optics and waves.pdf

Lecture 3

Superposition and Standing Waves

Superposition and interference

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• Two traveling waves can pass through each other without being destroyed or altered (two particles cannot). This is a consequence of the superposition principle.

• The combination of separate waves in the same region of space to produce a resultant wave is called interference.

For mechanical waves, linear waves have

amplitudes much smaller than their

wavelengths.

• Waves can be combined in the same location in space. The superposition principle is used for wave combinations analysis :

if two or more traveling waves are moving through a medium, the resultant value of the wave function at any point is the algebraic sum of the values of the wave functions of the individual waves.

• Linear waves obey the superposition principle.

Superposition – 1

• Two pulses are traveling in opposite directions.

• The wave function of the pulse moving to the right is y1 and for the one moving to the left is y2.

• The pulses have different shapes.

• The displacement of the elements is positive for both.

• When the waves start to overlap (b), the resultant wave function is y1 + y2.

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Superposition – 2

• When crest meets crest (c) the resultant wave has a larger amplitude than either of the original waves.

• The two pulses then separate (d), and they continue moving in their original directions.

• The shapes of the pulses remain unchanged.

• This type of superposition is called constructive interference.

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Constructive interference

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Destructive interference

• Two pulses traveling in opposite directions

• Their displacements are inverted with respect to each other

• When they overlap, their displacements partially cancel each other

• When these pulses overlap, the resultant pulse is y1 + y2.

• This type of superposition is called destructive interference.

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Types of interference

• Constructive interference occurs when the displacements caused by the two pulses are in the same direction;

the amplitude of the resultant pulse is greater than either individual pulse.

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• Destructive interference occurs when the displacements caused by the two pulses are in opposite directions;

the amplitude of the resultant pulse is less than either individual pulse.

Superposition of sinusoidal waves

• Assume two waves are traveling in the same direction, with the same frequency, wavelength and amplitude.

• The waves differ only in phase:

y1 = A sin (kx wt) and y2 = A sin (kx wt + f)

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1 2

1 1Since sin sin 2sin cos

2 2

Let and , we have :

2 cos sin2 2

a b a b a b

a kx t b kx t

y y y A kx t

w w f

f fw

• The resultant wave function, y is also sinusoidal. • The resultant wave has the same frequency and wavelength

as the original waves. • The amplitude of the resultant wave is • The phase of the resultant wave is

2

f

2 cos2

Af

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Superposition of sinusoidal waves

Sinusoidal waves with constructive

interference

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• When f = 0, or f = n, where n is an even integer (n = 2, 4, 6,…) then cos (f /2) = 1

• The amplitude of the resultant wave is 2A.

The crests of one wave coincide with the crests of the other wave.

• The waves are everywhere in phase and the waves interfere constructively.

• In general, constructive interference occurs when cos (f /2) = 1. That is true, when f = 0, 2, 4,…, i.e. f is an even multiple of .

Sinusoidal waves with destructive

interference

• When f = n, where n = 1, 3, 5,…, then cos (f/2) = 0.

• The amplitude of the resultant wave is 0.

• Crests of one wave coincide with troughs of the other wave.

• The waves interfere destructively.

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General interference of sinusoidal waves

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• In general, when f is other than 0 or an even multiple of , the amplitude of the resultant is between 0 and 2A,

i.e. 0 f n, then, 0 amplitude 2A.

• The wave functions (linear) still add (superposition).

• The interference is neither constructive nor destructive.

Summary of interference of sinusoidal waves

• Constructive interference occurs when f = n where n is an even integer (including 0).

Amplitude of the resultant is 2A.

• Destructive interference occurs when f = n where n is an odd integer.

Amplitude is 0.

• General interference occurs when 0 < f < n .

Amplitude is 0 < Aresultant < 2A

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Interference in sound waves – 1

• Sound from S can reach R by two different paths.

• The distance along any path from speaker to receiver is called the path length, r.

• The lower path length, r1, is fixed. The upper path length, r2, can be varied.

• Whenever r = |r2 – r1| = n, constructive interference occurs, where n = 0, 1, 2, 3, …

• A maximum in sound intensity is detected at the receiver.

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An acoustical system for demonstrating interference of sound waves

adjustable

Interference in sound waves – 2

• Whenever r = |r2 – r1| = (n is odd), destructive interference occurs. Hence, no sound is detected at the receiver.

• A phase difference may arise between two waves generated by the same source when they travel along paths of unequal lengths.

