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

13.5 VELOCITY AND ACCELERATION IN SIMPLE HARMONIC MOTION The speed of a particle v in uniform circular motion is its angular speed ω times the radius of the circle A. v = ω A (13.8) The direction of velocity v at a time t is along the tangent to the circle at the point where the particle is located at that instant. From the Fig. 13.12 The acceleration, a(t), of the particle P′ isgeometry of Fig. 13.11, it is clear that the velocity the projection of the acceleration a of theof the projection particle P′ at time t is reference particle P. v(t) = –ωA sin (ωt + φ ) (13.9) Eq. (13.11) gives the acceleration of a particle in SHM. The same equation can again be obtained directly by differentiating velocity v(t) given by Eq. (13.9) with respect to time: d a (t ) = v (t ) (13.12) d t We note from Eq. (13.11) the important Fig. 13.11 The velocity, v (t), of the particle P′ is property that acceleration of a particle in SHM the projection of the velocity v of the is proportional to displacement. For x(t) > 0, reference particle, P. a(t) < 0 and for x(t) < 0, a(t) > 0. Thus, whatever Reprint 2025-26 OSCILLATIONS 267 the value of x between –A and A, the acceleration (b) Using Eq. (13.9), the speed of the body a(t) is always directed towards the centre. = – (5.0 m)(2π s–1) sin [(2π s–1) ×1.5 s For simplicity, let us put φ = 0 and write the + π/4] expression for x (t), v (t) and a(t) = – (5.0 m)(2π s–1) sin [(3π + π/4)] x(t) = A cos ωt, v(t) = – ω Asin ωt, a(t)=–ω2 A cos ωt = 10π × 0.707 m s–1 The corresponding plots are shown in Fig. 13.13. = 22 m s–1 All quantities vary sinusoidally with time; only (c) Using Eq.(13.10), the acceleration of the their maxima differ and the different plots differ body in phase. x varies between –A to A; v(t) varies = –(2π s–1)2 × displacement from –ωA to ωA and a(t) from –ω2A to ω2A. With = – (2π s–1)2 × (–3.535 m) respect to displacement plot, velocity plot has a = 140 m s–2 ⊳ phase difference of π/2 and acceleration plot has a phase difference of π. 13.6 FORCE LAW FOR SIMPLE HARMONIC MOTION Using Newton’s second law of motion, and the expression for acceleration of a particle undergoing SHM (Eq. 13.11), the force acting on a particle of mass m in SHM is F (t ) = ma = –mω2 x (t) i.e., F (t) = –k x (t) (13.13) where k = mω2 (13.14a) k or ω = (13.14b) m Like acceleration, force is always directed towards the mean position—hence it is sometimes called the restoring force in SHM. To summarise the discussion so far, simple harmonic motion can be defined in two equivalent ways, either by Eq. (13.4) for displacement or by Eq. (13.13) that gives Fig. 13.13 Displacement, velocity and acceleration of its force law. Going from Eq. (13.4) to Eq. (13.13) a particle in simple harmonic motion have required us to differentiate two times. Likewise, the same period T, but they differ in phase by integrating the force law Eq. (13.13) two times, we can get back Eq. (13.4). u Example 13.5 A body oscillates with SHM Note that the force in Eq. (13.13) is linearly according to the equation (in SI units), proportional to x(t). A particle oscillating under x = 5 cos [2π t + π/4]. such a force is, therefore, calling a linear harmonic oscillator. In the real world, the force At t = 1.5 s, calculate the (a) displacement, may contain small additional terms proportional (b) speed and (c) acceleration of the body. to x2, x3, etc. These then are called non-linear oscillators. Answer The angular frequency ω of the body = 2π s–1 and its time period T = 1 s. u Example 13.6 Two identical springs of At t = 1.5 s spring constant k are attached to a block (a) displacement = (5.0 m) cos [(2π s–1)× of mass m and to fixed supports as shown 1.5 s + π/4] in Fig. 13.14. Show that when the mass is = (5.0 m) cos [(3π + π/4)] displaced from its equilibrium position on = –5.0 × 0.707 m either side, it executes a simple harmonic = –3.535 m motion. Find the period of oscillations. Reprint 2025-26 268 PHYSICS 13.7 ENERGY IN SIMPLE HARMONIC MOTION Both kinetic and potential energies of a particle in SHM vary between zero and their maximum values. In section 13.5 we have seen that the velocity of a particle executing SHM, is a periodic Fig. 13.14 function of time. It is zero at the extreme positions of displacement. Therefore, the kinetic energy (K) Answer Let the mass be displaced by a small of such a particle, which is defined as distance x to the right side of the equilibrium position, as shown in Fig. 13.15. Under this 1 2 K = mv situation the spring on the left side gets 2 1 2 2 2 = m ω A sin (ωt + φ) 2 1 2 2 = k A sin (ωt + φ) (13.15) 2 is also a periodic function of time, being zero when the displacement is maximum and maximum when the particle is at the mean Fig. 13.15 position. Note, since the sign of v is immaterial in K, the period of K is T/2. elongated by a length equal to x and that on What is the potential energy (U) of a particle the right side gets