som no ar (T = 20ºC):

vsom ≈ 343m/s

13

20.1 Efeito Doppler

Quando um emissor (fonte) de ondas e um recetor estiverem parados, a frequência recebida é a mesma que a emitida.

14

20.1 Efeito Doppler

Quando um emissor (fonte) de ondas e um recetor se moverem, relativamente um ao outro, a frequência recebida não é a mesma que a emitida.

So a listener moving toward a source as in Fig. 16.26, hears a higherfrequency (higher pitch) than does a stationary listener. A listener moving awayfrom the source hears a lower frequency (lower pitch).

Moving Source and Moving ListenerNow suppose the source is also moving, with velocity (Fig. 16.27). The wavespeed relative to the wave medium (air) is still it is determined by the propertiesof the medium and is not changed by the motion of the source. But the wavelengthis no longer equal to Here’s why. The time for emission of one cycle of thewave is the period During this time, the wave travels a distance

and the source moves a distance The wavelength is thedistance between successive wave crests, and this is determined by the relativedisplacement of source and wave. As Fig. 16.27 shows, this is different in front ofand behind the source. In the region to the right of the source in Fig. 16.27 (thatis, in front of the source), the wavelength is

(wavelength in front of a moving source) (16.27)

In the region to the left of the source (that is, behind the source), it is

(wavelength behind a moving source) (16.28)

The waves in front of and behind the source are compressed and stretched out,respectively, by the motion of the source.

To find the frequency heard by the listener behind the source, we substituteEq. (16.28) into the first form of Eq. (16.25):

(Doppler effect, moving source and moving listener) (16.29)

This expresses the frequency heard by the listener in terms of the frequency of the source.

Although we derived it for the particular situation shown in Fig. 16.27, Eq. (16.29) includes all possibilities for motion of source and listener (relative to

ƒSƒL

ƒL =v + vL

v + vSƒS

ƒL =v + vL

lbehind=

v + vL1v + vS2>ƒS

lbehind =v + vS

ƒS

lin front = vƒS

-vS

ƒS=

v - vS

ƒS

vST = vS>ƒS.vT = v>ƒS

T = 1>fS.v>ƒS.

v;vS

1vL 6 02 1vL 7 02,

534 CHAPTER 16 Sound and Hearing

• Velocity of listener (L) 5 vL• Velocity of source (S) 5 vS• Speed of sound wave 5 v• Positive direction: from listener

to source

v

L to S1

vL

vv

v

v

v

v

vSvSa bS S

L

lin frontlbehind

16.27 Wave crests emitted by a movingsource are crowded together in front of thesource (to the right of this source) andstretched out behind it (to the left of thissource).

15

20.1 Efeito Doppler

Quando um emissor (fonte) de ondas e um recetor se moverem, relativamente um ao outro, a frequência recebida não é a mesma que a emitida.

16.8 The Doppler Effect 533

16.8 The Doppler EffectYou’ve probably noticed that when a car approaches you with its horn sounding,the pitch seems to drop as the car passes. This phenomenon, first described by the19th-century Austrian scientist Christian Doppler, is called the Doppler effect.When a source of sound and a listener are in motion relative to each other, thefrequency of the sound heard by the listener is not the same as the source fre-quency. A similar effect occurs for light and radio waves; we’ll return to this laterin this section.

To analyze the Doppler effect for sound, we’ll work out a relationship betweenthe frequency shift and the velocities of source and listener relative to the medium(usually air) through which the sound waves propagate. To keep things simple,we consider only the special case in which the velocities of both source and lis-tener lie along the line joining them. Let and be the velocity componentsalong this line for the source and the listener, respectively, relative to the medium.We choose the positive direction for both and to be the direction from thelistener L to the source S. The speed of sound relative to the medium, is alwaysconsidered positive.

Moving Listener and Stationary SourceLet’s think first about a listener L moving with velocity toward a stationarysource S (Fig. 16.26). The source emits a sound wave with frequency andwavelength The figure shows four wave crests, separated by equal dis-tances The wave crests approaching the moving listener have a speed of propa-gation relative to the listener of So the frequency with which thecrests arrive at the listener’s position (that is, the frequency the listener hears) is

(16.25)

or

(moving listener, stationary source)

(16.26)ƒL = av + vL

vbƒS = a1 +

vL

vbƒS

ƒL =v + vL

l=

v + vL

v>ƒS

ƒL1v + vL2.l.l = v>ƒS.

