Topic 4: Oscillations and waves 4.5 Wave properties

4.5.1 Describe the reflection and transmission of waves at the boundary between two media.

4.5.2 State and apply Snell’s law.

4.5.3 Explain and discuss qualitatively the diffraction of waves at apertures and obstacles.

4.5.4 Describe examples of diffraction.

4.5.5 State the principle of superposition and explain what is meant by constructive and destructive interference.

4.5.6 State and apply the conditions for constructive and destructive interference in terms of path difference and phase difference.

4.5.7 Apply the principle of superposition to determine the resultant of two waves.

Topic 4: Oscillations and waves4.5 Wave properties

Describe the reflection and transmission of waves at the boundary between two media.

Suppose a traveling wave approaches a boundary between two media. Reflection is where some of the wave’s energy bounces back at the boundary without passing through.

Consider a single wave pulse sent along a tight rope fixed to the wall as shown:


FYI

Waves reflected from a point at which the medium CANNOT move are reflected WITH A PHASE SHIFT of 180°.


Suppose a traveling wave approaches a boundary between two media. Reflection is where some of the wave’s energy bounces back at the boundary without passing through. Now consider a single wave pulse sent along a tight rope free to move on a pipe as shown:


FYI

Waves reflected from a point at which the medium CAN move are reflected WITHOUT A PHASE SHIFT.



PRACTICE:

(a) A wave pulse is shown in a cable whose right end is fixed securely to a wall. Which pulse configuration is possible later?

(b) A wave pulse is shown in a cable whose right end is free to slide on a pipe. Which pulse configuration is possible later?

A.B. C.

A. B. C.


Wave reflection occurs when a wave meets a boundary, such as a solid object, or a change in the medium (for example, a thermal layer), and is at least partially diverted backward.

For two-dimensional waves we look at wave rays which are wave velocity directions.

The angles of the rays are always measured with respect to the normal to the surface.

The relationship between the angle of incidence Incident and the angle of reflection Reflect is very simple:


Reflective

surface

Incident ray

Reflected

ray

IncidentReflect

Incident = Reflect reflected waves

normal


Wave reflection occurs when a wave meets a boundary, such as a solid object, or a change in the medium (for example, a thermal layer), and is at least partially diverted backward.

We can also look at wave fronts: Observe…


Reflective

surface

Reflective

surface

Spherical wavefront

Flat or straight wavefront


Wave refraction occurs when a wave meets a boundary, such as a solid object, or a change in the medium (for example, a thermal layer), and is at least partially allowed through the boundary.


CLEAR WATER MUDDY WATER

BOUN DARY

INCIDENT WAVE

REFRACTED WAVE

normal

Incidence

angle of incidence

Refraction angle of refraction


Imagine the ranks of a marching band. Obviously the cadence does not change.

Thus the period and the frequency do not change.

But speed and wavelength do change.

CONCRETE

DEEP MUD



During refraction the frequency and the period do not change.

It should be clear that the incident wave is faster than the refracted wave in this example.


CLEAR WATER MUDDY WATER

BOUN DARY

INCIDENT WAVE

REFRACTED WAVE

normal

Incidence

angle of incidence

Refraction angle of refraction


Snell’s law relates the wave’s velocities to their angles:


BOUN DARY

normal

12

v2/v1 = sin 2/sin 1 Snell’s law for refracted waves

EXAMPLE:

A wave traveling at 20 ms-1 hits a medium boundary at an angle of 28° relative to the normal to the boundary, and is refracted at 32°. What is the wave’s speed in the new medium?

From Snell’s law we have v2/v1 = sin 2/sin 1 so

v2/20 = sin 32°/sin 28°

v2 = 23 m s-1.



PRACTICE: A wave traveling at 50 m s-1 strikes a boundary as shown. Part of the wave energy is reflected and part is refracted.

(a) What is the angle of reflection?

Since all angles are measured wrt the normal, the incident angle is 90° – 68° = 22°. This is also the reflected angle since Incident = Reflect.

(b) What is the speed of the transmitted wave?

The angle of refraction is 90° – 75° = 15°. v2/v1 = sin 2/sin 1 v2/50 = sin 15°/sin 22° or v2 = 35 m s-1.