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2

n

Constructive interference occurs when the path difference is 0, , 2, 3, …

Destructive interference occurs when the path difference is /2, 3/2, 5/2, …

Example 3.1

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QUESTION: Two small loudspeakers, A and B, are driven by the same amplifier and emit pure sinusoidal waves in phase. (a) For what frequencies does constructive interference occur at point P? (b) For what frequencies does destructive interference occur?

SOLUTION: Using Pythagorean theorem, 4.47 m and BP 4.12 m

The path difference is therefore 4.47 m 4.12 m 0.35 m

AP

d

Constructive interference occurs when 0, , 2 ,... or 0, / , 2 / ,... /

350 m/sSo, the possible frequencies are 1, 2,3,...

0.35 m

n

d d v f v f nv f

nvf n n

d

1000 Hz, 2000 Hz, 3000 Hz,...

3 3 5Destructive interference occurs when , ,... or , , ,...

2 2 2 2 2

350 m/sSo, the possible frequencies are 1,3,5,...

2 2 0.35 m

n

v v vd d

f f f

nvf n n

d

500 Hz, 1500 Hz, 2500 Hz,...

Example 3.2

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QUESTION:

Two identical loudspeakers placed 3.00 m apart are driven by the same oscillator. A listener is originally at point O, located 8.00 m from the centre of line connecting the two speakers. The listener then moves to point P, which is a perpendicular distance 0.350 m from O, and she experiences the first minimum in sound intensity. What is the frequency of the oscillator?

Example 3.2 (continued)

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SOLUTION:

Because the sound waves from two separate sources combine, we apply the waves in interference analysis model.

Figure shows the physical arrangement of the speakers, along with two shaded right triangles that can be drawn on the basis of the lengths described. The first minimum occurs when the two waves reaching the listener at point P are 180° out of phase, in other words, when their path difference r equals /2. The path lengths: Hence, the path difference is r2 r1, = 0.13 m. Because this path difference must equal /2 for the first minimum, = 0.26 m. To obtain the oscillator frequency, we use v=f, where v is the speed of sound in air, 343 m/s:

2 2

1

2 2

2

8.00 m 1.15 m 8.08 m

8.00 m 1.85 m 8.21 m

r

r

343 m/s1.3 kHz

0.26 m

vf

Standing waves

• Assume two waves with the same amplitude, frequency and wavelength, traveling in opposite directions in a medium

y1 = A sin (kx – wt) and

y2 = A sin (kx + wt)

• They interfere according to the superposition principle.

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• The resultant wave is y = (2A sin kx) cos wt.

• This is the wave function of a standing wave. There is no (kx – wt) term, and therefore it is not a traveling wave.

• In observing a standing wave, there is no sense of motion in the direction of propagation of either of the original waves.

When the sound waves overlap, identical waves travelling in opposite directions will combine to form standing waves.

(3.1)

Standing wave example

• An oscillation pattern with a stationary outline that results from the superposition of two identical waves traveling in opposite directions.

• The amplitude of the simple harmonic motion of a given element is 2A sin(kx). This depends on the location x of the element in the medium.

• Each individual element vibrates at w.

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Multiflash photography of a standing wave on a string.

Nodes and antinodes – 1

• A node occurs at a point of zero amplitude.

This corresponds to positions x where kx = 2 x/ = n

• An antinode occurs at a point of maximum displacement, 2A.

This corresponds to positions x where kx = 2x/ (m+1/2); m = 0,1,2,…; or kx = 2x/ n/2, n =1,3,5,…

0,1, 2, 3,2

nx n

1, 3, 5,4

nx n

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(3.2)

(3.3)

2 2

Amplitude: 2 sin 2 sin , where x

A kx A k

Nodes and antinodes – 2

• The distance between adjacent antinodes is /2.

• The distance between adjacent nodes is /2.

• The distance between a node and an adjacent antinode is /4.

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• The diagrams below show standing-wave patterns produced at various times by two waves of equal amplitude traveling in opposite directions.

Standing wave patterns produced at various times by two waves of equal amplitude traveling in opposite directions

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Standing waves

Standing waves in a string – 1

• Consider a string fixed at both ends, with a length L.

• Standing waves are set up by a continuous superposition of waves incident on and reflected from the ends.

• There is a boundary condition on the waves: the ends of the strings must necessarily be nodes. They are fixed and therefore must have zero displacement.

• The boundary condition results in the string having a set of normal modes of vibration.

The normal modes of oscillation for the string can be described by imposing the requirements that the ends be nodes and that the separation of nodes and antinodes is /4.

• We identify an analysis model called waves under boundary conditions model.

