14.1 Introduction
Chapter 14 — Waves
In the year 1687, Isaac Newton did something no one had done before: he calculated the speed of sound using pure physics. He reasoned that sound is a wave of compressions and rarefactions moving through air, and that the speed of such a wave should depend on how “stiff” the air is (its pressure) and how heavy it is (its density). He worked out the numbers and arrived at about 280 metres per second.
There was only one problem. When people actually measured the speed of sound — by timing echoes and distant cannon fire — they got about 331 metres per second. Newton’s answer was nearly 15% too small. For a scientist of his stature, this was an uncomfortable gap, and it stayed unexplained for over a century.
The mistake was finally traced by Pierre-Simon Laplace in the early 1800s. Newton had quietly assumed that as sound passes, the air has time to stay at a steady temperature. Laplace realised this was wrong: a sound wave squeezes and stretches the air so rapidly that heat cannot flow out in time. Once he accounted for this, the corrected value came out to 331 m/s — matching the measurements almost exactly.
This chapter is the story of how disturbances travel through matter, and how one careful correction turned a good guess into an exact law.
Figure to come
Fig. 14.0 – A pebble dropped in still water sending circular ripples outward, with a small cork floating on the surface bobbing up and down in place; beside it, a small inset showing a sound wave as bands of compression and rarefaction spreading from a speaker.
By the end of this chapter, you will be able to answer each of these questions using the physics of waves.
In the previous chapter, we studied objects that oscillate on their own — a single pendulum swinging, or a single mass bobbing on a spring. Each of these moved in isolation, without affecting anything around it. But in the real world, oscillating objects are rarely alone. What happens when we have a whole system of such objects linked together?
A material medium — a solid, a liquid, or a gas — is exactly such a system. Inside it, the particles are not free and separate. They are bound to one another by elastic forces, the same kind of “springy” forces that pull a stretched spring back to its natural length. Because of this binding, when one particle is disturbed and starts to move, it tugs on its neighbours, and their motion in turn affects the particles next to them. The disturbance does not stay in one place; it is passed along.
A disturbance that travels, while matter stays put
The simplest way to see this is to drop a small pebble into a still pond. The moment it hits, the water surface is disturbed, and this disturbance does not remain at that one spot — it spreads outward as a growing circle. If you keep dropping pebbles, you see circle after circle racing outward from the point of impact. It genuinely looks as if the water itself is flowing outward from the centre.
But watch more carefully. Place a few small pieces of cork on the surface. As the ripples pass under them, the cork pieces simply bob up and down — they do not travel outward with the circles. This is the key observation: the water as a whole is not moving away from the centre. What moves outward is only the disturbance, while the water itself stays where it is and merely oscillates in place.
The same thing happens with sound. When we speak, the sound moves outward from us and reaches a listener across the room — yet there is no wind, no bulk flow of air travelling from our mouth to their ear. The disturbances produced in air are far less obvious than ripples on water, and only our ears or an instrument like a microphone can detect them. But the principle is identical: a pattern of disturbance moves through the medium while the medium itself stays put.
These patterns, which travel from one place to another without the actual physical transfer or bulk flow of matter, are called waves. This chapter is a study of such waves.
What waves carry
A wave transports two things: energy, and the information contained in the pattern of the disturbance. This is why almost all of our communication depends on waves. When we speak, we produce sound waves in the air; when someone hears us, their ears detect those same waves. Signals of every kind — spoken words, television pictures, phone calls — are sent from one place to another by waves.
Do all waves need a medium?
Not every wave requires a material medium to travel through. Light is the clearest example. Light waves can travel through a vacuum — empty space with no particles at all. The light from distant stars, some of them hundreds of light years away, reaches us after crossing the vast emptiness of interstellar space, which is very nearly a perfect vacuum. If light needed a medium, this could never happen.
This gives us a natural way to sort waves into three broad families.
The first and most familiar family is the mechanical waves — waves on a string, water waves, sound waves, seismic waves inside the Earth, and so on. These waves need a medium to travel through; they cannot pass through a vacuum. They work by making the particles of the medium oscillate, and their behaviour depends on the elastic properties of that medium.
The second family is the electromagnetic waves, which you will study in detail in a later class. Unlike mechanical waves, these do not necessarily require a medium — they can travel through a vacuum. Light, radio waves, and X-rays are all electromagnetic waves. In vacuum, all electromagnetic waves travel at the same fixed speed, denoted by \(c\), whose value is
\[c = 299{,}792{,}458 \ \text{m s}^{-1} \tag{14.1}\]
Here \(c\) is the speed of light in vacuum, measured in metres per second (m s\(^{-1}\)).
The third family is the matter waves. These are associated with the tiny constituents of matter — electrons, protons, neutrons, atoms, and molecules. They appear in the quantum-mechanical description of nature, which you will meet in later studies. Though they are more abstract than mechanical or electromagnetic waves, they are already put to practical use.
In this chapter, we will study only mechanical waves — the kind that need a material medium to propagate.
A chain of springs
To see how oscillations lead to waves, imagine a long row of springs joined end to end, as shown in Fig. 14.1. Suppose the spring at one end is suddenly pulled and released. What happens? That first spring is disturbed from its natural (equilibrium) length. Since the second spring is attached to the first, it too gets stretched or compressed, and this passes on to the third, and so on down the line. In this way the disturbance travels steadily from one end to the other — yet each individual spring only makes small oscillations about its own equilibrium position. It never travels along the chain itself.
Figure to come
Fig. 14.1 – A horizontal row of identical springs connected end to end; the free end A is being pulled sideways, generating a disturbance shown moving along the chain toward the far end.
A familiar everyday version of this is a stationary train at a railway station. The bogies of the train are coupled to one another through spring couplings. When an engine is attached at one end and gives a push to the bogie next to it, that push is handed on from bogie to bogie all along the train — without the entire train sliding bodily along the track. The push (the disturbance) travels, while each bogie barely moves.
How sound travels through air
Now consider how sound moves through air using the same idea. As a sound wave passes, it squeezes together (compresses) or spreads apart (expands) a small region of air. Compressing a region changes its density by a small amount, written \(\delta\rho\), and this change in density produces a small change in pressure, written \(\delta p\).
Since pressure is force per unit area, this pressure change acts as a restoring force — a force that tries to push the air back to its normal state, just as a stretched spring is pushed back to its natural length. Here, the quantity that plays the role of the “stretch or squeeze” of a spring is the change in the density of the air.
The mechanism then works step by step. When a region of air is compressed, its molecules are packed closer together, so they tend to spread out into the neighbouring region. This raises the density there — creating a new compression in the adjoining region — while the first region, having lost molecules, becomes rarefied (thinner than normal). Where the air is rarefied, the surrounding air rushes in to fill it, so the rarefaction itself shifts to the next region along. In this way, compressions and rarefactions move steadily from one region to the next, and the disturbance travels through the air.
The same reasoning works inside solids. In a crystalline solid, the atoms (or groups of atoms) are arranged in a regular, repeating pattern called a lattice. Each atom sits in equilibrium, held in place by forces from the atoms surrounding it. If you displace one atom while keeping the others fixed, restoring forces immediately pull it back — behaving exactly like a spring. So we can picture the atoms in a lattice as points joined to their neighbours by tiny springs, allowing a disturbance to be passed from atom to atom.
In the sections that follow, we will build on this simple spring picture to study the various characteristic properties of waves — their shapes, speeds, and the ways they combine, reflect, and interfere.
14.2 Transverse and Longitudinal Waves
We have already seen that a mechanical wave is carried by the oscillations of the particles of a medium. But not all waves make their particles oscillate in the same way. The natural way to sort mechanical waves into types is to ask a simple question: in which direction do the particles oscillate, compared with the direction in which the wave travels?
There are two basic answers, and they give us the two main kinds of mechanical waves.
If the particles of the medium oscillate at right angles (perpendicular) to the direction in which the wave moves, the wave is called a transverse wave.
If instead the particles oscillate along the same line as the direction in which the wave moves — that is, back and forth in the direction of travel — the wave is called a longitudinal wave.
A pulse on a string — a transverse wave
To picture a transverse wave, imagine a very long, stretched string. Give one end a single quick up-and-down jerk. A single hump — called a pulse — runs along the string, as shown in Fig. 14.2. A pulse is just one short-lived disturbance, not a repeating wave.
Figure to come
Fig. 14.2 – A long horizontal stretched string lying along the x-direction with a single raised hump (pulse) travelling to the right; a small arrow on the string shows a point moving up and down in the y-direction.
As the pulse passes any point on the string, that point rises up and then comes back down, returning to its original rest position. It does not travel along with the pulse. If the string is very long compared with the size of the pulse, the pulse simply dies out (damps out) before it reaches the far end, so we need not worry about any reflection from that end.
Now suppose that, instead of a single jerk, an external agent moves the end of the string up and down continuously and smoothly (sinusoidally). The result is no longer a single pulse but a repeating sinusoidal wave running along the string, as shown in Fig. 14.3.
Figure to come
Fig. 14.3 – A stretched string carrying a smooth repeating (sinusoidal) wave travelling to the right along the x-axis, with the string elements displaced up and down along the y-axis.
In both cases — the single pulse and the continuous wave — each small element of the string oscillates up and down about its own equilibrium (rest) position as the disturbance passes. These oscillations are directed at right angles to the direction in which the wave travels along the string. This is exactly why a wave on a string is a clear example of a transverse wave.
