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IOP Concise Physics

Musical Sound, Instruments, and Equipment

Panos Photinos

Chapter 1

Properties of waves

1.1 IntroductionIn everyday language the term wave has several uses; for example, a wave of e-mails,a wave of enthusiasm, a wave of applause, a heat wave, and so on. The general ideais that something is suddenly going above or below normal. In more technicallanguage, a wave indicates a repeating pattern of highs and lows in some quantity.In terms of sound, what is ‘waving’ is the air pressure, going higher and lower thanthe ambient atmospheric pressure. This chapter will introduce the basic conceptsthat are commonly used to characterize waves, and sound waves in particular.

1.2 Periodic wavesA most familiar wave is the pattern of circles generated by dropping a coin in a stillpond, shown in figure 1.1. The pattern consists of highs and lows (crests and troughs,

doi:10.1088/978-1-6817-4680-7ch1 1-1 ª Morgan & Claypool Publishers 2017

respectively) traveling outwards from the center. At each point of the surface of thepond, the water level oscillates in cycles, above and below the undisturbed level ofthe pond. We will refer to the undisturbed level as the equilibrium level of the watersurface. The pattern propagates away from the point of impact (the source of thewave) and at each point the propagation is along the line of sight to the source. In ashallow flat-bottomed pond, the crests (and troughs) travel at the same speed. As thespeed is the same for all crests, the distance between successive crests remains thesame as the wave travels. The distance between successive crests (or successivetroughs) is defined as the wavelength. The time elapsed between the crossings of twosuccessive crests through a given point is the period of the wave. In one period, thewave travels a distance of one wavelength; in other words, the speed (i.e. distancetraveled divided by time of travel) is the ratio of the wavelength divided by theperiod.

If we count the number of crests crossing through one point in a given timeinterval, say in one second, then we have a very important concept in the study ofsound, namely the frequency. The frequency is the inverse of the period, i.e.

frequency = 1/(period) and therefore period = 1/(frequency).

If four successive crests cross a given point in 1 s (i.e. if the frequency is 4 crestsper second) then the time elapsed between two successive crests (i.e. the period) is 1/4of a second. As the speed of the wave equals the wavelength divided by the period,and since frequency is the inverse of the period, it follows that the speed of the waveis equal to the product of the wavelength times the frequency:

Speed of wave = (frequency) × (wavelength).

The frequency is fundamental in characterizing sound tones, and is a measure ofwhat we call the pitch. High pitch tones correspond to high frequencies, and lowpitch tones correspond to low frequencies. The frequency is measured in units ofHertz (Hz for short).

The difference in height between the top of a wave crest and the undisturbedwater level is the amplitude of the wave. The amplitude of the wave depends on theweight and speed of the impacting object. A small coin will cause a smaller

Figure 1.1. A wave in a pond.

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amplitude than a huge rock. The amplitude relates to the intensity of the wave. Notethat as the wave pattern expands, the amplitude diminishes, and eventually the wavedies out. With sound, this observation relates to everyday experience: the farther weare from the source, the weaker it sounds.

The succession of crests and troughs in the water pond example occurs because ofgravity. Water that happens to be above the equilibrium level of the surface is pulleddown by gravity. The downward speed builds up, and that amount of water fallsbelow the equilibrium level of the surface and becomes part of a trough. Whilemoving down it pushes adjacent parts of the water upward, which become part of acrest, and so forth. The entire cycle can be viewed as an attempt of gravity to restorethe water level back to the equilibrium level, as it was before the coin was dropped.In the process of restoring the equilibrium water level, it keeps overshooting thetarget. Thus, in our example, gravity is the restoring force. The overshoot occursbecause of the energy imparted by the impacting coin.

All waves require a restoring force. For example, in a vibrating string, the tensionof the string acts as the restoring force. There is a relation between the frequency ofthe wave and the strength of the restoring force. A stronger restoring force makes theup-and-down oscillation faster, which means that the frequency will be higher. Thisrelation will be discussed in more detail in connection with strings and stringinstruments.

