VCE PHYSICS Unit 4 Topic 3 SOUND
UNIT OUTLINE • To achieve this outcome students should demonstrate the knowledge and skills to; • explain sound as the transmission of energy via longitudinal pressure waves; mathematically model the relationship between wavelength, frequency and speed of propagation of sound waves using v = fλ explain the difference between sound intensity (Wm-2) and sound intensity level (dB) calculate sound intensity at different distances from a source using an inverse square law. explain resonance in terms of superposition of a travelling sound wave and its reflection analyse, for strings and open and closed resonant tubes the fundamental as the 1st harmonic and subsequent harmonics explain qualitatively, in terms of electrical and electromagnetic effects the operation of microphones, including electret - condenser, crystal, dynamic and velocity microphones dynamic loudspeakers explain qualitatively the effects of baffles and enclosures for loudspeakers interpret frequency response curves of microphones, speakers, simple sound systems and hearing including loudness (phon) evaluate the fidelity of microphones and loudspeakers in terms of purpose, frequency response and qualitatively construction interpret qualitatively the directional spread of various frequencies in terms of different gap width or obstacle size including the significance of the magnitude of the λ/w ratio use safe and responsible practices when working with sound sources and sound equipment
Chapter 1 Topics covered: • Wave nature of Sound. • Transverse Waves. • Longitudinal Waves. • Sound Production, Transmission, Detection and Absorption
The “WAVE” nature of SOUND 1.0 The Wave Nature of Sound • Waves are a method of TRANSFERRING ENERGY from one place to another. • Some waves (eg. Sound, Water Waves) need a MEDIUM through which to travel. • The MEDIUM (eg. air, water), although disturbed by the passage of the waves, does NOT suffer any PERMANENT DISTORTION due to the wave’s movement through it. Sound Waves require matter (either solid, liquid or gas) as their medium This means, of course, no one can hear you scream in space. There are two basic types of waves: TRANSVERSE WAVES. LONGITUDINAL WAVES.
Direction of motion of the medium’s particles Direction of motion of wave 1.1 Transverse Waves Transverse waves are characterised by having the individual particles of the medium through which the wave travels, moving perpendicular to the direction of motion of the wave. 1.1 Transverse Waves Notice the “medium” does not move along with the wave. Pick a spot and follow its motion.
Individual particles of Medium Direction of Wave Motion Direction of Motion of Particles of Medium 1.2 Longitudinal Waves LONGITUDINAL WAVES are characterised by having the individual particles which make up the medium through which the wave travels, moving parallel to the direction of motion of the wave. Sound is a LONGITUDINAL WAVE. Again, notice the “medium” does not move along with the wave. Pick a spot and follow its motion.
Question 1 In the sentences below, options are given within the brackets. Only one of the options will be correct. Circle the best option. A sound wave is a [torsional / transverse / longitudinal] wave in which the air particles move [at right angles to / parallel to / by spiralling around]the direction of propagation of the wave. The wave transmits[energy / air particles / wave maxima] from the source to the receiver.
Compression Rarefaction 1.3 Sound Production Sound is produced by making an object vibrate (move backward and forward). As the object vibrates back and forth, it pushes on the air particles immediately next to it, creating a series of COMPRESSIONS and RAREFACTIONS which move outward from the source. This moving chain of compressions and rarefactions form a Sound Wave. The faster the object vibrates, the higher the frequency of the sound.
Consider a dust particle one metre in front of a loudspeaker that is producing a constant tone sound wave. Question 2 Which one of the following statements and diagrams (A to D below) best describes the motion of the dustparticle? The dust particle oscillates in a vertical direction. The dust particle travels away from the speaker with the wave. The dust particle remains stationary as the wave passes. The dust particle oscillates in a horizontal direction
1.4 Sound Transmission • Sound is transmitted from one place to another through a MEDIUM. • The medium may be solid, liquid or gas. • Generally the DENSER the medium the FASTER the speed of sound. • Sound is transmitted through a medium by causing the particles of the medium to be disturbed from their mean or average positions as the wave passes by. • The particles making up the medium DO NOT move along with the sound wave. • The medium suffers no permanent “effect” from having a sound wave pass through it.
