10.7 POLARISATION Consider holding a long string that is held horizontally, the other end of which is assumed to be fixed. If we move the end of the string up and down in a periodic manner, we will generate a wave propagating in the +x direction (Fig. 10.17). Such a wave could be described by the following equation FIGURE 10.17 (a) The curves represent the displacement of a string at t = 0 and at t = Dt, respectively when a sinusoidal wave is propagating in the +x-direction. (b) The curve represents the time variation of the displacement at x = 0 when a sinusoidal wave is propagating in the +x-direction. At x = Dx, the time variation of the displacement will be slightly displaced to the right. y (x,t) = a sin (kx – wt) (10.15) where a and w(= 2pn) represent the amplitude and the angular frequency of the wave, respectively; further, 2 π λ = (10.16) k represents the wavelength associated with the wave. We had discussed propagation of such waves in Chapter 14 of Class XI textbook. Since the displacement (which is along the y direction) is at right angles to the direction of propagation of the wave, we have what is known as a transverse wave. Also, since the displacement is in the y direction, it is often referred to as a y-polarised wave. Since each point on the string moves on a straight line, the wave is also referred to as a linearly polarised 269 Reprint 2025-26 Physics wave. Further, the string always remains confined to the x-y plane and therefore it is also referred to as a plane polarised wave. In a similar manner we can consider the vibration of the string in the x-z plane generating a z-polarised wave whose displacement will be given by z (x,t) = a sin (kx – wt) (10.17) It should be mentioned that the linearly polarised waves [described by Eqs. (10.15) and (10.17)] are all transverse waves; i.e., the displacement of each point of the string is always at right angles to the direction of propagation of the wave. Finally, if the plane of vibration of the string is changed randomly in very short intervals of time, then we have what is known as an unpolarised wave. Thus, for an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation. Light waves are transverse in nature; i.e., the electric field associated with a propagating light wave is always at right angles to the direction of propagation of the wave. This can be easily demonstrated using a simple polaroid. You must have seen thin plastic like sheets, which are called polaroids. A polaroid consists of long chain molecules aligned in a particular direction. The electric vectors (associated with the propagating light wave) along the direction of the aligned molecules get absorbed. Thus, if an unpolarised light wave is incident on such a polaroid then the light wave will get linearly polarised with the electric vector oscillating along a direction perpendicular to the aligned molecules; this direction is known as the pass-axis of the polaroid. Thus, if the light from an ordinary source (like a sodium lamp) passes through a polaroid sheet P1, it is observed that its intensity is reduced by half. Rotating P1 has no effect on the transmitted beam and transmitted intensity remains constant. Now, let an identical piece of polaroid P2 be placed before P1. As expected, the light from the lamp is reduced in intensity on passing through P2 alone. But now rotating P1 has a dramatic effect on the light coming from P2. In one position, the intensity transmitted by P2 followed by P1 is nearly zero. When turned by 90° from this position, P1 transmits nearly the full intensity emerging from P2 (Fig. 10.18). The experiment at figure 10.18 can be easily understood by assuming that light passing through the polaroid P2 gets polarised along the pass- axis of P2. If the pass-axis of P2 makes an angle q with the pass-axis of P1, then when the polarised beam passes through the polaroid P2, the component E cos q (along the pass-axis of P2) will pass through P2. Thus, as we rotate the polaroid P1 (or P2), the intensity will vary as: I = I0 cos2q (10.18) where I0 is the intensity of the polarized light after passing through P1. This is known as Malus’ law. The above discussion shows that the Reprint 2025-26 Wave Optics FIGURE 10.18 (a) Passage of light through two polaroids P2 and P1. The transmitted fraction falls from 1 to 0 as the angle between them varies from 0° to 90°. Notice that the light seen through a single polaroid P1 does not vary with angle. (b) Behaviour of the electric vector when light passes through two polaroids. The transmitted polarisation is the component parallel to the polaroid axis. The double arrows show the oscillations of the electric vector. intensity coming out of a single polaroid is half of the incident