In the example of the water wave, the quantity that oscillates is the water level, ascompared to the equilibrium level. In the case of sound waves in air, the oscillatingquantity is the air pressure. The sound wave in air is a succession of layers of low andhigh pressures, the rarefactions and compressions, respectively. Low and highpressures are with reference to the undisturbed air pressure of the atmosphere.Note that, in the pond example, as the wave travels in the horizontal direction, theoscillation is up and down. In other words, the oscillation is at a right angle to thedirection of travel. This is an example of a transverse wave: the oscillation istransverse to the direction of travel. In the case of a sound wave in air, the pressureoscillates back and forth, along the direction of travel. This is an example of alongitudinal wave.

It is convenient to use graphs to represent waves. The simplest periodic waveformis the sinusoidal wave, i.e. described by the sin (sine) or cos (cosine) functions knownfrom trigonometry. Three cycles of a sinusoidal are shown in figure 1.2. The cycle or

Figure 1.2. Three cycles of a sinusoidal waveform.

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periodicity of the repeating pattern is equal to the distance between two successiveequivalent points; for example, two successive highs or two successive lows. Theinterval between adjacent highs and lows is half a cycle, and so on. The phase at anypoint of the graph is the fraction of the cycle elapsed from the starting point of thewaveform. For example, in figure 1.2, the phase at the first high on the left is one-quarter of a cycle; the phase at the first low is three-quarters of a cycle.

When using graphs to represent waves, the vertical axis of the graph is used toshow the value of the oscillating quantity (e.g. displacement or pressure). In thehorizontal axis we usually have two options. We can choose the horizontal axis toshow distance or time, as indicated in figure 1.3.

With reference to the water wave in a pond, the vertical axis shows the height ofthe water level. From the graph, we see that the amplitude of the wave is 1 m. If wechoose to show distance in the horizontal axis, then the graph will give the profile ofthe height at a given instant. If the horizontal axis is chosen to indicate time, then thegraph will give the variation of the height at a given point. It is important to notethat the two graphs provide different information about the wave. Recalling thedefinitions of the wavelength and period, we see that the distance between successivecrests in figure 1.3(a) is equal to the wavelength. In this example, the first crest occursat a distance of about 1.8 m, and the second crest occurs at a distance of about 7.8 m.The wavelength is found by taking the difference of the locations of the two crests,i.e. 7.8−1.8 = 6 m. In figure 1.3(b), the first crest occurs at time = 1 s, and the secondcrest at time = 5 s. The interval between successive crests is the period of the wave,and in this example it is 5−1 = 4 s. As the frequency is the inverse of the period, itfollows that the frequency of this waveform is (1/4)= 0.25 Hz.

Figure 1.3. Sinusoidal waveform. (a) Horizontal axis is distance. (b) Horizontal axis is time.

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1.3 Addition of waveformsIn this section we discuss simple ways in which waves can combine with each other.Comparison of the phase of the interacting waves is the key concept in under-standing the outcome. Figure 1.4 shows two identical waveforms, i.e. they have thesame amplitude and the same periodicity. The horizontal axis is not labeled, and canbe either distance or time without affecting the conclusions. The graphs are offsetvertically, and the horizontal lines represent zero displacement for each wave. Tofind the waveform resulting from combining waves A and B, we add the displace-ments at each point of the horizontal axis. The result is shown in the bottom graph,and is simply the sum of the two waves, i.e. the amplitude of the resulting wave isdoubled, and the periodicity (which is the wavelength or the period depending on thechoice of the horizontal axis) remains the same. In this case, we combined two wavesthat are in step, or in-phase. This means at each point of the horizontal axis, the twowaves have the same phase, and the phase difference between them is zero.

In figure 1.5, wave B is displaced to the left by about one quarter of a cycle. In thiscase the two waves are not in-phase, and there is a phase difference of one-quarter ofa cycle. If the horizontal axis were distance, the two crests of the two waves would beseparated by one quarter of a wavelength. In the same way, if the horizontal axisindicated time, then the crests of the two waves would be separated by one quarter ofthe period. To find the waveform resulting from combining waves A and B, we addthe displacements at each point of the horizontal axis. The result is shown in thebottom of figure 1.5. We note that the sum of waves A and B has the sameperiodicity, but the amplitude is smaller than the sum of the amplitudes of A and B.The crest of the resulting wave occurs somewhere in between the crests of waves Aand B.

