Chapter 8: Sound

Stand a few hundred metres from a construction site and watch a hammer strike a steel girder. You see the strike, and only a moment later does the clang reach your ear. The gap is short, but it is real, and it carries a clue: sound takes time to travel from one place to another.

Early scientists turned that everyday gap into a measurement. In the early eighteenth century, the English clergyman and natural philosopher William Derham studied the sound of cannon fired several kilometres away. Watching through a telescope from a church tower, he saw the flash of the gunpowder almost at the instant it was produced, and then timed, with a pendulum, the seconds that passed before the bang arrived. Dividing the known distance by that delay gave him a speed for sound in air close to the value we accept today — a little over 300 metres every second. He also noticed something that puzzled him: the number was not perfectly fixed. Wind and the warmth of the day nudged it up and down.

That one experiment holds this whole chapter in miniature. Sound is a disturbance that travels through a material medium at a definite speed, carrying energy from place to place without carrying the medium along with it — and that speed depends on what the medium is made of and on the conditions it is in.

Diagram

Fig. 8.1 – A distant cannon firing on a hillside, with an observer on a tower seeing the flash while the sound wave (drawn as spreading arcs) is still on its way.

NoteCuriosity Corner

Q1. Lightning and thunder are produced at the same instant, so why do we always see the flash first and hear the rumble afterwards? Q2. Steel is thousands of times denser than air, yet sound races through steel far faster than through air. How can a heavier medium carry sound more quickly? Q3. Newton worked out a formula for the speed of sound in air, but his answer came out about 16% too small — and it was not an error in measurement. What had he overlooked? Q4. Why do you hear a clear echo when you shout in a hilly valley, but never inside your own classroom? Q5. Why does the whistle of an approaching train sound shrill, and then suddenly drop to a lower pitch the moment the train passes you? Q6. A 40 dB sound is ten thousand times more intense than a 10 dB sound. Why, then, does it not seem ten thousand times louder to your ears?

By the end of this chapter, you will be able to answer each of these questions for yourself.

8.1 Introduction

Before you begin, recall from your earlier classes: (1) What type of wave is a sound wave? (2) Can sound travel through vacuum? (3) What are reverberation and echo? (4) What is meant by the pitch of a sound?

Drop a stone into a still pond and you see rings of ripples spreading outward from the point where it fell, as shown in Fig. 8.2. Something is clearly travelling across the water surface — but if you look closely at a dry leaf floating some distance away, you notice that the leaf does not get carried outward with the ripple. It simply bobs up and down and stays where it is.

Diagram

Fig. 8.2 – Circular ripples spreading outward from the point where a stone falls into water, with a floating leaf shown bobbing up and down in the same spot.

This is the central idea of wave motion. The water molecules oscillate up and down about their equilibrium positions — their normal resting positions — but they do not travel from one point to another along the surface. What travels outward is the disturbance, not the water.

Q. If the water itself does not move outward, then what exactly is moving outward?

A. The disturbance moves — that is, the pattern of displacement, and with it energy and momentum. Each particle of water passes the disturbance on to its neighbour and then returns to its own place. The pattern travels; the matter does not.

So a wave is a periodic and regular disturbance in a medium that causes no flow of material, but does cause a flow of energy and momentum from one point to another.

NoteDefinition

A wave is a periodic and regular disturbance travelling through a medium, which does not transport any material of the medium, but transports energy and momentum from one point to another.

Not every wave, however, needs a material medium to travel through. Light, as you already know, is a type of wave, and it reaches us from the Sun across the near-vacuum of space. So before we study sound waves in particular, it is useful to sort waves into types, and then look at the properties that all of them share.

Types of Waves

(i) Mechanical waves. A wave is said to be mechanical if a material medium is essential for its propagation. Without matter to be disturbed, such a wave simply cannot exist. Water waves, waves along a stretched string, seismic waves and sound waves are all mechanical waves.

NoteDefinition

A wave whose propagation requires a material medium is called a mechanical wave.

In the 1660s, working with the newly invented air pump, Robert Boyle placed a ringing bell inside a glass jar and slowly pumped the air out. As the air thinned, the sound grew fainter and fainter, until almost nothing could be heard — even though the bell could still be seen ringing. The light from the bell crossed the empty jar without difficulty; the sound did not. This was direct evidence that sound is a mechanical wave and needs a material medium, while light does not.

(ii) Electromagnetic (EM) waves. These are generated by periodic vibrations in electric and magnetic fields. They can pass through material media, but a material medium is not essential for their propagation — which is why they can travel through vacuum. Light is one example. EM waves are studied in detail in Chapter 13.

NoteDefinition

Waves produced by periodic vibrations of electric and magnetic fields, which do not require a material medium for their propagation, are called electromagnetic waves.

(iii) Matter waves. There is always a wave associated with any object that is in motion. Such waves are called matter waves, and they are studied in quantum mechanics.

NoteDefinition

A wave associated with any moving object is called a matter wave.

The electron microscope is built on the idea of matter waves. Because a fast-moving electron behaves as a wave of very short wavelength — much shorter than the wavelength of visible light — an instrument that “illuminates” a specimen with electrons can reveal detail far finer than any optical microscope can, which is how we obtain images of viruses and of the fine structure of materials.

Progressive Waves and Stationary Waves

Waves are also classified by what the disturbance does over time. In a travelling (or progressive) wave, a disturbance created at one place travels onwards to distant points, and keeps travelling until it is stopped by some external agency. Energy is carried from one point to another as the wave advances.

NoteDefinition

A wave in which the disturbance created at one place travels to distant points, and keeps travelling unless stopped by some external agency, is called a travelling or progressive wave.

The ripples we began with are travelling waves. They keep moving outward from the point where the stone was dropped until they meet the walls of the container or the boundary of the water body, and are stopped there.

There is another kind of wave, in which the disturbance does not advance at all but stays confined between fixed boundaries — for example, the vibration of a plucked guitar string held at both ends. These are called stationary waves, and you will study them in Class XII. Throughout this chapter, unless it is said otherwise, we deal with progressive waves.

In the sections that follow, we shall first look at the properties common to all waves, then at the two ways in which a wave can be organised (transverse and longitudinal), and only then turn to sound itself.

8.2 Common Properties of all Waves

The properties described below hold for waves of every kind. Here, however, we shall describe them for mechanical waves, since those are the easiest to picture: a rope, a water surface, a column of air — something you can point at and watch vibrate.

Keep one picture in mind throughout. Every particle of the medium is doing nothing more exciting than swinging back and forth about its own equilibrium position. The wave is the pattern those swings make when you look at all the particles together.

Amplitude (A)

The amplitude of a wave motion is the largest displacement of a particle of the medium, measured from its rest position, as the wave passes through it. It tells us how violently the particles are being shaken, and therefore how much energy the wave carries. In SI units it is measured in metre (m).

NoteDefinition

The amplitude (A) of a wave is the largest displacement of a particle of the medium, from its rest position, during the passage of the wave. SI unit: metre (m).

Wavelength (λ)

The wavelength is the distance between two successive particles which are in the same state of vibration. “Same state of vibration” is a phrase we shall sharpen a little later in this section, once phase has been introduced. For the moment, read it as: the particles are doing exactly the same thing at the same instant. Wavelength is measured in metre (m).

NoteDefinition

The wavelength (λ) of a wave is the distance between two successive particles of the medium that are in the same state of vibration. SI unit: metre (m).

Period (T)

The period is the time required by a particle of the medium to complete one full vibration. It is measured in second (s). Notice that the period belongs to the particle’s motion, not to the wave’s travel — every particle in the path of the wave takes the same time T to complete one oscillation.

NoteDefinition

The period (T) of a wave is the time required by a particle of the medium to complete one vibration. SI unit: second (s).

Double Periodicity

Waves possess a property called double periodicity. At any one fixed location, the wave motion repeats itself at equal intervals of time — so the motion is periodic in time. And at any one fixed instant, if you freeze the wave and walk along it, the form of the wave repeats itself at equal distances — so the motion is also periodic in space.

Wave motion is therefore a doubly periodic phenomenon: periodic in time and periodic in space.

Q. What are the “repeat intervals” of these two periodicities?

A. In time, the pattern repeats after every period T. In space, it repeats after every wavelength λ. This is why T and λ are the two basic yardsticks of any wave: one measures the repetition of the wave in time, the other in space.

Frequency (n)

The frequency of a wave is the number of vibrations performed by a particle during each second. Its SI unit is the hertz (Hz), where 1 Hz means one vibration per second.

Since the particle needs a time T to complete one vibration, the number of vibrations it completes in one second must be the reciprocal of T. Hence frequency is the reciprocal of the time period:

\[n = \frac{1}{T}\]

where \(n\) is the frequency in hertz (Hz) and \(T\) is the period in second (s).

NoteDefinition

The frequency (n) of a wave is the number of vibrations performed by a particle of the medium in one second. It is the reciprocal of the period, \(n = 1/T\). SI unit: hertz (Hz).

Frequency is the quantity a musician actually tunes. Orchestras and instrument makers across the world agree on a standard reference note — the A above middle C — fixed at 440 Hz. When a violinist tightens a string until it matches that reference, she is adjusting the number of vibrations the string makes per second, and every other note on the instrument is then set relative to it.

Velocity of a Wave (v)

The distance covered by a wave per unit time is called the velocity of the wave. Here we must be careful to separate two motions that students often mix up: the particles of the medium only oscillate about their fixed positions, while the wave pattern advances steadily through the medium. It is the advance of the pattern whose speed we are describing.

During one period T, the wave pattern advances by exactly one wavelength λ, because after a time T the whole pattern has shifted forward by one full repeat. Therefore,

\[\text{Magnitude of velocity} = \frac{\text{distance}}{\text{time}}\]

\[\text{v} = \frac{\text{wavelength}}{\text{period}}\]

\[\text{v} = \frac{\lambda}{T} \qquad \text{--- (8.1)}\]

but

\[\frac{1}{T} = n \ \text{(frequency)} \qquad \text{--- (8.2)}\]

\[\therefore \ \text{v} = n\lambda \qquad \text{--- (8.3)}\]

Here \(\text{v}\) is the magnitude of the wave velocity in metre per second (m/s), \(n\) is the frequency in hertz (Hz), and \(\lambda\) is the wavelength in metre (m).

NoteDefinition

The velocity of a wave is the distance covered by the wave per unit time. Its magnitude is given by \(\text{v} = n\lambda\). SI unit: metre per second (m/s).

Equation (8.3) carries a message that is easy to state and easy to misuse, so read it slowly. In a given medium, the magnitude of the wave velocity is constant — it is fixed by the medium, not by the source. So if the frequency of the wave is increased, its wavelength must decrease in exactly the same proportion, to keep the product \(n\lambda\) unchanged.

Now suppose the wave passes from one medium into another. The frequency does not change, because frequency is decided by the source that is shaking the medium, and the source has not changed. What must change, then, are the speed and the wavelength together.

Q. A wave slows down as it enters a new medium. Does its frequency drop as well?

A. No. Frequency is set by the source, and the first layer of the new medium is being shaken at exactly the rate at which the wave arrives — so the same number of vibrations per second is passed on. Since v = nλ and v has fallen while n is fixed, it is the wavelength that shrinks.

A sound wave of frequency 500 Hz travels in air, where the speed of sound is 340 m/s, and then passes into water, where the speed of sound is 1500 m/s. Find (a) the wavelength of the wave in air, (b) its frequency in water, and (c) its wavelength in water.

Properties the Medium Must Possess

A mechanical wave cannot travel through just any material. For a medium to carry such a wave, it must possess the following properties.

  1. The medium should be continuous and elastic, so that it regains its original state after the deforming force is removed. Elasticity is what supplies the restoring force that pulls a displaced particle back — without it, a displaced particle would simply stay displaced and no oscillation, and hence no wave, could occur.

  2. The medium should possess inertia. The medium must be capable of storing energy and of transferring it in the form of waves. Inertia is what makes a particle overshoot its equilibrium position instead of stopping there, which is what keeps the oscillation going.

  3. The frictional resistance of the medium must be negligible, so that the oscillations are not damped — that is, so that the wave does not die out quickly as it travels.

Elasticity and inertia are thus a pair: elasticity brings the particle back, inertia carries it past. Between them they produce the endless to-and-fro that a mechanical wave depends on.

Phase and Phase Difference

In Fig. 8.3 (a), the displacements of various particles along a sinusoidal wave travelling along the + x-axis are plotted against their respective distances from the source (at O), all at one given instant. This plot is valid for a transverse as well as a longitudinal wave.

Diagram

Fig. 8.3 (a) – Displacement as a function of distance along the wave, a sine curve with labelled particles A, B, C, D, E, F, G, H, P and Q at various points on it.

The state of oscillation of a particle is called its phase. To describe the phase of a particle at a place completely, we need to know three things: (a) its displacement, (b) the direction of its velocity, and (c) its oscillation number — that is, which oscillation it is currently performing.

NoteDefinition

The phase of a particle in a wave is its state of oscillation, described by its displacement, the direction of its velocity, and its oscillation number.

Why are all three needed? Look at Fig. 8.3 (a). Particles P and Q (and likewise the pairs E and C, or B and D) have the same displacement — but the directions of their velocities are opposite. One is on its way up while the other is on its way down. Displacement alone therefore does not fix the state of a particle.

Now compare particles B and F. They have the same magnitude of displacement and the same direction of velocity. Such particles are said to be in phase during their respective oscillations. Moreover, they are successive particles with this property — the nearest such pair. The separation between these two particles is the wavelength λ.

These two successive particles differ by ‘1’ in their oscillation number: if particle B is at its nth oscillation, particle F will be at its (n + 1)th oscillation, since the wave is travelling along the + x direction and reached B before it reached F.

The most convenient way to understand phase is in terms of an angle. For a sinusoidal wave, the variation in displacement is a ‘sine’ function of the distance from the source and of the time, as we shall see in Section 8.4. For such waves it is possible for us to assign angles corresponding to the displacement (or to the time).

At the instant at which the graph of Fig. 8.3 (a) is drawn, the disturbance (the energy) has just reached particle A. The phase angle corresponding to this particle A can therefore be taken as 0°.

At this same instant, particle B has completed a quarter of an oscillation and reached its positive maximum (sin θ = +1). The phase angle θ of this particle B is therefore \(\pi^c/2 = 90^\circ\).

Similarly, the phase angles of particles C and E are \(\pi^c\) (180°) and \(2\pi^c\) (360°) respectively. Particle F has completed one full oscillation and is at its positive maximum during its second oscillation. Hence its phase angle is

\[2\pi^c + \frac{\pi^c}{2} = \frac{5\pi^c}{2}\]

B and F are the successive particles in the same state — same displacement and same direction of velocity — during their respective oscillations. The separation between these two is the wavelength (λ). The phase angle between these two differs by \(2\pi^c\).

This gives us a sharper and more useful way of defining wavelength than the one we started with.

NoteDefinition

The wavelength of a wave is the separation between two successive particles that differ in phase by \(2\pi^c\).

Q. Two particles in a wave are separated by half a wavelength. What is the phase difference between them?

A. A separation of one full wavelength corresponds to a phase difference of \(2\pi^c\). Half a wavelength therefore corresponds to \(\pi^c\), or 180°. Such particles have equal displacements in opposite directions and are said to be exactly out of phase.

As noted earlier in this section, waves possess double periodicity. This means the displacements of the particles are periodic in space, as shown in Fig. 8.3 (a), as well as periodic in time. Figure 8.3 (b) shows the displacement of one particular particle as a function of time.

Diagram

Fig. 8.3 (b) – Displacement of a single particle as a function of time, drawn as a sine curve on displacement–time axes.

The two graphs look alike, and that is precisely the point — but do not confuse them. Fig. 8.3 (a) is a snapshot of many particles at one instant, and the repeat distance along its horizontal axis is the wavelength λ. Fig. 8.3 (b) is the life story of one particle over time, and the repeat interval along its horizontal axis is the period T.

  1. Using axes of displacement and distance, sketch two waves A and B such that A has twice the wavelength and half the amplitude of B. (2) Determine the wavelength and amplitude of each of the two waves P and Q shown in Fig. 8.4.

Diagram

Fig. 8.4 – Two sinusoidal waves P and Q of different wavelengths and amplitudes plotted on the same displacement–distance axes, with the displacement axis marked in units up to 4.

Characteristics of a Progressive Wave
  1. All vibrating particles of the medium have the same amplitude, the same period and the same frequency.

  2. The state of oscillation, that is the phase, changes from particle to particle. This is what makes the wave appear to travel: each particle starts its oscillation a little later than the one before it.

The speed of sound in air is 330 m/s and that in glass is 4500 m/s. What is the ratio of the wavelength of sound of a given frequency in the two media?

Solution: The phrase “of a given frequency” is the key. The same source is producing the sound in both media, so the frequency \(n\) is the same in air and in glass. We can therefore use \(\text{v} = n\lambda\) separately for each medium and divide.

\[\text{v}_{air} = n \lambda_{air}\]

\[\text{v}_{glass} = n \lambda_{glass}\]

\[\therefore \frac{\lambda_{air}}{\lambda_{glass}} = \frac{\text{v}_{air}}{\text{v}_{glass}} = \frac{330}{4500} = 7.33 \times 10^{-2}\]

\[= 0.0733 \approx 7.33 \times 10^{-2}\]

Notice that the common factor \(n\) cancels out on dividing — which is exactly why the answer can be found without ever being told what the frequency is.