intensity. By putting a second polaroid, the intensity can be further controlled from 50% to zero of the incident intensity by adjusting the angle between the pass-axes of two polaroids. Polaroids can be used to control the intensity, in sunglasses, windowpanes, etc. Polaroids are also used in photographic cameras and 3D movie cameras. Example 10.2 Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids? Solution Let I0 be the intensity of polarised light after passing through the first polariser P1. Then the intensity of light after passing through second polariser P2 will be I = I 0cos 2θ, where q is the angle between pass axes of P1 and P2. Since P1 and P3 are crossed the angle between the pass axes of P2 and P3 will be (p/2–q). Hence the intensity of light emerging from P3 will be  π  – θ I = I 0 cos 2θ cos 2  2  EXAMPLE = I0 cos2q sin2q =(I0/4) sin22q Therefore, the transmitted intensity will be maximum when q = p/4. 10.2 271 Reprint 2025-26 Physics SUMMARY 1. Huygens’ principle tells us that each point on a wavefront is a source of secondary waves, which add up to give the wavefront at a later time. 2. Huygens’ construction tells us that the new wavefront is the forward envelope of the secondary waves. When the speed of light is independent of direction, the secondary waves are spherical. The rays are then perpendicular to both the wavefronts and the time of travel is the same measured along any ray. This principle leads to the well known laws of reflection and refraction. 3. The principle of superposition of waves applies whenever two or more sources of light illuminate the same point. When we consider the intensity of light due to these sources at the given point, there is an interference term in addition to the sum of the individual intensities. But this term is important only if it has a non-zero average, which occurs only if the sources have the same frequency and a stable phase difference. 4. Young’s double slit of separation d gives equally spaced interference fringes. 5. A single slit of width a gives a diffraction pattern with a central λ 2λ maximum. The intensity falls to zero at angles of ± , ± , etc., a a with successively weaker secondary maxima in between. 6. Natural light, e.g., from the sun is unpolarised. This means the electric vector takes all possible directions in the transverse plane, rapidly and randomly, during a measurement. A polaroid transmits only one component (parallel to a special axis). The resulting light is called linearly polarised or plane polarised. When this kind of light is viewed through a second polaroid whose axis turns through 2p, two maxima and minima of intensity are seen. POINTS TO PONDER 1. Waves from a point source spread out in all directions, while light was seen to travel along narrow rays. It required the insight and experiment of Huygens, Young and Fresnel to understand how a wave theory could explain all aspects of the behaviour of light. 2. The crucial new feature of waves is interference of amplitudes from different sources which can be both constructive and destructive, as shown in Young’s experiment. 3. Diffraction phenomena define the limits of ray optics. The limit of the ability of microscopes and telescopes to distinguish very close objects is set by the wavelength of light. 4. Most interference and diffraction effects exist even for longitudinal waves like sound in air. But polarisation phenomena are special to transverse waves like light waves. Reprint 2025-26 Wave Optics EXERCISES

14.7 Beats the cork pieces move up and down but do not move away Summary from the centre of disturbance. This shows that the water Points to ponder mass does not flow outward with the circles, but rather a Exercises moving disturbance is created. Similarly, when we speak, the sound moves outward from us, without any flow of air from one part of the medium to another. The disturbances produced in air are much less obvious and only our ears or a microphone can detect them. These patterns, which move without the actual physical transfer or flow of matter as a whole, are called waves. In this Chapter, we will study such waves. Waves transport energy and the pattern of disturbance has information that propagate from one point to another. All our communications essentially depend on transmission of sig- nals through waves. Speech means production of sound waves in air and hearing amounts to their detection. Often, communication involves different kinds of waves. For exam- ple, sound waves may be first converted into an electric cur- rent signal which in turn may generate an electromagnetic wave that may be transmitted by an optical cable or via a Reprint 2025-26 WAVES 279 satellite. Detection of the original signal will usu- We shall