BOUN DARY

normal68°

75°

22°22° 15°



PRACTICE: The refractive index n is often used when dealing with light waves. The index for air and empty space is 1.00. The relationship between the indices of refraction and the angles is given by n1sin 1 = n2sin 2.

(a) Light enters a glass sample and is refracted as in the figure. What is the index of refraction of the glass sample?

From n1sin 1 = n2sin 2 we obtain

1(sin 40°) = n2(sin 36°)

n2 = 1.1 (unitless)

12n1 n2

40°36°

air

glass



PRACTICE: The refractive index n is often used when dealing with light waves. The index for air and empty space is 1.00. The relationship between the indices of refraction and the angles is given by n1sin 1 = n2sin 2.

(b) What is the speed of light in the glass sample? The speed of light in air is 3.0108 m s-1.

From Snell’s law: v2/v1 = sin 2/sin 1 v2/3.0108 = sin 36°/sin 40°

v2 = 2.7108 m s-1.

12n1 n2

40°36°

air

glass


Recall Snell’s law:

From n1sin 1 = n2sin 2 we get

n1/n2 = sin 2/sin 1.

By inspection of Snell’s law we see that it can be expanded:


v2/v1 = sin 2/sin 1 Snell’s law for refracted waves

n1/n2 = v2/v1 = sin 2/sin 1 Snell’s law for refracted wavesn = 1 for air and free space

EXAMPLE: The index of refraction for a particular glass is 1.65. What is the speed of light in it?

SOLUTION:

n1/n2 = v2/v1

1/1.65 = v2/3.0108 so that v2 = 1.8108 m s-1.

Explain and discuss qualitatively the diffraction of waves at apertures and obstacles.

If a wave meets a hole in a wall that is of comparable size to its wavelength, the wave will be bent through a process called diffraction.

If the aperture (hole, opening, etc.) is much larger than the wavelength, diffraction will be minimal to nonexistent.


INCIDENT WAVE

DIFFRACTED WAVE

REFLECTED WAVE

FYI

Diffraction is caused by objects within the medium that interact with the wave. It is not caused by two mediums and their boundary.


The reason waves can turn corners is that the incoming wave transmits a disturbance by causing the medium to vibrate.

And wherever the medium vibrates it becomes the center of a new wave front as illustrated below.

Note that the smaller the aperture b the more pronounced the diffraction effect.


FYI

The aperture size must be of the order of a wavelength in order for diffraction to occur. b = 12

b = 6b = 2

b b b

PRACTICE:

If d is too small the bat’s sound waves will not even be disturb- ed enough for the bat to detect the insect.



Describe examples of diffraction.


PRACTICE: Classify each property of a sound wave.

(a) What wave property is being demonstrated here?

Diffraction.




(b) What wave property is being demonstrated here?

Reflection. This is the same principle behind fiber optics. Light is used instead of sound.




(c) What wave property is being demonstrated here?

Refraction.(d) T or F: The period changes in the different media.

(e) T or F: The frequency changes in the different media.

(f) T or F: The wavelength changes in the different media.

(g) T or F: The wave speed changes in the different media.

(h) T or F: The sound wave is traveling fastest in the warm air.

A sound pulse entering and leaving a pocket of cold air (blue).




(i) What wave property is being demonstrated here?

Reflection.This is an echo from a wall.(j) T or F: The frequency of the reflected wave is different from the frequency of the incident wave?

(k) Fill in the blanks: The angle of ___________ equals the angle of ___________.

(l) Fill in the blank: In this example, both angles equal ___°.

Remember that angles are measured wrt the normal.

incidencereflection

0

State the principle of superposition and explain what is meant by constructive and destructive interference.

Wave superposition is simply the addition of two or more waves.

Superposition is also called interference and can be constructive or destructive.

Consider two in-phase pulses coming from each end of a taught rope.

The amplitudes x0 of the two pulses add together, producing a momentary pulse of amplitude 2x0.


0x0

2x0 Constructive interference


Wave superposition is simply the addition of two or more waves.

Superposition is also called interference and can be constructive or destructive.

Consider two 180° out-of-phase pulses coming from each end of a taught rope.

The amplitudes x0 of the two pulses cancel, producing a momentary pulse of amplitude 0.