Two ways to look at a wave
It helps to notice that a wave can be viewed in two different ways.
One way is to freeze a single instant of time, like a photograph, and look at the wave spread out in space. This shows us the shape of the whole wave at that moment.
The other way is to fix our attention on one particular element of the medium — say one point on the string — and watch how it moves up and down over time. This shows us the oscillation in time of a single point. Both views describe the same wave, and we will use each of them later.
Sound in a pipe — a longitudinal wave
The most familiar longitudinal wave is sound. Consider a long pipe filled with air, with a movable piston at one end, as shown in Fig. 14.4. Give the piston a single sudden push forward and then pull it back. This creates a pulse made of a condensation — a region where the air is squeezed to higher density — followed by a rarefaction, a region where the air is thinned to lower density. These travel down the pipe through the air.
Figure to come
Fig. 14.4 – A long vertical pipe filled with air with a piston at the top; below it, alternating dark bands (condensations, higher density) and light bands (rarefactions, lower density) travel along the pipe in the same direction as the piston’s push, with an arrow showing a volume element of air oscillating parallel to the direction of propagation.
If the piston is pushed and pulled continuously and periodically (sinusoidally), instead of just once, a repeating sinusoidal wave is generated and travels along the length of the pipe. Here the air elements oscillate back and forth along the pipe — in the same direction as the wave travels — so this is clearly a longitudinal wave.
Travelling (progressive) waves — the medium does not travel
The waves we have described so far, whether transverse or longitudinal, are called travelling waves or progressive waves, because the disturbance travels from one part of the medium to another.
This last point is worth stressing. A flowing stream is water moving as a whole from one place to another. A water wave is completely different: here it is only the disturbance that moves across the surface, while the water itself stays roughly in place. In the same way, wind is air moving as a whole, but a sound wave is not wind — it is a disturbance in the pressure and density of the air, travelling through the air without the air itself being carried along.
Which media can carry which wave?
There is an important physical reason why transverse and longitudinal waves need different kinds of media.
In a transverse wave, the particles move sideways, at right angles to the direction of travel. As the wave passes, neighbouring layers of the medium slide past one another. This kind of deformation — where layers slide over each other and the shape (but not the volume) changes — is called a shearing strain, and the internal force that resists it is called shearing stress. To carry a transverse wave, a medium must be able to resist such sliding — that is, it must be able to sustain shearing stress.
Solids can resist shearing; they have a definite shape and springy internal forces that push the sliding layers back. Fluids — liquids and gases — cannot resist shearing, because they simply flow when their layers are made to slide. For this reason, transverse waves can travel only through media that can sustain shearing stress, namely solids, and not through fluids.
Both solids and fluids, however, can resist being compressed — they can sustain compressive strain, where the volume is squeezed. Because longitudinal waves rely on compression and rarefaction, they can travel through all elastic media. For example, in a solid like steel, both transverse and longitudinal waves can travel, whereas air can carry only longitudinal waves.
Waves on the surface of water
Water waves are a special case, and are of two kinds. The first are capillary waves — small ripples of fairly short wavelength, usually not more than a few centimetres. The restoring force that produces them is the surface tension of water. The second kind are gravity waves, with much longer wavelengths, ranging from several metres to several hundred metres. Here the restoring force is the pull of gravity, which always tends to bring the water surface back to its lowest, flat level.
In these water waves, the motion of the particles is more complicated than a simple up-and-down. The particles not only move up and down but also back and forth, and their oscillation is not confined to the surface — it extends downward with steadily decreasing amplitude, reaching all the way to the bottom. Because of this, ocean waves are actually a combination of both longitudinal and transverse motion.
Finally, it is found that, in general, transverse and longitudinal waves travel at different speeds in the same medium. We will see in a later section exactly what decides the speed of each kind of wave.
14.3 Displacement Relation in a Progressive Wave
So far we have described waves in words and pictures. To do real physics, we now need a mathematical description of a travelling wave — a formula that tells us the exact displacement of any particle, at any position, at any moment.
Such a formula must depend on two quantities at once: the position \(x\) (where along the medium we are looking) and the time \(t\) (when we are looking). This is because a wave has two faces. At any fixed instant, the formula should give us the shape of the whole wave spread out in space. And at any fixed location, the same formula should describe how that one particle moves up and down as time passes.
Building the wave function
If we want to describe a smooth, repeating (sinusoidal) travelling wave — like the one on the string in Fig. 14.3 — then the function we use must itself be sinusoidal, built from sine or cosine. For convenience, let us take the wave to be transverse. Then the position of a particle along the medium is \(x\), and its displacement away from its rest (equilibrium) position is \(y\).
A sinusoidal travelling wave is then described by
\[y(x,t) = a\sin(kx - \omega t + \phi) \tag{14.2}\]
Here \(y(x,t)\) is the displacement (in metres, m) of the particle located at position \(x\) (in m) at time \(t\) (in seconds, s). The constant \(a\) is the amplitude, \(k\) the angular wave number, \(\omega\) the angular frequency, and \(\phi\) the initial phase — quantities we will interpret one by one in the coming subsections.
The extra term \(\phi\) inside the sine is important. Including it is the same as writing the wave as a mixture (a linear combination) of a pure sine and a pure cosine:
\[y(x,t) = A\sin(kx - \omega t) + B\cos(kx - \omega t) \tag{14.3}\]
Comparing Equations (14.2) and (14.3), the two forms describe the same wave provided
\[a = \sqrt{A^2 + B^2} \quad \text{and} \quad \phi = \tan^{-1}\left(\frac{B}{A}\right)\]
So the single “phase-shifted sine” form of Eq. (14.2) and the “sine-plus-cosine” form of Eq. (14.3) are just two ways of writing the same thing.
Why Equation (14.2) really is a travelling wave
To see that Eq. (14.2) describes a wave that moves, look at it in the two ways we discussed earlier.
First, freeze the clock at some fixed instant \(t = t_0\). Then \(\omega t_0\) is just a constant, and the argument of the sine becomes \(kx + \text{constant}\). As a function of \(x\), this is simply a sine curve. So at any fixed instant, the shape of the wave along the medium is a sine wave.
Next, stand at one fixed location \(x = x_0\). Now \(kx_0\) is a constant, and the argument becomes \(-\omega t + \text{constant}\). As a function of \(t\), this varies sinusoidally. So the particle at a fixed position moves up and down sinusoidally with time — meaning that the particles of the medium at every position are performing simple harmonic motion (SHM), the back-and-forth motion you studied in oscillations.
Finally, ask what happens as time goes on. To keep the same point of the wave (the same value of the phase \(kx - \omega t + \phi\)), whenever \(t\) increases, \(x\) must also increase. In other words, the pattern shifts toward larger \(x\). This is exactly why Eq. (14.2) represents a sinusoidal (harmonic) wave travelling along the positive direction of the \(x\)-axis.
By the same reasoning, a function with a plus sign,
\[y(x,t) = a\sin(kx + \omega t + \phi) \tag{14.4}\]
represents a wave travelling along the negative direction of the \(x\)-axis.
Naming the quantities in the wave equation
Each symbol in Eq. (14.2) has a physical meaning, summarised in Fig. 14.5. We will interpret them fully in the next three subsections, but here are their names at a glance.
Figure to come
Fig. 14.5 – A labelled table listing the standard symbols in Eq. (14.2) and their meanings.
The quantity \(y(x,t)\) is the displacement, written as a function of position \(x\) and time \(t\). The quantity \(a\) is the amplitude of the wave. The quantity \(\omega\) is the angular frequency of the wave. The quantity \(k\) is the angular wave number. The full expression \(kx - \omega t + \phi\) is called the phase of the wave, and its value at \(x = 0\) and \(t = 0\) is simply \(\phi\), which is therefore called the initial phase angle.
Watching the wave move: crests, troughs, and two markers
To picture how the wave changes, Fig. 14.6 shows the plots of Eq. (14.2) at several instants of time, each separated by an equal time interval.
Figure to come
Fig. 14.6 – A stack of sine-wave plots of the same wave at equal time intervals, showing a crest marked with a cross (×) shifting steadily to the right, and a particle at the origin marked with a solid dot (•) moving up and down in place.
In any wave, the point of maximum upward (positive) displacement is called a crest, and the point of maximum downward (negative) displacement is called a trough.
To follow how the wave travels, we can fix our attention on one crest and watch how it advances with time. In Fig. 14.6 this chosen crest is marked with a cross (×). At the same time, we can watch one particular particle at a fixed location — say the particle at the origin — marked with a solid dot (•).
The plots reveal two things clearly. The solid dot (•) at the origin moves up and down periodically: the particle there oscillates about its mean position as the wave passes, and the same is true for a particle at any other location. Meanwhile, the cross (×) marking the crest steadily advances. In fact, in exactly the time the dot takes to complete one full oscillation, the crest has moved forward by a certain distance. This neatly separates the two ideas we will need: how far the pattern moves, and how long each particle takes to oscillate.
Using these plots, we now go on to define carefully the various quantities of Eq. (14.2) — the amplitude and phase, the wavelength and angular wave number, and the period, angular frequency and frequency — in the subsections that follow.
14.3.1 Amplitude and Phase
Now that we have the wave equation, let us interpret its two most basic features — how big the oscillation is, and where in its cycle each particle happens to be. These are captured by the amplitude and the phase.
