PH 201-4A spring 2007PH 201 4A spring 2007

Interference PhenomenaInterference Phenomena

Lecture 28

Chapter 17(Cutnell & Johnson Physics 7th edition)(Cutnell & Johnson, Physics 7 edition)

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The Superposition of Waves

A iti i i l h t i t i i tA 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 displacementsdisplacements

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)

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Consider two waves with the following properties:

• propagate in the same direction The superposition yields a wave of twice the amplitude of the individual• same frequency

• same amplitude• in phase (wave crests and troughs coincide)

twice the amplitude of the individual waves.

Reinforcement of one wave by another is called constructive interferenceis called constructive interference.

y1 = Asin(Kx – ωt)y2 = Asin(Kx – ωt)y = y1 + y2 = 2Asin(Kx – ωt)

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• propagate in the same direction• same frequency

lit d

Consider two waves with the following properties:

• same amplitude• out of phase (wave crests of one wave match the wave troughs of the other)

The superposition yields a wave of zero amplitude. A cancellation of one wave b th i ll d d t ti i t fby another is called destructive interference.

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• propagate in the same direction• same frequency

Consider two waves with the following properties:

• same frequency• amplitudes are not equal• out of phase

The superposition yields a wave, amplitude does not equal zero

Destructive interference – cancellation is not total

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InterferenceLike waves of any kind, sound waves interfere when they meet. 2 y , yloudspeakers emit sound waves in phase.

The phase difference at particular points in space Δθ is related to the path difference (PD) by a simple relationship:p ( ) y p p

y = AcosKx; θ = Kx; Δθ = kΔx=2π/λ * (PD)

Constructive interference – when Δθ=2πn; or 2πn=2π/λ * (PD). or; ( )PD=nλ is a multiple of the wavelength

Δθ = 2π/λ (AC – BC) = 2π/λ(4λ - 3λ) = 2πPD = nλ n = 0, 1, 2, 3, , ,

Δθ = 2π/λ (3λ – 2.5λ) = π out of phase

Destructive interference – when the path difference is an odd pnumber of half-wavelengths

PD = m(λ/2) m = 1, 3, 5, …

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At an open-air concert on a hot day (Tc = 25°C, Vs = 346.5 m/s), a person sits at a location that 7.0 m and 9.1 m respectively from speakers at each side of the stage. A musician, warming up, plays a single 494 Hz tone. What does the spectator hear?

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Constructive and Destructive Interference of Sound WavesAssume that two loudspeakers in the figure are vibrating out of phase instead of in phase. (see example 2 from the text) The speed of sound is 343 m/s. What is the smallest f th t ill d d t ti i t f t i t C?frequency that will produce destructive interference at point C?

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Constructive and Destructive Interference of Sound WavesSpeakers A and B are vibrating in phase. They are directly facing each other, are 7.80 m apart, and are each playing a 73.0 Hz tone. The speed of sound is 343 m/s. On the line between the speakers there are three points where constructive interference occurs. What are the distances of these three points from speaker A?

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Consider two waves:• propagate in the same direction• same amplitude• f1 does not equal f2; λ1 does not equal λ2; f1 – f2 -> small difference• x = 0 - in phase => interfere constructively• x = p - out of phase => interfere destructively

• x = Q phase difference one• x = Q - phase difference – one cycle crests coincide => interfere constructively

Superposition of the two waves displays regularly alternating regions of large and small amplitude. Amplitude – pulsates.

The periodic variations in the amplitude of the wave pattern gives rise to the phenomenon – beats.

The frequency with which the amplitude pulsates is called the beat frequencyf f ffbeat = f2 – f1

Beats are a sensitive indication of small frequency differences.10

BeatsProblem 48A tuning fork vibrates at a frequency of 534 Hz An out-of-tune piano string vibrates atA tuning fork vibrates at a frequency of 534 Hz. An out of tune piano string vibrates at 529 Hz. How much time separates successive beats?

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One car is approaching and another is moving away from a bystander. Each car is moving with a speed of 6.00 m/s, and each blows a horn that has a frequency of 392 Hz. The speed of sound is 343 m/s. What is the beat frequency heard by the bystander?

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The superposition of two waves traveling in opposite directions is a standing wave

t = 0waves in phase

Points P, Q, R at which the sum of the waves ist = ½ period

red wave moved left ½ of the λblue – right ½ λ

the sum of the waves is zero are called nodes

Nodes are λ/2 apartg

t = ¼ of a periodred moved left λ/4bl i ht λ/4

Antinodes – points at which sum of the two waves is max.

blue - right λ/4Out of phase -> they cancel everywhere.

t = 3/8 of a period

t = ½ periodagain in phase

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The standing wave travels neither to the right nor to the left; its wave crests remain at fixed positions, while the entire wave increases and decreases in unisonf = f1 = f2fr f1 f2

Standing wave on a tightly stretched string with fixed ends.

In contrast to the case of traveling wave, where the amplitudes of the harmonic oscillations of all the particles are the same, the amplitudes of oscillation depend (for standing wave) on position. Anodes = min; Aantinodes = max

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Standing Waves – the Vibrating Strings. Math Description.

If a string is fixed at both ends and a wave train gis sent down the string, then the wave is reflected from the fixed ends.