ƒS

vL

v,vLvS

vLvS

Test Your Understanding of Section 16.7 One tuning fork vibrates at 440 Hz,while a second tuning fork vibrates at an unknown frequency. When both tuning forks aresounded simultaneously, you hear a tone that rises and falls in intensity three times persecond. What is the frequency of the second tuning fork? (i) 434 Hz; (ii) 437 Hz; (iii) 443 Hz; (iv) 446 Hz; (v) either 434 Hz or 446 Hz; (vi) either 437 Hz or 443 Hz. ❙

vL to S

1

vL

v v

v

v

v

v

l

LS

• Velocity of listener (L) 5 vL• Velocity of source (S) 5 0 (at rest)• Speed of sound wave 5 v• Positive direction: from listener

to source

16.26 A listener moving toward a sta-tionary source hears a frequency that ishigher than the source frequency. This isbecause the relative speed of listener andwave is greater than the wave speed v.

ActivPhysics 10.8: Doppler Effect: ConceptualIntroductionActivPhysics 10.9: Doppler Effect: Problems

Fonte em movimento

animação

Velocidade do som no ar (T = 20ºC):

vsom ≈ 343m/s

Doppler 2

animação

16

20.1 Efeito Doppler

O emissor emite uma onda, com frequência fE e comprimento de onda λE , que se propaga com velocidade v = fE λE . Se o recetor se aproximar do emissor, com velocidade vR , a frente de onda que se aproxima do recetor tem uma velocidade de propagação, relativamente ao recetor, igual a v + vR . Assim, a frequência que o recetor vai receber é

fR =v + vRλE

=v + vRvfE

fR =v + vRv

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE = 1+

vRv

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE

16.8 The Doppler Effect 533

16.8 The Doppler EffectYou’ve probably noticed that when a car approaches you with its horn sounding,the pitch seems to drop as the car passes. This phenomenon, first described by the19th-century Austrian scientist Christian Doppler, is called the Doppler effect.When a source of sound and a listener are in motion relative to each other, thefrequency of the sound heard by the listener is not the same as the source fre-quency. A similar effect occurs for light and radio waves; we’ll return to this laterin this section.

To analyze the Doppler effect for sound, we’ll work out a relationship betweenthe frequency shift and the velocities of source and listener relative to the medium(usually air) through which the sound waves propagate. To keep things simple,we consider only the special case in which the velocities of both source and lis-tener lie along the line joining them. Let and be the velocity componentsalong this line for the source and the listener, respectively, relative to the medium.We choose the positive direction for both and to be the direction from thelistener L to the source S. The speed of sound relative to the medium, is alwaysconsidered positive.

Moving Listener and Stationary SourceLet’s think first about a listener L moving with velocity toward a stationarysource S (Fig. 16.26). The source emits a sound wave with frequency andwavelength The figure shows four wave crests, separated by equal dis-tances The wave crests approaching the moving listener have a speed of propa-gation relative to the listener of So the frequency with which thecrests arrive at the listener’s position (that is, the frequency the listener hears) is

(16.25)

or

(moving listener, stationary source)

(16.26)ƒL = av + vL

vbƒS = a1 +

vL

vbƒS

ƒL =v + vL

l=

v + vL

v>ƒS

ƒL1v + vL2.l.l = v>ƒS.

ƒS

vL

v,vLvS

vLvS

Test Your Understanding of Section 16.7 One tuning fork vibrates at 440 Hz,while a second tuning fork vibrates at an unknown frequency. When both tuning forks aresounded simultaneously, you hear a tone that rises and falls in intensity three times persecond. What is the frequency of the second tuning fork? (i) 434 Hz; (ii) 437 Hz; (iii) 443 Hz; (iv) 446 Hz; (v) either 434 Hz or 446 Hz; (vi) either 437 Hz or 443 Hz. ❙

vL to S

1

vL

v v

v

v

v

v

l

LS

• Velocity of listener (L) 5 vL• Velocity of source (S) 5 0 (at rest)• Speed of sound wave 5 v• Positive direction: from listener

to source

16.26 A listener moving toward a sta-tionary source hears a frequency that ishigher than the source frequency. This isbecause the relative speed of listener andwave is greater than the wave speed v.

ActivPhysics 10.8: Doppler Effect: ConceptualIntroductionActivPhysics 10.9: Doppler Effect: Problems

(emissor em repouso, recetor a aproximar-se) Doppler 2 animação

Quando um emissor (fonte) de ondas e um recetor se moverem, relativamente um ao outro, a frequência recebida não é a mesma que a emitida.