Amplitude
Look again at the wave equation, \(y(x,t) = a\sin(kx - \omega t + \phi)\). The sine function can never be larger than \(+1\) or smaller than \(-1\). Since \(y\) equals \(a\) multiplied by this sine, the displacement \(y(x,t)\) must swing between a largest value of \(+a\) and a smallest value of \(-a\), and never beyond.
We always choose \(a\) to be a positive constant — this can be done without any loss of generality. With that choice, \(a\) is exactly the maximum displacement of the particles of the medium away from their equilibrium (rest) position. In Fig. 14.6, it is the height of a crest above the central rest line.
Notice the difference carefully: the displacement \(y\) itself may be positive or negative (the particle moves above and below its rest position), but the amplitude \(a\) is always taken as a single positive number — the size of the largest swing. It is called the amplitude of the wave.
Phase
The whole quantity inside the sine, \((kx - \omega t + \phi)\), is called the phase of the wave. Once the amplitude \(a\) is fixed, it is the phase that decides the actual displacement of the wave at any chosen position \(x\) and any chosen instant \(t\). In other words, the phase tells you exactly where in its up-and-down cycle a particle is.
Now set \(x = 0\) and \(t = 0\) in the phase. Everything except \(\phi\) vanishes, so the phase at the origin and at the starting instant is simply \(\phi\). For this reason, \(\phi\) is called the initial phase angle.
The initial phase angle is not a fixed property of the wave — it depends on where we choose to place the origin of \(x\) and when we choose to start our clock. By a suitable choice of the origin on the \(x\)-axis and of the initial time, we can always arrange for \(\phi\) to be zero. Because of this freedom, there is no loss of generality in dropping \(\phi\) altogether and writing Eq. (14.2) with \(\phi = 0\) whenever it is convenient.
14.3.2 Wavelength and Angular Wave Number
The amplitude and phase told us how far a particle swings and where it is in its cycle. Now we ask a question about the wave spread out in space: how long is one complete wave? This is measured by the wavelength, and it is closely tied to a quantity called the angular wave number.
Wavelength
Look along a wave frozen at one instant. The minimum distance between two points that are in exactly the same state of oscillation — the same phase — is called the wavelength of the wave, usually written as \(\lambda\) (the Greek letter lambda).
For simplicity, we can pick the “same-phase” points to be two crests, or two troughs, since these are easy to spot. The wavelength is then just the distance between two consecutive crests, or between two consecutive troughs, as marked in Fig. 14.6.
Figure to come
Fig. 14.6 – A sine-shaped wave frozen in space, with the wavelength λ marked as the distance between two successive crests (and equivalently between two successive troughs).
Finding wavelength from the wave equation
We can extract the wavelength directly from the wave formula. Take \(\phi = 0\) for convenience, and freeze the wave at \(t = 0\). Then Eq. (14.2) becomes
\[y(x,0) = a\sin kx \tag{14.5}\]
Now recall a basic property of the sine function: it repeats its value every time its angle changes by \(2\pi\). So
\[\sin kx = \sin(kx + 2n\pi) = \sin k\!\left(x + \frac{2n\pi}{k}\right)\]
This tells us that the displacement at a point \(x\) is exactly the same as the displacement at the point \(x + \dfrac{2n\pi}{k}\), where \(n = 1, 2, 3, \dots\) In other words, the wave looks identical at these points — they are all in the same phase.
The least such distance (the wavelength) is found by taking the smallest value, \(n = 1\). This gives
\[\lambda = \frac{2\pi}{k} \quad \text{or} \quad k = \frac{2\pi}{\lambda} \tag{14.6}\]
Angular wave number
The quantity \(k\) appearing here is called the angular wave number, or the propagation constant. From \(k = 2\pi/\lambda\), its SI unit is radian per metre (rad m\(^{-1}\)).
14.3.3 Period, Angular Frequency and Frequency
In the previous subsection we looked at the wave spread out in space and found its wavelength. Now we switch to the other point of view: we stand at one fixed spot and watch a single particle bob up and down over time. This gives us three closely related quantities — the period, the angular frequency, and the frequency.
A plot in time, not in space
Fig. 14.7 again shows a sinusoidal curve, but this time it means something different from the earlier picture. It is not the shape of the wave frozen in space; instead it shows the displacement of one element of the medium, at a fixed location, as time goes on.
Figure to come
Fig. 14.7 – A sinusoidal graph of displacement versus time for a single particle at a fixed location, with the amplitude a marked vertically and the time period T marked as the length of one full cycle along the time axis.
To describe this motion, take Eq. (14.2) with \(\phi = 0\) and watch the particle at the origin, \(x = 0\). Substituting these in gives
\[y(0,t) = a\sin(-\omega t) = -a\sin\omega t\]
So the particle at a fixed location simply oscillates sinusoidally with time, exactly like a body in simple harmonic motion.
Period and angular frequency
The period of oscillation is the time taken by an element of the medium to complete one full oscillation. If \(T\) is this period, then the displacement must return to the same value after a time \(T\):
\[-a\sin\omega t = -a\sin\omega(t + T) = -a\sin(\omega t + \omega T)\]
For the sine to repeat itself, its angle must increase by exactly \(2\pi\). Therefore \(\omega T = 2\pi\), which gives
\[\omega T = 2\pi \quad \text{or} \quad \omega = \frac{2\pi}{T} \tag{14.7}\]
The quantity \(\omega\) here is called the angular frequency of the wave. From \(\omega = 2\pi/T\), its SI unit is radian per second (rad s\(^{-1}\)).
Frequency
The frequency \(\nu\) (the Greek letter nu) is simply the number of oscillations completed in one second. Since one oscillation takes a time \(T\), the number per second is \(1/T\). Combining this with Eq. (14.7),
\[\nu = \frac{1}{T} = \frac{\omega}{2\pi} \tag{14.8}\]
Frequency is usually measured in hertz (Hz), where one hertz means one oscillation per second.
The same equation for longitudinal waves
All of the discussion above has referred to a wave on a string — that is, a transverse wave, where the displacement \(y\) is at right angles to the direction of travel. But everything carries over to longitudinal waves as well, where the displacement of a particle is along the direction the wave travels.
For a longitudinal wave, we simply rename the displacement. Instead of \(y\), we write \(s\), and the wave is described by
\[s(x,t) = a\sin(kx - \omega t + \phi) \tag{14.9}\]
Here \(s(x,t)\) is the displacement of an element of the medium in the direction of propagation of the wave, at position \(x\) and time \(t\). The constant \(a\) is now called the displacement amplitude. Every other quantity — \(k\), \(\omega\), \(\phi\), \(T\), \(\nu\) — carries exactly the same meaning as for a transverse wave; the only change is that the displacement function \(y(x,t)\) is replaced by \(s(x,t)\).
14.4 The Speed of a Travelling Wave
Before we work out how fast a wave travels, it helps to consolidate the quantities from the previous section with a worked example. The following NCERT example uses the amplitude, wavelength, period and frequency we have just defined.
Now, the main question of this section: how fast does the wave pattern itself move?
To find the speed at which a travelling wave moves, we fix our attention on any one particular point of the wave — a point recognised by a definite value of its phase — and watch how that point moves as time passes. The most convenient point to track is a crest of the wave.
Fig. 14.8 shows the shape of the wave at two instants of time separated by a small time interval \(\Delta t\). In that short time, the whole wave pattern shifts to the right (the positive \(x\)-direction) by a distance \(\Delta x\). In particular, the crest marked by a dot (\(\bullet\)) moves forward by \(\Delta x\) in the time \(\Delta t\).
Figure to come
Fig. 14.8 – The same sinusoidal wave drawn at time \(t\) and again at time \(t + \Delta t\), shifted to the right; a marked crest (dot) has moved forward by a small distance \(\Delta x\), with arrows indicating the rightward shift.
The speed of the wave is therefore \(\Delta x/\Delta t\). We could just as well have placed the dot on a point of any other phase — it would move with the same speed \(v\). (If different points moved at different speeds, the wave pattern would not keep its fixed shape as it travelled.)
Deriving the wave speed
A “fixed phase point” is a point whose phase does not change as it moves along. So its motion is described by keeping the phase constant:
\[kx - \omega t = \text{constant} \tag{14.10}\]
As time \(t\) changes, the position \(x\) of this point must change so that the phase stays the same. Comparing the phase at time \(t\) (position \(x\)) with the phase a moment later at time \(t + \Delta t\) (position \(x + \Delta x\)):
\[kx - \omega t = k(x + \Delta x) - \omega(t + \Delta t)\]
Cancelling the common terms leaves
\[k\,\Delta x - \omega\,\Delta t = 0.\]
Taking \(\Delta x\) and \(\Delta t\) vanishingly small, this becomes the speed of the point:
\[\frac{dx}{dt} = \frac{\omega}{k} = v \tag{14.11}\]
Now express this in terms of the more familiar quantities. Using \(\omega = 2\pi\nu\) and \(k = 2\pi/\lambda\),
\[v = \frac{\omega}{k} = \frac{2\pi\nu}{2\pi/\lambda} = \lambda\nu = \frac{\lambda}{T} \tag{14.12}\]
Here \(v\) is the wave speed (m s\(^{-1}\)), \(\lambda\) the wavelength (m), \(\nu\) the frequency (Hz), and \(T\) the period (s).