8.3 Transverse Waves and Longitudinal Waves

Progressive waves can be of two types: transverse waves and longitudinal waves. The difference between them lies in a single question — in which direction do the particles of the medium vibrate, compared with the direction in which the wave itself advances?

That one question decides almost everything else about the wave: which media it can travel through, what pattern it forms in the medium, and even whether it can be polarised.

Transverse Waves

A wave in which the particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called a transverse wave.

NoteDefinition

A wave in which the particles of the medium vibrate in a direction perpendicular to the direction of propagation of the wave is called a transverse wave.

Water waves are transverse waves. The water molecules vibrate perpendicular to the surface of the water — up and down — while the wave itself propagates along the surface. This is exactly the behaviour we noticed with the floating leaf in Section 8.1: the leaf bobbed vertically while the ripple ran horizontally away from it.

The same picture appears if you tie one end of a rope to a wall and shake the free end up and down, as shown in Fig. 8.5. Your hand moves vertically; the hump you create travels horizontally.

Diagram

Fig. 8.5 – A rope fixed at one end and shaken vertically at the other, showing crests and troughs travelling horizontally while an arrow marks the vertical vibration of one particle.

Characteristics of Transverse Waves
  1. All particles of the medium in the path of the wave vibrate in a direction perpendicular to the direction of propagation of the wave, with the same period and the same amplitude.

  2. When a transverse wave passes through a medium, the medium is divided into alternate crests, i.e., regions of positive displacements, and troughs, i.e., regions of negative displacements.

NoteDefinition

A crest is a region of positive (upward) displacement of the medium, and a trough is a region of negative (downward) displacement, produced by a transverse wave.

  1. A crest and an adjacent trough together form one cycle of a transverse wave. The distance measured along the wave between any two consecutive points in the same phase — two consecutive crests, or two consecutive troughs — is called the wavelength of the wave.

  2. Crests and troughs advance in the medium, and they are responsible for the transfer of energy.

  3. Transverse waves can travel through solids and on the surfaces of liquids only. They can not travel through liquids and gases. EM waves are transverse waves, but they do not require a material medium for propagation.

The reason behind point 5 is worth spelling out, because it is a favourite of examiners. To send a transverse wave through a material, one layer of the material must be able to drag the next layer sideways along with it. That sideways grip is called elasticity of shape, or rigidity, and only solids possess it. The interior of a liquid or a gas offers no resistance to being sheared sideways — the layers simply slide past one another — so no transverse wave can be sustained there.

  1. When transverse waves advance through a medium there is no change in the pressure and density at any point of the medium; however, the shape changes periodically.

  2. If the vibrations of all the particles along the path of a wave are constrained to be in a single plane, then the wave is called a polarised wave. A transverse wave can be polarised.

NoteDefinition

A wave in which the vibrations of all the particles are confined to a single plane is called a polarised wave.

  1. The medium conveying a transverse wave must possess elasticity of shape.

Polaroid sunglasses work only because light is a transverse wave. Sunlight reflected off a road surface or a water surface vibrates mostly in the horizontal plane, and that is what produces the harsh glare. The polaroid film in the lens is built to transmit vibrations in only one plane and block the perpendicular one, so the horizontally vibrating glare is largely cut out while the rest of the scene comes through. No such trick is possible with sound, for the reason we shall see in a moment.

Longitudinal Waves

A wave in which the particles of the medium vibrate in a direction parallel to the direction of propagation of the wave is called a longitudinal wave. Sound waves are longitudinal waves.

NoteDefinition

A wave in which the particles of the medium vibrate in a direction parallel to the direction of propagation of the wave is called a longitudinal wave.

Picture a long, loosely coiled spring lying on a table, as in Fig. 8.6. Push one end sharply along the length of the spring and let go. The coils near your hand crowd together, then spread apart, and this pattern of crowding and spreading runs along the spring. Each coil moves only a little, back and forth along the spring’s own length — parallel to the direction in which the disturbance travels.

Diagram

Fig. 8.6 – A long coiled spring pushed along its length, showing alternate regions of crowded coils (compressions) and spread-out coils (rarefactions) travelling along the spring.

Air behaves in exactly the same way when a loudspeaker cone or a vibrating tuning fork pushes it. That is why sound is a longitudinal wave, and why sound in air travels as a pattern of changing pressure rather than as a pattern of changing shape.

Characteristics of Longitudinal Waves
  1. All the particles of the medium along the path of the wave vibrate in a direction parallel to the direction of propagation of the wave, with the same period and the same amplitude.

  2. When a longitudinal wave passes through a medium, the medium is divided into regions of alternate compressions and rarefactions. A compression is the region where the particles of the medium are crowded together (a high pressure zone), while a rarefaction is the region where the particles of the medium are more widely separated, i.e., the medium gets rarefied (a low pressure zone).

NoteDefinition

A compression is a region of a longitudinal wave where the particles of the medium are crowded together, forming a high pressure zone; a rarefaction is a region where the particles are more widely separated, forming a low pressure zone.

  1. A compression and an adjacent rarefaction together form one cycle of a longitudinal wave. The distance measured along the wave between any two consecutive points having the same phase is the wavelength of the wave.

Compression and rarefaction are therefore, for a longitudinal wave, exactly what crest and trough are for a transverse wave. The vocabulary changes, but the geometry does not.

A sound wave of frequency 500 Hz travels through air in which the speed of sound is 340 m/s. Calculate (a) the distance between two successive compressions, and (b) the distance between a compression and the rarefaction immediately next to it.

  1. For the propagation of longitudinal waves, the medium should possess the property of elasticity of volume. Thus longitudinal waves can travel through solids, liquids and gases. A longitudinal wave can not travel through vacuum or free space.

Elasticity of volume simply means that the material resists being squeezed and springs back when the squeeze is released. Solids, liquids and gases all resist being compressed, which is why all three can carry sound — and why the empty space of a vacuum, having nothing to compress, cannot.

  1. The compressions and rarefactions advance in the medium, and they are responsible for the transfer of energy.

  2. When a longitudinal wave advances through a medium there are periodic variations in the pressure and the density along the path of the wave, and also with time.

  3. Longitudinal waves can not be polarised, because the direction of vibration of the particles and the direction of propagation of the wave are the same, or parallel.

Q. Why can a transverse wave be polarised while a longitudinal wave cannot?

A. A transverse wave has a genuine choice of vibration direction — the particles may vibrate up-down, left-right, or in any direction in the plane perpendicular to the wave’s travel. Polarisation means restricting that choice to one plane. A longitudinal wave has no such choice: its particles must vibrate along the direction of travel, and there is only one such direction. With nothing to restrict, there is nothing to polarise.

Comparing the Two

It is worth pausing to line up the two kinds of wave against each other, because most examination questions on this section are really asking you to make this comparison.

In a transverse wave, the vibration is perpendicular to the propagation, the medium shows crests and troughs, the shape of the medium changes but its pressure and density do not, elasticity of shape is required, only solids and liquid surfaces can carry it, and the wave can be polarised.

In a longitudinal wave, the vibration is parallel to the propagation, the medium shows compressions and rarefactions, the pressure and density vary periodically, elasticity of volume is required, solids and liquids and gases can all carry it, and the wave cannot be polarised.

Sound, the subject of this chapter, sits firmly in the second column: it is a longitudinal, progressive, mechanical wave.

8.4 Mathematical Expression of a Wave

So far we have described waves in words and in pictures. We now describe a progressive wave mathematically — that is, we write down a single equation that contains the whole moving pattern.

Since it is a progressive wave, we require a function of both the position \(x\) and the time \(t\). Two demands are made of this function. First, it must describe the shape of the wave at any instant of time — freeze the clock, and the function should hand you the snapshot. Second, it must describe the motion of the particle of the medium at any chosen point — stand at one place, and the function should hand you that particle’s oscillation.

A sinusoidal progressive wave can be described by a sinusoidal function. Let us assume that the progressive wave is transverse, so that the position of a particle of the medium is described by a fixed value of \(x\), and its displacement from the equilibrium position can be described by \(y\). Such a sinusoidal wave can be written as follows:

\[y(x,t) = a \sin (kx - \omega t + \phi) \qquad \text{--- (8.4)}\]

Here \(a\), \(k\), \(\omega\) and \(\phi\) are constants.

Why This Equation Describes a Wave

An equation is not a wave merely because we call it one. Let us see the justification for writing Eq. (8.4), by testing it against the two demands listed above.

Test 1 — freeze the time. At a particular instant, say \(t = t_0\),

\[y(x, t_0) = a \sin (kx - \omega t_0 + \phi)\]

\[= a \sin (kx + \text{constant})\]

because \(t_0\) is fixed, so the whole term \((-\omega t_0 + \phi)\) is just a fixed number. Thus the shape of the wave at \(t = t_0\), as a function of \(x\), is a sine wave. The snapshot is a sine curve, exactly as in Fig. 8.7.

Diagram

Fig. 8.7 – Snapshot of a sinusoidal wave on displacement–distance axes, with the amplitude a and the wavelength λ marked on it.

Test 2 — freeze the position. At a fixed location, say \(x = x_0\),

\[y(x_0, t) = a \sin (kx_0 - \omega t + \phi)\]

\[= a \sin (\text{constant} - \omega t)\]

Hence the displacement \(y\), at \(x = x_0\), varies as a sine function of time.

This means that the particles of the medium, through which the wave travels, execute simple harmonic motion around their equilibrium position. Both demands are met, and Eq. (8.4) does describe a wave.

Which Way Is the Wave Travelling?

There remains one more thing to check: the direction in which this wave advances. Look at the argument of the sine, \((kx - \omega t + \phi)\).

A particular point of the wave pattern — say the top of a particular crest — is a point where the argument of the sine keeps a fixed value. As time \(t\) increases, the term \(\omega t\) grows larger. To keep \((kx - \omega t + \phi)\) constant, the term \(kx\) must grow larger by the same amount. In other words, \(x\) must increase in the positive direction as \(t\) increases. The crest therefore moves towards larger \(x\).

Thus Eq. (8.4) represents a wave travelling along the positive \(x\) axis.

A wave represented by

\[y(x, t) = a \sin (kx + \omega t + \phi) \qquad \text{--- (8.5)}\]

is a wave travelling in the direction of the negative \(x\) axis. Here the same reasoning runs the other way: as \(t\) increases, \(\omega t\) grows, so \(kx\) must decrease to keep the argument constant, and the pattern moves towards smaller \(x\).

Q. How can I remember which sign goes with which direction?

A. A minus sign between \(kx\) and \(\omega t\) means the wave travels along \(+x\); a plus sign means it travels along \(-x\). The signs feel reversed, and that is the trap. The reason is that the two terms must cancel each other’s growth to keep the phase constant, so they must have opposite signs for the wave to move forward.

The Symbols in Eq. (8.4)

\(y(x, t)\) is the displacement as a function of position (\(x\)) and time (\(t\)). It is measured in metre (m).

\(a\) is the amplitude of the wave, measured in metre (m).

\(\omega\) is the angular frequency of the wave, measured in radian per second (rad/s). It is related to the frequency \(n\) and the period \(T\) of Section 8.2 by \(\omega = 2\pi n = 2\pi/T\), so it simply counts the phase angle swept per second rather than the oscillations completed per second.

\(k\) is the angular wave number, measured in radian per metre (rad/m). It is related to the wavelength by \(k = 2\pi/\lambda\), so it counts the phase angle gained per metre travelled along the wave.

\(\phi\) is a constant, called the initial phase or epoch. It fixes what the wave was doing at \(x = 0\) at the instant \(t = 0\).

\((kx_0 - \omega t + \phi)\) is the argument of the sinusoidal wave, and is the phase of the particle at \(x\) at time \(t\). It is the mathematical form of the “state of oscillation” we described in words in Section 8.2.

Reading the Speed out of the Equation

Since \(\omega = 2\pi n\) and \(k = 2\pi/\lambda\), dividing one by the other gives

\[\frac{\omega}{k} = \frac{2\pi n}{2\pi / \lambda} = n\lambda = \text{v}\]

which is nothing but Eq. (8.3) in a new dress. So if a wave equation is handed to you in the form of Eq. (8.4), you can read off its amplitude directly as the number in front of the sine, its wavelength from \(k\), its frequency from \(\omega\), and its speed as the ratio \(\omega/k\) — without any further information. Numerical questions in this section almost always turn on exactly this skill.

A progressive wave travelling in air is described by \(y(x,t) = 0.03 \sin (4\pi x - 1360\pi t)\), where \(y\) and \(x\) are in metre and \(t\) is in second. Find (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed of the wave, and (e) the direction in which it travels.

8.5 The Speed of Travelling Waves

The speed of a mechanical wave depends upon the elastic properties and the density of the medium — that is, on how strongly the medium springs back when disturbed, and on how much matter has to be moved.

The same medium can support both transverse and longitudinal waves, and these two travel at different speeds through it. We therefore treat them one at a time in the subsections that follow, and then come to the speed of sound in particular.

Q. Why should the speed of a wave be decided by the medium at all, rather than by the source that produced it?

A. Because once the first layer of the medium has been disturbed, the source has no further say in the matter. From that point on, the disturbance is handed from layer to layer by the medium’s own restoring forces, working against the medium’s own inertia. How quickly it is handed on therefore depends only on elasticity and density — the two properties named above.

8.5.1 The Speed of Transverse Waves

The speed of a wave is determined by the restoring force produced in the medium when it is disturbed. A stronger restoring force snaps the displaced particles back more sharply, and so the disturbance is handed on to the next layer more quickly.

But the restoring force is only half the story. The speed also depends on the inertial properties of the medium, such as its mass density. Heavier particles are more sluggish to accelerate, and so they pass the disturbance on more slowly. Speed, then, is always a contest between these two — the elastic push-back and the inertial reluctance.

The waves produced on a string are transverse waves, and they are the simplest case in which to see this contest at work, as shown in Fig. 8.8. In this case the restoring force is provided by the tension \(T\) in the string: the tighter the string, the harder it pulls a displaced element back towards the straight line.

Diagram

Fig. 8.8 – A string of mass M and length L stretched under tension T, with a transverse pulse travelling along it.

The inertial property here is the linear mass density \(m\), i.e., the mass carried by each metre of the string. It can be determined from the mass of the string \(M\) and its length \(L\) as \(m = M/L\), and is measured in kilogram per metre (kg/m).

The formula for the speed of a transverse wave on a stretched string is given by

\[\text{v} = \sqrt{\frac{T}{m}} \qquad \text{--- (8.6)}\]

where \(\text{v}\) is the speed of the transverse wave in metre per second (m/s), \(T\) is the tension in the string in newton (N), and \(m\) is the linear mass density of the string in kilogram per metre (kg/m). The derivation of this formula is beyond the scope of this book.

NotePrinciple / Law

The speed of a transverse wave on a stretched string is \(\text{v} = \sqrt{T/m}\), where \(T\) is the tension in the string and \(m = M/L\) is its linear mass density.

Notice how the formula matches the physical argument exactly. Tension — the elastic property — sits in the numerator, so more tension means more speed. Linear mass density — the inertial property — sits in the denominator, so a heavier string means less speed. The square root simply tells us the dependence is not proportional but gentler: to double the wave speed you must make the tension four times as large.

The important point here is that the speed of a transverse wave depends only on the properties of the string, \(T\) and \(m\). It does not depend on the wavelength or the frequency of the wave.

Q. If I pluck a stretched string harder, the wave I create is bigger. Does it also travel faster?

A. No. Plucking harder increases the amplitude, not the speed. Equation (8.6) contains only \(T\) and \(m\) — neither amplitude, nor frequency, nor wavelength appears in it. The louder pulse carries more energy, but it races along the string at exactly the same speed as a gentle one.

Look at the strings of a guitar and you will see that they are not all alike: the strings meant for the low, deep notes are visibly thicker than those meant for the high, bright ones. A thicker string has a much larger linear mass density \(m\), so by Eq. (8.6) the wave speed along it is lower, and the note it sounds is lower in pitch. The tuning pegs then let the player adjust \(T\) for each string separately — the same equation, used in the other direction.

A stretched wire of length 2 m has a mass of 20 g and is kept under a tension of 100 N. Calculate the linear mass density of the wire and the speed of a transverse wave travelling along it.

8.5.2 The Speed of Longitudinal Waves

In the case of longitudinal waves, the particles of the medium oscillate forward and backward along the direction of wave propagation. This causes compression and rarefaction, which travel through the medium because the medium possesses elastic property — it resists being squeezed, and springs back when the squeeze is released.

The mechanism is worth stating plainly, because it is the reason a longitudinal wave moves at all. A layer of the medium is pushed into its neighbour, crowding it; the crowded neighbour, resisting the squeeze, pushes outward on the layer beyond it; that layer is crowded in turn. The compression is thus handed forward from layer to layer, with a rarefaction following behind it, as shown in Fig. 8.9.

Diagram

Fig. 8.9 – A vibrating tuning fork sending compressions and rarefactions through a column of air, with a graph of pressure versus distance drawn below it.

The speed of sound in liquids and solids is higher than that in gases. The speed of sound as a longitudinal wave in an ideal gas is given by Newton’s formula, as discussed in the next subsection. The speed of sound in different media is given in Table 8.1 below.