17

20.1 Efeito Doppler

Quando um emissor (fonte) de ondas e um recetor se movem, relativamente um ao outro, a frequência recebida não é a mesma que é emitida.

Se o recetor se afastar do emissor, a frequência que vai receber é

fR =v − vRλE

=v − vRvfE

fR =v − vRv

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE = 1−

vRv

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE

16.8 The Doppler Effect 533

16.8 The Doppler EffectYou’ve probably noticed that when a car approaches you with its horn sounding,the pitch seems to drop as the car passes. This phenomenon, first described by the19th-century Austrian scientist Christian Doppler, is called the Doppler effect.When a source of sound and a listener are in motion relative to each other, thefrequency of the sound heard by the listener is not the same as the source fre-quency. A similar effect occurs for light and radio waves; we’ll return to this laterin this section.

To analyze the Doppler effect for sound, we’ll work out a relationship betweenthe frequency shift and the velocities of source and listener relative to the medium(usually air) through which the sound waves propagate. To keep things simple,we consider only the special case in which the velocities of both source and lis-tener lie along the line joining them. Let and be the velocity componentsalong this line for the source and the listener, respectively, relative to the medium.We choose the positive direction for both and to be the direction from thelistener L to the source S. The speed of sound relative to the medium, is alwaysconsidered positive.

Moving Listener and Stationary SourceLet’s think first about a listener L moving with velocity toward a stationarysource S (Fig. 16.26). The source emits a sound wave with frequency andwavelength The figure shows four wave crests, separated by equal dis-tances The wave crests approaching the moving listener have a speed of propa-gation relative to the listener of So the frequency with which thecrests arrive at the listener’s position (that is, the frequency the listener hears) is

(16.25)

or

(moving listener, stationary source)

(16.26)ƒL = av + vL

vbƒS = a1 +

vL

vbƒS

ƒL =v + vL

l=

v + vL

v>ƒS

ƒL1v + vL2.l.l = v>ƒS.

ƒS

vL

v,vLvS

vLvS

Test Your Understanding of Section 16.7 One tuning fork vibrates at 440 Hz,while a second tuning fork vibrates at an unknown frequency. When both tuning forks aresounded simultaneously, you hear a tone that rises and falls in intensity three times persecond. What is the frequency of the second tuning fork? (i) 434 Hz; (ii) 437 Hz; (iii) 443 Hz; (iv) 446 Hz; (v) either 434 Hz or 446 Hz; (vi) either 437 Hz or 443 Hz. ❙

vL to S

1

vL

v v

v

v

v

v

l

LS

• Velocity of listener (L) 5 vL• Velocity of source (S) 5 0 (at rest)• Speed of sound wave 5 v• Positive direction: from listener

to source

16.26 A listener moving toward a sta-tionary source hears a frequency that ishigher than the source frequency. This isbecause the relative speed of listener andwave is greater than the wave speed v.

ActivPhysics 10.8: Doppler Effect: ConceptualIntroductionActivPhysics 10.9: Doppler Effect: Problems

Portanto, para um emissor em repouso e um recetor em movimento, a relação entre as frequências é

fR =v ± vRv

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE

Doppler 3 animação

So a listener moving toward a source as in Fig. 16.26, hears a higherfrequency (higher pitch) than does a stationary listener. A listener moving awayfrom the source hears a lower frequency (lower pitch).

Moving Source and Moving ListenerNow suppose the source is also moving, with velocity (Fig. 16.27). The wavespeed relative to the wave medium (air) is still it is determined by the propertiesof the medium and is not changed by the motion of the source. But the wavelengthis no longer equal to Here’s why. The time for emission of one cycle of thewave is the period During this time, the wave travels a distance

and the source moves a distance The wavelength is thedistance between successive wave crests, and this is determined by the relativedisplacement of source and wave. As Fig. 16.27 shows, this is different in front ofand behind the source. In the region to the right of the source in Fig. 16.27 (thatis, in front of the source), the wavelength is

(wavelength in front of a moving source) (16.27)

In the region to the left of the source (that is, behind the source), it is

(wavelength behind a moving source) (16.28)

The waves in front of and behind the source are compressed and stretched out,respectively, by the motion of the source.

To find the frequency heard by the listener behind the source, we substituteEq. (16.28) into the first form of Eq. (16.25):

(Doppler effect, moving source and moving listener) (16.29)

This expresses the frequency heard by the listener in terms of the frequency of the source.