Equation (14.12) is a general relation that holds for all progressive waves — transverse or longitudinal, on strings, in air, or anywhere else. It carries a simple physical meaning: in the time required for one full oscillation of any particle of the medium (that is, one period \(T\)), the wave pattern advances by exactly one wavelength \(\lambda\).
What decides the wave speed?
An important point: the speed of a mechanical wave is not set by the source that makes it. It is fixed by the properties of the medium the wave travels through. Two kinds of properties matter:
The inertial property — how much mass the medium has to be moved. For a string this is the linear mass density (mass per unit length), and for a bulk medium it is the ordinary mass density.
The elastic property — how strongly the medium springs back when disturbed. Depending on the situation this is the Young’s modulus, the shear modulus, or the bulk modulus of the medium.
The medium fixes the speed; then Eq. (14.12) simply relates the wavelength to the frequency for that given speed. As noted earlier, one medium can carry both transverse and longitudinal waves, and these generally travel at different speeds in the same medium. In the next two subsections we will derive specific formulas for the wave speed in particular media.
14.4.1 Speed of a Transverse Wave on Stretched String
In the previous section we said that the speed of a mechanical wave is fixed by two kinds of properties of the medium — its elastic (restoring) property and its inertial property. Let us now apply this idea to a specific case we can picture easily: a transverse wave running along a stretched string.
What controls the speed on a string?
The speed of the wave is decided by two competing factors. One is the restoring force in the medium — how strongly the medium pulls a disturbed part back toward its rest position. The other is the inertial property — how much mass has to be set moving. We expect the speed to be larger when the restoring force is stronger, and smaller when there is more mass to move. So the speed should be directly related to the restoring force and inversely related to the inertia.
For a stretched string, the restoring force is provided by the tension \(T\) in the string — the pull along the string that tries to straighten out any disturbance. (The tension is a property of the stretched string, set up by an external stretching force.)
The inertial property here is the linear mass density \(\mu\), which is the mass \(m\) of the string divided by its length \(L\):
\[\mu = \frac{m}{L}.\]
Figure to come
Fig. 14.8a – A horizontal stretched string pulled tight by a tension \(T\) at both ends, carrying a small transverse hump (pulse) travelling along it, with a label showing linear mass density \(\mu = m/L\).
Finding the formula by dimensional analysis
Using Newton’s laws of motion, one can derive an exact formula for the wave speed on a string, but that full derivation is beyond the level of this book. Instead, we use a shortcut you met earlier — dimensional analysis. Recall its one limitation: dimensional analysis can tell us how the quantities must combine, but it can never fix the value of a pure dimensionless number (a constant) out front. That constant is always left undetermined.
Let us set up the dimensions. Writing \([\text{M}]\) for mass, \([\text{L}]\) for length, and \([\text{T}]\) for time:
The dimension of \(\mu\) (mass per length) is \([\text{ML}^{-1}]\).
The dimension of tension \(T\) is the same as that of force, namely \([\text{MLT}^{-2}]\).
We want to combine these to get the dimension of speed, which is \([\text{LT}^{-1}]\). A quick inspection shows that the combination \(T/\mu\) does the job:
\[\frac{[\text{MLT}^{-2}]}{[\text{ML}^{-1}]} = [\text{L}^2\text{T}^{-2}]\]
This is the square of a speed. So \(\sqrt{T/\mu}\) has exactly the dimension of speed.
Assuming \(T\) and \(\mu\) are the only physical quantities that matter, the wave speed must therefore be
\[v = C\sqrt{\frac{T}{\mu}} \tag{14.13}\]
where \(C\) is the undetermined dimensionless constant. The exact derivation (using Newton’s laws) shows that \(C = 1\). Hence the speed of transverse waves on a stretched string is
\[v = \sqrt{\frac{T}{\mu}} \tag{14.14}\]
Here \(v\) is the wave speed (m s\(^{-1}\)), \(T\) the tension (newton, N), and \(\mu\) the linear mass density (kg m\(^{-1}\)).
Speed depends on the medium, not on the wave
Notice the key point in Eq. (14.14): the speed \(v\) depends only on the properties of the medium, \(T\) and \(\mu\). It does not depend on the wavelength or the frequency of the wave itself. However you shake the string — fast or slow, gently or strongly — the wave still travels at the same speed set by the string.
(In higher studies you will meet special “dispersive” media, where the wave speed does depend on frequency. But for a simple stretched string this is not the case.)
This clears up how \(\lambda\) and \(\nu\) are fixed. Of the two, it is the source of the disturbance that decides the frequency \(\nu\) of the wave produced. Once the medium fixes the speed \(v\), and the source fixes the frequency \(\nu\), then Eq. (14.12) fixes the wavelength:
\[\lambda = \frac{v}{\nu} \tag{14.15}\]
14.4.2 Speed of a Longitudinal Wave (Speed of Sound)
We now turn to the other great class of waves — longitudinal waves, of which sound is the most important example. In a longitudinal wave, the particles of the medium oscillate forward and backward along the same direction the wave travels. As we saw earlier, sound travels through air as a series of compressions and rarefactions of small volume elements of the medium (recall Fig. 14.4).
Which elastic property matters here?
For a string, the relevant elastic property was its tension. For sound, the disturbance squeezes and stretches the volume of small parcels of the medium. The elastic property that measures how a medium resists being compressed is its bulk modulus \(B\), which you met earlier while studying the mechanical properties of matter. It is defined by
\[B = -\frac{\Delta P}{\Delta V / V} \tag{14.16}\]
Here a change in pressure \(\Delta P\) produces a fractional change in volume \(\Delta V/V\), called the volumetric strain. The minus sign appears because an increase in pressure (\(\Delta P\) positive) causes the volume to shrink (\(\Delta V\) negative), and the bulk modulus is defined to be a positive quantity. \(B\) has the same dimensions as pressure and is measured in pascals (Pa) in SI units.
The inertial property relevant to sound is simply the mass density \(\rho\) of the medium — its mass per unit volume — which has dimensions \([\text{ML}^{-3}]\).
Finding the formula by dimensional analysis
Just as with the string, we combine the elastic property \(B\) and the inertial property \(\rho\) to build a speed. Checking dimensions, the bulk modulus has the dimensions of pressure, \([\text{ML}^{-1}\text{T}^{-2}]\), so
\[\frac{[\text{ML}^{-1}\text{T}^{-2}]}{[\text{ML}^{-3}]} = [\text{L}^2\text{T}^{-2}] \tag{14.17}\]
which is again the square of a speed. Taking \(B\) and \(\rho\) to be the only relevant quantities,
\[v = C\sqrt{\frac{B}{\rho}} \tag{14.18}\]
where \(C\) is the undetermined dimensionless constant. The exact derivation again gives \(C = 1\), so the general formula for the speed of a longitudinal wave in a medium is
\[v = \sqrt{\frac{B}{\rho}} \tag{14.19}\]
Longitudinal waves in a solid bar
There is one special case worth noting. For a thin solid bar, when a longitudinal wave passes, the bar can be treated as being squeezed and stretched only along its length; its sideways (lateral) expansion is negligible. In this situation the relevant elastic constant is not the bulk modulus but Young’s modulus \(Y\) — the modulus that governs stretching and compressing along one direction. Young’s modulus has the same dimensions as the bulk modulus, so the dimensional analysis is identical, and again the constant works out to unity. The speed of longitudinal waves in a solid bar is therefore
\[v = \sqrt{\frac{Y}{\rho}} \tag{14.20}\]
where \(Y\) is the Young’s modulus of the material of the bar.
Table 14.1 lists the measured speed of sound in some common media.
Table 14.1 — Speed of Sound in some Media (speed in m s\(^{-1}\))
| Medium | Speed (m s\(^{-1}\)) |
|---|---|
| Gases | |
| Air (0 °C) | 331 |
| Air (20 °C) | 343 |
| Helium | 965 |
| Hydrogen | 1284 |
| Liquids | |
| Water (0 °C) | 1402 |
| Water (20 °C) | 1482 |
| Seawater | 1522 |
| Solids | |
| Aluminium | 6420 |
| Copper | 3560 |
| Steel | 5941 |
| Granite | 6000 |
| Vulcanised Rubber | 54 |
Why sound is faster in solids and liquids
The table shows that solids and liquids generally carry sound much faster than gases. (For solids, the value quoted is the speed of longitudinal waves in the solid.) At first this seems strange: solids and liquids are far denser than gases, and Eq. (14.19) has density \(\rho\) in the denominator, which should slow the wave down.
The resolution is that both \(B\) and \(\rho\) increase as we go from gases to liquids to solids — but \(B\) increases far more. Solids and liquids are enormously harder to compress than gases, so their bulk modulus \(B\) is very much larger. In the ratio \(B/\rho\), the big rise in \(B\) easily beats the smaller rise in \(\rho\), so \(\sqrt{B/\rho}\) — and hence the wave speed — comes out larger. This is why sound races through steel and water far faster than through air.
Estimating the speed of sound in a gas: Newton’s formula
We can go further and estimate the speed of sound in a gas from theory. For an ideal gas, the pressure \(P\), volume \(V\) and absolute temperature \(T\) are related by
\[PV = N k_B T \tag{14.21}\]
where \(N\) is the number of molecules in volume \(V\), \(k_B\) is the Boltzmann constant, and \(T\) is the temperature in kelvin.