Table 8.1: Speed of Sound in Gas, Liquids, and Solids

Medium Speed (m/s)
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
Vulcanised Rubber 54
Copper 3560
Steel 5941
Granite 6000
Aluminium 6420

Read the table carefully and a puzzle jumps out at once. Steel is thousands of times denser than air, and denser matter is more sluggish — yet sound travels through steel roughly seventeen times faster than through air. Density alone clearly cannot be deciding the outcome.

The resolution lies in the contest we described in Section 8.5.1. Speed is set by elasticity working against inertia, and in going from a gas to a solid, the elasticity rises far more steeply than the density does. Air is extremely easy to compress; steel is enormously stiff, so the restoring push it delivers when squeezed is immense. The gain in stiffness overwhelms the penalty of the extra mass, and the wave races through. The next subsection puts this argument into a formula.

Q2. Steel is thousands of times denser than air, yet sound races through steel far faster than through air. How can a heavier medium carry sound more quickly?

A. The speed of sound is not decided by density alone. It is decided by a competition between the elasticity of the medium, which speeds the wave up, and the density, which slows it down. Steel is denser than air, but it is very much stiffer than air — its elastic modulus is larger by a far bigger factor than its density is. The stiffness wins the contest, and so sound travels through steel at 5941 m/s while managing only 343 m/s in air at 20°C.

Notice also the odd entry in the solids list: vulcanised rubber, at 54 m/s, is slower than air. Rubber is a solid, but it is a very soft one — its elasticity is feeble, so it loses the same contest that steel wins. Being a solid is not, by itself, a guarantee of high speed.

There is an old railway practice of placing one’s ear against the steel rail to tell whether a train is approaching. The sound of the wheels reaches the listener through the rail long before the same sound arrives through the air, because steel carries it at 5941 m/s while air manages only a few hundred metres per second. The same principle is used in the laboratory to test metal castings for hidden cracks: a pulse of sound sent through the metal arrives late, or does not arrive cleanly, if there is a flaw in its path.

Always remember: when a sound wave goes from one medium to another, its velocity changes, and so does its wavelength. Its frequency, which is decided by the source, remains constant.

A source produces a sound of frequency 1000 Hz. Using the values in Table 8.1, calculate the wavelength of this sound (a) in air at 20°C and (b) in water at 20°C. State which quantity has remained the same in the two media.

8.5.3 Newton’s Formula for Velocity of Sound

The propagation of longitudinal waves was studied by Newton. Sound waves travel through a medium in the form of compressions and rarefactions. The density of the medium is greater at a compression, while it is smaller at a rarefaction. Hence the velocity of sound depends on the elasticity and the density of the medium — the very contest we described qualitatively in the previous subsection.

Newton formulated the relation as

\[\text{v} = \sqrt{\frac{E}{\rho}} \qquad \text{--- (8.7)}\]

where \(E\) is the proper modulus of elasticity of the medium, measured in newton per square metre (N/m²), and \(\rho\) is the density of the medium, measured in kilogram per cubic metre (kg/m³). The speed \(\text{v}\) then comes out in metre per second (m/s).

NotePrinciple / Law

(Newton’s formula for the velocity of sound.) The velocity of sound in a medium is \(\text{v} = \sqrt{E/\rho}\), where \(E\) is the appropriate modulus of elasticity of the medium and \(\rho\) is its density.

The word “proper” in the definition of \(E\) is doing quiet but important work here. Which modulus of elasticity is the appropriate one depends on how the compressions and rarefactions are actually produced — and it is precisely on this point that Newton was to be caught out.

Newton’s Assumption: An Isothermal Process

Newton assumed that, during the propagation of sound, there is no change in the average temperature of the medium. On this assumption, sound wave propagation in air is an isothermal process — a process in which the temperature remains constant — and isothermal elasticity should be considered.

The volume elasticity of air determined under an isothermal change is called the isothermal bulk modulus, and it is equal to the atmospheric pressure ‘\(P\)’. Hence Newton’s formula for the speed of sound in air is given by

\[\text{v} = \sqrt{\frac{P}{\rho}} \qquad \text{--- (8.8)}\]

where \(P\) is the atmospheric pressure in N/m² and \(\rho\) is the density of air in kg/m³.

Putting Numbers into the Formula

To use Eq. (8.8) we need the atmospheric pressure in SI units. Atmospheric pressure is given by \(P = hdg\), the pressure at the base of a mercury column of height \(h\) and density \(d\), as in the barometer of Fig. 8.10.

Diagram

Fig. 8.10 – A simple mercury barometer, with the column height h = 0.76 m marked, illustrating P = hdg.

At NTP,

\(h\) = 0.76 m of Hg

\(d\) = 13600 kg/m³ — density of mercury

\(\rho\) = 1.293 kg/m³ — density of air

and \(g\) = 9.8 m/s²

\[\text{v} = \sqrt{\frac{0.76 \times 13600 \times 9.8}{1.293}}\]

\[\text{v} = 279.9 \ \text{m/s at NTP}\]

The Discrepancy Newton Could Not Explain

This is the value of the velocity of sound according to Newton’s formula. But the experimental value of the velocity of sound at 0°C, as determined earlier by a number of scientists, is 332 m/s.

The difference between the value predicted by Newton’s formula and the experimental value is large, and it is not due to experimental error. The experimental value is 16% greater than the value given by the formula. A gap of that size cannot be blamed on sloppy measurement; it means that something in the theory is wrong.

Newton could not give a satisfactory explanation of this discrepancy. It was resolved much later by the French physicist Pierre Simon Laplace (1749–1827), whose correction forms the subject of the next subsection.

Newton’s calculation, published in the Principia in the late seventeenth century, was the first serious attempt in history to predict the speed of sound from theory rather than simply measure it — and that alone was a remarkable achievement. But his prediction of about 280 m/s stubbornly refused to match the measurements, which clustered around 330 m/s. For nearly a century and a half the gap remained an open embarrassment for physics, an example of a beautiful derivation whose one hidden assumption was quietly wrong.

Q. Newton’s formula gives 279.9 m/s while experiment gives 332 m/s. Does this mean Eq. (8.7) itself is wrong?

A. No. Equation (8.7), \(\text{v} = \sqrt{E/\rho}\), is perfectly correct. What is wrong is the choice of \(E\) — Newton put in the isothermal bulk modulus, and that turns out to be the wrong modulus for the way sound actually compresses air. Fix the modulus and the same formula gives the right answer, as we shall see.

Suppose you are listening to an open-air live concert, sitting at a distance of 150 m from the speakers. Your friend is listening to a live broadcast of the same concert in another country, and the radio signal has to travel 3000 km to reach him. Who will hear the music first, and what will be the time difference between the two? (Velocity of light = 3 × 10⁸ m/s and that of sound is 330 m/s.)

Solution: The two paths are travelled by two completely different kinds of wave. The sound reaching you crosses 150 m of air as a mechanical wave at 330 m/s. The signal reaching your friend crosses 3000 km as an EM wave at 3 × 10⁸ m/s. So we simply compute time = distance ÷ speed for each.

Time taken by sound to reach you

\[= \frac{150}{330} \ s = 0.4546\]

Time taken by the broadcasted sound (which is done by EM waves having velocity = 3 × 10⁸ m/s)

\[= \frac{3000 \ \text{km}}{3 \times 10^{5} \ \text{km/s}} = \frac{3 \times 10^{3}}{3 \times 10^{5}} = 10^{-2} \ s\]

\(\therefore\) your friend will hear the sound first. The time difference will be

\[= 0.4546 - 0.01\]

\[= 0.4446 \ s\]

The result is worth pausing over: the listener thousands of kilometres away hears the note before the listener sitting in the audience. Sound is slow, and light — along with every other EM wave — is very nearly instantaneous over ordinary distances.

Q1. Lightning and thunder are produced at the same instant, so why do we always see the flash first and hear the rumble afterwards?

A. Because light and sound travel at hugely different speeds. Light is an EM wave and reaches you at 3 × 10⁸ m/s, so from a storm a few kilometres away it arrives in a few millionths of a second — effectively at once. Thunder is a sound wave and crawls through the air at only about 330 m/s, so it needs roughly 3 seconds to cover each kilometre. The flash and the bang begin together; they simply do not finish the journey together.

8.5.4 Laplace’s Correction

According to Laplace, the generation of compression and rarefaction is not a slow process but is a rapid process. That single observation is the key to the whole puzzle left behind by Newton.

Think about what a sound wave actually does to a small parcel of air. If the frequency is 256 Hz, the air is compressed and rarefied 256 times in a second. Such a process must be a rapid process — each squeeze and each release lasts only a few thousandths of a second.

Now, heat is produced during compression, because squeezing a gas warms it, and heat is lost during rarefaction, because letting a gas expand cools it. But this heat does not get sufficient time to dissipate — it cannot flow away into the surrounding air in the tiny fraction of a second available before the next rarefaction arrives, as suggested in Fig. 8.11. Due to this, the total heat content remains the same.

Diagram

Fig. 8.11 – A sound wave in air with a compression marked slightly warmer and the adjacent rarefaction slightly cooler, and crossed-out arrows showing that heat has no time to flow out of either region.

Such a process — one in which no heat enters or leaves — is called an adiabatic process. Hence the elasticity to be used must be adiabatic elasticity, and not isothermal elasticity, as was assumed by Newton.

Always remember: in an isothermal process the temperature remains constant, while in an adiabatic process there is neither transfer of heat nor of mass.

Q. Newton assumed the temperature of the air stays constant. Is that not exactly what we observe — the air in a room does not get hot when someone speaks?

A. The average temperature of the room certainly does not change. But that is not the question. Inside each compression the air is momentarily a little warmer, and inside each rarefaction a little cooler, and these tiny local temperature differences never get the chance to even out before the wave has moved on. It is these local, un-cancelled temperature changes that Newton overlooked, and they are what stiffen the air and speed the sound up.

The correction was supplied in the early nineteenth century by Pierre Simon Laplace, who realised that Newton’s error lay not in his mathematics but in his thermodynamics — a subject that did not properly exist in Newton’s day. By replacing the isothermal bulk modulus with the adiabatic one, Laplace closed a gap that had stood open in physics for well over a century, and did so without changing a single step of Newton’s derivation.

The Corrected Formula

The adiabatic modulus of elasticity of air is given by

\[E = \gamma P \qquad \text{--- (8.9)}\]

where \(P\) is the pressure of the medium (air), measured in N/m², and \(\gamma\) is the ratio of the specific heat of air at constant pressure (\(C_p\)) to the specific heat of air at constant volume (\(C_v\)), called the adiabatic ratio,

i.e.,

\[\gamma = \frac{C_p}{C_v} \qquad \text{--- (8.10)}\]

Being a ratio of two like quantities, \(\gamma\) has no unit.

For air the ratio of \(C_p / C_v\) is 1.41

i.e. \(\gamma\) = 1.41

Newton’s formula for the speed of sound in air, as modified by Laplace, gives

\[\text{v} = \sqrt{\frac{\gamma P}{\rho}} \qquad \text{--- (8.11)}\]

NotePrinciple / Law

(Newton’s formula with Laplace’s correction.) The speed of sound in air is \(\text{v} = \sqrt{\gamma P / \rho}\), where \(P\) is the pressure of the air, \(\rho\) its density, and \(\gamma = C_p/C_v\) is the adiabatic ratio of the gas, because the compressions and rarefactions of a sound wave take place adiabatically and not isothermally.

Compare Eq. (8.11) with Newton’s Eq. (8.8) and you will see that the whole of Laplace’s correction amounts to inserting the single factor \(\gamma\) under the square root. Since \(\gamma\) = 1.41 for air, the corrected speed is larger than Newton’s by a factor of \(\sqrt{1.41}\), which is about 1.19 — an increase of roughly 19%, which is just what was needed to close the 16% shortfall.

According to this formula, the velocity of sound at NTP is

\[\text{v} = \sqrt{\frac{1.41 \times 0.76 \times 13600 \times 9.8}{1.293}}\]

\[= 332.3 \ \text{m/s}\]

This value is in close agreement with the experimental value of 332 m/s. As seen above, the velocity of sound depends on the properties of the medium.

Q3. Newton worked out a formula for the speed of sound in air, but his answer came out about 16% too small — and it was not an error in measurement. What had he overlooked?

A. Newton assumed that the compressions and rarefactions of a sound wave happen slowly enough for the air to stay at a constant temperature — an isothermal process. Laplace pointed out that they happen hundreds of times a second, far too fast for the heat produced in a compression to flow away. The process is therefore adiabatic, and the correct elasticity to use is the adiabatic bulk modulus \(\gamma P\), not the isothermal one \(P\). Inserting the factor \(\gamma\) = 1.41 raises the predicted speed from 279.9 m/s to 332.3 m/s, in close agreement with experiment.

The density of helium at NTP is 0.179 kg/m³ and its adiabatic ratio is \(\gamma\) = 1.67. Taking the atmospheric pressure as 1.01 × 10⁵ N/m², use Laplace’s corrected formula to calculate the speed of sound in helium, and compare your answer with the value listed in Table 8.1.

8.5.5 Factors Affecting Speed of Sound

As sound waves travel through the atmosphere (open air), some factors related to the air affect the speed of sound. Laplace’s formula, Eq. (8.11), contains the pressure \(P\), the density \(\rho\) and the adiabatic ratio \(\gamma\) of the air, so it is natural to ask what happens to the speed when the pressure changes, when the temperature changes, and when the air becomes moist. We take the three questions in turn.

Do you know? \(C_p\), the specific heat of a gas at constant pressure, is defined as the quantity of heat required to raise the temperature of unit mass of the gas through 1 K when the pressure remains constant. \(C_v\), the specific heat of a gas at constant volume, is defined as the quantity of heat required to raise the temperature of unit mass of the gas through 1 K when the volume remains constant. When the pressure is kept constant, the volume of the gas increases with an increase in temperature. Thus additional heat is required to increase the volume of the gas against the external pressure. Therefore the heat required to raise the temperature of unit mass of a gas through 1 K when the pressure is kept constant is greater than the heat required when the volume is kept constant, i.e., \(C_p > C_v\).

Effect of Pressure on Velocity of Sound

According to Laplace’s formula, the velocity of sound in air is

\[\text{v} = \sqrt{\frac{\gamma P}{\rho}}\]

At first sight this seems to say that raising the pressure must raise the speed. But that reading forgets something: you cannot squeeze air into a smaller space without also making it denser. \(P\) and \(\rho\) are not independent, and we must handle them together.

If \(M\) is the mass and \(V\) is the volume of air, then

\[\rho = \frac{M}{V}\]

\[\therefore \text{v} = \sqrt{\frac{\gamma P V}{M}} \qquad \text{--- (8.12)}\]

At constant temperature, \(PV\) = constant, according to Boyle’s law. Also \(M\) and \(\gamma\) are constant, hence v = constant.

Therefore, at constant temperature, a change in pressure has no effect on the velocity of sound in air. The increase in \(P\) is exactly cancelled by the increase in \(\rho\), and their ratio — which is all the formula cares about — stays put.

This can be seen in another way. For a gaseous medium, \(PV = nRT\), \(n\) being the number of moles.

\[\therefore \text{v} = \sqrt{\frac{\gamma n R T}{M}} \qquad \text{--- (8.13)}\]

Here \(R\) is the universal gas constant, \(T\) is the absolute temperature in kelvin (K), \(n\) is the number of moles, and \(M\) is the mass of the gas in kilogram (kg). Pressure has disappeared from the expression altogether — the only quantity on the right that can vary is \(T\).

Hence, for a gaseous medium obeying the ideal gas equation, a change in pressure has no effect on the velocity of sound unless there is a change in temperature.

Q. If pressure has no effect, why is the speed of sound lower high up in the atmosphere, where the pressure is clearly much lower?

A. Because the pressure is not the thing that has changed on its own. As you climb, the pressure falls — but so does the density, in nearly the same proportion, so the ratio \(P/\rho\) is barely affected. What really changes with altitude is the temperature, and it is the drop in temperature, acting through Eq. (8.13), that slows the sound down.

Consider a closed box of rigid walls, so that the density of the air inside it is constant. On heating, the pressure of this enclosed air is increased from \(P_0\) to \(P\). It is now observed that sound travels 1.5 times faster than at pressure \(P_0\). Calculate \(P/P_0\).

Solution: Note carefully why this example does not contradict the result we have just proved. There, the pressure changed at constant temperature and the density changed with it. Here the walls are rigid, so the density is forced to stay constant while the temperature — and hence the pressure — rises. The two situations are physically different.

\[\text{v}_P = \sqrt{\frac{\gamma P_o}{\rho}}\]

\[\text{v}_{Po} = \sqrt{\frac{\gamma P_o}{\rho}}\]

\[\text{v}_P = 1.5 \ \text{v}_{Po}\]

\[\sqrt{\frac{\gamma P}{\rho}} = 1.5 \sqrt{\frac{\gamma P_o}{\rho}}\]

Squaring both sides, \(\gamma\) cancels and, since the density is unchanged, so does \(\rho\):

\[\frac{P}{\rho} = 2.25 \frac{P_o}{\rho}\]

\[P = 2.25 \ P_o\]

So the pressure must be raised to 2.25 times its original value, since \(1.5^2 = 2.25\) and the speed depends on the square root of the pressure.