compressed by the same executing simple harmonic motion? In length. The forces acting on the mass are Chapter 6, we have seen that the concept of then, potential energy is possible only for conservative forces. The spring force F = –kx is a conservative F1 = –k x (force exerted by the spring on force, with associated potential energy the left side, trying to pull the mass towards the mean 1 2 U = k x position) (13.16) 2 F2 = –k x (force exerted by the spring on Hence the potential energy of a particle the right side, trying to push the executing simple harmonic motion is, mass towards the mean position) 1 2 The net force, F, acting on the mass is then U(x) = k x given by, 2 F = –2kx 1 2 2 Hence the force acting on the mass is = k A cos (ωt + φ) (13.17) 2proportional to the displacement and is directed towards the mean position; therefore, the motion Thus, the potential energy of a particle executed by the mass is simple harmonic. The executing simple harmonic motion is also time period of oscillations is, periodic, with period T/2, being zero at the mean m position and maximum at the extreme T = 2 π displacements. 2k ⊳ Reprint 2025-26 OSCILLATIONS 269 It follows from Eqs. (13.15) and (13.17) that Observe that both kinetic energy and the total energy, E, of the system is, potential energy in SHM are seen to be always positive in Fig. 13.16. Kinetic energy can, of E = U + K course, be never negative, since it is proportional to the square of speed. Potential 1 1 = k A 2 cos 2 (ωt + φ) + k A 2 sin 2 (ωt + φ) energy is positive by choice of the undermined 2 2 constant in potential energy. Both kinetic energy and potential energy peak twice during each period of SHM. For x = 0, the energy is 1 2 2 2 = k A cos (ωt + φ) + sin (ωt + φ)  kinetic; at the extremes x = ±A, it is all   2 potential energy. In the course of motion Using the familiar trigonometric identity, the between these limits, kinetic energy increases value of the expression in the brackets is unity. at the expense of potential energy or Thus, vice-versa. 1 2 E = k A (13.18) u Example 13.7 A block whose mass is 1 kg 2 is fastened to a spring. The spring has a The total mechanical energy of a harmonic spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from itsoscillator is thus independent of time as equilibrium position at x = 0 on a frictionlessexpected for motion under any conservative surface from rest at t = 0. Calculate the force. The time and displacement dependence kinetic, potential and total energies of the of the potential and kinetic energies of a block when it is 5 cm away from the mean linear simple harmonic oscillator are shown position. in Fig. 13.16. Answer The block executes SHM, its angular frequency, as given by Eq. (13.14b), is k ω = m 1 50 N m– = 1kg = 7.07 rad s–1 Its displacement at any time t is then given by, x(t) = 0.1 cos (7.07t) Fig. 13.16 Kinetic energy, potential energy and total energy as a function of time [shown in (a)] Therefore, when the particle is 5 cm away from and displacement [shown in (b)] of a particle in SHM. The kinetic energy and potential the mean position, we have energy both repeat after a period T/2. The total energy remains constant at all t or x. 0.05 = 0.1 cos (7.07t) Reprint 2025-26 270 PHYSICS let it go. The stone executes a to and fro motion,Or cos (7.07t) = 0.5 and hence it is periodic with a period of about two seconds. 3 We shall show that this periodic motion is sin (7.07t) = = 0.866 2 simple harmonic for small displacements from Then, the velocity of the block at x = 5 cm is = 0.1 × 7.07 × 0.866 m s–1 = 0.61 m s–1 Hence the K.E. of the block, 1 2 = m v 2 = ½[1kg × (0.6123 m s–1 )2 ] = 0.19 J (a) The P.E. of the block, 1 2 = k x 2 = ½(50 N m–1 × 0.05 m × 0.05 m) = 0.0625 J The total energy of the block at x = 5 cm, = K.E. + P.E. = 0.25 J we also know that at maximum displacement, K.E. is zero and hence the total energy of the (b) system is equal to the P.E. Therefore, the total Fig. 13.17 (a) A bob oscillating about its mean position. (b) The radial force T-mg cosθ energy of the system, provides centripetal force but no torque = ½(50 N m–1 × 0.1 m × 0.1 m ) about the support. The tangential force mg sinθ provides the restoring torque. = 0.25 J which is same as the sum of the two energies at the mean position. Consider simple pendulum a displacement of 5 cm. This is in conformity — a small bob of mass m tied to an inextensible with the principle of conservation of energy. ⊳ massless string of length L. The other end of the string is fixed to a rigid support. The bob13.8 The Simple Pendulum oscillates in a plane about the vertical lineIt is said that Galileo measured the periods of a through the support. Fig. 13.17(a) shows thisswinging chandelier in a church by his pulse system. Fig. 13.17(b) is a kind of ‘free-body’beats. He observed that the motion of the chandelier was periodic. The system is a kind diagram of the simple pendulum showing the of pendulum. You can