Newton assumed that as sound passes, the compressions and rarefactions happen slowly enough that the temperature of each region stays constant — that is, the changes are isothermal (constant temperature). For an isothermal change, \(T\) is fixed, so \(PV\) stays constant. Taking small changes in Eq. (14.21):
\[V\Delta P + P\Delta V = 0 \quad \Rightarrow \quad -\frac{\Delta P}{\Delta V/V} = P.\]
Comparing this with the definition of bulk modulus, Eq. (14.16), we see that under this assumption the bulk modulus of the gas is simply
\[B = P.\]
Substituting into Eq. (14.19), the speed of sound in an ideal gas becomes
\[v = \sqrt{\frac{P}{\rho}} \tag{14.22}\]
This result was first given by Newton and is known as Newton’s formula.
14.5 The Principle of Superposition of Waves
Up to now we have followed a single wave travelling through a medium. But in the real world, many waves often pass through the same region at once — think of several instruments playing together, or ripples from two stones dropped in a pond. What happens where two waves meet and overlap?
Two pulses that pass through each other
Consider two wave pulses travelling in opposite directions along a string and about to cross each other, as shown in Fig. 14.9. A remarkable thing happens: after they have crossed, each pulse continues on its way completely unchanged, keeping its original shape and size — as if the other had never been there. Only during the moment of overlap is the wave pattern different from either pulse alone.
Figure to come
Fig. 14.9 – A sequence of snapshots (at successive instants) of two pulses of equal and opposite shape moving toward each other along a string; in the middle frame they overlap and momentarily cancel to give zero displacement everywhere, then re-emerge unchanged on the far sides.
Fig. 14.9 shows the special case of two pulses of equal size but opposite shape (one up, one down) approaching each other. While they overlap, the displacement at each point of the string is just the sum of the displacements the two pulses would produce separately. Since one pulse pushes the string up and the other pushes it down, they partly or wholly cancel. In the middle frame, graph (c), the effect is dramatic: the two displacements cancel exactly, and for that instant the string is completely flat, with zero displacement everywhere.
The principle of superposition
This adding-up rule is called the principle of superposition of waves. According to it, each wave travels as if the others were not present, and the actual displacement of any particle of the medium is the algebraic sum (taking signs into account) of the displacements each wave would give on its own.
The word “algebraic” is important: because displacements can be positive (up) or negative (down), overlapping waves can add up to something larger, or cancel to something smaller — even to zero.
Writing superposition mathematically
To state this in symbols, let \(y_1(x,t)\) and \(y_2(x,t)\) be the displacements that two separate wave disturbances would produce in the medium. If both waves reach a region at the same time and overlap, the net displacement \(y(x,t)\) is simply their sum:
\[y(x,t) = y_1(x,t) + y_2(x,t) \tag{14.25}\]
The same idea extends to any number of waves. If several waves, all travelling with speed \(v\) in the \(+x\) direction, have wave functions
\[y_1 = f_1(x - vt), \quad y_2 = f_2(x - vt), \quad \dots, \quad y_n = f_n(x - vt),\]
then the total disturbance in the medium is the sum of all of them:
\[y = f_1(x - vt) + f_2(x - vt) + \dots + f_n(x - vt) = \sum_{i=1}^{n} f_i(x - vt) \tag{14.26}\]
The principle of superposition is the foundation of a very important phenomenon called interference — the combining of waves to give a resultant that may be stronger or weaker than the individual waves.
Superposing two harmonic waves
Let us now apply superposition to two neat harmonic waves and see what comes out. Take two harmonic waves travelling on a stretched string, both with the same angular frequency \(\omega\) and the same angular wave number \(k\) — and therefore the same wavelength \(\lambda\) and the same speed. Let them also have equal amplitudes \(a\), and let both travel in the \(+x\) direction. The only difference between them is their initial phase.
Using the standard form of Eq. (14.2), the two waves are
\[y_1(x,t) = a\sin(kx - \omega t) \tag{14.27}\] \[y_2(x,t) = a\sin(kx - \omega t + \phi) \tag{14.28}\]
where \(\phi\) is the phase difference between them. By the principle of superposition, the net displacement is
\[y(x,t) = a\sin(kx - \omega t) + a\sin(kx - \omega t + \phi) \tag{14.29}\]
We now use the familiar trigonometric identity \(\sin A + \sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\). With \(A = kx - \omega t\) and \(B = kx - \omega t + \phi\), this gives
\[y(x,t) = a\left[2\sin\!\left(\frac{(kx - \omega t) + (kx - \omega t + \phi)}{2}\right)\cos\frac{\phi}{2}\right] \tag{14.30}\]
Simplifying the average inside the sine leads to
\[y(x,t) = 2a\cos\frac{\phi}{2}\,\sin\!\left(kx - \omega t + \frac{\phi}{2}\right) \tag{14.31}\]
Look closely at Eq. (14.31). It is itself a harmonic travelling wave in the \(+x\) direction, with the same frequency and wavelength as the originals. Its initial phase angle is \(\phi/2\). But the most important feature is its amplitude, which is not simply \(2a\) — it depends on the phase difference \(\phi\) between the two waves:
\[A(\phi) = 2a\cos\tfrac{1}{2}\phi \tag{14.32}\]
Here \(A(\phi)\) is the amplitude of the resultant wave, \(a\) the amplitude of each original wave, and \(\phi\) their phase difference.
Two special cases: constructive and destructive interference
The formula \(A(\phi) = 2a\cos\tfrac{1}{2}\phi\) tells the whole story, but two cases stand out.
When \(\phi = 0\), the two waves are exactly in phase — crest falls on crest and trough on trough. Then \(\cos 0 = 1\), so the resultant amplitude is the largest possible, \(A = 2a\):
\[y(x,t) = 2a\sin(kx - \omega t) \tag{14.33}\]
This is called constructive interference — the two waves reinforce each other, and their amplitudes add up.
When \(\phi = \pi\), the two waves are completely out of phase — crest falls on trough. Then \(\cos\tfrac{\pi}{2} = 0\), so the resultant amplitude is zero everywhere, at all times:
\[y(x,t) = 0 \tag{14.34}\]
This is called destructive interference — the two waves cancel each other out.
Fig. 14.10 shows both cases side by side — the reinforced wave for \(\phi = 0\) and the complete cancellation for \(\phi = \pi\).
Figure to come
Fig. 14.10 – Two panels showing the superposition of two equal harmonic waves: panel (a) with phase difference \(\phi = 0\) giving a resultant of amplitude \(2a\) (constructive), and panel (b) with \(\phi = \pi\) giving zero resultant (destructive).
14.6 Reflection of Waves
Until now we have imagined waves travelling through an unbounded medium — a string or a body of air with no edges. But every real medium ends somewhere. What happens when a pulse or a wave reaches a boundary?
Reflection at a rigid boundary
If the boundary is rigid — a hard, immovable wall — the pulse or wave bounces back. This bouncing-back is called reflection. A familiar example is an echo: when you shout toward a distant cliff or a large wall, the sound reflects off it and returns to your ears a moment later.
Reflection and refraction at an interface
If the boundary is not perfectly rigid, or is an interface between two different elastic media, the situation is a little more involved. Now a part of the incident wave is reflected back, and another part passes through into the second medium. The part that crosses into the second medium is the transmitted wave.
When a wave strikes such a boundary at an angle (obliquely) and passes into the second medium, its direction of travel bends. This transmitted, bent wave is called the refracted wave, and the process is called refraction. The incident and refracted waves obey Snell’s law of refraction (the same law you meet in optics), while the incident and reflected waves obey the usual laws of reflection — for instance, the angle of incidence equals the angle of reflection.
Why reflection at a rigid boundary flips the wave
Fig. 14.11 shows a pulse travelling along a stretched string whose far end is fixed to a rigid support, and being reflected there. If we assume the boundary absorbs no energy, the reflected pulse has the same shape and size as the incident pulse — but it comes back inverted. That is, it undergoes a phase change of \(\pi\) (equivalently, \(180^\circ\)) on reflection.
Figure to come
Fig. 14.11 – A sequence of frames showing a pulse travelling along a stretched string toward a rigid wall, reaching it, and returning inverted (flipped upside down) as the reflected pulse.
Why must the wave flip? Because the end is rigidly fixed, that point of the string can never move — its displacement must be zero at all times. By the principle of superposition, the total displacement at the wall is the sum of the incident and reflected waves. For this sum to stay zero at the wall, the reflected wave must always be the exact negative of the incident wave — which means the two differ in phase by exactly \(\pi\). Only an inverted reflection keeps the fixed end at rest.
There is also a nice way to see this using forces. As the pulse arrives at the wall, it exerts a force on the wall. By Newton’s third law, the wall pushes back on the string with an equal and opposite force. This oppositely directed push is what launches a reflected pulse that is inverted — differing in phase by \(\pi\) from the incident one.
Reflection at a free (open) boundary
Now suppose the boundary is the opposite kind — not rigid, but completely free to move. An example is a string tied to a light ring that can slide freely up and down a smooth rod. Here the reflected pulse comes back with the same phase and the same amplitude as the incident pulse (again assuming no energy is lost). Since the incident and reflected pulses now add up at the free end instead of cancelling, the net maximum displacement there is twice the amplitude of each pulse. A common example of such a free (open) boundary is the open end of an organ pipe.