Although we derived it for the particular situation shown in Fig. 16.27, Eq. (16.29) includes all possibilities for motion of source and listener (relative to

ƒSƒL

ƒL =v + vL

v + vSƒS

ƒL =v + vL

lbehind=

v + vL1v + vS2>ƒS

lbehind =v + vS

ƒS

lin front = vƒS

-vS

ƒS=

v - vS

ƒS

vST = vS>ƒS.vT = v>ƒS

T = 1>fS.v>ƒS.

v;vS

1vL 6 02 1vL 7 02,

534 CHAPTER 16 Sound and Hearing

• Velocity of listener (L) 5 vL• Velocity of source (S) 5 vS• Speed of sound wave 5 v• Positive direction: from listener

to source

v

L to S1

vL

vv

v

v

v

v

vSvSa bS S

L

lin frontlbehind

16.27 Wave crests emitted by a movingsource are crowded together in front of thesource (to the right of this source) andstretched out behind it (to the left of thissource).

18

20.1 Efeito Doppler

Nota: a frequência tende a aumentar quando a)  o emissor se aproxima do recetor b)  o recetor se aproxima do emissor

Se o emissor também estiver em movimento, com velocidade vE , a relação entre as frequências passa a ser

fR =v ± vRv ∓ vE

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE

fR =v + vRv − vE

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE

19

Exemplo Dois submarinos, imersos, dirigem-se um para o outro. O submarino A desloca-se a 8 m/s e emite uma onda (sonar) com a frequência de 1400 Hz. O submarino B move-se a 9 m/s e a velocidade de propagação do som na água é 1533 m/s. Determine: a)  a frequência detetada por um observador no submarino B; b)  a frequência da alínea anterior, depois dos submarinos terem passado um pelo

outro.

a)

b)

fR =v ± vRv ∓ vE

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fE

A B

vA = 8 m/s→← vB = 9 m/s

S E C T I O N 17. 4 • The Doppler Effect

523

vwaves

(b)

vwaves

(a)

vwaves

(c)

vboat

vboat

Figure 17.7(a) Waves mov-

ing toward a stationary boat.

The waves travel to the left,

and their source is far to the

right of the boat, out of the

frame of the photograph.

(b) The boat moving toward

the wave source. (c) The

boat moving away from the

wave source.

circles are the intersections of these three-dimensional wave fronts with the two-dimen-

sional paper.We take the frequency of the source in Figure 17.8 to be f, the wavelength to be !,

and the speed of sound to be v. If the observer were also stationary, he or she would

detect wave fronts at a rate f. (That is, when vO "0 and vS "

0, the observed frequency

equals the source frequency.) When the observer moves toward the source, the speed

of the waves relative to the observer is v# "v + vO , as in the case of the boat, but the

×

O

vO

S

Active Figure 17.8An observer O (the cyclist) moves with a speed vO toward a station-

ary point source S, the horn of a parked truck. The observer hears a frequency f # that is

greater than the source frequency.

At the Active Figures link

at http://www.pse6.com, you

can adjust the speed of the

observer.

fB =v + vBv − vA

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fA =

1533+91533−8

⎛

⎝⎜

⎞

⎠⎟1400 =1416 Hz

fB =v − vBv + vA

⎛

⎝⎜⎜

⎞

⎠⎟⎟ fA =

1533−91533+8

⎛

⎝⎜

⎞

⎠⎟1400 =1385 Hz

φ1 x,t( ) = Asen ω1t − k1x( )φ2 x,t( ) = Asen ω2t − k2x( )φ x,t( ) =ϕ1 x,t( )+ϕ2 x,t( ) = 2Acos Δω.t −Δk.x( )sen ωt − kx( )

20

20.2 Batimentos (interferência no tempo) Sobreposição de duas ondas com frequências muito próximas uma da outra (batimentos):

( )ff

kv .

2.2

λλπ

πω===Velocidade de fase:

Velocidade de grupo: k

uΔ

Δ=

ω

sena+ senb = 2sen a+b2

⎛

⎝⎜

⎞

⎠⎟cos

a−b2

⎛

⎝⎜

⎞

⎠⎟

Δω =ω1−ω22

Δk =k1− k22

ω =ω1+ω22

k =k1+ k22

Batimentos simulação

21

20.3 Interferência

máximosd senθ =mλ

mínimos

d senθ =mλ + λ2

22

20.3 Interferência

Interferência 2

simulação

Interferência 1 simulação

Mínimos de difração 1ª ordem

a2

senθ = λ2

⇒ a senθ = λ

Condição geral a senθ = nλ

Interferência

Difração

20.4 Difração

23