Effect of Temperature on Speed of Sound

Suppose \(\text{v}_0\) and \(\text{v}\) are the speeds of sound at \(T_0\) and \(T\) in kelvin respectively. Let \(\rho_0\) and \(\rho\) be the densities of the gas at these two temperatures. The velocity of sound at temperatures \(T_0\) and \(T\) can be written by using Eq. (8.13),

\[\text{v}_0 = \sqrt{\frac{\gamma R T_0}{M}} \qquad \text{--- } M \text{ is molar mass, } n = 1\]

\[\text{v} = \sqrt{\frac{\gamma R T}{M}}\]

Dividing one by the other, the constants \(\gamma\), \(R\) and \(M\) all cancel:

\[\therefore \frac{\text{v}}{\text{v}_0} = \sqrt{\frac{RT}{RT_0}}\]

\[\therefore \frac{\text{v}}{\text{v}_0} = \sqrt{\frac{T}{T_0}} \qquad \text{--- (8.14)}\]

This equation shows that the speed of sound in air is directly proportional to the square root of the absolute temperature. Thus, the speed of sound in air increases with an increase in temperature.

NotePrinciple / Law

The speed of sound in a gas is directly proportional to the square root of the absolute temperature of the gas, \(\text{v} \propto \sqrt{T}\), the pressure being immaterial for an ideal gas.

Equation (8.14) uses temperature in kelvin. In everyday work, however, temperatures are quoted in degrees celsius, so it is useful to convert it into a form we can use directly. Taking \(T_0\) = 273 K and writing \(T = (273 + t)\) K, where \(t\) is the temperature in degree celsius, the ratio of the velocity of sound in air at \(t\) °C to that at 0°C is given by,

\[\therefore \frac{\text{v}}{\text{v}_0} = \sqrt{\frac{273 + t}{273}}\]

\[\therefore \frac{\text{v}}{\text{v}_0} = \sqrt{1 + \frac{t}{273}}\]

\[\therefore \frac{\text{v}}{\text{v}_0} = \sqrt{1 + \alpha t} \quad \text{where} \quad \alpha = \frac{1}{273}\]

or,

\[\text{v} = \text{v}_0 (1 + \alpha t)^{\frac{1}{2}}\]

As \(\alpha\) is very small, we can write — using the binomial approximation \((1+x)^{1/2} \approx 1 + \tfrac{1}{2}x\) for small \(x\) —

\[\text{v} \simeq \text{v}_0 \left(1 + \frac{1}{2}\alpha t \right)\]

\[\text{v} = \text{v}_0 \left(1 + \frac{1}{2} \times \frac{1}{273} \ t \right)\]

\[\text{v} = \text{v}_0 \left(1 + \frac{t}{546} \right)\]

\[\text{v} = \text{v}_0 + \frac{\text{v}_0}{546} \ t\]

But \(\text{v}_0\) = 332 m/s at 0°C

\[\therefore \text{v} = \text{v}_0 + \frac{332}{546} \ t\]

\[\therefore \text{v} \simeq \text{v}_0 + (0.61) \ t \qquad \text{--- (8.15)}\]

i.e., for a 1°C rise in temperature the velocity increases by 0.61 m/s. Hence for small variations in temperature (< 50°C), the speed of sound changes linearly with temperature, as shown in Fig. 8.12.

Diagram

Fig. 8.12 – Graph of speed of sound in air versus temperature in °C, a straight line starting from 332 m/s at 0°C with a slope of 0.61 m/s per °C.

Two cautions are worth fixing in your mind here, because both are common sources of lost marks. Equation (8.14) needs the temperature in kelvin; Eq. (8.15) needs it in degree celsius. And Eq. (8.15) is an approximation valid only for modest temperature changes — for a large change, go back to Eq. (8.14).

Wind instruments go noticeably sharp as a concert proceeds. A flute or a trumpet sounds a note whose frequency is set by the speed of sound in the air column inside the instrument, and the player’s warm breath steadily raises the temperature of that air. By Eq. (8.15), a rise of 10°C lifts the speed of sound by about 6 m/s, and the pitch of every note rises with it — which is why wind players re-tune after warming up, while string players, whose pitch depends on tension rather than air temperature, do not have quite the same problem.

The speed of sound in air at 0°C is 332 m/s. Using Eq. (8.15), calculate the speed of sound in air on a summer afternoon when the temperature is 42°C, and state by how much it exceeds the value at 0°C.

Effect of Humidity on Speed of Sound

Humidity (moisture) in air depends upon the presence of water vapour in it. Let \(\rho_m\) and \(\rho_d\) be the densities of moist and dry air respectively. If \(\text{v}_m\) and \(\text{v}_d\) are the speeds of sound in moist air and dry air, then using Eq. (8.11),

\[\text{v}_m = \sqrt{\frac{\gamma P}{\rho_m}}\]

\[\text{and} \quad \text{v}_d = \sqrt{\frac{\gamma P}{\rho_d}}\]

\[\therefore \frac{\text{v}_m}{\text{v}_d} = \sqrt{\frac{\rho_d}{\rho_m}} \qquad \text{--- (8.16)}\]

Notice that the density of the dry air ends up in the numerator, because \(\rho\) sits in the denominator of each speed and the two fractions invert on dividing.

Moist air is always less dense than dry air, i.e.,

\[\rho_m < \rho_d\]

\[(\rho_m = 0.81 \ \text{kg/m}^3 \ \text{(at 0°C) and} \ \rho_d = 1.29 \ \text{kg/m}^3 \ \text{(at 0°C)})\]

\[\therefore \text{v}_m > \text{v}_d\]

Thus, the speed of sound in moist air is greater than the speed of sound in dry air, i.e., the speed increases with an increase in the moistness of the air.

Q. Water is far denser than air — so how can adding water vapour to air make it less dense?

A. Because the water vapour does not simply pile on top of the air; it displaces some of it. A water molecule is lighter than the nitrogen and oxygen molecules it pushes aside, so a given volume of moist air contains slightly less mass than the same volume of dry air at the same pressure and temperature. Less mass in the same volume means lower density — and by Eq. (8.16), a higher speed of sound.

8.6 Principle of Superposition of Waves

Waves do not display any repulsion towards each other. Two wave patterns can therefore overlap in the same region of space without affecting each other — neither is bounced away, and neither is destroyed. This is quite unlike two particles, which cannot occupy the same place at the same time.

When two waves overlap, their displacements add vectorially. This additive rule is referred to as the principle of superposition of waves. Two pulses meeting on a stretched string, as in Fig. 8.13, show it plainly: where they overlap, the string takes up the sum of the two displacements.

Diagram

Fig. 8.13 – Two pulses travelling towards each other on a string, shown before, during and after overlap, with the resultant displacement in the overlap region drawn as the sum of the two.

When two or more waves travelling through a medium arrive at a point of the medium simultaneously, each wave produces its own displacement at that point independent of the others. Hence the resultant displacement at that point is equal to the vector sum of the displacements due to all the waves.

NotePrinciple / Law

(Principle of superposition of waves.) When two or more waves arrive simultaneously at a point of a medium, each produces its own displacement at that point independently of the others, and the resultant displacement of the particle is the vector sum of the individual displacements.

The word “vector” earns its place in that statement. If both waves happen to displace the particle in the same direction at that instant, the displacements add and the resultant is larger than either. If they displace it in opposite directions, they subtract, and the resultant may be smaller than either — even zero, if the two are equal and opposite. The phenomenon of superposition will be discussed in detail in Class XII.

Q. When two pulses overlap and momentarily cancel each other, has their energy been destroyed?

A. No. At that instant the displacement of the string is zero, but its particles are moving fastest — the energy is present as kinetic energy rather than as visible displacement. A moment later the two pulses emerge from the overlap completely unchanged in shape, size and speed, and travel on as though nothing had happened.

Noise-cancelling headphones are superposition put to work. A small microphone on the outside of the earcup samples the steady drone of an aircraft cabin, and the electronics inside generate a second sound wave of the same amplitude but exactly opposite displacement — half a wavelength out of step. The two waves superpose at your eardrum, their displacements add vectorially to nearly zero, and the drone very largely disappears, while the speech you want to hear, which the system does not attempt to match, comes through.

8.7 Echo, Reverberation and Acoustics

Sound waves obey the same laws of reflection as those of light. When a sound wave strikes a hard, rigid surface, it bounces back with the angle of reflection equal to the angle of incidence, the incident wave, the reflected wave and the normal to the surface all lying in one plane, as shown in Fig. 8.14.

Diagram

Fig. 8.14 – A sound wave incident on a rigid wall and reflected from it, with the normal drawn and the angles of incidence and reflection marked equal.

NotePrinciple / Law

Sound obeys the laws of reflection: the angle of reflection is equal to the angle of incidence, and the incident wave, the reflected wave and the normal to the reflecting surface lie in the same plane.

Everything in the three subsections that follow — echo, reverberation and the design of auditoriums — is a consequence of this single fact.

Q. If sound reflects like light, why does a wall not act as a mirror for sound the way it does for light?

A. It does, but the wavelengths are wildly different. Light has a wavelength of less than a micrometre, so even a slightly rough wall is smooth compared with it only when polished. Sound in air has wavelengths of centimetres to metres, so an ordinary brick wall is effectively “smooth” to sound and reflects it cleanly — which is exactly why you can get an echo off a wall you could never see your face in.

8.7.1 Echo

An echo is the repetition of the original sound because of reflection from some rigid surface at a distance from the source of sound. If we shout in a hilly region, we are likely to hear an echo, as illustrated in Fig. 8.15.

Diagram

Fig. 8.15 – A person shouting towards a distant cliff, with the sound wave travelling to the cliff and reflecting back, and the distance d between the person and the cliff marked.

NoteDefinition

An echo is the repetition of the original sound, heard because of reflection of the sound from a rigid surface situated at a distance from the source of sound.

Why We Do Not Hear an Echo Everywhere

Why can’t we hear an echo at every place? Reflection, after all, happens at every wall. The answer lies not in the sound but in the listener.

At 22°C, the velocity of sound in air is 344 m/s. Our brain retains a sound for 0.1 second — that is, a sound sensation persists in the mind for about a tenth of a second after the sound itself has stopped. If the reflected sound arrives within that window, it merges into the original and is not perceived as a separate sound at all.

Thus, for us to hear a distinct echo, the sound should take more than 0.1 s after starting from the source (i.e., from us) to get reflected and come back to us.

In 0.1 s the sound covers

distance = speed × time

\[= 344 \times 0.1\]

\[= 34.4 \ \text{m}\]

But this 34.4 m is the total journey — out to the surface and back again. To be able to hear a distinct echo, the reflecting surface should therefore be at a minimum distance of half of the above distance, i.e. 17.2 m.

As the velocity depends on the temperature of the air, this distance will change with temperature. On a hotter day sound travels faster, covers more ground in the same 0.1 s, and so the minimum distance required for a distinct echo becomes a little larger.

Q4. Why do you hear a clear echo when you shout in a hilly valley, but never inside your own classroom?

A. Because your brain holds on to a sound for about 0.1 second. For the reflection to be heard as a separate sound, it must arrive later than that, and at 344 m/s this means the reflecting surface must be at least about 17.2 m away. A hillside easily satisfies this. A classroom wall is only a few metres away, so its reflection returns in a small fraction of a second and merges with the original sound instead of standing apart from it as an echo.

The Gol Gumbaz at Vijayapura (Bijapur) in Karnataka is famous precisely because of this effect. Its enormous dome sits far enough from a speaker standing beneath it, and its curved surface is hard enough, that a single clap returns as a series of clearly separated repetitions. Visitors come to test the effect for themselves — a seventeenth-century monument that has become an unintended demonstration of the reflection of sound.

A man shouts loudly close to a high wall. He hears an echo. If the man is at 40 m from the wall, how long after the shout will the echo be heard? (Speed of sound in air = 330 m/s.)

Solution: The first step is the one students most often skip: the sound does not travel 40 m, it travels 40 m out and 40 m back. So the distance travelled by the sound wave

= 2 × distance from man to wall

\[= 2 \times 40\]

\[= 80 \ \text{m}\]

\(\therefore\) Time taken to travel the distance \(= \dfrac{\text{distance}}{\text{speed}}\)

\[= \frac{80 \ \text{m}}{330 \ \text{m/s}}\]

\[= 0.24 \ \text{s}\]

\(\therefore\) The man will hear the echo 0.24 s after he shouts.

Notice that 0.24 s is comfortably more than 0.1 s, which is why the man hears the echo as a distinct sound at all — consistent with the 17.2 m minimum found above.

A boy standing in front of a cliff claps his hands and hears the echo 0.6 s later. If the speed of sound in air on that day is 340 m/s, calculate the distance of the cliff from the boy.

8.7.2 Reverberation

If the reflecting surface is nearer than 15 m from the source of sound, the echo joins up with the original sound, which then seems to be prolonged. The reflection has not vanished — it has simply arrived too soon to be heard as a separate event, and so it is heard as a stretching-out of the sound you were already hearing.

In a closed room, this happens not once but over and over again. Sound waves get reflected multiple times from the walls and the roof of a closed room, which are nearer than 15 m, as shown in Fig. 8.16. Each reflected wave sets off again, strikes another surface, and returns once more. This causes a single sound to be heard not just once but continuously. This is called reverberation.

Diagram

Fig. 8.16 – A source of sound inside a closed hall, with rays reflecting repeatedly off the walls, floor and ceiling before reaching a listener.

It is this — the persistence of sound after the source has switched off, as a result of repeated reflection from walls, ceilings and other surfaces — that we call reverberation. Reverberation characteristics are important in the design of concert halls, theatres and similar spaces.

NoteDefinition

Reverberation is the persistence of sound in a closed space after the source has stopped producing it, caused by repeated reflection of the sound from the walls, ceiling and other surfaces.

Q. Echo and reverberation are both reflections of sound. What exactly is the difference?

A. The difference is time, and therefore distance. In an echo, the reflected sound arrives more than 0.1 s after the original, from a surface more than about 17 m away, and is heard as a separate, repeated sound. In reverberation, the reflections come back from nearby surfaces in far less than 0.1 s, over and over, and are heard not as repetitions but as a prolonging of the original sound.

Why Too Much Reverberation Is a Problem

If the time between successive reflections of a particular sound wave reaching us is small, the reflected sound gets mixed up with the sound arriving directly from the source, and produces a continuous sound of increased loudness which can’t be heard clearly.

That last phrase contains an apparent contradiction worth resolving. Reverberation makes a hall louder but less clear. Each syllable a speaker utters is still ringing around the room when the next syllable arrives, so the two overlap and smear into each other. The sound is plentiful; the information carried by it is lost.

Reducing Reverberation

Reverberation can be decreased by making the walls and roofs rough and by using curtains in the hall, to avoid the clean reflection of sound. A rough surface scatters the sound in many directions instead of returning it as a single strong reflection.

Chairs and wall surfaces are covered with sound absorbing materials. Porous cardboard sheets, perforated acoustic tiles, gypsum boards, thick curtains and similar materials at the ceilings and at the walls are the most convenient means of reducing reverberation. What these have in common is porosity: sound entering the tiny air spaces is repeatedly reflected inside them and gradually converted into heat, so that less of it comes back into the hall.

Acoustic laboratories take this idea to its limit in an anechoic chamber — a room whose walls, ceiling and floor are covered entirely with deep foam wedges that absorb almost all the sound falling on them. With reverberation reduced practically to zero, engineers can measure the true sound output of a loudspeaker or a machine without any reflections contaminating the reading. Visitors usually find the room deeply unsettling: with no reverberation at all, voices sound flat and dead, because our ears have never in ordinary life heard a sound unaccompanied by its reflections.

A student claps once while standing at the centre of a hall whose walls are 12 m away on either side. Taking the speed of sound in air as 340 m/s, calculate the time after which the first reflection from a wall reaches her, and use your answer to explain whether she will hear an echo or reverberation.

8.7.3 Acoustics

The branch of physics which deals with the study of the production, transmission and reception of sound is called acoustics.

NoteDefinition

Acoustics is the branch of physics dealing with the production, transmission and reception of sound.

This is useful during the construction of theatres and auditoriums. While designing an auditorium, proper care for the absorption and reflection of sound should be taken. Otherwise the audience will not be able to hear the sound clearly. The two effects we have just studied — echo and reverberation — are the very things the designer must manage: too little reflection and the hall sounds dead and faint, too much and every word is smeared into the next.

Conditions for Good Acoustics in an Auditorium

For proper acoustics in an auditorium the following conditions must be satisfied.

  1. The sound should be heard sufficiently loudly at all the points in the auditorium. The surface behind the speaker should be parabolic, with the speaker at its focus, so that the distribution of sound is uniform in the auditorium. Reflection of sound is helpful in maintaining good loudness throughout the entire auditorium, as suggested in Fig. 8.17.

Diagram

Fig. 8.17 – A parabolic reflecting surface behind a speaker standing at its focus, with the reflected sound leaving as a uniform parallel beam into the hall.

Q. Why a parabolic surface behind the speaker, and why must the speaker stand at its focus?

A. A parabola has the special property that all rays leaving its focus are reflected into a parallel beam. Placing the speaker at the focus therefore sends the sound out evenly into the hall instead of letting it spread and weaken in all directions — so the listeners at the back receive nearly as much sound as those at the front.

  1. Echoes and reverberation must be eliminated or reduced. Echoes can be reduced by making the reflecting surfaces more absorptive. The echo will be less if the auditorium is full, since the clothing and bodies of the audience are themselves good absorbers of sound.

  2. Unnecessary focusing of sound should be avoided, and there should not be any zone of poor audibility or region of silence. For that purpose, curved surfaces of the wall or ceiling should be avoided. A concave wall can concentrate sound at one spot and starve another, so that two listeners a few seats apart hear quite different things.