Each mode has a characteristic frequency.

Standing waves in a string – 2

• This is the first normal mode that is consistent with the boundary conditions.

• There are nodes at both ends and one antinode in the middle.

• This is the longest wavelength mode,

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• Consecutive normal modes add an antinode at each step.

• The section of the standing wave from one node to the next is called a loop.

• The second mode corresponds to = L.

• The third mode corresponds to = 2L/3.

1 2 .

2L L

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String vibration

Standing waves on a string – 3

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2 2n

n

v v n Tf

L L

n

2 1,2,3,...n

Ln

n (3.4)

(3.5)

• The wavelengths of the normal modes for a string of length L fixed at both ends are:

where n is the nth normal mode of oscillation

• These are the possible modes for the string.

• The natural frequencies are

• fn are also called quantised frequencies

Waves on a string, harmonic series

• The fundamental frequency corresponds to n = 1. It is the lowest frequency, ƒ1

• The frequencies of the remaining natural modes are

integer multiples of the fundamental frequency, i.e.

• Frequencies of normal modes that exhibit this relationship form a harmonic series.

• The normal modes are called harmonics. 28

1

1

1

2 2

v v Tf

L L

1ƒnf n (3.6)

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Example 3.3

Figure below shows a pattern of resonant oscillation of a string of mass m = 2.5 g and length L = 0.8 m and that is under tension T = 325.0 N. (A) What is the wavelength of the transverse waves producing the

standing waves pattern, and what is the harmonic number n? (B) What is the frequency f of the transverse waves and of the

oscillations of the moving string elements? (C) What is the maximum magnitude of the transverse velocity, um of

the element oscillating at coordinate x = 0.18 m?

QUESTION:

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Example 3.3 (continued)

(A) Two full wavelengths fit into the length L = 0.8 m of the string, Thus, we have

SOLUTION:

0.8 m2 0.4 m

2 2

LL

By counting the number of loops (or half-wavelengths), we can obtain the harmonic number as n = 4.

3

325 N 0.8 m322.49 m/s

/ 2.50 10 kg

T T TLv

m L m

(B) The speed of wave

322.49 m/s 322.49 m/s

Frequency 806.2 Hz or 4 806 Hz0.4 m 2 2 0.8 m

v vf f n

L

(C) The displacement 2 sin cosmy y kx tw

Transverse speed 2 sin cos 2 sin sinm m

yu y kx t y kx t

t tw w w

3

2 0.18 mMaximum magnitude 2 2.0 10 m 2 806.2 Hz sin 6.26 m/s

0.4 mmu

Musical instrument

• The musical note is defined by its fundamental frequency.

• The frequency of the string can be changed by changing either its length or its tension.

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• For example, the tension in guitar and violin strings is varied by a screw adjustment mechanism or by tuning pegs located on the neck of the instrument.

• As the tension is increased, the frequency of the normal modes increases in accordance with eq (3.5). Once the instrument is "tuned," players vary the frequency by moving their fingers along the neck, thereby changing the length of the oscillating portion of the string.

• As the length is shortened, the frequency increases because, as eq (3.5) specifies, the normal mode frequencies are inversely proportional to string length.

The frequency of a string that defines the

musical note that it plays is that of the

fundamental. The string's frequency can be

varied by changing either the string's

tension or its length.

Resonance

• A system is capable of oscillating in one or more normal modes.

• If a periodic force is applied to such a system, the amplitude of the resulting motion is greatest when the frequency of the force is equal to one of the natural frequencies of the system. This phenomena is called resonance.

• Because an oscillating system exhibits a large amplitude when driven at any of its natural frequencies, these frequencies are referred to as resonance frequencies, f0.

• If the system is not driven at one of the natural frequencies, a standing wave is not setup, then only small oscillation is obtained.

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Example 3.4

QUESTION:

A violin string 15.0 cm long and fixed at both ends oscillates in its n = 1 mode. The speed of waves on the string is 250 m/s, and the speed of sound in air is 348 m/s. What are the (a) frequency and, (b) wavelength of the emitted sound wave?

SOLUTION: (1)(250m/s)

(a) Resonant frequency for violin, 833Hz2 2(0.150m)

(b) The frequency of the wave on the string is the same as the frequency

of the sound wave it produces during its vibration.

nvf

L

348 m/s Consequently, the wavelength in air is 0.418 m

833 Hz

v

f

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Example 3.5

High-frequency sound can be used to produce standing-wave vibrations in a wine glass. A standing-wave vibration in a wine glass is observed to have four nodes and four antinodes equally spaced around the 25.0-cm circumference of the rim of the glass. If transverse waves move around the glass at 940 m/s, an opera singer would have to produce a high harmonic with what frequency to shatter the glass with a resonant vibration?