Summarising with equations
To put all this in symbols, let the incident travelling wave be
\[y_i(x,t) = a\sin(kx - \omega t)\]
At a rigid boundary, the reflected wave carries the extra phase of \(\pi\):
\[y_r(x,t) = a\sin(kx - \omega t + \pi) = -a\sin(kx - \omega t) \tag{14.35}\]
At an open boundary, the reflected wave carries no extra phase:
\[y_r(x,t) = a\sin(kx - \omega t + 0) = a\sin(kx - \omega t) \tag{14.36}\]
From Eq. (14.35) we can immediately check the rigid-boundary condition: the total displacement there is \(y = y_i + y_r = a\sin(kx - \omega t) - a\sin(kx - \omega t) = 0\) at all times, exactly as required for a fixed end. (Here we have written the reflected wave in the same \((kx - \omega t)\) form to highlight the phase flip. In the next subsection we will treat the reflected wave properly as one travelling in the opposite direction, which is what leads to standing waves.)
14.6.1 Standing Waves and Normal Modes
In the previous section we looked at reflection at a single boundary. But many familiar systems have two boundaries — a string fixed at both ends, or an air column in a pipe. What happens then?
Two boundaries, and a wave that stops travelling
In a string fixed at both ends, a wave travelling one way gets reflected at one end, travels back, and is reflected again at the other end. This back-and-forth reflection continues, and the many reflected waves keep superposing on one another. Very quickly, a steady pattern establishes itself on the string — a pattern that no longer appears to travel in either direction. Such patterns are called standing waves or stationary waves.
The mathematics of a standing wave
To see how this works, take a wave travelling in the \(+x\) direction and an equal reflected wave travelling in the \(-x\) direction. Using Eqs. (14.2) and (14.4) with \(\phi = 0\):
\[y_1(x,t) = a\sin(kx - \omega t)\] \[y_2(x,t) = a\sin(kx + \omega t)\]
By the principle of superposition, the resultant on the string is
\[y(x,t) = y_1(x,t) + y_2(x,t) = a\big[\sin(kx - \omega t) + \sin(kx + \omega t)\big]\]
Now use the identity \(\sin(A - B) + \sin(A + B) = 2\sin A\cos B\), with \(A = kx\) and \(B = \omega t\):
\[y(x,t) = 2a\sin kx\,\cos\omega t \tag{14.37}\]
Look carefully at Eq. (14.37) and compare it with a travelling wave, Eq. (14.2). In a travelling wave, position and time always appear locked together as \((kx - \omega t)\). Here they appear separately: a \(\sin kx\) part that depends only on position, and a \(\cos\omega t\) part that depends only on time.
This separation changes everything. The quantity in front of \(\cos\omega t\), namely \(2a\sin kx\), acts as the amplitude — but it now depends on the position \(x\). So different points of the string oscillate with different amplitudes, yet all of them oscillate with the same angular frequency \(\omega\), and all in step (in phase) with one another. The pattern as a whole does not move left or right. This is exactly why these are called standing (stationary) waves.
Nodes and antinodes
Since the amplitude \(2a\sin kx\) depends on position, some points get a very special treatment. At points where \(\sin kx = 0\), the amplitude is zero — these points never move at all. They are called nodes. At points where \(\sin kx\) is largest, the amplitude reaches its maximum value \(2a\) — these are called antinodes.
Fig. 14.12 shows a standing-wave pattern built from two travelling waves moving in opposite directions, with the node positions staying fixed at all times.
Figure to come
Fig. 14.12 – A standing wave shown as several superposed snapshots in time, formed from two opposite travelling waves, with the fixed node points (zero displacement) marked N along the string.
Where the nodes and antinodes sit
The node positions come from setting the amplitude to zero:
\[\sin kx = 0 \quad \Rightarrow \quad kx = n\pi; \quad n = 0, 1, 2, 3, \dots\]
Using \(k = 2\pi/\lambda\), this gives the node positions
\[x = \frac{n\lambda}{2}; \quad n = 0, 1, 2, 3, \dots \tag{14.38}\]
So two successive nodes are separated by a distance of \(\lambda/2\). In the same way, antinodes occur where \(|\sin kx| = 1\):
\[kx = \left(n + \tfrac{1}{2}\right)\pi \quad \Rightarrow \quad x = \left(n + \tfrac{1}{2}\right)\frac{\lambda}{2}; \quad n = 0, 1, 2, 3, \dots \tag{14.39}\]
Two successive antinodes are also separated by \(\lambda/2\). (And a node and its neighbouring antinode are \(\lambda/4\) apart.)
Normal modes of a string fixed at both ends
Here is the most important consequence of standing waves: the boundaries do not allow just any wavelength or frequency. The system can only vibrate at a special set of allowed frequencies, called its natural frequencies or normal modes. This is quite unlike a travelling wave, which can have any frequency at all.
Let us find these modes for a string of length \(L\) fixed at both ends. Since both ends are clamped, they cannot move — so \(x = 0\) and \(x = L\) must both be nodes. Taking the node condition from Eq. (14.38), the end \(x = 0\) is automatically a node. The end \(x = L\) being a node requires
\[L = \frac{n\lambda}{2}; \quad n = 1, 2, 3, \dots \tag{14.40}\]
So the only wavelengths that can form standing waves are
\[\lambda = \frac{2L}{n}; \quad n = 1, 2, 3, \dots \tag{14.41}\]
Using \(v = \lambda\nu\), the corresponding allowed frequencies are
\[\nu = \frac{nv}{2L}; \quad n = 1, 2, 3, \dots \tag{14.42}\]
where \(v\) is the wave speed set by the string’s properties (its tension and linear mass density).
Harmonics
These allowed frequencies are the normal modes of the string. The lowest of them, with \(n = 1\), is called the fundamental mode or the first harmonic:
\[\nu_1 = \frac{v}{2L}.\]
The next frequency (\(n = 2\)) is the second harmonic, \(n = 3\) gives the third harmonic, and so on. We label them \(\nu_n\) for \(n = 1, 2, 3, \dots\). Fig. 14.13 shows the first six harmonics of a string fixed at both ends.
Figure to come
Fig. 14.13 – The first six harmonics of a stretched string fixed at both ends, each labelled (fundamental, second harmonic, … sixth harmonic) with nodes N and antinodes A marked.
A real string need not vibrate in a single pure mode. In general its motion is a superposition of several modes at once — some strongly present, some weakly. Musical instruments work on exactly this idea.
Normal modes of an air column: one end closed, one open
Now consider an air column in a pipe closed at one end and open at the other. A glass tube partly filled with water is a good example — the air column sits above the water.
At the closed end (in contact with the water), the air cannot move, so it is a node of displacement. At the open end, the air is free to move the most, so it is an antinode. (In terms of pressure, it is the reverse: the closed end, a displacement node, is a point of largest pressure change, while the open end, a displacement antinode, has the least pressure change.)
Taking the closed end as \(x = 0\), the node condition Eq. (14.38) is already satisfied there. Requiring the open end \(x = L\) to be an antinode, we use Eq. (14.39):
\[L = \left(n + \tfrac{1}{2}\right)\frac{\lambda}{2}, \quad n = 0, 1, 2, 3, \dots\]
This restricts the possible wavelengths to
\[\lambda = \frac{2L}{\left(n + \tfrac{1}{2}\right)}, \quad n = 0, 1, 2, 3, \dots \tag{14.43}\]
and the natural frequencies to
\[\nu = \left(n + \tfrac{1}{2}\right)\frac{v}{2L}; \quad n = 0, 1, 2, 3, \dots \tag{14.44}\]
The fundamental frequency, at \(n = 0\), is \(\dfrac{v}{4L}\). The higher frequencies turn out to be the odd harmonics only — odd multiples of the fundamental: \(3\left(\dfrac{v}{4L}\right)\), \(5\left(\dfrac{v}{4L}\right)\), and so on. Fig. 14.14 shows the first six odd harmonics of such a closed-open air column.
Figure to come
Fig. 14.14 – Normal modes of an air column closed at one end and open at the other, showing the fundamental and the higher odd harmonics (third, fifth, …) with a node at the closed end and an antinode at the open end.
Pipe open at both ends
For a pipe open at both ends, each end is an antinode. Fitting an antinode at both ends allows all the harmonics — first, second, third, and so on — so an air column open at both ends generates the complete set of harmonics, as shown in Fig. 14.15.
Figure to come
Fig. 14.15 – Standing waves in a pipe open at both ends, showing the first four harmonics, each with an antinode at both open ends.
Resonance
The strings and air columns above can also be driven by an external periodic force — this is a forced oscillation. If the driving frequency is close to one of the system’s natural frequencies, the amplitude of vibration grows very large. This effect is called resonance.
Finally, the same ideas extend to two-dimensional systems. The normal modes of a circular membrane clamped at its rim, as in a tabla, are set by the condition that no point on the rim can vibrate. Working out these frequencies is more complicated because the wave spreads in two dimensions, but the underlying physics — standing waves fixed by boundary conditions — is exactly the same.
14.7 Beats
We end our study of waves with a striking effect that comes straight out of the principle of superposition — the phenomenon of beats.
What beats are
When two harmonic sound waves of close, but not exactly equal, frequencies reach our ears at the same time, we hear two things at once. First, we hear a single steady note whose pitch is roughly the average of the two frequencies. But on top of that, we hear the loudness of this note rise and fall, over and over — a slow throbbing or pulsing of the sound. This regular rise and fall in intensity is called beats.