  3. Echelon effect: it is due to the mixing of sound produced in the hall by the echoes of sound produced in front of a regular structure like a set of stairs. Because the steps are evenly spaced, the reflections from them return at evenly spaced intervals, and the ear picks up this regular train of echoes as an unpleasant musical note that was never in the original sound. To avoid this, stair-type construction must be avoided in the hall.

  4. The auditorium should be sound-proof when closed, so that stray sound can not enter from outside.

  5. For proper acoustics no sound should be produced from the inside fittings, seats, etc. Instead of fans, air conditioners may be used. Soft-action door closers should be used.

Acoustics Observed in Nature

The importance of acoustic principles goes far beyond human hearing. Several animals use sound for navigation.

  1. Bats depend on sound rather than light to locate objects. So they can fly in the total darkness of caves. They emit short ultrasonic pulses of frequency 30 kHz to 150 kHz. The resulting echoes give them information about the location of the obstacle.

  2. Dolphins use an analogous system for underwater navigation. The frequencies are subsonic, about 100 Hz. They can sense an object of about the size of a wavelength, i.e., 1.4 m or larger.

Q. How can an echo tell a bat where an obstacle is, and not merely that one exists?

A. From the time delay. The pulse travels out at the speed of sound and returns at the same speed, so the delay between the squeak and its echo is proportional to the distance of the obstacle — exactly the calculation you performed for the man and the wall in Section 8.7.1. The direction is judged from the slight difference between the echoes arriving at the bat’s two ears.

Medical Applications of Acoustics
  1. Shock waves, which are high pressure and high amplitude waves, are used to split kidney stones into smaller pieces without invasive surgery. A shock wave is produced outside the body and is then focused by a reflector or acoustic lens, so that as much of its energy as possible converges on the stone. When the resulting stresses in the stone exceed its tensile strength, it breaks into small pieces which can be removed easily. Notice that the focusing here is deliberate — the very effect that condition (3) above forbids in an auditorium is exactly what is wanted in this treatment.

  2. Reflection of ultrasonic waves from regions in the interior of the body is used for ultrasonic imaging. It is used for prenatal (before the birth) examination, for the detection of anomalous conditions like a tumour, and for the study of heart valve action. The principle is once again the echo: a pulse is sent in, and the time taken by each reflected pulse to return tells the instrument how deep the reflecting boundary lies.

  3. At very high power levels, ultrasound is a selective destroyer of pathological tissues, in the treatment of arthritis and certain types of cancer.

Other Applications of Acoustics
  1. SONAR is an acronym for Sound Navigational Ranging. This is a technique for locating objects underwater by transmitting a pulse of ultrasonic sound and detecting the reflected pulse. The time delay between the transmission of a pulse and the reception of the reflected pulse indicates the depth of the object. This system is useful to measure the motion and position of submerged objects like a submarine, as shown in Fig. 8.18.

Diagram

Fig. 8.18 – A ship emitting an ultrasonic pulse downwards through water, the pulse reflecting off a submarine and returning to the ship’s detector.

A SONAR system on a ship sends an ultrasonic pulse straight down and receives the echo from the sea bed 0.8 s later. Taking the speed of sound in seawater from Table 8.1, calculate the depth of the sea bed below the ship.

  1. The acoustic principle has an important application to environmental problems like noise control. The design of a quiet mass transit vehicle involves the study of the generation and propagation of sound in the motor’s wheels and supporting structures.

  2. We can study the properties of the Earth by measuring the reflected and refracted elastic waves passing through its interior. It is useful for geological studies, to detect local anomalies like oil deposits, etc.

Taken together, these examples make one point worth carrying away from the section. Every application listed here — the bat in the cave, the ultrasound scanner, the SONAR pulse, the geologist’s survey — rests on the same simple fact with which Section 8.7 opened: sound reflects, and the time its reflection takes to return carries information about whatever it bounced off.

8.8 Qualities of Sound

Audible Sound, or Human Response to Sound

Whenever we talk about audible sound, what matters is how we perceive it. This is purely a subjective attribute of sound waves. The wave itself is described completely by its amplitude, frequency and speed; but the sensation it produces in a listener is a separate matter, and it is that sensation we now examine.

The major qualities of sound that are of our interest are (i) pitch, (ii) timbre or quality, and (iii) loudness. Each of them is a response, and each is connected to a measurable property of the wave — but never in a simple, proportional way.

Pitch

This aspect refers to the sharpness or shrillness of the sound. If the frequency of a sound is increased, what we perceive is an increase in the pitch, or we feel the sound to be sharper.

Two related words need to be kept apart. Tone refers to the single frequency of that wave, while a note may contain one or more than one tone. We use the words high pitch or high tones if the frequency is higher.

NoteDefinition

Pitch is the subjective sensation of the sharpness or shrillness of a sound, and it increases as the frequency of the sound increases.

As sharpness is a subjective term, sentences like “a sound of double frequency is doubly sharp” make no sense. Frequency can be doubled; a sensation cannot be. Also, a high pitch sound need not be louder — pitch and loudness are independent qualities, and a shrill whistle may be very faint.

The tones of a guitar are sharper than those of a bass guitar; the sound of a tabla is sharper than that of a dagga; and, in general, a female voice is sharper than a male voice, and so on. In every one of these cases, the sharper source is the one vibrating at the higher frequency.

For a sound amplifier (or equaliser), when we raise the treble knob (or treble button), high frequencies are boosted, and if we raise the bass knob, low frequencies are boosted.

Q. Two sounds have the same loudness, but one is a shrill whistle and the other a deep drum. What is physically different about the two waves?

A. Their frequencies. Loudness is governed mainly by the intensity of the wave — that is, by its amplitude — while pitch is governed by frequency. Since these are two independent properties of the same wave, a sound can be loud and deep, loud and shrill, faint and deep, or faint and shrill, in any combination.

Timbre (Sound Quality)

During a telephonic conversation with a friend, you are usually able to know who is speaking at the other end even if you are not told who is speaking. Quite often we say, “Couldn’t you recognise the voice?” The sound quality in this context is called timbre.

The same song played on a guitar, a violin, a harmonium or a piano feels significantly different, and we can easily identify the instrument. Yet all four instruments may be sounding the very same note, at the very same loudness. Something else must be distinguishing them, and that something is timbre.

NoteDefinition

Timbre, or quality, is the characteristic of a sound that allows us to distinguish between two sounds of the same pitch and the same loudness produced by different sources.

The quality of sound of any sound instrument (including our vocal organ) depends upon the mixture of tones and overtones in the sound generated by that instrument. A real instrument never produces a single pure frequency; it produces a principal tone together with a set of higher tones, and the recipe in which these are mixed is the instrument’s signature, as suggested by the differing waveforms in Fig. 8.19.

Diagram

Fig. 8.19 – Waveforms of the same musical note played on a tuning fork, a guitar and a violin, drawn one below the other, showing the same repeat interval but different shapes.

Even our own sound quality during the morning (just after we get up) and in the evening is different. It is drastically affected if we are suffering from a cold or cough — which is precisely why a friend can often tell over the phone that you are unwell, without your having said so. The concept of overtones will be discussed in Class XII.

Loudness

The intensity of a wave is a measurable quantity which is proportional to the square of the amplitude (\(I \propto A^2\)), and is measured in the SI unit of W/m². Human perception of the intensity of sound is loudness. Obviously, if the intensity is more, the loudness is more.

Keep the distinction firmly in mind: intensity is a property of the wave and can be read off an instrument; loudness is a sensation in a listener and cannot.

NoteDefinition

The intensity of a sound wave is the energy carried by the wave per unit time per unit area held perpendicular to the direction of propagation. It is proportional to the square of the amplitude and is measured in W/m².

NoteDefinition

Loudness is the human perception of the intensity of a sound. It increases with intensity, but not in direct proportion to it.

The human response to intensity is not linear, i.e., a sound of double intensity is louder but not doubly loud. This is also valid for the brightness of light. In both cases, the response is approximately logarithmic. Using this property, loudness (and brightness) can be measured.

The Bel and the Decibel

Under ideal conditions, for a perfectly healthy human ear, the least audible intensity is \(I_0 = 10^{-12}\) W/m². This is taken as the reference against which all other intensities are compared. The loudness of a sound of intensity \(I\), measured in the unit bel, is given by

\[L_{bel} = log_{10} \left( \frac{I}{I_0} \right) \qquad \text{--- (8.17)}\]

where \(I\) is the intensity of the sound in W/m² and \(I_0 = 10^{-12}\) W/m² is the least audible intensity.

The popular or commonly used unit for loudness is the decibel. We know that 1 decimetre, or 1 dm, = 0.1 m. Similarly, 1 decibel, or 1 db, = 0.1 bel. \(\therefore\) 1 bel = 10 db. Thus, loudness expressed in db is 10 times the loudness in bel:

\[\therefore L_{db} = 10 \ L_{bel} = 10 \ log_{10} \left( \frac{I}{I_0} \right)\]

For a sound of the least audible intensity \(I_0\),

\[L_{db} = 10 \ log_{10} \left( \frac{I_0}{I_0} \right) = 10 \ log_{10} (1) = 0 \qquad \text{--- (8.18)}\]

This corresponds to the threshold of hearing. A reading of 0 db therefore does not mean “no sound at all” — it means sound at the faintest level a healthy ear can just detect.

For a sound of 10 db,

\[10 = 10 \ log_{10} \left( \frac{I}{I_0} \right) \ \therefore \left( \frac{I}{I_0} \right) = 10^{1} \ \text{or} \ I = 10 \ I_0\]

For a sound of 20 db,

\[20 = 10 \ log_{10} \left( \frac{I}{I_0} \right) \ \therefore \left( \frac{I}{I_0} \right) = 10^{2} \ \text{or} \ I = 100 \ I_0\]

and so on.

Hence, the loudness of a 20 db sound is felt to be double that of a 10 db sound, but its intensity is 10 times that of the 10 db sound. Now, we feel a 40 db sound to be twice as loud as a 20 db sound, but its intensity is 100 times that of the 20 db sound, and 10000 times that of the 10 db sound. This is the power of a logarithmic or exponential scale: enormous ranges of intensity are compressed into a comfortable range of numbers, because that is exactly what the ear itself does.

Q6. A 40 dB sound is ten thousand times more intense than a 10 dB sound. Why, then, does it not seem ten thousand times louder to your ears?

A. Because the ear does not respond to intensity in direct proportion — its response is approximately logarithmic. Each step of 10 db multiplies the intensity by 10, but the ear registers each such step as roughly a doubling of loudness. Going from 10 db to 40 db is three such steps, so the intensity is multiplied by 10 × 10 × 10 = 10000, while the sensation of loudness only about doubles three times over. The ear compresses a huge range of intensities into a manageable range of sensations.

The loudness of a sound at a certain point is 50 db. The intensity of the sound is then increased 100 times. Taking \(I_0 = 10^{-12}\) W/m², calculate (a) the original intensity of the sound in W/m², and (b) the new loudness in db.

Intensity and Distance

If we move away from a (practically) point source, the intensity of its sound varies inversely with the square of the distance, i.e.,

\[I \propto \frac{1}{r^2}\]

where \(r\) is the distance from the source in metre (m). The reason is a matter of geometry rather than of sound: the energy leaving the source spreads over the surface of an expanding sphere, and the area of that sphere grows as \(r^2\), so the energy falling on each square metre must fall off as \(1/r^2\).

Whenever you are using earphones, or jam your mobile against your ear, the distance from the source is too small. Obviously, such a habit for a long time can affect your normal hearing.

Do you know? According to the World Health Organisation, a billion young people could be at risk of hearing loss due to unsafe listening practices. Among teenagers and young adults aged 12–35 years, (i) about 50% are exposed to unsafe levels of sound from the use of personal audio devices, and (ii) about 40% are exposed to potentially damaging sound levels at clubs, discotheques and bars.

When heard independently, two sound waves produce sensations of 60 db and 55 db respectively. How much will the sensation be if those are sounded together, perfectly in phase?

Solution: Loudness in db cannot simply be added — 60 db and 55 db do not make 115 db. We must convert each loudness back into an intensity, combine the intensities correctly, and then convert the result back into db.

\[L_1 = 60 \ db = 10 \ log_{10} \frac{I_1}{I_0} \ \therefore \frac{I_1}{I_0} = 10^{6} \ \text{or} \ I_1 = 10^{6} I_0\]

Similarly, \(I_2 = 10^{5.5} I_0\)

As the waves combine perfectly in phase, the vector addition of their amplitudes will be given by

\[A^2 = (A_1 + A_2)^2 = A_1^2 + A_2^2 + 2A_1 A_2\]

As intensity is proportional to the square of the amplitude, the cross term \(2A_1A_2\) becomes \(2\sqrt{I_1 I_2}\):

\[\therefore I = I_1 + I_2 + 2\sqrt{I_1 I_2}\]

\[= 10^{5} I_0 \left( 10^{1} + 10^{0.5} + 2\sqrt{10^{1.5}} \right)\]

\[= 10^{5} I_0 \left( 10 + 3.1623 + 2 \times 10^{0.75} \right)\]

\[= 24.41 \times 10^{5} I_0 = 2.441 \times 10^{6} I_0\]

\[\therefore L = 10 \ log_{10} \left( \frac{I}{I_0} \right) = 10 \ log_{10} \left( 2.441 \times 10^{6} \right)\]

\[= 10 \left[ log_{10} (2.441) + log_{10} \left( 10^{6} \right) \right]\]

\[= 10 \left( 0.3876 + 6 \right)\]

\[= 63.876 \ db \approx 64 \ db\]

It is interesting to note that there is only a marginal increase in the loudness. Two fairly loud sounds, added together, gain only about four decibels between them — another consequence of the logarithmic response of the ear.

Table 8.2: Approximate Decibel Ratings of Some Audible Sounds

Source or description of noise | Loudness, L_db | Effect |

|—|—|—|

Extremely loud | 160 | Immediate ear damage |
Jet aeroplane, near 25 m | 150 | Rupture of eardrum |
Auto horn, one metre; Aircraft take off, 60 m | 110 | Strongly painful |
Diesel train, 30 m; Average factory | 80 | |
Highway traffic, 8 m | 70 | Uncomfortable |
Conversation at a restaurant | 60 | |
Conversation at home | 50 | |
Quiet urban background sound | 40 | |
Quiet rural area | 30 | Virtual silence |
Whispering of leaves, 5 m | 20 | |
Normal breathing | 10 | |
Threshold of hearing | 0 | |

Read Table 8.2 alongside Eq. (8.18) and it becomes far more informative than it first appears. The jet aeroplane at 150 db is not “about twice” as intense as normal conversation at 50 db — it is \(10^{10}\) times as intense, and that is precisely why it damages tissue while conversation does not.

8.9 Doppler Effect

Have you ever heard an approaching train and noticed the distinct change in the pitch of the sound of its whistle when it passes away? While the train comes towards you the whistle sounds shrill; the instant it has gone past, the same whistle sounds noticeably deeper — even though the driver has done nothing at all to change it.

Something similar happens when a listener moves towards or away from a stationary source of sound. What matters, as we shall see, is not who is moving but that there is relative motion at all.

Such a phenomenon was first identified in 1842 by the Austrian physicist Christian Doppler (1803–1853), and is known as the Doppler effect.

Doppler’s claim was not accepted quietly, because in the 1840s there was no easy way to make a sound source travel fast enough to test it. The Dutch scientist Christophorus Buys Ballot found one. In the mid-1840s he hired a locomotive, placed a group of trumpeters on an open railway carriage, and had them hold a steady note as the train rushed past a second group of musicians standing beside the track, whose task was to identify the note they heard. The listeners reported a note that was higher as the train approached and lower as it receded, exactly as Doppler had predicted.

When a source of sound and a listener are in motion relative to each other, the frequency of sound heard by the listener is not the same as the frequency emitted by the source.

NoteDefinition

The Doppler effect is the apparent change in the frequency of sound heard by a listener, due to relative motion between the source of sound and the listener.

It is important to be clear about what is not happening. The source has not changed its frequency; it is emitting exactly as before. The speed of sound in the air has not changed either — that, as Section 8.5 established, is fixed by the medium. What changes is the rate at which the crests of the wave arrive at the listener, as suggested by the crowded wavefronts in Fig. 8.20.

Diagram

Fig. 8.20 – Circular wavefronts emitted by a moving source, crowded together in front of it and spread apart behind it, with a listener shown on each side.

The Doppler effect is a wave phenomenon. It holds for sound waves and also for EM waves. But here we shall consider it for sound waves only.

The change in frequency can be studied under 3 different conditions:

  1. When the listener is stationary but the source is moving.

  2. When the listener is moving but the source is stationary.

  3. When the listener and the source both are moving.

We take these three cases one by one in the subsections that follow. They lead to formulae that look very similar to one another, and the commonest mistake in this whole topic is to reach for the wrong one — so it is worth noticing, as we go, exactly which quantity ends up in the numerator and which in the denominator, and why.

Q. If the source is receding, is the listener actually hearing “less sound”?

A. No — the loudness is a separate matter. What the listener receives is a lower frequency: fewer crests arrive per second than the source is sending out per second, because each successive crest is emitted from a point slightly further away and so has slightly further to travel. The pitch drops; the sound has not become weaker on that account.

8.9.1 Source Moving and Listener Stationary

Consider a source of sound S, moving away from a stationary listener L — a situation called relative recede — with velocity \(\text{v}_s\). The speed of the sound waves with respect to the medium is \(\text{v}\), which is always positive: sound, once emitted, travels through the air at the speed the air dictates, no matter how the source is moving.