- x0

0x0 Destructive

interference

Apply the principle of superposition to determine the resultant of two waves.


PRACTICE: Two pulses are shown in a string approaching each other at 1 m s-1.

Sketch diagrams to show each of the following:

(a) The shape of the string at exactly t = 0.5 s later.

(b) The shape of the string at exactly t = 1.0 s later.

1 m 1 m 1 m

Since the pulses are 1 m apart and approaching each other at 1 ms-1 they will overlap at this time canceling out.

Since the pulses are 1 m apart and approaching each other at 1 m s-1, at this time they will each have moved ½ m and their leading edges will just meet.



EXAMPLE:

Two waves P and Q reach the same point at the same time, as shown in the graph.

The amplitude of the resulting wave is

A. 0.0 mm. B. 1.0 mm. C. 1.4 mm. D. 2.0 mm.

In the orange regions P and Q do not cancel.In the purple regions P and Q do cancel.Focus on orange. Try various combos until…Blue (0.7 mm) plus green (0.7 mm) = 1.4 mm.Note that the crests to not coincide, so NOT 2.0.



EXAMPLE: Fourier series are examples of the superposition principle. You can create any waveform by summing up sine waves!

T 2Tt

y

0

14

12

14

12

-

-

y1 = - sin t11

y2 = - sin 2t12

y3 = - sin 3t13 y4 = - sin 4t

14

y5 = - sin 5t15y = yn

n=1

5



PRACTICE: Two apertures in a sea wall produce two diffraction patterns as shown in the animation.

(a) Which letter represents a maximum displacement above equilibrium? __(b) Which letter represents a maximum displacement below equilibrium? __(c) Which letter represents a minimum displacement from equilibrium? __Crest-crest = max high. Trough-trough = max low.Crest-trough = minimum displacement.

A

B

C

D

C (crest-crest)

B (trough-trough)

A (crest-trough)



EXAMPLE: Rogue waves are examples of superposition. Which type of interference is illustrated?

Constructive interference.SIGNAL

WAVE

IRREGULAR SEA WAVE

SUPERPOSITION OF ABOVE TWO

WAVESSOURCE: http://www.math.uio.no/~karstent/waves/index_en.html


S1S2

P1

L 1 = 4

L 2 =

2

L 1 =

4

L 2 =

3

State and apply the conditions for constructive and destructive interference in terms of path difference and phase difference.

The following animation showing two coherent (in-phase and same frequency) wave sources S1 and S2 should show thatpath difference = n condition for

constructive interference n is an integer


S1S2

P2

L 1 =

3

L2 =

2.5L 1

= 3

.5

L2 = 4

State and apply the conditions for constructive and destructive interference in terms of path difference and phase difference.

The following animation showing two coherent wave sources S1 and S2 should show that

path difference = (n + ½) condition for destructive interferencen is an integer


State and apply the conditions for constructive and destructive interference in terms of path difference and phase difference.PRACTICE: Two identical wave sources in a ripple tank produce waves having a wavelength of . The interference pattern is shown. Four reference lines are labeled A through D.

(a) Which reference line or lines represent constructive interference? ________

(b) Which reference line or lines represent destructive interference? ________

D

CB

A

A and B

C and D


State and apply the conditions for constructive and destructive interference in terms of path difference and phase difference.PRACTICE: Two identical wave sources in a ripple tank produce waves having a wavelength of . The interference pattern is shown. Four reference lines are labeled A through D.

(c) Which reference line or lines represent a path difference of 2.5? ________2.5 is a condition for destructive interference.These two represent the smallest difference 0.5.C represents 1.5 difference, and D is thus 2.5.

D

CB

A

D


State and apply the conditions for constructive and destructive interference in terms of path difference and phase difference.PRACTICE: Observe the 2D sound wave animation. Sketch in the regions where there is complete destructive interference.

Note that the red regions and the blue regions are 180° out of phase.

Where they meet we have destructive interference in purple.

In the purple region there is no displacement.

Apply the principle of superposition to determine the resultant of two waves.The principle of superposition yields a surprising resultant for two identical waves traveling in opposite directions.

Snapshots of the blue and the green waves and their red resultant follow:

Because the resultant red wave appears to not be traveling it is called a standing wave.