also make your own forces acting on the bob. pendulum by tying a piece of stone to a long Let θ be the angle made by the string with unstretchable thread, approximately 100 cm the vertical. When the bob is at the mean long. Suspend your pendulum from a suitable position, θ = 0 support so that it is free to oscillate. Displace There are only two forces acting on the bob; the stone to one side by a small distance and the tension T along the string and the vertical Reprint 2025-26 OSCILLATIONS 271 force due to gravity (=mg). The force mg can be Table 13.1 sin θ as ma function of angle θ resolved into the component mg cosθ along the string and mg sinθ perpendicular to it. Since (degrees) (radians) sin the motion of the bob is along a circle of length L and centre at the support point, the bob has a radial acceleration (ω2L) and also a tangental acceleration; the latter arises since motion along the arc of the circle is not uniform. The radial acceleration is provided by the net radial force T –mg cosθ, while the tangential acceleration is provided by mg sinθ. It is more convenient to Equation (13.24) is mathematically, identical towork with torque about the support since the radial force gives zero torque. Torque τ about Eq. (13.11) except that the variable is angular displacement. Hence we have proved that forthe support is entirely provided by the tangental small q, the motion of the bob is simple harmonic.component of force From Eqs. (13.24) and (13.11), τ = –L (mg sinθ ) (13.19) This is the restoring torque that tends to reduce mgL ω =angular displacement — hence the negative Isign. By Newton’s law of rotational motion, τ = I α (13.20) and where I is the moment of inertia of the system about the support and α is the angular I T = 2π (13.25)acceleration. Thus, mgL I α = –m g sin θ L (13.21) Now since the string of the simple pendulum is massless, the moment of inertia I is simply mL2. Eq. (13.25) then gives the well-known Or, formula for time period of a simple pendulum. m g L α = − sin θ (13.22) L I T = 2π (13.26) g We can simplify Eq. (13.22) if we assume that the displacement θ is small. We know that sin θ u Example 13.8 What is the length of a can be expressed as, simple pendulum, which ticks seconds ? θ3 θ5 Answer From Eq. (13.26), the time period of a sin θ = θ− + ± ... (13.23) 3! 5! simple pendulum is given by, L where θ is in radians. T = 2π Now if θ is small, sin θ can be approximated g by θ and Eq. (13.22) can then be written as, From this relation one gets, gT 2 mgL L = 2 θ α = − (13.24) 4π I The time period of a simple pendulum, which In Table 13.1, we have listed the angle θ in ticks seconds, is 2 s. Therefore, for g = 9.8 m s–2 degrees, its equivalent in radians, and the value of the function sin θ . From this table it and T = 2 s, L is can be seen that for θ as large as 20 degrees, 9.8(m s –2 ) × 4(s 2 ) = sin θ is nearly the same as θ expressed 2 4π in radians. = 1 m ⊳ Reprint 2025-26 272 PHYSICS SUMMARY 1. The motion that repeats itself is called periodlic motion. 2. The period T is the time reequired for one complete oscillation, or cycle. It is related to the frequency v by, 1 T = v The frequency ν of periodic or oscillatory motion is the number of oscillations per unit time. In the SI, it is measured in hertz : 1 hertz = 1 Hz = 1 oscillation per second = 1s–1 3. In simple harmonic motion (SHM), the displacement x (t) of a particle from its equilibrium position is given by, x (t) = A cos (ωt + φ ) (displacement), in which A is the amplitude of the displacement, the quantity (ωt + φ ) is the phase of the motion, and φ is the phase constant. The angular frequency ω is related to the period and frequency of the motion by, 2π ω= = 2πν (angular frequency). T 4. Simple harmonic motion can also be viewed as the projection of uniform circular motion on the diameter of the circle in which the latter motion occurs. 5. The particle velocity and acceleration during SHM as functions of time are given by, v (t) = –ωA sin (ωt + φ ) (velocity), a (t) = –ω2A cos (ωt + φ ) = –ω2x (t) (acceleration), Thus we see that both velocity and acceleration of a body executing simple harmonic motion are periodic functions, having the velocity amplitude vm=ω A and acceleration amplitude am =ω 2A, respectively. 6. The force acting in a simple harmonic motion is proportional to the displacement and is always directed towards the centre of motion. 7. A particle executing simple harmonic motion has, at any time, kinetic energy K = ½ mv2 and potential energy U = ½ kx2. If no friction is present the mechanical energy of the system, E = K + U always remains constant even though K and U change with time. 8. A particle of mass m oscillating under the influence of Hooke’s law restoring force given by F = – k x exhibits simple harmonic motion with k ω = (angular frequency) m m T = 2π (period) k Such a system is also called a linear oscillator. 