A particle of dust is floating at rest 10 cm directly in front of a loudspeaker that is not operating. The loudspeaker then emits sound of frequency of 10 Hz and speed of 330 ms–1. Question 3 Which one of the following statements best describes the motion of the dust particle? A. It vibrates vertically up and down at 10 Hz remaining on average 10 cm in front of the loudspeaker. B. It vibrates horizontally backwards and forwards at 10 Hz remaining on average 10 cm in front of the loudspeaker. C. It travels away from the loudspeaker at 330 ms–1 while moving horizontally backwards and forwards at 10 Hz. D. It remains at rest.
1.5 Sound Detection • Sound is detected by making a receiver vibrate. • Sound detection occurs in devices such as microphones where incoming sound waves cause the production of an electrical signal. • Sound is detected by humans using our ears, in particular our Cochlea, a circular canal lined with clumps of hairs. Each hair clump is designed to react to a particular frequency. • Sound Level Meters are used to measure Sound Intensity Levels, which are displayed in decibels (dB). • The output from most sound level meters is adjusted to mirror the ear’s response by using the so called dB(A) scale. (see Slide 3.4 - Frequency Response Graphs)
1.6 Sound Absorption • When a sound strikes a barrier, it is either reflected off, transmitted through or absorbed by, that barrier. • The amount of reflection, transmission or absorption depends upon the nature of the barrier. • The physical absorption of sound, as measured by the ABSORPTION COEFFICIENT (A.C.), occurs when the energy of the wave is transformed into other forms of energy (eg. Heat) within the absorbing material. • The A.C. varies with frequency. • Hard, rigid, non-porous materials have low A.C.’s • Soft, pliable, porous materials have high A.C.’s
Chapter 2 Topics covered: • Amplitude. • Period. • Frequency. • Wavelength. • Wave Speed. • Sound Waves in Air.
Point of Max. Pressure above Atmospheric, a COMPRESSION P Amplitude Time Point of Min. Pressure below Atmospheric, a RAREFACTION Atmospheric Pressure 2.0 Amplitude Amplitude is a measure of the size of a disturbance above or below a mean or average value. In sound wave representations, the amplitude is measured as a variation in air pressure (P), above or below the normal atmospheric pressure. This method allows sound to be presented as a transverse rather than a longitudinal wave. The unit for P is the PASCAL (1 Pa = 1 Nm-2). Human ears interpret Amplitude as the “Loudness” of a sound: Large amplitude = loud sound, Small amplitude = soft sound
P Low Frequency Time High Frequency 2.1 Frequency • Frequency (symbol f ) is most generally defined as the number of events which occur during a time interval. • In terms of Sound Waves it represents the number of complete sound waves passing a given point in a given time. • In the SI system, frequency is defined as the number of events or cycles per second. • The UNIT for frequency is the HERTZ (Hz), where 1 Hz = 1 cycle per second Human ears interpret frequency as the “pitch” of a sound: High frequency = high pitch, Low frequency = low pitch
P Time 0.02 0.04 2.2 Period • Period (symbol T) is defined as the time it takes for one event to occur. • It is the time it takes for one complete sound wave to pass a given point. • Period is the measure of a time interval, thus has the unit seconds (s). • Period and frequency are the inverse of one another thus: Period (T) = 0.02 s f = 1/T = 1/0.02 = 50 Hz Thus, a wave of period 0.02 s has a frequency of 50 Hz
λ Compression Rarefaction λ λ 2.3 Wavelength Wavelength, (symbol , Greek Letter LAMBDA), is a measure of the distance between two adjacent points on a wave undergoing similar motions. Thus the distance between two adjacent compressions or two adjacent rarefactions would be 1 wavelength. Wavelength is a distance measure, hence the unit for is metres (m).