Apply the principle of superposition to determine the resultant of two waves.The standing wave has two important properties. It does not travel to the left or the right as the blue and the green wave do.Its “lobes” grow and shrink and reverse, but do not go to the left or the right.Any points where the standing wave has no displacement is called a node (N).The lobes that grow and shrink and reverse are called antinodes (A).


N N N N N N N

A A A A A A

Apply the principle of superposition to determine the resultant of two waves.You may be wondering how a situation could ever develop in which two identical waves come from opposite directions. Well, wonder no more.When you pluck a stringed instrument, waves travel to the ends of the string and reflect at each end, and return to interfere under precisely the conditions needed for a standing wave.

Note that there are two nodes and one antinode.Why must there be a node at each end of the string?


L

N NA

Because it is fixed at each end.


Observe that precisely half a wavelength fits along the length of the string.Thus we see that = 2L.Since v = f we see that f = v/(2L) for a string.This is the lowest frequency you can possibly get from this string configuration, so we call it the fundamental frequency f1.The fundamental frequency of any system is called the first harmonic.


L

N NA

fundamental frequency f1 = v/(2L)

1st harmonic


The next higher frequency has another node and another antinode.

We now see that = L.Since v = f we see that f = v/L.This is the second lowest frequency you can possibly get and since we called the fundamental frequency f1, we’ll name this one f2.This frequency is also called the second harmonic.


L

N NA

f2 = v/LA

N 2nd harmonic

PRACTICE: Complete the table below with both sketch and formula.

Remember that there are always nodes on each end of a string.

Add a new well-spaced node each time.Decide the relationship between and L.We see that = (2/3)L.Since v = f we see that f = v/(2/3)L = 3v/2L.



f2 = v/L

f1 = v/2L

f3 = 3v/2L

Apply the principle of superposition to determine the resultant of two waves.We can also set up standing waves in pipes.In the case of pipes, longitudinal waves are created (instead of translational waves), and these waves are reflected from the ends of the pipe.Consider a closed pipe of length L which gets its wave energy from a mouthpiece on the left side.

Why must the mouthpiece end be an antinode?Why must the closed end be a node?


(1/4)1 = L (3/4)2 = L (5/4)2 = Lf1 = v/4L f2 = 3v/4L f3 = 5v/4L

Air can’t move.

Apply the principle of superposition to determine the resultant of two waves.In an open-ended pipe you have an antinode at the open end because the medium can vibrate there (and, of course, at the mouthpiece).


(1/2)1 = L 2 = L (3/2)2 = Lf1 = v/2L f2 = 2v/2L f3 = 3v/2L

FYI

The IBO requires you to be able to make sketches of string and pipe harmonics and find wavelengths and frequencies.

PRACTICE: A tube is filled with water and a vibrating tuning fork is held above the open end. As the water runs out the of the tap at the bottom sound is loudest when the water level is a distance x from the top. The next loudest sound comes when the water level is at a distance y from the top. Which expression for is correct, if v is the speed of sound in air?

A. = x B. = 2x C. = y-x D. = 2(y-x)v = f and since v and f are constant, so is .The first possible standing wave is sketched.The sketch shows that = 4x, not a choice.



PRACTICE: A tube is filled with water and a vibrating tuning fork is held above the open end. As the water runs out the of the tap at the bottom sound is loudest when the water level is a distance x from the top. The next loudest sound comes when the water level is at a distance y from the top. Which expression for is correct, if v is the speed of sound in air?

A. = x B. = 2x C. = y-x D. = 2(y-x)The second possible standing wave is sketched.Notice that y – x is half a wavelength.Thus the answer is = 2(y - x).



y-x



PRACTICE: This drum head set to vibrating at different resonant frequencies has black sand on it, which reveals 2D standing waves.

Does the sand reveal nodes, or does it reveal antinodes?

Why does the edge have to be a node?

Nodes, because there is no displacement to throw the sand off.

The drumhead cannot vibrate at the edge.



EXAMPLE:

These images are of hologram interferograms of 3D standing waves on a vibrating c5 handbell. The nodes appear bright white, and the antinodes appear patterned.

Topic 4: Oscillations and waves 4.5 Wave properties

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Transcript of Topic 4: Oscillations and waves 4.5 Wave properties