illustrate this connection through ally involve these steps in reverse order. simple examples. Not all waves require a medium for their Consider a collection of springs connected to propagation. We know that light waves can one another as shown in Fig. 14.1. If the spring travel through vacuum. The light emitted by at one end is pulled suddenly and released, the stars, which are hundreds of light years away, disturbance travels to the other end. What has reaches us through inter-stellar space, which is practically a vacuum. The most familiar type of waves such as waves on a string, water waves, sound waves, seismic waves, etc. is the so-called mechanical waves. Fig. 14.1 A collection of springs connected to each These waves require a medium for propagation, other. The end A is pulled suddenly they cannot propagate through vacuum. They generating a disturbance, which then involve oscillations of constituent particles and propagates to the other end. depend on the elastic properties of the medium. The electromagnetic waves that you will learn happened? The first spring is disturbed from its in Class XII are a different type of wave. equilibrium length. Since the second spring is Electromagnetic waves do not necessarily require connected to the first, it is also stretched or a medium - they can travel through vacuum. compressed, and so on. The disturbance moves Light, radiowaves, X-rays, are all electromagnetic from one end to the other; but each spring only waves. In vacuum, all electromagnetic waves executes small oscillations about its equilibrium have the same speed c, whose value is : position. As a practical example of this situation, consider a stationary train at a railway station. c = 299, 792, 458 ms–1. (14.1) Different bogies of the train are coupled to each A third kind of wave is the so-called Matter other through a spring coupling. When an waves. They are associated with constituents of engine is attached at one end, it gives a push to matter : electrons, protons, neutrons, atoms and the bogie next to it; this push is transmitted from molecules. They arise in quantum mechanical one bogie to another without the entire train description of nature that you will learn in your being bodily displaced. later studies. Though conceptually more abstract Now let us consider the propagation of sound than mechanical or electro-magnetic waves, they waves in air. As the wave passes through air, it have already found applications in several compresses or expands a small region of air. This devices basic to modern technology; matter causes a change in the density of that region, waves associated with electrons are employed say δρ, this change induces a change in pressure, in electron microscopes. δp, in that region. Pressure is force per unit area, In this chapter we will study mechanical so there is a restoring force proportional to waves, which require a material medium for the disturbance, just like in a spring. In this their propagation. case, the quantity similar to extension or The aesthetic influence of waves on art and compression of the spring is the change in literature is seen from very early times; yet the density. If a region is compressed, the molecules first scientific analysis of wave motion dates back in that region are packed together, and they tend to the seventeenth century. Some of the famous to move out to the adjoining region, thereby scientists associated with the physics of wave increasing the density or creating compression motion are Christiaan Huygens (1629-1695), in the adjoining region. Consequently, the air Robert Hooke and Isaac Newton. The in the first region undergoes rarefaction. If a understanding of physics of waves followed the region is comparatively rarefied the surrounding physics of oscillations of masses tied to springs air will rush in making the rarefaction move to and physics of the simple pendulum. Waves in the adjoining region. Thus, the compression or elastic media are intimately connected with rarefaction moves from one region to another, harmonic oscillations. (Stretched strings, coiled making the propagation of a disturbance springs, air, etc., are examples of elastic media). possible in air. Reprint 2025-26 280 PHYSICS In solids, similar arguments can be made. In a crystalline solid, atoms or group of atoms are arranged in a periodic lattice. In these, each atom or group of atoms is in equilibrium, due to forces from the surrounding atoms. Displacing one atom, keeping the others fixed, leads to restoring forces, exactly as in a spring. So we can think of atoms in a lattice as end points, with springs between pairs of them. In the subsequent sections of this chapter we are going to discuss various characteristic Fig. 14.3 A