In figure 1.6, wave B is displaced to the left by half a cycle, and the result ofadding these two waves is total cancellation. As the amplitudes of A and B are thesame and the oscillations are in opposite directions, the cancellation is complete. If

Figure 1.4. Adding two identical waveforms that are in-phase.

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the amplitudes were not the same, the resulting wave would have amplitude equal tothe difference between the amplitudes of A and B. For instance, if A had amplitude 3and B had amplitude 1, the resulting wave would have amplitude equal to thedifference 3 − 1 = 2.

Using similar diagrams, we can add waves of the same periodicity. The generalconclusions are:

• The resulting wave will always have the same wavelength and frequency.• The amplitude will be equal to the sum of the amplitudes, if the phasedifference is zero.

• The amplitude will be equal to the difference between amplitudes, if the phasedifference is 1/2 cycle.

Figure 1.6. Adding two identical waveforms that are 1/2 cycle out of phase. The result is completecancellation.

Figure 1.5. Adding two identical waveforms that are 1/4 of a cycle out of phase.

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• If the phase difference between waves A and B has value between 0 and 1/2cycle, the resulting amplitude will have value between the amplitudes A + Band A − B1.

Cancellation of waves as shown in figure 1.6 is the basic principle behind noisecancelling devices. For example, noise cancellation headphones use electronics togenerate replicas of the sound before it reaches each ear. The replicas generateddiffer in phase from the incoming wave by half a cycle. Inside each ear, the incomingsound and the out-of-phase replica cancel each other.

1.4 BeatsA very interesting situation occurs when adding two waves of slightly differentfrequencies. Figure 1.7 shows two waves, A and B, of equal amplitude. Thehorizontal axis shows time. By counting crests, we see that in the time it takeswave A to complete 10 cycles, wave B completes 8 cycles. In other words, wave Ahas a shorter period, therefore higher frequency than wave B. The result of addingwaves A and B is shown at the bottom of figure 1.7. For our purposes, the importantfeature of the combined wave is that the amplitude has cycles of highs and lows, thatoccur approximately every 30 s. This change in the amplitude is referred to as beats.Thus, the amplitude of the resulting waveform A + B is modulated, i.e. changes withtime. The period of this modulation is about 30 s. A more detailed analysis2 showsthat the result of adding two waves of different frequencies is a wave of frequencyequal to the average frequency of the two waves. The amplitude of the resultingwave is not constant, but modulated at a frequency equal to the difference betweenthe frequencies of the two waves.

1Or B − A if the amplitude of B is larger. This is so because by convention, the amplitude is always a positivenumber.2Details are provided in appendix B.3.

Figure 1.7. Adding two waveforms of the same amplitude but different frequencies.

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So, if we add a wave of 400 Hz and a wave of 402 Hz, the result will be a wave of401 Hz, with amplitude that is modulated with a frequency of 402 − 400= 2 Hz,which is the beat frequency. Therefore, the highs of the amplitude will repeat every1/(2 Hz) = 0.5 s, i.e. the beats will occur every 0.5 s. The phenomenon of beats playsa significant role in music and musical instruments, particularly those with double ortriple strings, such as the piano. Each string in a pair should produce sound of thesame frequency. If there is a mismatch of 2 Hz between the frequencies of the twostrings, then from our calculation above we find that the intensity of the sound willfluctuate up and down every 0.5 s, producing a tone of poor quality.

Figure 1.8 shows the waveform of the sound produced by an out-of-tune piano.The entire waveform lasts about 2 s, from which we can estimate that the pattern ofbeats shown repeats about every 0.5 s. Therefore, the beat frequency is about 1/0.5 =2 Hz, meaning that the two strings are off by 2 Hz, which, as will be discussed insection 5.6 is quite noticeable and unpleasant to the ear. Experienced tuners andstring instrument players can use beats to tune their instruments very precisely.