The throbbing repeats at a rate exactly equal to the difference between the two frequencies. Musicians make direct use of this. When tuning two instruments to each other, they listen for these beats and keep adjusting one instrument until the beats disappear completely — at which point the two are perfectly in tune.
This is exactly the throbbing you would hear from two guitar strings tuned slightly apart: each string produces a steady note, but together they produce a pulsing “wah–wah–wah” whose rate equals the gap between their two frequencies. Tighten one string until the pulsing stops, and the strings are in tune.
The mathematics of beats
Let us see where the throbbing comes from. Consider two harmonic sound waves of nearly equal angular frequencies \(\omega_1\) and \(\omega_2\), and to keep things simple, fix our attention at one location, \(x = 0\). Choosing the phase conveniently (\(\phi = \pi/2\) for each turns the sine into a cosine) and taking equal amplitudes, the two waves at that point are
\[s_1 = a\cos\omega_1 t \quad \text{and} \quad s_2 = a\cos\omega_2 t \tag{14.45}\]
Here we write the displacement as \(s\) rather than \(y\), because sound is a longitudinal wave. Let \(\omega_1\) be the slightly larger of the two frequencies. By the principle of superposition, the resultant displacement is
\[s = s_1 + s_2 = a(\cos\omega_1 t + \cos\omega_2 t).\]
Using the trigonometric identity \(\cos A + \cos B = 2\cos\dfrac{A+B}{2}\cos\dfrac{A-B}{2}\),
\[s = 2a\cos\!\left(\frac{(\omega_1 - \omega_2)t}{2}\right)\cos\!\left(\frac{(\omega_1 + \omega_2)t}{2}\right) \tag{14.46}\]
To make this easier to read, define two new frequencies — the average and the half-difference:
\[\omega_a = \frac{\omega_1 + \omega_2}{2} \quad \text{and} \quad \omega_b = \frac{\omega_1 - \omega_2}{2}.\]
Then the result takes the compact form
\[s = \big[\,2a\cos\omega_b t\,\big]\cos\omega_a t \tag{14.47}\]
Reading the result
Because the two frequencies are nearly equal, \(\omega_b\) (built from their tiny difference) is very small, while \(\omega_a\) (their average) is large — so \(\omega_a \gg \omega_b\).
This lets us read Eq. (14.47) as an ordinary oscillation at the fast average frequency \(\omega_a\), but with an amplitude given by the slowly-changing term in square brackets, \(2a\cos\omega_b t\). Unlike a pure harmonic wave, this amplitude is not constant — it swells and shrinks slowly with time.
The sound is loudest whenever the amplitude term reaches its extreme value, that is, whenever \(\cos\omega_b t = +1\) or \(-1\). Now, \(|\cos\omega_b t|\) reaches its maximum value of \(1\) twice in each full cycle of \(\cos\omega_b t\). So the loudness peaks at a rate of \(2\omega_b = \omega_1 - \omega_2\). Since \(\omega = 2\pi\nu\), this rate of throbbing — the beat frequency \(\nu_{beat}\) — is simply
\[\nu_{beat} = \nu_1 - \nu_2 \tag{14.48}\]
Fig. 14.16 shows this for two waves of frequency \(11\ \text{Hz}\) and \(9\ \text{Hz}\): the resultant swells and fades at a beat frequency of \(11 - 9 = 2\ \text{Hz}\).
Figure to come
Fig. 14.16 – Three stacked graphs: (a) a wave of \(11\ \text{Hz}\), (b) a wave of \(9\ \text{Hz}\), and (c) their superposition, whose amplitude envelope rises and falls twice per second, showing beats at \(2\ \text{Hz}\).
14.8 Summary
Mechanical waves can exist only in a material medium and are governed by Newton’s laws.
Transverse waves are waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation.
Longitudinal waves are waves in which the particles of the medium oscillate along the direction of wave propagation.
A progressive wave is a wave that moves from one point of the medium to another.
The displacement in a sinusoidal wave propagating in the positive \(x\) direction is given by
\[y(x,t) = a\sin(kx - \omega t + \phi)\]
where \(a\) is the amplitude of the wave, \(k\) is the angular wave number, \(\omega\) is the angular frequency, \((kx - \omega t + \phi)\) is the phase, and \(\phi\) is the phase constant or phase angle.
Wavelength \(\lambda\) of a progressive wave is the distance between two consecutive points of the same phase at a given time. In a stationary wave, it is twice the distance between two consecutive nodes or antinodes.
Period \(T\) of oscillation of a wave is the time any element of the medium takes to move through one complete oscillation. It is related to the angular frequency \(\omega\) by
\[T = \frac{2\pi}{\omega}\]
- Frequency \(\nu\) of a wave is \(1/T\) and is related to angular frequency by
\[\nu = \frac{\omega}{2\pi}\]
- Speed of a progressive wave is given by
\[v = \frac{\omega}{k} = \frac{\lambda}{T} = \lambda\nu\]
- The speed of a transverse wave on a stretched string is set by the properties of the string. For tension \(T\) and linear mass density \(\mu\),
\[v = \sqrt{\frac{T}{\mu}}\]
- Sound waves are longitudinal mechanical waves that can travel through solids, liquids, or gases. The speed \(v\) of a sound wave in a fluid of bulk modulus \(B\) and density \(\rho\) is
\[v = \sqrt{\frac{B}{\rho}}\]
The speed of longitudinal waves in a metallic bar is
\[v = \sqrt{\frac{Y}{\rho}}\]
For gases, since \(B = \gamma P\), the speed of sound is
\[v = \sqrt{\frac{\gamma P}{\rho}}\]
- When two or more waves traverse simultaneously in the same medium, the displacement of any element of the medium is the algebraic sum of the displacements due to each wave. This is the principle of superposition of waves:
\[y = \sum_{i=1}^{n} f_i(x - vt)\]
- Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two travel in the same direction with the same amplitude \(a\) and frequency but differ in phase by a phase constant \(\phi\), the result is a single wave of the same frequency \(\omega\):
\[y(x,t) = \left[2a\cos\tfrac{1}{2}\phi\right]\sin\!\left(kx - \omega t + \tfrac{1}{2}\phi\right)\]
If \(\phi = 0\) or an integral multiple of \(2\pi\), the waves are exactly in phase and the interference is constructive; if \(\phi = \pi\), they are exactly out of phase and the interference is destructive.
- A travelling wave, at a rigid boundary or a closed end, is reflected with a phase reversal, but reflection at an open boundary takes place without any phase change.
For an incident wave \(y_i(x,t) = a\sin(kx - \omega t)\), the reflected wave at a rigid boundary is
\[y_r(x,t) = -a\sin(kx + \omega t)\]
For reflection at an open boundary,
\[y_r(x,t) = a\sin(kx + \omega t)\]
- The interference of two identical waves moving in opposite directions produces standing waves. For a string with fixed ends, the standing wave is
\[y(x,t) = [2a\sin kx]\cos\omega t\]
Standing waves are characterised by fixed locations of zero displacement called nodes and fixed locations of maximum displacement called antinodes. The separation between two consecutive nodes or antinodes is \(\lambda/2\).
A stretched string of length \(L\) fixed at both ends vibrates with frequencies
\[\nu = \frac{nv}{2L}, \quad n = 1, 2, 3, \dots\]
The set of frequencies given by this relation are the normal modes of oscillation of the system. The mode with the lowest frequency is the fundamental mode or the first harmonic. The second harmonic is the mode with \(n = 2\), and so on.
A pipe of length \(L\) with one end closed and the other open (such as an air column) vibrates with frequencies
\[\nu = \left(n + \tfrac{1}{2}\right)\frac{v}{2L}, \quad n = 0, 1, 2, 3, \dots\]
The set of frequencies represented by this relation are the normal modes of oscillation of such a system. The lowest frequency, \(v/4L\), is the fundamental mode or the first harmonic.
A string of length \(L\) fixed at both ends, or an air column closed at one end and open at the other, or open at both ends, vibrates with certain frequencies called its normal modes. Each of these frequencies is a resonant frequency of the system.
Beats arise when two waves of slightly different frequencies, \(\nu_1\) and \(\nu_2\), and comparable amplitudes are superposed. The beat frequency is
\[\nu_{beat} = \nu_1 \sim \nu_2\]
14.9 Points to Ponder
A wave is not the motion of matter as a whole in a medium. A wind is different from the sound wave in air. The wind involves the motion of air from one place to the other. The sound wave involves compressions and rarefactions of layers of air.
In a wave, energy and not the matter is transferred from one point to the other.
In a mechanical wave, energy transfer takes place because of the coupling through elastic forces between neighbouring oscillating parts of the medium.
Transverse waves can propagate only in a medium with a shear modulus of elasticity. Longitudinal waves need a bulk modulus of elasticity and are therefore possible in all media — solids, liquids, and gases.
In a harmonic progressive wave of a given frequency, all particles have the same amplitude but different phases at a given instant of time. In a stationary wave, all particles between two nodes have the same phase at a given instant but have different amplitudes.
Relative to an observer at rest in a medium, the speed of a mechanical wave in that medium (\(v\)) depends only on the elastic and other properties (such as mass density) of the medium. It does not depend on the velocity of the source.