harmonic (sinusoidal) wave travelling properties of waves. along a stretched string is an example of a transverse wave. An element of the string 14.2 TRANSVERSE AND LONGITUDINAL in the region of the wave oscillates about WAVES its equilibrium position perpendicular to the direction of wave propagation. We have seen that motion of mechanical waves involves oscillations of constituents of the position as the pulse or wave passes through medium. If the constituents of the medium them. The oscillations are normal to the oscillate perpendicular to the direction of wave direction of wave motion along the string, so this propagation, we call the wave a transverse wave. is an example of transverse wave. If they oscillate along the direction of wave We can look at a wave in two ways. We can fix propagation, we call the wave a longitudinal an instant of time and picture the wave in space. wave. This will give us the shape of the wave as a Fig.14.2 shows the propagation of a single whole in space at a given instant. Another way pulse along a string, resulting from a single up is to fix a location i.e. fix our attention on a and down jerk. If the string is very long compared particular element of string and see its oscillatory motion in time. Fig. 14.4 describes the situation for longitudinal waves in the most familiar example of the propagation of sound waves. A long pipe filled with air has a piston at one end. A single sudden push forward and pull back of the piston will generate a pulse of condensations (higher density) and rarefactions (lower density) in the medium (air). If the push-pull of the piston is continuous and periodic (sinusoidal), a Fig. 14.2 When a pulse travels along the length of a stretched string (x-direction), the elements of the string oscillate up and down (y- direction) to the size of the pulse, the pulse will damp out before it reaches the other end and reflection from that end may be ignored. Fig. 14.3 shows a similar situation, but this time the external agent gives a continuous periodic sinusoidal up Fig. 14.4 Longitudinal waves (sound) generated in a and down jerk to one end of the string. The pipe filled with air by moving the piston up resulting disturbance on the string is then a and down. A volume element of air oscillates sinusoidal wave. In either case the elements of in the direction parallel to the direction of the string oscillate about their equilibrium mean wave propagation. Reprint 2025-26 WAVES 281 sinusoidal wave will be generated propagating u Example 14.1 Given below are some in air along the length of the pipe. This is clearly examples of wave motion. State in each case an example of longitudinal waves. if the wave motion is transverse, longitudinal The waves considered above, transverse or or a combination of both: longitudinal, are travelling or progressive waves (a) Motion of a kink in a longitudinal spring since they travel from one part of the medium produced by displacing one end of the to another. The material medium as a whole spring sideways. does not move, as already noted. A stream, for (b) Waves produced in a cylinder example, constitutes motion of water as a whole. containing a liquid by moving its piston back and forth.In a water wave, it is the disturbance that moves, (c) Waves produced by a motorboat sailing not water as a whole. Likewise a wind (motion in water. of air as a whole) should not be confused with a (d) Ultrasonic waves in air produced by a sound wave which is a propagation of vibrating quartz crystal. disturbance (in pressure density) in air, without the motion of air medium as a whole. Answer In transverse waves, the particle motion is (a) Transverse and longitudinal normal to the direction of propagation of the (b) Longitudinal wave. Therefore, as the wave propagates, each (c) Transverse and longitudinal element of the medium undergoes a shearing (d) Longitudinal ⊳ strain. Transverse waves can, therefore, be 14.3 DISPLACEMENT RELATION INpropagated only in those media, which can A PROGRESSIVE WAVE sustain shearing stress, such as solids and not in fluids. Fluids, as well as, solids can sustain For mathematical description of a travelling compressive strain; therefore, longitudinal wave, we need a function of both position x and time t. Such a function at every instant shouldwaves can be propagated in all elastic media. give the shape of the wave at that instant. Also,For example, in medium like steel, both at every given location, it should describe thetransverse and longitudinal waves can motion of the constituent of the medium at thatpropagate, while air can sustain only location. If we wish to describe a sinusoidal longitudinal waves. The waves on the surface travelling wave (such as the one shown in