Th 2=> There are 2 wavesOne is traveling to the right y1 = Asin(Kx – ωt)Reflected one -> y2 = Asin(Kx + ωt)

The resultant wave is the sum of these waves y = y1 + y2 = Asin(Kx – ωt) + Asin(Kx + ωt)

Use trigonometric identity sinB + sinC = 2sin( [B+C]/2 )*cos( [B C]/2)Use trigonometric identity sinB + sinC = 2sin( [B+C]/2 )*cos( [B-C]/2)where B = (Kx + ωt) ; C = (Kx – ωt)y = 2Asin[ { (Kx – ωt) + (Kx + ωt) }/2 ]*cos[ { (Kx + ωt) – (Kx – ωt) }/2 ] = Equation of standing wave y = 2Asin(Kx)cos(ωt)

The amplitude = 2Asin(Kx) node 2Asin(Kx) = 0Kx = nπ n = 0, 1, 2, 3=> x = nπ/K ; K = 2π/λ => x = nλ/2 nodes=> x = nπ/K ; K = 2π/λ => x = nλ/2 nodes

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Possible Standing Waves of the Stretched String with Fixed Ends (normal modes)

The fundamental modeλ1 = 2L ; f1 = v/2L – fundamental freq.

n =1

L = nλ/2λ = 2L/n n = 1λ = 2L

The first overtone

n = 2

λ2 = 2L/2 = L ; f2 = v/λ2 = v/L = 2v/2L = 2f1

The second overtoneλ = 2L/3 ; f = v/λ3 = 3v/3L = 3f

n = 3

λ3 = 2L/3 ; f3 = v/λ3 = 3v/3L = 3f1

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Transverse Standing WavesProblem 25 The G string on a guitar has a fundamental frequency of 196 Hz and a length of 0.62 m. This string is pressed against the proper fret to produce the note C, whose g p g p p p ,fundamental frequency is 262 Hz. What is the distance L2 between the fret and the end of the string at the bridge of the guitar?

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A steel wire having a mass of 4.0 g and length of 1.25 m is tightened to a tension of 1020 N. Find the wavelength and the frequency of the fundamental and third harmonic of the standing waves on this stringharmonic of the standing waves on this string.

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Longitudinal Standing WavesA cylindrical tube sustains standing waves at the following frequencies: 500, 700 and 900 Hz. There are no standing waves at frequencies of 600 and 800 Hz.g qA) What is the fundamental frequency?B) Is the tube open at both ends or at only one end?

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Longitudinal Standing WavesProblem 49 A person hums into the top of a well and finds that standing waves are established at frequencies of 42, 70.0 and 98 Hz. The frequency of 42 Hz is not

il th f d t l f Th d f d i 343 / H d i thnecessarily the fundamental frequency. The speed of sound is 343 m/s. How deep is the well?

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ReflectionReflection occurs when a wave

RefractionWhen a wave strikes a boundary, some of itsReflection occurs when a wave

strikes an object or comes to a boundary of another medium and is at least partly diverted backward.

When a wave strikes a boundary, some of its energy is reflected and some is transmitted or absorbed. When a wave crosses a boundary into another medium, its velocity h b th t i l h

An echo is an example of reflection of sound waves and mirrors reflect light waves.

changes because the new material has different characteristics. Thus the transmitted wave may move in a direction different from that of the incident wave. This phenomenon g pis called refraction.

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Diffraction

A deflection of waves by the edge of an obstacle is called diffractionA deflection of waves by the edge of an obstacle is called diffraction.

General rule – the amount of diffraction suffered by a wave passing through an aperture depends on the ratio of wavelength to the side of the aperture.p p g p

Increasing λ makes the diffraction more pronounced.

An important role of diffraction in the propagation of soundAn important role of diffraction in the propagation of sound.

- We can hear a person facing away from us or talking in an adjacent room (mouth is out of our direct line of sight) Why? – the sound waves diffract through the open mouth or the open door and spread to fill the entire room.

Application – design of loudspeakers for hi-fi systems

The choices of the sizes of the speaker ensure that all the waves are strongly diffracted.

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Diffraction

• Each vibrating molecule in the doorway generates a sound wave that expands outwards and bends, or diffracts, around the edges of the doorway.• Because of interference effects among the sound waves produced by all the molecules in the doorway. The sound intensity is mostly confined to the region defined by the angle θ on either side of the doorway.

Sinθ = λ/2 single slit – first min.Sinθ = 1.22λ/D circular opening – first min.

• because the dispersion of high frequencies is less than that of low frequencies, you should be directly in front of the speaker to hear both the high and low frequencies especially wellwell• for hi-fi loudspeakers small speakers are used to produce high frequency sound• large speakers are used for low frequency sounds

So the person will hear both frequencies equally well. 23

DiffractionProblem 15 A row of seats is parallel to a stage at a distance of 8.7 m from it. At the center and front of the stage is a diffraction horn loudspeakers that has a width D = 0.075

Th k i l i t th t h f f 1 0 104 H Th d f dm. The speaker is playing a tone that has a frequency of 1.0 x 104 Hz. The speed of sound is 343 m/s. What is the separation between two seats, located on opposite sides of the center of the row, at which the tone cannot be heard?

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DiffractionProblem 12 The width of D of a diffraction horn and the diameter of a circular speaker are equal. The sound produced by the two speakers has the same diffraction angle θ. q p y p gWhat is the ratio of the wavelength of the sound produced by the diffraction horn to that produced by the circular speaker?

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Fourier’s TheoremAny arbitrary periodic wave can be constructed by the superposition of a sufficiently large number of harmonic waves of different amplitudes and wavelengths.

ab c d

periodic wave (a) is constructed by superposition of three harmonic waves of wavelengths λ (b) ; λ/3 (c) and λ/5 (d) whose amplitude are in ratio 1 : 1/3 : 1/5

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