1.5 Energy and intensityWaves carry energy. So, when we drop a coin in a pond, some of the energy of thefalling coin (the so-called kinetic energy) is transferred to the water, and carriedaway by the wave generated by the impact. A heavier coin or faster moving coin willresult in a wave of larger amplitude, because it imparts a larger amount of energy atthe point of impact. In many cases, the important quantity is not the energyimparted at the point of impact, but how much energy is flowing at points away fromthe source of the wave. To understand the concept of energy flow, it would be helpfulto make an analogy to collecting rainwater in a tub. The more rain is coming down,the more raindrops will be caught in the tub, but the amount of water collected willalso depend on the size of the tub because a wider tub has larger collecting area. Theamount of water collected will depend on time as well: the longer we wait, the morewater will be collected. We can characterize the intensity of the rain by referring to acommonly agreed upon tub area (e.g. 1 m2) in a specific time interval (e.g. theamount collected in 24 h). Also note that the amount collected will depend on the tiltof the tub. A vertical tub will not catch much water! The maximum amount of waterwill be collected if the tub is perpendicular to the direction of the falling drops.

Figure 1.8. Sound produced by an out-of-tune piano. Note the four cycles of the beat pattern.

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For waves, we can define the intensity as the rate at which energy is flowingthrough an area of 1 m2. The area must be oriented perpendicular to the direction ofenergy flow. The rate of energy flow (i.e. amount of energy flowing per unit time) isdefined as the power, and it is measured in Watts (W). Therefore, the intensity ismeasured in Watts per square meter (W m−2). The intensity is related to the squareof the amplitude of the wave. This means that if we double the amplitude of the wavethe intensity is not doubled but quadrupled (following the square 22). If theamplitude is tripled, the intensity increases 32 = 9 times, and so on.

As the wave propagates outward from the source, the energy that started at thesource is spread over a larger and larger surface. Consequently, the intensity (andamplitude) will decrease more and more the further we move away from the source.In addition, some of the energy that started at the source is absorbed, usuallyconverted to heat as a result of friction. We refer generically to the decrease inintensity as attenuation. Attenuation will be discussed in detail in section 2.7.

1.6 Further discussionExamples of waves

Electromagnetic (EM) waves include light, radio waves, microwaves, ultravioletradiation, x-rays, and more. The oscillating quantity in EM waves is an electric fieldcombined with a magnetic field. In empty space, all EM waves travel at the speed oflight. EM waves can be produced by the oscillation of electrical charges.

WiFi is a wireless connection between devices based on transmitting and receivingEM waves of frequency 2.4 billion Hz (GHz for short) and 5 GHz. Similarly, an AMradio is a wireless connection, using EM waves of frequency in the range of about500–1600 thousand Hz (kHz). And an FM radio uses EM waves of frequency in therange from 87.5 to 108 million Hz (MHz). All these waves carry the soundinformation, but what we hear is not the EM wave itself. The EM wave is onlythe carrier, the so-called carrier wave. The sound information is ‘imprinted’ on thecarrier signal as a modulation of the amplitude (AM = Amplitude Modulation) orthe frequency (FM = FrequencyModulation) of the carrier wave. The bottom graphin figure 1.7 is an amplitude-modulated signal.

One of the most exciting discoveries in recent years was the detection ofgravitational waves. In principle, gravitational waves can be produced by oscillationsof huge amounts of mass (like stars) by analogy to EM waves that can result fromoscillating electric charges. The oscillation quantity in the case of gravitationalwaves is the ‘fabric’ (or the curvature) of space, as predicted by Einstein’s generaltheory of relativity.

1.7 EquationsFrequency, period, wavelength, and speed

The following symbols are commonly used:c = speedλ = wavelengthf = frequency

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T = periodf is defined as f = 1/TFor sinusoidal waves we have:c = f λ and by the definition of f we have c = λ/T.

1.8 Questions1. A person’s heart rate is 120 beats per minute.

(a) Find the period of the heart rate in seconds.(b) Find the frequency of the heart rate in Hz.

2. A wave has frequency 20 Hz and wavelength 200 meters. Find the speed ofthe wave.

3. Suppose we have two sound waves, A and B, traveling in air simultaneously.The amplitude of wave B is twice the amplitude of wave A.

(a) Which of the two waves has higher intensity?(b) Which of the two waves travels faster?

4. Suppose we combine two sound waves of frequencies 100 and 106 Hz,respectively. Find the beat frequency resulting from the combination of thetwo waves.

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