Suppose the listener uses a detector for counting each wave crest that reaches it. Counting crests is the whole of this derivation: the frequency the listener records is simply the number of crests his detector registers per second.

Diagram

Fig. 8.21 (a) – Doppler effect detected when the source is moving and the listener is at rest in the medium: source at S₁ and later at S₂, separated by v_s T₀, with the listener L at a distance d from S₁.

Following the Crests

Initially (at \(t = 0\)), the source, which is at point S₁, emits a crest when at a distance \(d\) from the listener; see Fig. 8.21 (a). This crest travels the distance \(d\) at speed \(\text{v}\), and so reaches the listener at time \(t_1 = d/\text{v}\).

Let \(T_0\) be the time period at which the waves are emitted. Thus, at \(t = T_0\), the source moves a distance \(= \text{v}_s T_o\) and reaches the point S₂. The distance of S₂ from the listener is \((d + \text{v}_s T_o)\) — larger than before, because the source has retreated.

When at S₂, the source emits the second crest. This crest has the longer distance \((d + \text{v}_s T_o)\) to cover, and it sets out a time \(T_0\) after the first one. It therefore reaches the listener at

\[t_2 = T_0 + \left[ \frac{d + \text{v}_s T_0}{\text{v}} \right] \qquad \text{--- (8.19)}\]

Similarly, at time \(pT_o\), the source emits its \((p+1)^{th}\) crest (where \(p\) is an integer, \(p\) = 1, 2, 3, …). By then the source has retreated a distance \(p\,\text{v}_s T_0\), so this crest reaches the listener at time

\[t_{p+1} = pT_o + \left[ \frac{d + p\text{v}_s T_0}{\text{v}} \right]\]

From Crest Count to Recorded Period

Hence the listener’s detector counts \(p\) crests in the time interval

\[t_{p+1} - t_1 = pT_o + \left[ \frac{d + p\text{v}_s T_0}{\text{v}} \right] - \frac{d}{\text{v}}\]

Hence the period of the wave as recorded by the listener is that whole interval divided by the number of crests counted in it:

\[T = \frac{(t_{p+1} - t_1)}{p}\]

or

\[T = \frac{\left[ pT_0 + \dfrac{d + p\text{v}_s T_0}{\text{v}} - \dfrac{d}{\text{v}} \right]}{p}\]

The two terms containing \(d\) cancel — the initial distance of the source drops out, which is why the answer does not depend on how far away the source happened to start. What survives is

\[T = T_0 + \frac{\text{v}_s T_0}{\text{v}}\]

\[T = T_0 \left[ 1 + \frac{\text{v}_s}{\text{v}} \right]\]

\[T = T_0 \left[ \frac{\text{v} + \text{v}_s}{\text{v}} \right]\]

Taking the reciprocal of both sides converts periods into frequencies:

\[\therefore \frac{1}{T} = \frac{1}{T_0} \left[ \frac{\text{v}}{\text{v} + \text{v}_s} \right]\]

\[\therefore n = n_0 \left[ \frac{\text{v}}{\text{v} + \text{v}_s} \right] \qquad \text{--- (8.20)}\]

where \(n\) is the frequency recorded by the listener and \(n_o\) is the frequency emitted by the source, both in hertz (Hz), while \(\text{v}\) is the speed of sound in the medium and \(\text{v}_s\) the speed of the source, both in m/s.

NotePrinciple / Law

When the source of sound recedes from a stationary listener with speed \(\text{v}_s\), the frequency recorded by the listener is \(n = n_0 \left[ \dfrac{\text{v}}{\text{v} + \text{v}_s} \right]\), which is lower than the emitted frequency \(n_0\).

Since \((\text{v} + \text{v}_s)\) is greater than \(\text{v}\), the bracket in Eq. (8.20) is less than 1, so \(n < n_0\). A receding source is heard flat. This is exactly what the crest-counting picture predicted: each successive crest is emitted from further away and has further to travel, so the crests arrive spaced out in time.

Source Approaching the Listener

If the source of sound is moving towards the listener with speed \(\text{v}_s\) — called relative approach — the source is now closing the gap rather than opening it, so each successive crest has a shorter distance to travel than the one before. The second term from Eq. (8.19) onwards will therefore be negative (or will be subtracted).

Thus, in this case,

\[n = n_0 \left[ \frac{\text{v}}{\text{v} - \text{v}_s} \right] \qquad \text{--- (8.21)}\]

NotePrinciple / Law

When the source of sound approaches a stationary listener with speed \(\text{v}_s\), the frequency recorded by the listener is \(n = n_0 \left[ \dfrac{\text{v}}{\text{v} - \text{v}_s} \right]\), which is higher than the emitted frequency \(n_0\).

Now the denominator is smaller than \(\text{v}\), the bracket exceeds 1, and \(n > n_0\): an approaching source is heard sharp.

Q. How do I decide whether to write \(\text{v} + \text{v}_s\) or \(\text{v} - \text{v}_s\) in the denominator without memorising two formulae?

A. Decide the answer first, then choose the sign that produces it. If the source is approaching, the pitch must rise, so the fraction must come out greater than 1 — which requires the smaller denominator, \(\text{v} - \text{v}_s\). If the source is receding, the pitch must fall, so the fraction must be less than 1 — which requires the larger denominator, \(\text{v} + \text{v}_s\). Physics first, algebra second.

Q5. Why does the whistle of an approaching train sound shrill, and then suddenly drop to a lower pitch the moment the train passes you?

A. This is the Doppler effect for a moving source. While the train approaches, each successive crest of the whistle’s sound is emitted from a point nearer to you than the last, so the crests reach your ear more frequently than they were emitted, and by Eq. (8.21) you hear a raised frequency, \(n = n_0 [\text{v}/(\text{v} - \text{v}_s)]\). The instant the train passes, the source begins to recede: each crest is now emitted from further away, the crests arrive spaced out, and by Eq. (8.20) you hear a lowered frequency, \(n = n_0 [\text{v}/(\text{v} + \text{v}_s)]\). The driver has changed nothing — the whistle’s frequency \(n_0\) is the same throughout.

An ambulance sounding a siren of frequency 640 Hz travels at 20 m/s along a straight road, while a pedestrian stands still at the roadside. Taking the speed of sound in air as 340 m/s, calculate the frequency heard by the pedestrian (a) while the ambulance approaches him, and (b) after it has passed and is moving away. Also state the difference between the two frequencies.

8.9.2 Listener Approaching a Stationary Source with Velocity v_L

Consider a listener approaching with velocity \(\text{v}_L\) towards a stationary source S, as shown in Fig. 8.21 (b). The source now stays put and emits crests at perfectly regular intervals; it is the listener who is closing the gap, running into the crests before they would otherwise have reached him.

Diagram

Fig. 8.21 (b) – Doppler effect detected when the listener is moving and the source is at rest in the medium: source S at rest, listener moving from L₁ towards the source, with the distances d, v_L t₁ and (d − v_L t₁) marked.

The First Wave

Let the first wave be emitted by the source at \(t = 0\), when the listener was at L₁ at an initial distance \(d\) from the source. Let \(t_1\) be the instant when the listener receives this wave, his position being L₂.

During time \(t_1\), the listener travels a distance \(\text{v}_L t_1\) towards the stationary source. In this same time, the sound wave travels the distance \(\left( d - \text{v}_L t_1 \right)\) with speed \(\text{v}\) — the sound has to cover only what is left of the original gap after the listener has eaten into it.

\[\therefore t_1 = \frac{d - \text{v}_L t_1}{\text{v}}\]

Solving this for \(t_1\) by taking all the \(t_1\) terms to one side gives

\[\therefore t_1 = \frac{d}{\text{v} + \text{v}_L}\]

The combination \((\text{v} + \text{v}_L)\) appearing here has a clear meaning: the gap between the wave and the listener is closing at the combined speed of the two, since they are approaching each other head-on.

The Second Wave, and the Rest

The second wave is emitted by the source at \(t = T_0\), the time period of the waves emitted by the source. Let \(t_2\) be the instant when the listener receives this second wave.

During time \(t_2\), the distance travelled by the listener is \(\text{v}_L t_2\). Thus, the distance to be travelled by the sound to reach the listener is then \(d - \text{v}_L t_2\).

\(\therefore\) The sound (the second wave) travels this distance with speed \(\text{v}\) in time \(= \dfrac{d - \text{v}_L t_2}{\text{v}}\)

However, this time should be counted after \(T_0\), as the second wave was emitted at \(t = T_0\) and not at \(t = 0\).

\[\therefore t_2 = T_0 + \frac{d - \text{v}_L t_2}{\text{v}} \qquad \therefore t_2 = \frac{\text{v}T_0 + d}{\text{v} + \text{v}_L}\]

Similarly, for the third wave, emitted at \(t = 2T_0\),

\[t_3 = 2T_0 + \frac{d - \text{v}_L t_3}{\text{v}} \qquad \therefore t_3 = \frac{2\text{v}T_0 + d}{\text{v} + \text{v}_L}\]

Extending this argument to the \((p+1)^{th}\) wave, we can write,

\[t_{p+1} = pT_0 + \frac{d - \text{v}_L t_{p+1}}{\text{v}} \qquad \therefore t_{p+1} = \frac{p\text{v}T_0 + d}{\text{v} + \text{v}_L}\]

The Recorded Period and Frequency

The time duration between the instances of receiving successive waves is the observed, or recorded, period \(T\). Over \(p\) intervals, that duration is \(pT\), so

\[\therefore pT = t_{p+1} - t_1 = \frac{p\text{v}T_0 + d}{\text{v} + \text{v}_L} - \frac{d}{\text{v} + \text{v}_L} = \frac{p\text{v}T_0}{\text{v} + \text{v}_L}\]

Once again the initial distance \(d\) cancels out, and dividing through by \(p\) gives

\[\therefore T = T_0 \left( \frac{\text{v}}{\text{v} + \text{v}_L} \right) \qquad \text{--- (8.22)}\]

\[\therefore \frac{1}{T} = \frac{1}{T_0} \left( \frac{\text{v} + \text{v}_L}{\text{v}} \right)\]

\[\therefore n = n_0 \left( \frac{\text{v} + \text{v}_L}{\text{v}} \right) \qquad \text{--- (8.23)}\]

where \(n\) is the frequency recorded by the moving listener and \(n_0\) the frequency emitted by the stationary source, both in hertz (Hz), \(\text{v}\) is the speed of sound and \(\text{v}_L\) the speed of the listener, both in m/s.

NotePrinciple / Law

When a listener approaches a stationary source with speed \(\text{v}_L\), the frequency he records is \(n = n_0 \left( \dfrac{\text{v} + \text{v}_L}{\text{v}} \right)\), which is higher than the emitted frequency \(n_0\). If the listener recedes from the stationary source, \(\text{v}_L\) is taken as negative and the recorded frequency is lower.

Since \((\text{v} + \text{v}_L) > \text{v}\), the bracket exceeds 1 and \(n > n_0\): a listener running towards a source hears it sharp, just as he does when the source runs towards him.

Q. In Section 8.9.1 the speed of the source appeared in the denominator; here the speed of the listener appears in the numerator. Why the difference, when both cases raise the pitch?

A. Because the two motions interfere with the wave in different ways. A moving source alters the spacing of the crests as they are laid down in the medium — it changes the wavelength, and wavelength sits in the denominator of \(n = \text{v}/\lambda\). A moving listener does not alter the crests at all; they are already in the air, evenly spaced. He simply sweeps through them faster, which changes the effective speed of approach from \(\text{v}\) to \(\text{v} + \text{v}_L\) — and speed sits in the numerator. The two effects raise the pitch by slightly different amounts, which is why the formulae are not identical.

A factory siren of frequency 500 Hz sounds continuously. A cyclist rides straight towards the siren at 10 m/s. Taking the speed of sound in air as 340 m/s, calculate the frequency the cyclist hears while approaching, and the frequency she hears after passing the siren and riding away from it at the same speed.

8.9.3 Both Source and Listener are Moving

In general, when both the source and the listener are in motion, we can write the observed frequency as

\[n = n_0 \left[ \frac{\text{v} \pm \text{v}_L}{\text{v} \mp \text{v}_s} \right] \qquad \text{--- (8.24)}\]

where \(n\) is the recorded frequency and \(n_0\) the emitted frequency in hertz (Hz), \(\text{v}\) is the speed of sound in the medium, \(\text{v}_L\) the speed of the listener and \(\text{v}_s\) the speed of the source, all in m/s.

Equation (8.24) is not a new result. It is Eq. (8.21) and Eq. (8.23) written as a single expression: the listener’s motion sits in the numerator, exactly as it did in Section 8.9.2, and the source’s motion sits in the denominator, exactly as it did in Section 8.9.1.

Here the upper signs (in both the numerator and the denominator) should be chosen during relative approach, while the lower signs should be chosen during relative recede.

It must be remembered that ‘when you are deciding the sign for any one of these, the other should be considered to be at rest’. This single sentence is the key to the whole formula, and it is worth reading twice. The listener’s sign is decided by asking whether he is moving towards or away from the source, pretending for the moment that the source is standing still — and the source’s sign is decided the same way, pretending the listener is standing still.

NotePrinciple / Law

When both source and listener are in motion, the recorded frequency is \(n = n_0 \left[ \dfrac{\text{v} \pm \text{v}_L}{\text{v} \mp \text{v}_s} \right]\), where the upper signs are used for relative approach and the lower signs for relative recede, the sign for each of the two being decided by treating the other as being at rest.

Illustration

Consider an observer or listener and a source moving with respective velocities \(\text{v}_L\) and \(\text{v}_S\) along the same direction — one chasing the other, as shown in Fig. 8.22.

Diagram

Fig. 8.22 – A listener and a source moving along the same straight line in the same direction, the listener behind with velocity v_L and the source ahead with velocity v_S, both arrows pointing the same way.

In this case, the listener is approaching the source with \(\text{v}_L\) (irrespective of whether the source is moving or not) — treat the source as at rest, and the listener is clearly closing in on it. Thus, the upper, i.e., positive, sign should be chosen for the numerator.

However, the source is moving with \(\text{v}_S\) away from the listener, irrespective of the listener’s motion — treat the listener as at rest, and the source is clearly running away. Thus the lower sign in the denominator, which is positive, has to be chosen.

\[\therefore n = n_o \left[ \frac{\text{v} + \text{v}_L}{\text{v} + \text{v}_s} \right] \qquad \text{--- (8.25)}\]

Notice the result: the numerator carries a plus sign and so does the denominator, which is not what a careless glance at “upper signs for approach” would have produced. The two signs are chosen independently, and here they happen to differ in their effect.

Three Cases Worth Knowing

Case (I) If \(|\text{v}_L| = |\text{v}_s|\), then \(n = n_o\). Thus there is no Doppler shift, as there is no relative motion — even if both are moving. The listener chases the source at exactly the speed at which it flees, the gap between them never changes, and the pitch is unaltered.

Case (II) If \(|\text{v}_L| > |\text{v}_s|\), the numerator will be greater than the denominator, so \(n > n_o\). This is because there is a relative approach: the listener approaches the source faster than the source is receding from him, so the gap is shrinking.

Case (III) If \(|\text{v}_L| < |\text{v}_s|\), then \(n < n_o\), as now there is relative recede — the source recedes faster than the listener approaches, so the gap is growing.

Q. Case (I) says a chase at equal speeds produces no Doppler shift at all. But both are moving through the air — surely the air must notice?

A. The air notices, but it makes no difference to the pitch. The moving source lays down crests that are bunched closer together, which by itself would raise the pitch; the listener, fleeing at the same speed, sweeps through them more slowly, which by itself would lower it. In this particular case the two effects cancel exactly, and the listener hears the source’s true frequency \(n_0\).

A train sounding a whistle of frequency 600 Hz moves along a straight track at 30 m/s. A car travels along a road parallel to the track, moving directly towards the train at 20 m/s. Taking the speed of sound in air as 340 m/s, calculate the frequency of the whistle as heard by the driver of the car while the two are approaching each other.

8.9.4 Common Properties between Doppler Effect of Sound and Light

The Doppler effect is a property of waves in general, not of sound in particular. Light, being an EM wave, shows it too — and much of what we have derived for sound carries over unchanged. The following properties are common to both.

A) Wherever there is relative motion between the listener (or observer) and the source (of sound or light waves), the recorded frequency is different from the emitted frequency.

B) The recorded frequency is higher (than the emitted frequency), if there is relative approach.

C) The recorded frequency is lower, if there is relative recede.

D) If \(\text{v}_L\) or \(\text{v}_s\) are much smaller than the wave speed (the speed of sound or of light), we can use \(\text{v}_r\) as the relative velocity — that is, when the motions are slow, only the relative velocity of source and observer matters, and it no longer matters which of them is doing the moving. In this case, using Eq. (8.24),

\[\frac{\Delta n}{n} \simeq \frac{\text{v}_r}{\text{v}} \simeq \frac{\Delta \lambda}{\lambda} \qquad \text{--- (8.26)}\]

where \(\Delta n\) is the Doppler shift, or the change in the recorded frequency, i.e., \(|n - n_o|\), and \(\Delta \lambda\) is the recorded change in wavelength. Here \(\text{v}_r\) is the relative velocity of source and observer in m/s and \(\text{v}\) is the speed of the wave in m/s.

\[\therefore \frac{|n - n_0|}{n} \simeq \frac{\text{v}_r}{\text{v}}\]

\[\therefore n = n_0 \left( 1 \pm \frac{\text{v}_r}{\text{v}} \right) \qquad \text{--- (8.27)}\]

Once again, the upper sign is to be used during relative approach, while the lower sign is to be used during relative recede.

