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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.
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Topic 4: Oscillations and waves4.5 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°.
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. 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.
Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. 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. B. C. A.
reflected waves Incident = Reflect Reflective surface Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. 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: Incident ray Incident normal Reflect Reflected ray
Reflective surface Reflective surface Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. 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… Flat or straight wavefront Spherical wavefront
Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. 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. BOUN DARY REFRACTED WAVE angle of incidence Incidence Refraction normal angle of refraction INCIDENT WAVE CLEAR WATER MUDDY WATER
CONCRETE DEEP MUD Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. Imagine the ranks of a marching band. Obviously the cadence does not change. Thus the period and the frequen- cy do not change. But speed and wave- length do change.
BOUN DARY REFRACTED WAVE angle of incidence Incidence Refraction normal angle of refraction INCIDENT WAVE CLEAR WATER MUDDY WATER Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. 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.
BOUN DARY 1 2 Snell’s law for refracted waves v2/v1 = sin2/sin1 normal Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. Snell’s law relates the wave’s velocities to their angles: 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 = sin2/sin1 so v2/20 = sin32°/sin28° v2 = 23 ms-1.
BOUN DARY 68° normal 75° Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. • PRACTICE: A wave travel- ing at 50 ms-1 strikes a boundary as shown. Part of the wave energy is reflected and part is refracted. • 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 = sin2/sin1 • v2/50 = sin15°/sin22° or v2 = 35 ms-1. 22° 22° 15°
1 2 n2 n1 glass 40° 36° air Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. PRACTICE: The refractive indexn 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 n1sin1 = n2sin2. (a) Light enters a glass sample and is refracted as in the figure. What is the index of refraction of the glass sample? From n1sin1 = n2sin2 we obtain 1(sin40°) = n2(sin36°) n2 = 1.1 (unitless)
1 2 n2 n1 glass 40° 36° air Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. PRACTICE: The refractive indexn 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 n1sin1 = n2sin2. (b) What is the speed of light in the glass sample? The speed of light in air is 3.0108 ms-1. From Snell’s law: v2/v1 = sin2/sin1 v2/3.0108 = sin36°/sin40° v2 = 2.7108 ms-1.
Snell’s law for refracted waves v2/v1 = sin2/sin1 Snell’s law for refracted waves n1/n2 = v2/v1 = sin2/sin1 n = 1 for air and free space Topic 4: Oscillations and waves4.5 Wave properties Describe the reflection and transmission of waves at the boundary between two media. Recall Snell’s law: From n1sin1 = n2sin2 we get n1/n2 =sin2/sin1. By inspection of Snell’s law we see that it can be expanded: 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 ms-1.
Topic 4: Oscillations and waves4.5 Wave properties 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. REFLECTED WAVE DIFFRACTED WAVE INCIDENT 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.
b b b Topic 4: Oscillations and waves4.5 Wave properties Explain and discuss qualitatively the diffraction of waves at apertures and obstacles. 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 = 2 b = 6 b = 12
Topic 4: Oscillations and waves4.5 Wave properties Explain and discuss qualitatively the diffraction of waves at apertures and obstacles. 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.
Topic 4: Oscillations and waves4.5 Wave properties Describe examples of diffraction. PRACTICE: Classify each property of a sound wave. (a) What wave property is being demonstrated here? Diffraction.
Topic 4: Oscillations and waves4.5 Wave properties Describe examples of diffraction. PRACTICE: Classify each property of a sound wave. (b) What wave property is being demonstrated here? Reflection. This is the same principle behind fiber optics. Light is used instead of sound.
Topic 4: Oscillations and waves4.5 Wave properties Describe examples of diffraction. PRACTICE: Classify each property of a sound wave. (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).
Topic 4: Oscillations and waves4.5 Wave properties Describe examples of diffraction. PRACTICE: Classify each property of a sound wave. (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. incidence reflection 0
Topic 4: Oscillations and waves4.5 Wave properties 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. 2x0 Constructive interference x0 0
Topic 4: Oscillations and waves4.5 Wave properties 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 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 Destructive interference 0 - x0
1 m 1 m 1 m Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. PRACTICE: Two pulses are shown in a string approach- ing each other at 1 ms-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. Since the pulses are 1 m apart and approach- ing each other at 1 ms-1, at this time they will each have moved ½ m and their leading edges will just meet. Since the pulses are 1 m apart and approaching each other at 1 ms-1 they will overlap at this time canceling out.
Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. • 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 • 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.
5 y =yn n=1 y 1 2 1 4 t 0 T 2T 1 4 - 1 2 - 1 1 y1 = - sin t 1 2 1 3 1 4 1 5 y2 = - sin 2t y3 = - sin 3t y4 = - sin 4t y5 = -sin 5t Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. EXAMPLE: Fourier series are examples of the superposition principle. You can create any waveform by summing up sine waves!
Topic 4: Oscillations and waves4.5 Wave properties State the principle of superposition and explain what is meant by constructive and destructive interference. 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. C (crest-crest) A B B (trough-trough) C D A (crest-trough)
SIGNAL WAVE IRREGULAR SEA WAVE SUPERPOSITION OF ABOVE TWO WAVES Topic 4: Oscillations and waves4.5 Wave properties State the principle of superposition and explain what is meant by constructive and destructive interference. EXAMPLE: Rogue waves are examples of superposition. Which type of interference is illustrated? Constructive interference. SOURCE: http://www.math.uio.no/~karstent/waves/index_en.html
condition for constructive interference path difference = n n is an integer Topic 4: Oscillations and waves4.5 Wave properties 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 that L1 = 4 L2 = 3 P1 L1 = 4 L2 = 2 S2 S1
condition for destructive interference path difference = (n + ½) n is an integer Topic 4: Oscillations and waves4.5 Wave properties 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 L2 = 4 P2 L1 = 3.5 L1 = 3 L2 = 2.5 S2 S1
Topic 4: Oscillations and waves4.5 Wave properties 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? ________ B C A D A and B C and D
Topic 4: Oscillations and waves4.5 Wave properties 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. B C A D D
Topic 4: Oscillations and waves4.5 Wave properties 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.
Topic 4: Oscillations and waves4.5 Wave properties 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.
N N N N N N N Topic 4: Oscillations and waves4.5 Wave properties 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). A A A A A A
Topic 4: Oscillations and waves4.5 Wave properties 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 N A Because it is fixed at each end.
L N N A Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. 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 frequencyf1. The fundamental frequency of any system is called the first harmonic. 1st harmonic fundamental frequency f1 = v/(2L)
L Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. 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. N N N 2nd harmonic A A f2 = v/L
f1 = v/2L f2 = v/L f3 = Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. 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. 3v/2L
Topic 4: Oscillations and waves4.5 Wave properties 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 = L f3 = 5v/4L f2 = 3v/4L f1 = v/4L Air can’t move.
Topic 4: Oscillations and waves4.5 Wave properties 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 = L f3 = 3v/2L f2 = 2v/2L f1 = v/2L FYI The IBO requires you to be able to make sketches of string and pipe harmonics and find wavelengths and frequencies.
Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. 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.
y-x Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. 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).
Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. 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.
Topic 4: Oscillations and waves4.5 Wave properties Apply the principle of superposition to determine the resultant of two waves. 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.