Rachel and Bruce have assembled some laboratory equipment and are planning a series of sound-relatedexperiments. An audio-signal generator is used to drive a small loudspeaker, which emits sound uniformally in all directions.The audio power from the loudspeaker is kept constant at all frequencies used in the experiments. A sound level meter is used to measure sound intensity. This is shown in Figure 2. Initially, the frequency of thesignal generator is set to 476 Hz. The speed of sound at the time of the experiment was 340 ms-1. Question 4 Calculate the wavelength of the 476 Hz sound wave. Include a unit in your answer. λ = v/f = 340/476 = 0.71 m
2.4 Wave Speed The relation is summarised in the so called “WAVE EQUATION”, v = f Wave Speed (symbol v) is a measure of how quickly a “wave train” is moving. where; v = Speed (ms-1), f = Frequency (Hz) = Wavelength (m). The wave speed is dependent on the frequency and wavelength of the wavetrain. The speed of Sound in Air is temperature dependent and is approx 340 ms-1 at 200C Sound travels faster through denser mediums
Roger, an instrument maker, is constructing and testing pipes for a pipe organ. He measures the speed of sound in air at the time of the test to be 333 ms–1. Question 5 One pipe is designed to produce the note middle C (256 Hz). Which one of the following best gives the wavelength corresponding to middle C? A. 0.38 m B. 0.77 m C. 1.3 m D. 2.6 m
2.5 Sound Waves in Air This means the air particles must vibrate back and forth around their MEAN, AVERAGE or CENTRAL POSITION. In air, the passage of a sound wave causes a series of COMPRESSIONS and RAREFACTIONS In areas of above average air pressure (Compressions), the particles are packed CLOSE TOGETHER. So only small scale vibrations are needed for them to transfer their “information” (sound wave energy), to adjacent particles. In areas of below average air pressure (Rarefactions) the particles are SPREAD APART. So large scale vibrations are needed for information to be transferred to adjacent particles. The energy lost per transfer is high and so sound travels a lesser distance than at normal air pressure. Energy lost per transfer is low and the sound travels a greater distance than at normal air pressure
Chapter 3 Topics covered: • Sound Intensity. • Sound Intensity Level. • The Decibel Scale. • Frequency Response Graphs. • Sound Intensity versus Distance • Human Response
Area (A) Sound Energy (P) 3.0 Sound Intensity The INTENSITY of a sound is DEFINED as THE RATE OF FLOW OF ENERGY through an area perpendicular to the direction of travel of the sound wave. • Mathematically: I = P/A Where, I = Sound Intensity (Wm-2) P = Total Acoustical Power (W) A = Area (m2) The rate of flow of energy is the definition of POWER. In this case the power is ACOUSTICAL POWER. Thus SOUND INTENSITY is defined as POWER/AREA.
It is a cold, windless morning and three hot-air balloons hover above a park. Each balloon is stationary and in direct line of sight, with no obstacles near them, as shown in Figure 3. Balloon A is equipped with a 100 W siren, which emits a 2000 Hz tone uniformally in all directions. On board balloons B and C are students with sound measuring equipment. Question 6 Which of the following is the best estimate of the sound intensity of the siren as measured at balloon B? A. 0.5 Wm-2 B. 2.5 × 10-2 Wm-2 C. 8.0 × 10-4 Wm-2 D. 2.5 × 10-5 Wm-2 100 W spread over a sphere of radius 100 m gives a sound intensity of 100/(4π(100)2) = 8.0 x 10-4 Wm-2
Alexander Graham Bell. 3.1 Bels The Bel (symbol B) is a unit of measurement of ratios, such as power levels and voltage levels. It is mostly used in telecommunications, electronics and acoustics. It was invented by engineers at the Bell Telephone Laboratory to quantify (give a number to) the reduction in audio level over a 1 mile length of standard telephone cable. It was named in honour of Alexander Graham Bell. The bel was too large for everyday use, so the decibel (dB), equal to 0.1 Bel, became the more commonly used unit.
4.7 x 1012 Power difference = log 2.3 x 101 3.2 Decibels We could use scientific notation, but a comparison between 2.3 x 101 and 4.7 x 1012 is still awkward. For convenience, we find the RATIO between the two numbers and convert that into a logarithm. The decibel is not a unit in the sense that a metre or a kilogram is. Metres and kilograms are defined quantities of distance and mass. They never change. A decibel is a RELATIONSHIP between two values of POWER. = 11.3 B Decibels are designed for talking about numbers of vastly different magnitudes, eg., 23 Watts vs. 4,700,000,000,000 Watts. With such vast differences, the most difficult problem is getting the number of zeros right. To make life a little easier, we can get rid of the decimal by multiplying the result by 10, so; So comparing the numbers above on this basis, you find that the larger number is 113dB bigger than the smaller number.