Fig. of water are of two kinds: capillary waves and 14.3) the corresponding function must also be gravity waves. The former are ripples of fairly sinusoidal. For convenience, we shall take the short wavelength—not more than a few wave to be transverse so that if the position of centimetre—and the restoring force that the constituents of the medium is denoted by x, produces them is the surface tension of water. the displacement from the equilibrium position Gravity waves have wavelengths typically may be denoted by y. A sinusoidal travelling ranging from several metres to several hundred wave is then described by: meters. The restoring force that produces these y ( x , t ) = a sin( kx −ωt + φ) (14.2)waves is the pull of gravity, which tends to keep The term φ in the argument of sine functionthe water surface at its lowest level. The means equivalently that we are considering aoscillations of the particles in these waves are linear combination of sine and cosine functions:not confined to the surface only, but extend with diminishing amplitude to the very bottom. The y ( x , t ) = A sin(kx − ωt ) + B cos(kx − ωt ) (14.3) particle motion in water waves involves a From Equations (14.2) and (14.3), complicated motion—they not only move up and  B down but also back and forth. The waves in an a = A 2 + B 2 and φ= tan −1ocean are the combination of both longitudinal  A  and transverse waves. To understand why Equation (14.2) It is found that, generally, transverse and represents a sinusoidal travelling wave, take a longitudinal waves travel with different speed fixed instant, say t = t0. Then, the argument of in the same medium. the sine function in Equation (14.2) is simply Reprint 2025-26 282 PHYSICS kx + constant. Thus, the shape of the wave (at any fixed instant) as a function of x is a sine wave. Similarly, take a fixed location, say x = x0. Then, the argument of the sine function in Equation (14.2) is constant -ωt. The displacement y, at a fixed location, thus, varies sinusoidally with time. That is, the constituents of the medium at different positions execute simple harmonic motion. Finally, as t increases, x must increase in the positive direction to keep kx – ωt + φ constant. Thus, Eq. (14.2) represents a sinusiodal (harmonic) wave travelling along the positive direction of the x-axis. On the other hand, a function y ( x , t ) = a sin( kx + ω t + φ ) (14.4) represents a wave travelling in the negative direction of x-axis. Fig. (14.5) gives the names of the various physical quantities appearing in Eq. (14.2) that we now interpret. y(x,t) : displacement as a function of position x and time t a : amplitude of a wave ω : angular frequency of the wave Fig. 14.6 A harmonic wave progressing along the k : angular wave number positive direction of x-axis at different times. kx–ωt+φ : initial phase angle (a+x = 0, t = 0) Using the plots of Fig. 14.6, we now define Fig. 14.5 The meaning of standard symbols in the various quantities of Eq. (14.2). Eq. (14.2) 14.3.1 Amplitude and Phase Fig. 14.6 shows the plots of Eq. (14.2) for In Eq. (14.2), since the sine function varies different values of time differing by equal between 1 and –1, the displacement y (x,t) varies intervals of time. In a wave, the crest is the between a and –a. We can take a to be a positive point of maximum positive displacement, the constant, without any loss of generality. Then, trough is the point of maximum negative a represents the maximum displacement of the displacement. To see how a wave travels, we constituents of the medium from their can fix attention on a crest and see how it equilibrium position. Note that the displacement progresses with time. In the figure, this is y may be positive or negative, but a is positive. shown by a cross (×) on the crest. In the same It is called the amplitude of the wave. manner, we can see the motion of a particular The quantity (kx – ωt + φ) appearing as the constituent of the medium at a fixed location, argument of the sine function in Eq. (14.2) issay at the origin of the x-axis. This is shown called the phase of the wave. Given theby a solid dot (•). The plots of Fig. 14.6 show amplitude a, the phase determines thethat with time, the solid dot (•) at the origin displacement of the wave at any position andmoves periodically, i.e., the particle at the origin oscillates about its mean position as at any instant. Clearly φ is the phase at x = 0 the wave progresses. This is true for any other and t = 0. Hence, φ is called the initial phase location also. We also see that during the time angle. By suitable choice of origin on the x-axis the solid dot (•) has completed one full and the intial time, it is possible to have φ = 0. oscillation, the crest has moved further by a Thus