Equation (8.26) is worth dwelling on, because it says something remarkably useful: the fractional change in frequency equals the fractional change in wavelength, and both equal the ratio of the relative speed to the wave speed. Measure the shift in a wavelength, and you have measured a speed.

NotePrinciple / Law

For relative speeds much smaller than the wave speed, the fractional Doppler shift is given by \(\dfrac{\Delta n}{n} \simeq \dfrac{\text{v}_r}{\text{v}} \simeq \dfrac{\Delta \lambda}{\lambda}\), where \(\text{v}_r\) is the relative velocity of source and observer, and the frequency recorded is \(n = n_0 \left(1 \pm \dfrac{\text{v}_r}{\text{v}}\right)\), the upper sign for relative approach and the lower for relative recede.

E) If the velocities of the source and the observer (listener) are not along the same line, their respective components along the line joining them should be chosen for the longitudinal Doppler effect, and the same mathematical treatment is applicable, as illustrated in Fig. 8.23.

Diagram

Fig. 8.23 – A source and an observer moving in different directions, with the components of each velocity resolved along the straight line joining them.

The physical reason is straightforward. Only motion along the line joining the two can change the distance between them, and only a changing distance can bunch or stretch the arriving crests. A component of velocity perpendicular to that line changes nothing, and is simply discarded.

Q. A car passes you at high speed on a straight road, sounding its horn. At the exact instant it is level with you, is the pitch you hear raised, lowered, or unchanged?

A. Unchanged. At that instant the car’s velocity is entirely perpendicular to the line joining you to it, so its component along that line is zero. Its distance from you is momentarily neither shrinking nor growing, and you hear the horn’s true frequency — the crossover point between the raised pitch of approach and the lowered pitch of recede.

The Doppler effect for light is how we know that the universe is expanding. Every element emits light at its own set of sharply defined wavelengths, and these can be identified in the light of a distant galaxy. In almost every galaxy examined, the whole set is found shifted towards longer wavelengths — a “redshift” — which by Eq. (8.26) means the galaxy is receding from us, and the size of \(\Delta\lambda/\lambda\) tells us how fast. The further away a galaxy is, the larger its redshift, and it was this pattern, read off photographic plates, that first revealed an expanding universe.

A spectral line emitted by hydrogen has a wavelength of 500 nm in the laboratory. In the light received from a distant star, the same line is observed at 500.05 nm. Taking the speed of light as 3 × 10⁸ m/s, use Eq. (8.26) to calculate the speed of the star relative to the Earth, and state whether the star is approaching us or receding from us.

8.9.5 Major Differences between Doppler Effects of Sound and Light

The similarities listed in the previous subsection should not tempt us into thinking that sound and light behave identically. They do not, and the differences all trace back to one fact: sound needs a medium and light does not.

A) As the speed of light is absolute, only the relative velocity between the observer and the source matters, i.e., who is in motion is not relevant.

B) Classical and relativistic Doppler effects are different in the case of light, while in the case of sound, it is only classical.

C) For obtaining the exact Doppler shift for sound waves, it is absolutely important to know who is in motion.

Points A and C are two sides of the same coin, and they deserve a moment’s thought, because they explain why we needed three separate subsections for sound. Sound travels through air, and the air provides a fixed background against which “moving” and “at rest” have real meanings. A source moving through that air genuinely bunches up the crests it lays down in it; a listener moving through the air does not. That is why Eq. (8.21) and Eq. (8.23) are not the same formula, and why a source approaching at 30 m/s and a listener approaching at 30 m/s do not give quite the same recorded frequency. Light has no medium and hence no such background — only the relative velocity survives, and one formula suffices.

A traffic police speed gun is a Doppler instrument working with EM waves. It sends out a radio wave, the moving vehicle reflects it, and the shift in frequency of the returned wave gives the vehicle’s speed through Eq. (8.26). Because light and radio waves are involved, the reading does not depend at all on whether the officer’s vehicle is parked or itself moving in traffic — only the relative velocity enters. The same measurement made with sound would require the officer to know exactly who was moving, and how fast, with respect to the air.

D) If wind is present, its velocity alters the speed of sound and hence affects the Doppler shift. In this case, the component of the wind velocity (\(\text{v}_w\)) is chosen along the line joining the source and the observer. This is to be algebraically added with the velocity of sound. Hence ‘\(\text{v}\)’ is to be replaced by (\(\text{v} \pm \text{v}_w\)) in all the above expressions.

The positive sign is to be used if \(\text{v}\) and \(\text{v}_w\) are along the same direction (remember that \(\text{v}\) is always positive and always from source to listener). The negative sign is to be used if \(\text{v}\) and \(\text{v}_w\) are oppositely directed. Wind is simply the medium itself in motion, so it carries the sound along with it — which is why it changes \(\text{v}\) and not the velocity of either the source or the listener.

Q. Does a following wind change the pitch of a sound, then?

A. On its own, no. A wind blowing from source to listener raises the effective speed of sound to \((\text{v} + \text{v}_w)\), but if neither source nor listener is moving there is still no Doppler shift — the sound merely arrives sooner. The wind matters only when a Doppler shift is already present: it changes the value of \(\text{v}\) to be used in Eqs. (8.20) to (8.25), and so alters the size of the shift.

A rocket is moving at a speed of 220 m/s towards a stationary target. It emits a wave of frequency 1200 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of sound detected by the target and (2) the frequency of the echo detected by the rocket. (Velocity of sound = 330 m/s.)

Solution: This problem must be done in two stages, because the roles of source and listener swap over halfway through. First the rocket is the source and the target the listener; then the target becomes the source (of the reflected sound) and the rocket becomes the listener.

Given, the target is stationary, i.e.,

\(\text{v}_L\) = 0, \(\text{v}_s\) = 220 m/s, \(\text{v}\) = 330 m/s

\(n_0\) = 1200 Hz

To find the frequency of sound detected by the target, we have to use Eq. (8.25). The source is approaching, so the lower sign in the denominator gives \((\text{v} - \text{v}_s)\).

\[n = n_0 \left[ \frac{\text{v}}{\text{v} - \text{v}_s} \right]\]

\[n = 1200 \left[ \frac{330}{330 - 220} \right]\]

\[n = 3600 \ \text{Hz}\]

The frequency of sound detected by the target = 3600 Hz.

When the echo is heard by the rocket’s detector, the target is considered as the source.

\(\therefore \text{v}_s = 0\)

The frequency of sound emitted by the source (i.e. the target) is \(n_0\) = 3600 Hz — the target reflects back exactly the frequency it received — and the frequency detected by the rocket is \(n'\). Now the listener is approaching the source, and so we have to use

\[n' = n \left[ \frac{\text{v} + \text{v}_L}{\text{v}} \right]\]

\[n = 3600 \left[ \frac{330 + 220}{330} \right]\]

\[n = 6000 \ \text{Hz}\]

The frequency of the echo detected by the rocket = 6000 Hz.

A bat, flying at velocity \(V_B\) = 12.5 m/s, is followed by a car running at velocity \(V_C\) = 50 m/s. The actual directions of the velocities of the car and the bat are as shown in Fig. 8.24, both being in the same horizontal plane (the plane of the figure). To detect the car, the bat radiates ultrasonic waves of frequency 36 kHz. The speed of sound at the surrounding temperature is 350 m/s.

There is an ultrasonic frequency detector fitted in the car. Calculate the frequency recorded by this detector.

The ultrasonic waves radiated by the bat are reflected by the car. The bat detects these waves and, from the detected frequency, it knows about the speed of the car. Calculate the frequency of the reflected waves as detected by the bat. (sin 37° = cos 53° ≈ 0.6, sin 53° = cos 37° ≈ 0.8)

Diagram

Fig. 8.24 – A bat moving with velocity V_B = 12.5 m/s at 37° to the line joining it to the car, and a car moving with velocity V_C = 50 m/s at 53° to that same line, both in one horizontal plane, with the components along the joining line marked as V_C cos 53° = 30 m/s and V_B cos 37° = 10 m/s.

Solution: Neither velocity lies along the line joining the bat and the car, so by property (E) of Section 8.9.4 we must first resolve each velocity along that line, and use only those components.

As shown in the figure, the components of the velocities of the bat and the car, along the line joining them, are

\[V_C \cos 53^0 \cong 50 \times 0.6 = 30 \ \text{m s}^{-1} \ \text{and}\]

\[V_B \cos 37^0 \cong 12.5 \times 0.8 = 10 \ \text{m s}^{-1}\]

These should be used while calculating the Doppler shifted frequencies.

Doppler shifted frequency, \(n = n_0 \left( \dfrac{\text{v} \pm \text{v}_L}{\text{v} \mp \text{v}_s} \right)\); the upper signs are to be used during approach, the lower signs during recede.

Part I: Frequency radiated by the bat \(n_0\) = 36 × 10³ Hz. Frequency detected by the detector in the car = \(n\) = ?

In this case, the bat is the source, which is moving away from the car (receding), while the detector in the car is the listener, who is approaching the source (the bat).

\[\text{v}_s = V_B \cos 37^0 = 10 \ \text{m/s} \quad \text{and} \quad \text{v}_L = V_C \cos 53^0 = 30 \ \text{m/s}\]

The source (the bat) is receding, while the listener (the car) is approaching

\[\therefore n = n_0 \left( \frac{\text{v} + \text{v}_L}{\text{v} + \text{v}_s} \right)\]

\[\therefore n = 36 \times 10^3 \left( \frac{350 + 30}{350 + 10} \right)\]

\[= 38 \times 10^3 \ \text{Hz} = 38 \ \text{kHz}\]

Part II: Reflected frequency, as detected by the bat. The frequency reflected by the car is the Doppler shifted frequency as detected at the car. Thus, this time, the car is the source, with emitted frequency

\[n_0 = 38 \times 10^3 \ \text{Hz}, \ n = ?\]

The car, the source, is approaching the listener (the bat).

Thus, \(\text{v}_S = \text{v}_C \cos 53^0 = 30\) m/s

Thus, \(\text{v}_L = \text{v}_B \cos 37^0 = 10\) m/s

Now the bat — the listener — is receding, while the car — the source — is approaching

\[\therefore n = n_0 \left( \frac{\text{v} - \text{v}_L}{\text{v} - \text{v}_s} \right)\]

\[\therefore n = 38 \times 10^3 \left( \frac{350 - 10}{350 - 30} \right)\]

\[= 38 \times 10^3 \times \frac{34}{32}\]

\[= 40.375 \times 10^3 \ \text{Hz}\]

\[= 40.375 \ \text{kHz}\]

Notice how the signs in Part II are the reverse of those in Part I, even though nothing physical has changed direction. That is entirely because the roles have swapped: the same bat that was the receding source is now the receding listener, and its speed component has moved from the denominator to the numerator. This is exactly the situation the rule in Section 8.9.3 was written for — decide each sign separately, treating the other party as at rest.

Internet my friend: https://hyperphysics.phys-astr.gsu.edu/hbase/hframe.html

8.10 Summary

Waves and Their Types
  1. A wave is a periodic and regular disturbance in a medium which causes no flow of material, but does cause a flow of energy and momentum from one point to another.
  2. Mechanical waves require a material medium (water waves, waves on a string, seismic waves, sound waves). EM waves are produced by periodic vibrations of electric and magnetic fields and do not require a material medium. Matter waves are associated with any moving object and are studied in quantum mechanics.
  3. In a travelling (progressive) wave, a disturbance created at one place travels to distant points and keeps travelling unless stopped by an external agency. Stationary waves, in which the disturbance does not advance, are studied in Class XII.
Common Properties of All Waves
  1. Amplitude (A): the largest displacement of a particle of the medium from its rest position. SI unit: metre (m).
  2. Wavelength (λ): the distance between two successive particles in the same state of vibration, i.e., two successive particles differing in phase by \(2\pi^c\). SI unit: metre (m).
  3. Period (T): the time required by a particle to complete one vibration. SI unit: second (s).
  4. Frequency (n): the number of vibrations performed by a particle per second; \(n = \dfrac{1}{T}\). SI unit: hertz (Hz).
  5. Double periodicity: wave motion is periodic in time (repeating after every T) and periodic in space (repeating after every λ).
  6. Velocity: \(\text{v} = \dfrac{\lambda}{T}\) — (8.1), and since \(\dfrac{1}{T} = n\) — (8.2), \(\text{v} = n\lambda\) — (8.3).
  7. In a given medium the magnitude of the wave velocity is constant, so a higher frequency means a shorter wavelength. When a wave passes from one medium to another, its frequency does not change (it is decided by the source); its speed and wavelength both change.
Requirements of the Medium
  1. The medium must be continuous and elastic, so that it regains its original state after the deforming force is removed.
  2. The medium must possess inertia, so that it can store energy and transfer it as a wave.
  3. The frictional resistance of the medium must be negligible, so that the oscillations are not damped.
Phase
  1. The phase of a particle is its state of oscillation, described by (a) its displacement, (b) the direction of its velocity, and (c) its oscillation number.
  2. Particles in the same state of oscillation are in phase; two successive such particles are separated by one wavelength and differ in phase by \(2\pi^c\).
Transverse and Longitudinal Waves
  1. Transverse wave: particles vibrate perpendicular to the direction of propagation. The medium shows crests (positive displacement) and troughs (negative displacement). Pressure and density do not change; the shape changes periodically. Requires elasticity of shape, so it travels through solids and on the surfaces of liquids only. It can be polarised — vibrations confined to a single plane. EM waves are transverse.
  2. Longitudinal wave: particles vibrate parallel to the direction of propagation. The medium shows compressions (crowded particles, high pressure) and rarefactions (widely separated particles, low pressure). Pressure and density vary periodically with position and time. Requires elasticity of volume, so it travels through solids, liquids and gases, but not through vacuum. It cannot be polarised. Sound waves are longitudinal.
Mathematical Expression of a Wave
  1. A sinusoidal progressive wave travelling along the positive x-axis: \[y(x,t) = a \sin (kx - \omega t + \phi) \qquad \text{--- (8.4)}\]
  2. A wave travelling along the negative x-axis: \[y(x,t) = a \sin (kx + \omega t + \phi) \qquad \text{--- (8.5)}\]
  3. Here \(y(x,t)\) is the displacement, \(a\) the amplitude, \(\omega\) the angular frequency, \(k\) the angular wave number, and \((kx_0 - \omega t + \phi)\) is the argument of the sine — the phase of the particle at \(x\) at time \(t\).
  4. The particles of the medium execute simple harmonic motion about their equilibrium positions.
Speed of Travelling Waves
  1. The speed of a mechanical wave depends on the elastic properties and the density of the medium.
  2. Transverse wave on a stretched string: \[\text{v} = \sqrt{\frac{T}{m}} \qquad \text{--- (8.6)}\] where \(T\) is the tension (N) and \(m = M/L\) is the linear mass density (kg/m). The speed does not depend on the wavelength or frequency of the wave.
  3. Longitudinal wave (Newton’s formula): \[\text{v} = \sqrt{\frac{E}{\rho}} \qquad \text{--- (8.7)}\] where \(E\) is the proper modulus of elasticity and \(\rho\) the density of the medium.
  4. Newton assumed the process to be isothermal, giving \[\text{v} = \sqrt{\frac{P}{\rho}} \qquad \text{--- (8.8)}\] At NTP, with \(h\) = 0.76 m of Hg, \(d\) = 13600 kg/m³, \(\rho\) = 1.293 kg/m³ and \(g\) = 9.8 m/s², this gives v = 279.9 m/s, whereas the experimental value at 0°C is 332 m/s — 16% higher.
Laplace’s Correction
  1. The compressions and rarefactions are produced rapidly; the heat produced in a compression has no time to dissipate. The process is therefore adiabatic, not isothermal.
  2. Adiabatic modulus of elasticity: \(E = \gamma P\) — (8.9), where \(\gamma = \dfrac{C_p}{C_v}\) — (8.10); for air \(\gamma\) = 1.41.
  3. Corrected formula: \[\text{v} = \sqrt{\frac{\gamma P}{\rho}} \qquad \text{--- (8.11)}\] which gives v = 332.3 m/s at NTP, in close agreement with experiment.
  4. In an isothermal process the temperature remains constant; in an adiabatic process there is neither transfer of heat nor of mass. Also, \(C_p > C_v\).
Factors Affecting the Speed of Sound
  1. Pressure: \(\text{v} = \sqrt{\dfrac{\gamma P V}{M}}\) — (8.12). At constant temperature \(PV\) = constant (Boyle’s law), so v = constant. Using \(PV = nRT\), \(\text{v} = \sqrt{\dfrac{\gamma n R T}{M}}\) — (8.13). A change in pressure has no effect on the speed of sound unless the temperature changes.
  2. Temperature: \(\dfrac{\text{v}}{\text{v}_0} = \sqrt{\dfrac{T}{T_0}}\) — (8.14): the speed is directly proportional to the square root of the absolute temperature. For small changes, \(\text{v} \simeq \text{v}_0 + (0.61) t\) — (8.15), where \(t\) is in °C and \(\text{v}_0\) = 332 m/s at 0°C — a rise of 0.61 m/s per °C, valid for temperature variations below about 50°C.
  3. Humidity: \(\dfrac{\text{v}_m}{\text{v}_d} = \sqrt{\dfrac{\rho_d}{\rho_m}}\) — (8.16). Moist air is less dense than dry air (\(\rho_m\) = 0.81 kg/m³ and \(\rho_d\) = 1.29 kg/m³ at 0°C), so \(\text{v}_m > \text{v}_d\): sound travels faster in moist air.
Principle of Superposition
  1. When two or more waves arrive simultaneously at a point of a medium, each produces its own displacement independently of the others, and the resultant displacement is the vector sum of the individual displacements.
Echo, Reverberation and Acoustics
  1. Sound obeys the same laws of reflection as light.
  2. Echo: the repetition of the original sound due to reflection from a rigid surface at a distance. The brain retains a sound for 0.1 s; at 22°C (v = 344 m/s) the sound covers 344 × 0.1 = 34.4 m in that time, so the reflecting surface must be at least 17.2 m away for a distinct echo. This minimum distance changes with temperature.
  3. Reverberation: the persistence of sound after the source has stopped, caused by repeated reflections from walls and ceilings nearer than 15 m. It is reduced by rough surfaces, curtains, porous cardboard sheets, perforated acoustic tiles and gypsum boards.
  4. Acoustics is the branch of physics dealing with the production, transmission and reception of sound. Good auditorium design requires uniform loudness (a parabolic surface behind the speaker, at whose focus the speaker stands), reduction of echoes and reverberation, avoidance of focusing and of the echelon effect, sound-proofing, and silent fittings.
  5. Applications: bats (ultrasonic pulses of 30 kHz to 150 kHz) and dolphins (about 100 Hz) for navigation; shock waves to break kidney stones; ultrasonic imaging; ultrasound therapy; SONAR (Sound Navigational Ranging); noise control; and the study of the Earth’s interior.
Qualities of Sound
  1. Pitch: the subjective sharpness or shrillness of a sound; it increases with frequency. A tone is a single frequency; a note may contain one or more tones. A high pitch sound need not be louder.
  2. Timbre (quality): allows us to distinguish two sounds of the same pitch and loudness from different sources; it depends on the mixture of tones and overtones.
  3. Loudness: the human perception of intensity. Intensity \(I \propto A^2\), measured in W/m². The response of the ear is approximately logarithmic, so a sound of double intensity is louder but not doubly loud.
  4. Least audible intensity \(I_0 = 10^{-12}\) W/m². Loudness in bel: \[L_{bel} = log_{10} \left( \frac{I}{I_0} \right) \qquad \text{--- (8.17)}\]
  5. 1 bel = 10 db, so \(L_{db} = 10 \ log_{10} \left( \dfrac{I}{I_0} \right)\). For \(I = I_0\): \[L_{db} = 10 \ log_{10} (1) = 0 \qquad \text{--- (8.18)}\] which is the threshold of hearing. A 10 db sound has \(I = 10 I_0\); a 20 db sound has \(I = 100 I_0\); each step of 10 db multiplies the intensity by 10 while roughly doubling the perceived loudness.
  6. For a point source, \(I \propto \dfrac{1}{r^2}\).
Doppler Effect
  1. The Doppler effect is the apparent change in the frequency of sound due to relative motion between the source and the listener. It was identified in 1842 by Christian Doppler (1803–1853). It applies to sound waves and to EM waves.
  2. Source moving, listener stationary: \[n = n_0 \left[ \frac{\text{v}}{\text{v} + \text{v}_s} \right] \ \text{(recede)} \qquad \text{--- (8.20)} \qquad n = n_0 \left[ \frac{\text{v}}{\text{v} - \text{v}_s} \right] \ \text{(approach)} \qquad \text{--- (8.21)}\]
  3. Listener moving, source stationary: \[T = T_0 \left( \frac{\text{v}}{\text{v} + \text{v}_L} \right) \qquad \text{--- (8.22)} \qquad n = n_0 \left( \frac{\text{v} + \text{v}_L}{\text{v}} \right) \qquad \text{--- (8.23)}\]
  4. Both moving: \[n = n_0 \left[ \frac{\text{v} \pm \text{v}_L}{\text{v} \mp \text{v}_s} \right] \qquad \text{--- (8.24)}\] Upper signs for relative approach, lower signs for relative recede; while deciding the sign for one of them, the other is considered to be at rest. For a listener and a source moving along the same direction, \(n = n_o \left[ \dfrac{\text{v} + \text{v}_L}{\text{v} + \text{v}_s} \right]\) — (8.25).
  5. If \(|\text{v}_L| = |\text{v}_s|\) there is no Doppler shift; if \(|\text{v}_L| > |\text{v}_s|\) then \(n > n_0\); if \(|\text{v}_L| < |\text{v}_s|\) then \(n < n_0\).
  6. For speeds much smaller than the wave speed, with \(\text{v}_r\) the relative velocity: \[\frac{\Delta n}{n} \simeq \frac{\text{v}_r}{\text{v}} \simeq \frac{\Delta \lambda}{\lambda} \qquad \text{--- (8.26)} \qquad \therefore n = n_0 \left( 1 \pm \frac{\text{v}_r}{\text{v}} \right) \qquad \text{--- (8.27)}\]
  7. If the velocities are not along the line joining source and listener, only their components along that line are used.
  8. Sound versus light: for light, only the relative velocity matters and who is moving is irrelevant; for sound, it is essential to know who is in motion. Classical and relativistic Doppler effects differ for light; for sound only the classical effect applies. Wind changes the speed of sound, so \(\text{v}\) is replaced by \((\text{v} \pm \text{v}_w)\) — positive when the wind is along the direction of \(\text{v}\) (always taken from source to listener), negative when opposite.