SIL’s for various objects or events 3.3 Sound Intensity Level • Sound Intensity measured in Wm-2 and Sound Intensity Level measured in decibels (dB) are NOT the same. • Sound Intensity Level is DEFINED as the Logarithm of the ratio of the intensity of a sound to that of a reference sound. • The intensity of the reference sound has a value of 1 x 10-12 Wm-2, and is the minimum audible sound intensity at 3000 Hz. Corresponds to displacement of air particles by 100 billionth of a metre. • The decibel scale ranges from 0 dB (the softest audible sound) to approx 140 dB (sound causing pain/ear damage) Mathematically: S.I.L. = 10 Log I Io where: S.I.L. = Sound Intensity Level (dB) I = Sound Intensity (Wm-2) Io = 1 x 10-12 Wm-2
An isolated siren emits sound of 3000 Hz uniformly in all directions. At a point 20 m from the siren, the soundintensity is measured to be 1.0 × 10–3 Wm–2. Question 7 Which one of the following best gives the sound intensity level (in dB) at this point? A. 1.0 × 10–3 dB B. 9.0 dB C. 90 dB D. 100 dB The sound intensity is measured at a distance of 60 m from the siren. Question 8 Which one of the following best gives the sound intensity (in Wm–2) at 60 m? A. 3.3 × 10–3 Wm–2 B. 1.1 × 10–4 Wm–2 C. 3.0 × 10–2 Wm–2 D. 9.0 × 10–3 Wm–2
3.4 Comparing Sound Intensity Levels Let I1 = 1 Wm-2 and I2 = 10 Wm-2 S.I.L. = 10 Log I2/I1 = 10 Log 10/1 = 10 dB • The Sound Intensity Level formula can also be used to determine CHANGES in dB levels between two intensities labelled I1 and I2. • Thus the equation becomes: SIL = 10 Log I2/I1 • When used in this form, the reference term (Io) is not used, and I1 and I2 are the two sound intensities being compared. This 10 dB increase in S.I.L. is perceived by the Human Ear as a Doubling in the LOUDNESS of the sound. In fact, every 10 dB increase leads to a doubling of the loudness. So an 80 dB sound will be perceived as twice as loud as a 70 dB sound NOTE: Loudness is a subjective, non measurable quantity, used by humans to characterise and compare sounds.
3.5 The Decibel Scale Remember when comparing S.I.L’s we use SIL = 10 Log I2/I1 Let I1 = 100 Wm-2 and I2 = 200 Wm-2 S.I.L. = 10 Log I2/I1 = 10 Log 200/100 = 3 dB • The decibel scale is used for a number of reasons: 1. The human ear responds to a vast range of sound intensities (from 10-12 Wm-2 to 102 Wm-2 - a range of 1014 or one hundred thousand billion units). 2. In order to bring this range to a more manageable size, the log of intensities is used, so the range now becomes 0 dB to 140 dB. 3. Luckily the ear also responds to sound intensities in a logarithmic rather than a linear fashion, as shown in last section. Thus, if the sound intensity doubles, this leads to a 3 dB increase in S.I.L. This is about the smallest change in S.I.L. detectable by the human ear. So, if you replace your 100 W speakers with far more expensive 200 W ones, you will barely notice any difference !!!!!!
Question 9 By how many decibels will the sound intensity level at balloon C be lower than at balloon B? Doubling the distance quartered the intensity. Each time the intensity was halved the sound level reduced by 3 dB, so the total reduction was 6 dB. Balloons B and C move so that they are at equal distances from balloon A. The sound intensity at balloon C is now measured as 1.0 × 10-2 Wm-2. Question 10 What is the sound intensity level (dB) at balloon B? SIL = 10 log I/Io = 10 log (1.0 × 10-2 )/(1.0 × 10-12) = 100 dB
3.6 Frequency Response Graphs • The Human Ear and various Musical/Electrical devices (eg. Microphones) respond to different Audible Frequencies in different ways. • We don’t hear each frequency with equal loudness. • In order for the Ear to perceive various frequencies at the SAME LOUDNESS, they must be played at VARYING SOUND INTENSITY LEVELS. • This is best shown on a Frequency Response Graph Thus a 20 Hz sound needs to be played at 25 dB for the ear to hear it at the same loudness as a 4000 Hz sound played at 1 dB The ear is most sensitive at the “lowest” point on the graph in this case 4000 Hz.