14.10 Physical Quantities
| Physical quantity | Symbol | Dimensions | Unit | Remarks |
|---|---|---|---|---|
| Wavelength | \(\lambda\) | \([\text{L}]\) | m | Distance between two consecutive points with the same phase. |
| Propagation constant | \(k\) | \([\text{L}^{-1}]\) | m\(^{-1}\) | \(k = \dfrac{2\pi}{\lambda}\) |
| Wave speed | \(v\) | \([\text{LT}^{-1}]\) | m s\(^{-1}\) | \(v = \nu\lambda\) |
| Beat frequency | \(\nu_{beat}\) | \([\text{T}^{-1}]\) | s\(^{-1}\) | Difference of two close frequencies of superposing waves. |
14.11 NCERT Questions
A string of mass \(2.50\ \text{kg}\) is under a tension of \(200\ \text{N}\). The length of the stretched string is \(20.0\ \text{m}\). If a transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
A stone dropped from the top of a tower of height \(300\ \text{m}\) splashes into the water of a pond near the base of the tower. When is the splash heard at the top, given that the speed of sound in air is \(340\ \text{m s}^{-1}\)? (Take \(g = 9.8\ \text{m s}^{-2}\).)
A steel wire has a length of \(12.0\ \text{m}\) and a mass of \(2.10\ \text{kg}\). What should be the tension in the wire so that the speed of a transverse wave on the wire equals the speed of sound in dry air at \(20\,^\circ\text{C}\), which is \(343\ \text{m s}^{-1}\)?
Use the formula \(v = \sqrt{\dfrac{\gamma P}{\rho}}\) to explain why the speed of sound in air
- is independent of pressure,
- increases with temperature,
- increases with humidity.
You have learnt that a travelling wave in one dimension is represented by a function \(y = f(x, t)\) where \(x\) and \(t\) must appear in the combination \(x - vt\) or \(x + vt\), i.e. \(y = f(x \pm vt)\). Is the converse true? Examine if the following functions for \(y\) can possibly represent a travelling wave:
- \((x - vt)^2\)
- \(\log\left[(x + vt)/x_0\right]\)
- \(1/(x + vt)\)
A bat emits ultrasonic sound of frequency \(1000\ \text{kHz}\) in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, and (b) the transmitted sound? Speed of sound in air is \(340\ \text{m s}^{-1}\) and in water \(1486\ \text{m s}^{-1}\).
A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is \(1.7\ \text{km s}^{-1}\)? The operating frequency of the scanner is \(4.2\ \text{MHz}\).
A transverse harmonic wave on a string is described by \[y(x, t) = 3.0\sin\left(36\,t + 0.018\,x + \frac{\pi}{4}\right)\] where \(x\) and \(y\) are in cm and \(t\) in s. The positive direction of \(x\) is from left to right.
- Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation?
- What are its amplitude and frequency?
- What is the initial phase at the origin?
- What is the least distance between two successive crests in the wave?
For the wave described in Exercise A-Q8, plot the displacement (\(y\)) versus time (\(t\)) graphs for \(x = 0\), \(2\) and \(4\) cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in a travelling wave differ from one point to another: amplitude, frequency, or phase?
For the travelling harmonic wave \[y(x, t) = 2.0\cos 2\pi\left(10\,t - 0.0080\,x + 0.35\right)\] where \(x\) and \(y\) are in cm and \(t\) in s, calculate the phase difference between the oscillatory motion of two points separated by a distance of
- \(4\ \text{m}\),
- \(0.5\ \text{m}\),
- \(\lambda/2\),
- \(3\lambda/4\).
The transverse displacement of a string (clamped at both its ends) is given by \[y(x, t) = 0.06\sin\left(\frac{2\pi}{3}x\right)\cos(120\pi t)\] where \(x\) and \(y\) are in m and \(t\) in s. The length of the string is \(1.5\ \text{m}\) and its mass is \(3.0 \times 10^{-2}\ \text{kg}\). Answer the following:
- Does the function represent a travelling wave or a stationary wave?
- Interpret the wave as a superposition of two waves travelling in opposite directions. What are the wavelength, frequency, and speed of each wave?
- Determine the tension in the string.
- For the wave on the string described in Exercise A-Q11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point \(0.375\ \text{m}\) away from one end?
Given below are some functions of \(x\) and \(t\) to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave, or (iii) none at all:
- \(y = 2\cos(3x)\sin(10t)\)
- \(y = 2\sqrt{x - vt}\)
- \(y = 3\sin(5x - 0.5t) + 4\cos(5x - 0.5t)\)
- \(y = \cos x\sin t + \cos 2x\sin 2t\)
A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of \(45\ \text{Hz}\). The mass of the wire is \(3.5 \times 10^{-2}\ \text{kg}\) and its linear mass density is \(4.0 \times 10^{-2}\ \text{kg m}^{-1}\). What is (a) the speed of a transverse wave on the string, and (b) the tension in the string?
A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed-frequency source (a tuning fork of frequency \(340\ \text{Hz}\)) when the tube length is \(25.5\ \text{cm}\) or \(79.3\ \text{cm}\). Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected.
A steel rod \(100\ \text{cm}\) long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod is given to be \(2.53\ \text{kHz}\). What is the speed of sound in steel?
A pipe \(20\ \text{cm}\) long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a \(430\ \text{Hz}\) source? Will the same source be in resonance with the pipe if both ends are open? (Speed of sound in air is \(340\ \text{m s}^{-1}\).)
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency \(6\ \text{Hz}\). The tension in string A is slightly reduced and the beat frequency is found to reduce to \(3\ \text{Hz}\). If the original frequency of A is \(324\ \text{Hz}\), what is the frequency of B?
Explain why (or how):
- in a sound wave, a displacement node is a pressure antinode, and vice versa,
- bats can ascertain distances, directions, nature, and sizes of obstacles without any “eyes”,
- a violin note and a sitar note may have the same frequency, yet we can distinguish between the two notes,
- solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and
- the shape of a pulse gets distorted during propagation in a dispersive medium.
14.12 Check Your Concepts
Name the three broad families of waves discussed in this chapter, giving one example of each. State clearly which of them require a material medium to travel and which do not.
Transverse mechanical waves can travel through solids but not through liquids or gases, whereas longitudinal waves can travel through all three. Explain this difference in terms of the elastic property (shear or bulk) that each type of wave requires.
Newton’s formula for the speed of sound in air gave a value about 15% below the measured value. Identify the wrong assumption Newton made, state the correction introduced by Laplace, and write the corrected formula.
A wave on a string is produced by a source. State whether the speed of the wave depends on (a) the frequency of the source, (b) the amplitude of the wave, and (c) the tension in the string. Justify each answer.
Explain why a string fixed at both ends can vibrate only at a discrete set of frequencies (its normal modes), whereas a travelling wave can have any frequency. What role do the boundary conditions play?
A pipe closed at one end produces only the odd harmonics, while a pipe open at both ends produces all harmonics. Using the node/antinode conditions at the ends, explain why.
A transverse pulse travelling on a string is reflected at a boundary. State, with reasons, whether the reflected pulse is inverted or upright when the boundary is (a) rigidly fixed, and (b) free to move.
Two sound waves of slightly different frequencies are sounded together. Describe what a listener hears, explain the terms “waxing and waning”, and state the formula for the beat frequency. Why is the pitch heard the average of the two frequencies?
Distinguish between a progressive (travelling) wave and a stationary (standing) wave with respect to (a) transport of energy, (b) amplitude of different particles, and (c) the phase of particles at a given instant.
14.13 Practice with Numericals
A transverse wave is described by \(y(x, t) = 0.05\sin(10\pi x - 400\pi t)\), where \(x\) and \(y\) are in metres and \(t\) in seconds. Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the wave speed, and (e) the direction of propagation.
A string \(1.0\ \text{m}\) long has a mass of \(20\ \text{g}\) and is stretched to a tension of \(80\ \text{N}\). Calculate the speed of a transverse wave on the string.
The bulk modulus of a certain liquid is \(2.0 \times 10^{9}\ \text{Pa}\) and its density is \(800\ \text{kg m}^{-3}\). Estimate the speed of sound in the liquid using \(v = \sqrt{B/\rho}\).
Estimate the speed of sound in a gas for which \(\gamma = 1.4\), the pressure is \(1.0 \times 10^{5}\ \text{Pa}\), and the density is \(1.25\ \text{kg m}^{-3}\), using the Laplace formula \(v = \sqrt{\gamma P/\rho}\).
A string of length \(1.2\ \text{m}\) is fixed at both ends, and transverse waves travel on it at a speed of \(90\ \text{m s}^{-1}\). Find the fundamental frequency and the frequency of the second harmonic.
A pipe of length \(25\ \text{cm}\) is closed at one end. Taking the speed of sound in air as \(340\ \text{m s}^{-1}\), find its fundamental frequency and the frequency of its next possible harmonic.
Two tuning forks of frequencies \(340\ \text{Hz}\) and \(344\ \text{Hz}\) are sounded together. What beat frequency will be heard? How many beats will be heard in \(5\ \text{seconds}\)?
Two harmonic waves of equal amplitude \(4\ \text{cm}\), travelling in the same direction with the same frequency, superpose with a phase difference of \(\dfrac{\pi}{3}\). Using \(A(\phi) = 2a\cos\tfrac{1}{2}\phi\), find the amplitude of the resultant wave.
A source produces sound of frequency \(680\ \text{Hz}\) in air, where the speed of sound is \(340\ \text{m s}^{-1}\). Calculate the wavelength of the sound. If the same sound passes into water (speed of sound \(1486\ \text{m s}^{-1}\)), what is its wavelength there?