QUESTION:

SOLUTION: The distance between two nodes or antinodes is /2. Because four nodes are present, the wavelength around the circumference of the glass is 4×/2 = 2.

Since the circumference is 25.0 cm, we have 2=25.0 cm =12.5 cm From 940 m/s

7520 Hz0.125 m

vf

The singer must match this

frequency in order to crack the glass.

Standing waves in air columns – 1

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• Waves under boundary conditions model can be applied.

• Standing waves can be set up in air columns as the result of interference between longitudinal sound waves traveling in opposite directions.

• The phase relationship between the incident and reflected waves depends upon whether the end of the pipe is opened or closed.

Organ pipes

Standing waves in air columns – 2

• The open end of a pipe is a displacement antinode in the standing wave.

• As the compression region of the wave exits the open end of the pipe, the constraint of the pipe is removed and the compressed air is free to expand into the atmosphere.

• The open end corresponds with a pressure node. It is a point of no pressure variation.

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Open end

Standing waves in air columns – 3

• A closed end of a pipe is a displacement node in the standing wave.

• The rigid barrier at this end will not allow longitudinal motion in the air.

• The closed end corresponds with a pressure antinode.

• It is a point of maximum pressure variations.

• The pressure wave is 90o out of phase with the displacement wave.

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Closed end

Standing waves in air columns – 4

• Both ends are displacement antinodes.

• The fundamental frequency is v/2L. This corresponds to the first diagram.

• The higher harmonics are ƒn = nƒ1 = n (v/2L) where n = 1, 2, 3, …

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For the case of open tube

Standing waves in air columns – 5

• The closed end is a displacement node.

• The open end is a displacement antinode.

• The fundamental frequency is v/4L.

• The frequencies are

ƒn = nƒ = n (v/4L) where n = 1, 3, 5, …odd integers only.

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For the case of one end of the tube is close

Standing waves in air columns – 6

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How a sound can be reflected from an open end? • Sound can be represented as a pressure wave, however, and a compression region of the

sound wave is constrained by the sides of the pipe as long as the region is inside the pipe.

• As the compression region exits at the open end of the pipe, the constraint of the pipe is removed and the compressed air is free to expand into the atmosphere.

• Therefore, there is a change in the character of the medium between the of the pipe and the outside even though there is no change in the material of the medium. This change in character is sufficient to allow some reflection.

• In a pipe open (or closed) at both ends, the natural frequencies of oscillation form a harmonic series that includes all integral multiples of the fundamental frequency.

• In a pipe closed at one end and open at other end, the natural frequencies of oscillations form a harmonic series that includes only odd integral multiples of the fundamental frequency.

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Example 3.6

A section of drainage culvert 1.28 m in length makes a howling noise when the wind blows across its open ends. (A) Determine the frequencies of the first three harmonics of the culvert if it is cylindrical in shape and open at both ends. Take v = 343 m/s as the speed of sound in air. (B) What are the three lowest natural frequencies of the culvert if it is blocked at one end?

QUESTION:

SOLUTION:

1

2 1

3 1

343 m/s139 Hz

2 2 1.23 m

2 279 Hz

3 418 Hz

vf

L

f f

f f

1

3 1

5 1

343 m/s69.7 Hz

4 4 1.23 m

3 209 Hz

5 349 Hz

vf

L

f f

f f

(A) (B)

Multiplying by odd integers

Multiplying by integers

Find the first harmonic Find the first harmonic

Standing waves in rods – 1

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• A rod is clamped in the middle.

• The clamp forces a displacement node.

• The ends of the rod are free to vibrate and so will correspond to displacement antinodes.

Standing waves in rods – 2

• By clamping the rod at other points, other normal modes of oscillation can be produced.

• Here the rod is clamped at L/4 from one end.

• This produces the second normal mode.

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An aluminium rod is clamped one quarter of the way along its length and set into longitudinal vibration by a variable-frequency driving source. The lowest frequency that produces resonance is 4000 Hz. The speed of sound in an aluminium rod is 5100 m/s. Find the length of the rod.

Example 3.7

QUESTION:

SOLUTION:

When the rod is clamped at one-quarter of its length, the vibration pattern reads A-N-A-N-A and the rod length is

L

5100 m/sTherefore, 1.16 m

4400 Hz

vL

f