8.11 Check Your Concepts

  1. A leaf floating on a pond bobs up and down as ripples pass beneath it, but it does not drift outward with them. What, then, is actually travelling outward across the pond, and what is not?

  2. A sound wave passes from air into water. State which of its three quantities — frequency, speed and wavelength — change, and which does not, and give the reason in each case.

  3. Two particles on a wave have equal displacements at a given instant, yet they are said to be in different states of oscillation. Explain how this is possible, and what additional information is needed to describe the phase of a particle completely.

  4. Why is a medium’s inertia as necessary for wave propagation as its elasticity? Explain what would happen if a medium possessed elasticity but no inertia.

  5. Transverse waves can travel through a steel rod but not through the interior of a lake, while longitudinal waves travel through both. Explain the physical property of the medium that is responsible for this difference.

  6. A student writes the equation of a wave travelling along the positive x-axis as \(y = a \sin (kx + \omega t)\). Identify the mistake and justify the correct form using the idea that the phase of a chosen crest must remain constant.

  7. Vulcanised rubber is a solid, yet the speed of sound in it (54 m/s) is lower than the speed of sound in air (343 m/s at 20°C). Explain this apparent contradiction using the relation \(\text{v} = \sqrt{E/\rho}\).

  8. Newton’s formula for the speed of sound in air is mathematically correct, yet it gives a value 16% below the experimental one. What exactly was wrong with Newton’s assumption, and how does Laplace’s correction repair it without altering the form of the formula?

  9. “A change in atmospheric pressure has no effect on the speed of sound in air.” Explain why this is true for an ideal gas, even though pressure appears explicitly in the formula \(\text{v} = \sqrt{\gamma P/\rho}\).

  10. Why does the speed of sound in air increase on a humid day, even though water is far denser than air?

  11. Two pulses of equal size and opposite displacement meet on a stretched string and momentarily cancel each other completely. Has the energy of the pulses been destroyed at that instant? Justify your answer.

  12. Why can you hear a distinct echo from a distant cliff but not from the wall of your classroom, even though both surfaces reflect sound equally well?

  13. Distinguish between echo and reverberation. State one situation in which reverberation is deliberately reduced, and explain how it is done.

  14. In an auditorium, curved wall surfaces are avoided, but a parabolic surface is placed behind the speaker. Explain why one kind of curved surface is a problem and the other is a solution.

  15. A bat flying in complete darkness can locate a moth. Explain, using the ideas of this chapter, how the time delay of an echo and the Doppler shift of that echo together give the bat both the distance and the speed of its target.

  16. Two sounds of the same frequency and the same loudness are produced, one by a violin and one by a harmonium, and a listener has no difficulty telling them apart. Name the quality of sound involved and explain what physically differs between the two waves.

  17. A sound of 60 db is not “six times” as intense as a sound of 10 db. Explain what the decibel scale actually measures, and why the ear is served well by a logarithmic scale.

  18. Why does listening to music through earphones at high volume for long periods pose a greater risk to hearing than listening to the same music through a loudspeaker across the room?

  19. A source and a listener move in the same direction along a straight line with exactly equal speeds. Show, using Eq. (8.24), that the listener records no change in frequency, and explain this result physically.

  20. A car sounding its horn passes a stationary observer at high speed. At the exact instant the car is level with the observer, is the frequency heard higher than, lower than, or equal to the emitted frequency? Justify your answer.

  21. For light, it does not matter who is moving — only the relative velocity matters. For sound, it matters a great deal. Explain the physical reason for this difference.

  22. A steady wind blows from a stationary source of sound towards a stationary listener. Does the listener hear a Doppler-shifted frequency? Explain your answer, and state how wind must be accounted for when a Doppler shift is present.

  23. Wave motion is doubly periodic. Explain.

  24. What is Doppler effect?

  25. Describe a transverse wave.

  26. Define a longitudinal wave.

  27. What do you mean by an echo?

  28. State any two applications of acoustics.

  29. Define amplitude and wavelength of a wave.

  30. Draw a wave and indicate points which are (i) in phase (ii) out of phase (iii) have a phase difference of π/2.

  31. Define the relation between velocity, wavelength and frequency of wave.

  32. State and explain the principle of superposition of waves.

  33. State the expression for apparent frequency when source of sound and listener are (i) moving towards each other (ii) moving away from each other.

  34. State the expression for apparent frequency when source is stationary and listener is (1) moving towards the source (2) moving away from the source.

  35. State the expression for apparent frequency when listener is stationary and source is (i) moving towards the listener (ii) moving away from the listener.

  36. Explain what is meant by phase of a wave.

  37. Define progressive wave. State any four properties.

  38. Distinguish between transverse waves and longitudinal waves.

  39. Explain Newton’s formula for velocity of sound. What is its limitation?

8.12 Practice with Numericals

  1. A wave of frequency 250 Hz travels through a medium with a speed of 500 m/s. Calculate its wavelength and its period.

  2. The speed of sound in air is 340 m/s and in steel it is 5941 m/s. A source of frequency 850 Hz produces sound in each of the two media. Calculate the wavelength of the sound in air and in steel.

  3. A progressive wave is given by \(y(x,t) = 0.05 \sin (5\pi x - 1700\pi t)\), where \(y\) and \(x\) are in metre and \(t\) is in second. Find the amplitude, wavelength, frequency, period and speed of the wave, and state the direction in which it travels.

  4. Two particles on a progressive wave of wavelength 0.8 m are separated by 0.2 m along the direction of propagation. Calculate the phase difference between them.

  5. A wire of length 4 m has a mass of 40 g and is stretched under a tension of 250 N. Calculate the linear mass density of the wire and the speed of a transverse wave along it.

  6. A transverse wave travels along a stretched string at 120 m/s. By what factor must the tension in the string be changed, keeping the string the same, so that the wave speed becomes 240 m/s?

  7. The density of a gas at NTP is 1.25 kg/m³ and its adiabatic ratio is \(\gamma\) = 1.40. Taking the atmospheric pressure as 1.01 × 10⁵ N/m², calculate the speed of sound in the gas.

  8. The speed of sound in air at 0°C is 332 m/s. Calculate the speed of sound in air (a) at 27°C and (b) at 35°C, using the linear approximation valid for small temperature changes.

  9. At what temperature (in °C) will the speed of sound in air be 1.2 times its value at 0°C? Use \(\text{v} \propto \sqrt{T}\) and take the speed at 0°C as 332 m/s.

  10. The density of dry air at 0°C is 1.29 kg/m³ and that of moist air at the same temperature is 0.81 kg/m³. Calculate the ratio of the speed of sound in moist air to that in dry air at 0°C, the pressure being the same in both cases.

  11. A boy standing in front of a tall cliff fires a cracker and hears its echo after 1.2 s. Taking the speed of sound in air as 340 m/s, calculate the distance of the cliff from the boy.

  12. A man stands between two parallel walls and claps once. He hears the first echo after 0.5 s and the second echo after 0.9 s. If the speed of sound in air is 340 m/s, calculate the distance between the two walls.

  13. A SONAR device on a ship sends an ultrasonic pulse vertically downwards and receives the echo from the sea bed after 1.6 s. Taking the speed of sound in seawater as 1522 m/s, calculate the depth of the sea bed below the ship.

  14. The intensity of a sound at a certain point is 10⁻⁶ W/m². Taking \(I_0 = 10^{-12}\) W/m², calculate the loudness of the sound in decibels.

  15. The loudness of a sound at a point is 70 db. Calculate the intensity of the sound in W/m², taking \(I_0 = 10^{-12}\) W/m². If the intensity is then reduced to one-hundredth of this value, what is the new loudness in db?

  16. The intensity of the sound from a small loudspeaker is 4 × 10⁻⁴ W/m² at a distance of 2 m from it. Assuming the loudspeaker behaves as a point source, calculate the intensity of the sound at a distance of 8 m from it.

  17. A motorcycle sounding a horn of frequency 400 Hz moves directly towards a stationary listener at 25 m/s. Taking the speed of sound in air as 340 m/s, calculate the frequency heard by the listener while the motorcycle approaches, and the frequency heard after it has passed and is receding at the same speed.

  18. A stationary siren emits a note of frequency 720 Hz. A scooter rider travels straight towards the siren at 15 m/s. If the speed of sound in air is 340 m/s, find the frequency recorded by the rider.

  19. A train moving at 25 m/s sounds a whistle of frequency 660 Hz. A car on a parallel road moves in the same direction as the train at 15 m/s, ahead of it. Taking the speed of sound as 340 m/s, calculate the frequency of the whistle as heard by the driver of the car.

  20. A ship sends an ultrasonic signal of frequency 40 kHz towards a submarine that is approaching it head-on at 10 m/s, while the ship itself is stationary. Taking the speed of sound in seawater as 1500 m/s, calculate the frequency of the signal received by the submarine.

  21. A bat flying at 8 m/s directly towards a stationary wall emits an ultrasonic note of frequency 45 kHz. Taking the speed of sound in air as 350 m/s, calculate (a) the frequency of the sound striking the wall and (b) the frequency of the reflected sound as detected by the bat.

  22. A spectral line of wavelength 600 nm emitted by a laboratory source is observed at 600.12 nm in the light from a distant galaxy. Taking the speed of light as 3 × 10⁸ m/s, calculate the speed of the galaxy relative to the Earth and state whether it is approaching or receding.

  23. A certain sound wave in air has a speed 340 m/s and wavelength 1.7 m. For this wave, calculate a) the frequency b) the period.

  24. A tuning fork of frequency 170 Hz produces sound waves of wavelength 2 m. Calculate the speed of sound.

  25. An echo-sounder in a fishing boat receives an echo from a shoal of fish 0.45 s after it was sent. If the speed of sound in water is 1500 m/s, how deep is the shoal?

  26. A girl stands 170 m away from a high wall and claps her hands at a steady rate so that each clap coincides with the echo of the one before. a) If she makes 60 claps in 1 minute, what value should be the speed of sound in air? b) Now, she moves to another location and finds that she should now make 45 claps in 1 minute to coincide with successive echoes. Calculate her distance for the new position from the wall.

  27. Sound wave A has period 0.015 s, sound wave B has period 0.025 s. Which sound has greater frequency?

  28. At what temperature will the speed of sound in air be 1.75 times its speed at N.T.P?

  29. A man standing between 2 parallel cliffs fires a gun. He hears two echoes, one after 3 seconds and the other after 5 seconds. The separation between the two cliffs is 1360 m. What is the speed of sound?

  30. If the velocity of sound in air at a given place on two different days of a given week are in the ratio of 1:1.1, assuming the temperatures on the two days to be the same, what quantitative conclusion can you draw about the condition on the two days?

  31. A police car travels towards a stationary observer at a speed of 15 m/s. The siren on the car emits a sound of frequency 250 Hz. Calculate the recorded frequency. The speed of sound is 340 m/s.

  32. The sound emitted from the siren of an ambulance has a frequency of 1500 Hz. The speed of sound is 340 m/s. Calculate the difference in frequencies heard by a stationary observer if the ambulance initially travels towards and then away from the observer at a speed of 30 m/s.

8.13 Objective Questions

  1. A sound carried by air from a sitar to a listener is a wave of the following type.
    1. Longitudinal stationary
    2. Transverse progressive
    3. Transverse stationary
    4. Longitudinal progressive
  2. When sound waves travel from air to water, which of these remains constant?
    1. Velocity
    2. Frequency
    3. Wavelength
    4. All of above
  3. The Laplace’s correction in the expression for velocity of sound given by Newton is needed because sound waves
    1. are longitudinal
    2. propagate isothermally
    3. propagate adiabatically
    4. are of long wavelength
  4. Speed of sound is maximum in
    1. air
    2. water
    3. vacuum
    4. solid
  5. The walls of the hall built for music concerns should
    1. amplify sound
    2. reflect sound
    3. transmit sound
    4. absorb sound