there is no loss of generality in dropping certain distance. φ, i.e., in taking Eq. (14.2) with φ = 0. Reprint 2025-26 WAVES 283 14.3.2 Wavelength and Angular Wave Number The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by λ. For simplicity, we can choose points of the same phase to be crests or troughs. The wavelength is then the distance between two consecutive crests or troughs in a wave. Taking φ = 0 in Eq. (14.2), Fig. 14.7 An element of a string at a fixed location the displacement at t = 0 is given by oscillates in time with amplitude a and period T, as the wave passes over it. y ( x ,0) = a sin kx (14.5) Since the sine function repeats its value after Now, the period of oscillation of the wave is the every 2π change in angle, time it takes for an element to complete one full oscillation. That is − a sin ωt = −a sin ω(t + T) = − a sin(ωt + ωT) That is the displacements at points x and at Since sine function repeats after every 2π, 2nπx + 2π k ωT = 2π or ω = (14.7) T are the same, where n=1,2,3,... The 1east is called the angular frequency of the wave.distance between points with the same ω displacement (at any given instant of time) is Its SI unit is rad s –1. The frequency ν is the obtained by taking n = 1. λ is then given by number of oscillations per second. Therefore, 1 ω 2π 2π ν= = (14.8) λ= or k = (14.6) T 2π k λ ν is usually measured in hertz. k is the angular wave number or propagation In the discussion above, reference has always constant; its SI unit is radian per metre or been made to a wave travelling along a string or rad m−*1 a transverse wave. In a longitudinal wave, the displacement of an element of the medium is 14.3.3 Period, Angular Frequency and parallel to the direction of propagation of the Frequency wave. In Eq. (14.2), the displacement function Fig. 14.7 shows again a sinusoidal plot. It for a longitudinal wave is written as, describes not the shape of the wave at a certain s(x, t) = a sin (kx – ωt + φ) (14.9) instant but the displacement of an element (at any fixed location) of the medium as a function where s(x, t) is the displacement of an element of time. We may for, simplicity, take Eq. (14.2) of the medium in the direction of propagation with φ = 0 and monitor the motion of the element of the wave at position x and time t. In Eq. (14.9), a is the displacement amplitude; othersay at x = 0 . We then get quantities have the same meaning as in case y (0, t ) = a sin( −ωt ) of a transverse wave except that the displacement function y (x, t) is to be replaced = −a sin ωt by the function s (x, t). * Here again, ‘radian’ could be dropped and the units could be written merely as m–1. Thus, k represents 2π times the number of waves (or the total phase difference) that can be accommodated per unit length, with SI units m–1. Reprint 2025-26 284 PHYSICS the shape of the wave at two instants of time, u Example 14.2 A wave travelling along a which differ by a small time internal ∆t. The string is described by, entire wave pattern is seen to shift to the right (positive direction of x-axis) by a distance ∆x. In y(x, t) = 0.005 sin (80.0 x – 3.0 t), particular, the crest shown by a dot (• ) moves a in which the numerical constants are in SI units (0.005 m, 80.0 rad m–1, and 3.0 rad s–1). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s ? Answer On comparing this displacement equation with Eq. (14.2), Fig. 14.8 Progression of a harmonic wave from time y (x, t) = a sin (kx – ωt), t to t + ∆t. where ∆t is a small interval. The wave pattern as a whole shifts to the we find right. The crest of the wave (or a point with (a) the amplitude of the wave is 0.005 m = 5 mm. any fixed phase) moves right by the distance (b) the angular wave number k and angular ∆x in time ∆t. frequency ω are distance ∆x in time ∆t. The speed of the wave is k = 80.0 m–1 and ω = 3.0 s–1 then ∆x/∆t. We can put the dot (• ) on a point We, then, relate the wavelength λ to k through with any other phase. It will move with the same Eq. (14.6), speed v (otherwise the wave pattern will not λ = 2π/k remain fixed). The motion of a fixed phase point on the wave is given by 2π = −1 kx – ωt = constant (14.10) 80.0 m Thus, as time t changes, the position x of the = 7.85 cm fixed phase point must change so that the phase (c) Now, we relate T to ω by the relation remains constant. Thus, T = 2π/ω kx – ωt = k(x+∆x) – ω(t+∆t) 2π or k ∆x – ω∆t =0 = −1 3.0 s Taking ∆x, ∆t vanishingly small, this gives = 2.09 s dx ω and frequency, v = 1/T = 0.48 Hz = = v (14.11) dt k The displacement y at x = 30.0 cm and Relating ω to T and k to λ, we get time t = 20 s is given by 2πν λ y = (0.005 m) sin (80.0 × 0.3 – 3.0 × 20) v = = λν= (14.12) 2π/λ T = (0.005 m) sin (–36 + 12π) = (0.005 m) sin (1.699) Eq. (14.12), a general relation for all progressive = (0.005 m) sin (970) j 5 mm ⊳ waves, shows that in the time required for one full oscillation by any constituent of the medium, the 14.4 THE SPEED OF A TRAVELLING WAVE wave pattern travels a distance equal to the To determine the speed of propagation of a wavelength of the wave. It should be noted that travelling wave, we can fix our attention on any the speed of a mechanical wave is determined by particular point on the wave (characterised by the inertial (linear mass density for strings, mass some value of the phase) and see how that point density in general) and elastic properties (Young’s moves in time. It is convenient to look at the modulus for linear media/ shear modulus, bulk motion of the crest of the wave. Fig. 14.8 gives modulus) of the medium. The medium determines Reprint 2025-26 WAVES 285 the speed; Eq. (14.12) then relates wavelength to arising due to an external force). It does not frequency for the given speed. Of course, as depend on wavelength or frequency of the wave remarked earlier, the medium can support both itself. In higher studies, you will come across transverse and longitudinal waves, which will have waves whose speed is not independent of different speeds in the same medium. Later in this frequency of the wave. Of the two parameters λ chapter, we shall obtain specific expressions for and ν the source of disturbance determines the the speed of mechanical waves in some media. frequency of the wave generated. Given the speed of the wave in the medium and the 14.4.1 Speed of a Transverse Wave on frequency Eq. (14.12) then fixes the wavelength Stretched String v The speed of a mechanical wave is determined λ = (14.15) by the restoring force setup in the medium when ν it is disturbed and the inertial properties (mass u Example 14.3 A steel wire 0.72 m longdensity) of the medium. The speed is expected to has a mass of 5.0 ×10–3 kg. If the wire isbe directly related to the former and inversely to under a tension of 60 N, what is the speedthe latter. For waves on a string, the restoring of transverse waves on the wire ?force is provided by the tension T in the string. The inertial property will in this case be linear Answer Mass per unit length of the wire,mass density µ, which is mass m of the string divided by its length L. Using Newton’s Laws of −3 5.0 × 10 kgMotion, an exact formula for the wave speed on µ = 0. 72 m a string can be derived, but this derivation is outside the scope of this book. We shall, = 6.9 ×10–3 kg m–1 therefore, use dimensional analysis. We already know that dimensional analysis alone can never Tension, T = 60 N yield the exact formula. The overall The speed of wave on the wire is given by dimensionless constant is always left T 60 N −1undetermined by dimensional analysis. = v = = 93 m s ⊳ µ 6.9 × 10− 3 kg m −1 The dimension of µ is [ML–1] and that of T is like force, namely [MLT–2]. We need to combine 14.4.2 Speed of a Longitudinal Wavethese dimensions to get the dimension of speed (Speed of Sound)v [LT–1]. Simple inspection shows that the quantity T/µ has the relevant dimension In a longitudinal wave, the constituents of the − 2 medium oscillate forward and backward in the MLT   2 − 2   =  L T  direction of propagation of the wave. We have [ ML ]   already seen that the sound waves travel in the Thus if T and µ are assumed to be the only form of compressions and rarefactions of small volume elements of air. The elastic property thatrelevant physical quantities, determines the stress under compressional strain is the bulk modulus of the medium defined T v = C (14.13) by (see Chapter 8) µ ∆P where C is the undetermined constant of B = (14.16)dimensional analysis. In the exact formula, it −∆V/V turms out, C=1. The speed of transverse waves Here, the change in pressure ∆P produces a on a stretched string is given by ∆V volumetric strain . B has the same dimension V T v = (14.14) as pressure and given in SI units in terms of µ pascal (Pa). The inertial property relevant for the Note the important point that the speed v propagation of wave is the mass density ρ, with depends only on the properties of the medium T dimensions [ML–3]. Simple inspection reveals and µ (T is a property of the stretched string that quantity B/ρ has the relevant dimension: Reprint 2025-26 286 PHYSICS ML − 2 T − 2   2 − 2 Liquids and solids generally have higher speed    L T =  (14.17) of sound than gases. [Note for solids, the speed − 3    ML    being referred to is the speed of longitudinal Thus, if B and ρ are considered to be the only waves in the solid]. This happens because they are much more difficult to compress than gasesrelevant physical quantities, and so have much higher values of bulk modulus. B