I/4 Point Source I/4 I I/4 I/4 r 2r 3.7 Sound Intensity vs Distance If a sound source is small (a so called POINT SOURCE), the sound it produces radiates out equally in all directions. This has the effect of producing an expanding sphere of sound. (NB. The surface area of a Sphere = 4r2) If the source is operating at a fixed power level, the sound intensity/unit area will decrease as the area (of the expanding sphere) increases. The rate at which the intensity drops off is inversely proportional to the square of the distance from the source: Mathematically: I 1/r2 So doubling the distance from the source leads to the intensity dropping in to ¼ of its original value.
At a distance of 4.0 m from a loudspeaker, a sound intensity of 1.25 x 10-4 Wm-2 is detected. Question 11 What sound intensity would be detected at 1.0 m from the source? I ∝ 1/r2, therefore, decreasing distance to one quarter increases intensity by a factor of 16. 16 x 1.25 x 10-4 = 2.0 x 10-3 Wm-2 Question 12 What sound intensity level would be detected at 1.0 m from the source? SIL =10 log I/IO = 10 log (2.0 x 10-3)/(1.0 x 10-12) = 93 dB
Music covers a wider range of frequencies from about 50 Hz to 12 kHz Human speech ranges from about 100 Hz to 8 kHz 3.8 Human Response The human ear responds to sound in the range from about 20 Hz to 20 kHz. The ear is most sensitive at about 4 kHz and has the lowest threshold of pain at the same frequency The distance between the red and green lines represents the range of audible sound for each frequency
3.9 Phons The ear is not equally responsive to all frequencies. The curves represent equal loudness as perceived by the average human ear The Phon is defined as a unit of apparent loudness, equal in number to the intensity in dB of a 1 kHz tone judged to be as loud as the sound being measured. The ear is less sensitive to low frequencies and this discrimination against lows becomes steeper for softer sounds Sound intensity in dB does not reflect changes in the ear’s sensitivity with frequency and sound level Thus 50 phon means: “as loud as a 50 dB, 1kHz tone” Curve for the threshold of hearing and 100 phon means: “as loud as a 100 dB, 1 kHz tone
The graph in Figure 1 shows the relationship between sound intensity level (dB), frequency (Hz) and loudness. Sound intensity level (dB) of a note of 10 000 Hz is measured by a sound meter to be 60 dB. Question 13 Which one of the values below best gives the loudness in phon at this point? A. 20 phon B. 40 phon C. 60 phon D 80 phon Question 14 The loudness scale (phon) specifically takes account of which one of the following factors? A. Intensity of sound, as perceived by human hearing, is inversely proportional to distance from the source. B. The perception of sound by human hearing is logarithmic, rather than linear, compared to soundintensity. C. The perception of the intensity of sound by human hearing varies with frequency. D. Human hearing has a very limited range of frequencies that it can hear.
Chapter 4. Topics covered: • Reflection. • Refraction • Diffraction. • Superposition. • Interference.
Direction of incoming Sound Waves Direction of Reflected Sound Waves Normal Incoming Compression Angle of Incidence Angle of Reflection 4.0 Reflection • Sound (like any other wave) undergoes reflection when it strikes a wall or barrier. • It will follow the laws of reflection: i =r • where i is measured between the direction of the incoming wave train and the Normal and r is measured between the Normal and the direction of the reflected sound waves.
Two physicists are discussing the design of a new theatre for use by a school choir. The design requirement is for good acoustic properties; in particular, for even distribution of sound over the whole frequency range throughout the theatre. A plan of the theatre to be used is shown in Figure 2. One of the physicists wants to line the walls of the audience area of the theatre with heavy sound-absorbingcurtains. Question 15 Which one of the following states why this is a good idea? A. The curtains will reduce the effect of diffraction through the stage opening, hence producing better quality sound. B. Without the curtains, different frequencies will reflect differently from the walls, causing distortion dueto diffraction effects. C. Without the curtains, different frequencies will reflect differently from the walls, causing distortion dueto interference effects. D. Without the curtains, there would be multiple paths from the speaker to each member of the audience, thus causing distortion and sound loss due to interference effects in some parts of the theatre.