Now, see Eq. (14.19). Solids and liquids have v = C (14.18) higher mass densities ( ρ) than gases. But the ρ corresponding increase in both the modulus (B)where, as before, C is the undetermined constant of solids and liquids is much higher. This is thefrom dimensional analysis. The exact derivation reason why the sound waves travel faster inshows that C=1. Thus, the general formula for solids and liquids.longitudinal waves in a medium is: We can estimate the speed of sound in a gas B in the ideal gas approximation. For an ideal gas, v = (14.19) ρ the pressure P, volume V and temperature T are related by (see Chapter 10). For a linear medium, like a solid bar, the lateral expansion of the bar is negligible and we PV = NkBT (14.21) may consider it to be only under longitudinal where N is the number of molecules in volumestrain. In that case, the relevant modulus of V, kB is the Boltzmann constant and T theelasticity is Young’s modulus, which has the temperature of the gas (in Kelvin). Therefore, forsame dimension as the Bulk modulus. an isothermal change it follows from Eq.(14.21) Dimensional analysis for this case is the same that as before and yields a relation like Eq. (14.18), V∆P + P∆V = 0 with an undetermined C, which the exact derivation shows to be unity. Thus, the speed of ∆ P or − = Plongitudinal waves in a solid bar is given by ∆V/V Hence, substituting in Eq. (14.16), we have Y v = (14.20) B = P ρ Therefore, from Eq. (14.19) the speed of a where Y is the Young’s modulus of the material longitudinal wave in an ideal gas is given by, of the bar. Table 14.1 gives the speed of sound in some media. v = P (14.22) ρ Table 14.1 Speed of Sound in some Media This relation was first given by Newton and is known as Newton’s formula. u Example 14.4 Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 ×10–3 kg. Answer We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is: ρo = (mass of one mole of air)/ (volume of one mole of air at STP) −3 29.0 × 10 kg = −3 3 22.4 × 10 m = 1.29 kg m–3 Reprint 2025-26 WAVES 287 According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP, = 280 m s–1 (14.23) ⊳ The result shown in Eq.(14.23) is about 15% smaller as compared to the experimental value of 331 m s–1 as given in Table 14.1. Where did we go wrong ? If we examine the basic assumption made by Newton that the pressure variations in a medium during propagation of sound are isothermal, we find that this is not correct. It was pointed out by Laplace that the pressure variations in the propagation of sound waves are so fast that there is little time for the heat flow to maintain constant temperature. These variations, therefore, are adiabatic and not isothermal. For adiabatic processes the ideal Fig. 14.9 Two pulses having equal and opposite gas satisfies the relation (see Section 11.8), γ displacements moving in opposite PV = constant directions. The overlapping pulses add up i.e. ∆(PVγ ) = 0 to zero displacement in curve (c). or P γ V γ –1 ∆V + V γ ∆P = 0 pulses. Figure 14.9 shows the situation when where γ is the ratio of two specific heats, two pulses of equal and opposite shapes move Cp/Cv. towards each other. When the pulses overlap, Thus, for an ideal gas the adiabatic bulk the resultant displacement is the algebraic summodulus is given by, of the displacement due to each pulse. This is ∆P known as the principle of superposition of waves. Bad = − ∆V/V According to this principle, each pulse moves = γP as if others are not present. The constituents of The speed of sound is, therefore, from Eq. the medium, therefore, suffer displacments due (14.19), given by, to both and since the displacements can be positive and negative, the net displacement is γ P an algebraic sum of the two. Fig. 14.9 gives v = (14.24) ρ graphs of the wave shape at different times. Note the dramatic effect in the graph (c); the This modification of Newton’s formula is referred displacements due to the two pulses have exactly to as the Laplace correction. For air cancelled each other and there is zero γ = 7/5. Now using Eq. (14.24) to estimate the speed displacement throughout. of sound in air at STP, we get a value 331.3 m s–1, To put the principle of superposition which agrees with the measured speed. mathematically, let y1 (x,t) and y2 (x,t) be the displacements due to two wave disturbances in

14.1 Introduction such an example. Here, elastic forces bind the constituents 14.2 Transverse and to each other and, therefore, the motion of one affects that of longitudinal waves the other. If you drop a little pebble in a pond of still water, 14.3 Displacement relation in a the water surface gets disturbed. The disturbance does not progressive wave remain confined to one place, but propagates outward along