SOUND REFRACTION Sound Shadow Sound Shadow 4.1 Refraction Refraction is the bending of waves when they enter a medium where their speed is different More Dense Medium Sound waves, unlike light waves, travel faster in denser materials, such as solids and liquids, than they travel in air. When sound waves leave a solid, their velocity and wavelength decrease and they are bent towards the normal to the surface of the solid. Less Dense Medium For sound waves in air, their speed is temperature dependent. (v = 331 + 0.6T) During a temperature inversion, the sound will refract back toward the ground. For normal conditions sound will refract away from the ground, producing a sound shadow as shown
The easiest large scale diffraction effect observable is that of ocean waves bending as they pass around an island or headland. 4.2 Diffraction For visible light, λ is about 550 nm or 5.5 x 10-7 m, so it needs to pass through a VERY NARROW gap to produce a diffraction effect. Diffraction is a phenomenon demonstrated by all waves and is best described as the bending of waves as they pass around objects or through gaps or openings. The extent of diffraction depends on the ratio between the wavelength of the wave and the size of the object, gap or opening. This is called the λ/w ratio Diffraction is rarely seen or experienced in the visual world but is part of everyday experience in the aural (hearing) world. For sound of frequency 4000 Hz (when the ear is at its most sensitive), λ = 8.75 cm, so sound can (and does) diffract around everyday objects.
Sound Shadow Diffracted Sound Waves Building Approaching Sound Waves 4.3 Diffraction Around Corners Since sound waves have wavelengths in the centimetre to metre range, sound can, and does, suffer diffraction in the world in which we live. • This is because the houses we live in, and the objects we surround ourselves with, have similar dimensions to the wavelengths of sound. He can hear what the ladies are talking about without being able to see them We can hear mum shouting to turn down the stereo in part because her sound waves are diffracted around the house.
Barrier with Narrow Gap Approaching Sound Waves Diffracted Sound Waves Gap Width w 4.4 Diffraction through Gaps When a series of straight waves approaches a gap, such as a doorway, those waves will suffer a change in direction in passing through the gap, ie. they will suffer diffraction. The EXTENT OF DIFFRACTION depends upon the ratio of wavelength () to the gap width (w). If the wavelength and gap are about the same size, appreciable diffraction will occur. If is very much bigger (or smaller) than w, no diffraction effects will occur. Thus: 1. Maximum diffraction occurs when /w is between about 0.1 and 50 2. No diffraction effects occur when /w 1 or /w 1
Question 16 In the paragraph below, options to complete each sentence are given within the brackets. Circle the correctoption in each case. Jamie is listening to the sound of an orchestra through a small gap in a partly open sliding door.When the sound wave travels through the gap,[constructive interference / destructive interference / diffraction] occurs and spreadingof the wave results. High pitched (frequency) instruments such as flutes experience[more / the same / less] spreading than lower pitched instruments. As the size of the gap decreases,the angle of spreading will [increase / not change / decrease].
Short Wavelength (High Frequency) Sound Hall with open doors A B Band playing inside Long Wavelength (Low Frequency) Sound 4.5 Diffraction Effects An observer at A will hear both high and low frequency sounds Band An observer at B will hear low frequency sound ONLY High Frequency sounds suffering less diffraction are said to be much more “directional” or “line of sight”. This is one of reasons we can hear the low frequency (bass) sounds but not hear the high frequency (treble) sounds coming from a “party” a few streets away on a Saturday night. Since wavelength and frequency are so closely related, the extent of diffraction for sound can be thought of in terms of the frequency of sound. Low Frequency (Bass) sounds are generally diffracted by a greater amount than High Frequency (Treble) sounds.
Question 17 A Speaker system uses a single, wide-frequency response speaker. Explain why the quality (fidelity) will deteriorate asthe listener moves off the centreline. Hence explain why a multiple-loudspeaker system, as shown in Figure 1,would be more satisfactory. The amount of diffraction depends on the ratio λ/w. For a single speaker, the high frequencies would not diffract away from the centre line as much as the low frequencies. Using different speaker sizes for different frequency ranges would ensure that comparable spreading will occur for all frequencies.
Chapter 5 Topics Covered: • Superposition • Standing Waves • Standing Waves on Strings. • Standing Waves in Open and Closed Pipes. • Overtones & Harmonics. • Resonance.
5.0 Superposition Two or more waves occupying the same space will interact to form a single, composite or total wave which reflects the size and orientation of the individual waves making it up. This addition process is called SUPERPOSITION. Superposition is a VECTOR addition process, so wave orientation as well as amplitude are important Destructive Superposition Constructive Superposition DESTRUCTIVE SUPERPOSITION occurs when two waves with opposite orientations add. CONSTRUCTIVE SUPERPOSITION occurs when two waves with similar orientations add.