9. The motion of a simple pendulum swinging through small angles is approximately simple harmonic. The period of oscillation is given by, L T = 2π g Reprint 2025-26 OSCILLATIONS 273 POINTS TO PONDER 1. The period T is the least time after which motion repeats itself. Thus, motion repeats itself after nT where n is an integer. 2. Every periodic motion is not simple harmonic motion. Only that periodic motion governed by the force law F = – k x is simple harmonic. 3. Circular motion can arise due to an inverse-square law force (as in planetary motion) as well as due to simple harmonic force in two dimensions equal to: –mω2r. In the latter case, the phases of motion, in two perpendicular directions (x and y) must differ by π/2. Thus, for example, a particle subject to a force –mω2r with initial position (0, A) and velocity (ωA, 0) will move uniformly in a circle of radius A. 4. For linear simple harmonic motion with a given ω, two initial conditions are necessary and sufficient to determine the motion completely. The initial conditions may be (i) initial position and initial velocity or (ii) amplitude and phase or (iii) energy and phase. 5. From point 4 above, given amplitude or energy, phase of motion is determined by the initial position or initial velocity. 6. A combination of two simple harmonic motions with arbitrary amplitudes and phases is not necessarily periodic. It is periodic only if frequency of one motion is an integral multiple of the other’s frequency. However, a periodic motion can always be expressed as a sum of infinite number of harmonic motions with appropriate amplitudes. 7. The period of SHM does not depend on amplitude or energy or the phase constant. Contrast this with the periods of planetary orbits under gravitation (Kepler’s third law). 8. The motion of a simple pendulum is simple harmonic for small angular displacement. 9. For motion of a particle to be simple harmonic, its displacement x must be expressible in either of the following forms : x = A cos ωt + B sin ωt x = A cos (ωt + α ), x = B sin (ωt + β ) The three forms are completely equivalent (any one can be expressed in terms of any other two forms). Thus, damped simple harmonic motion is not strictly simple harmonic. It is approximately so only for time intervals much less than 2m/b where b is the damping constant. Reprint 2025-26 274 PHYSICS Exercises 13.1 Which of the following examples represent periodic motion? (a) A swimmer completing one (return) trip from one bank of a river to the other and back. (b) A freely suspended bar magnet displaced from its N-S direction and released. (c) A hydrogen molecule rotating about its centre of mass. (d) An arrow released from a bow. 13.2 Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) general vibrations of a polyatomic molecule about its equilibrium position. 13.3 Fig. 13.18 depicts four x-t plots for linear motion of a particle. Which of the plots represent periodic motion? What is the period of motion (in case of periodic motion) ? Fig. 18.18 Reprint 2025-26 OSCILLATIONS 275 13.4 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant): (a) sin ωt – cos ωt (b) sin3 ωt (c) 3 cos (π/4 – 2ωt) (d) cos ωt + cos 3ωt + cos 5ωt (e) exp (–ω2t2) (f) 1 + ωt + ω2t2 13.5 A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction from A to B as the positive direction and give the signs of velocity, acceleration and force on the particle when it is (a) at the end A, (b) at the end B, (c) at the mid-point of AB going towards A, (d) at 2 cm away from B going towards A, (e) at 3 cm away from A going towards B, and (f) at 4 cm away from B going towards A. 13.6 Which of the following relationships between the acceleration a and the displacement x of a particle involve simple harmonic motion? (a) a = 0.7x (b) a = –200x2 (c) a = –10x (d) a = 100x3 13.7 The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ ). If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle ? The angular frequency of the particle is π s–1. If instead of the cosine function, we choose the sine function to describe the SHM : x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions. 13.8 A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body ? 13.9 A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released. Fig. 13.19 Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass. Reprint 2025-26 276 PHYSICS 13.10 In Exercise 13.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase? 13.11 Figures 13.20 correspond to two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution (i.e. clockwise or anti-clockwise) are indicated on each figure. Fig. 13.20 Obtain the corresponding simple harmonic motions of the x-projection of the radius vector of the revolving particle P, in each case. 13.12 Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s). (a) x = –2 sin (3t + π/3) (b) x = cos (π/6 – t) (c) x = 3 sin (2πt + π/4) (d) x = 2 cos πt 13.13 Figure 13.21(a) shows a spring of force constant k clamped rigidly at one end and a mass m attached to its free end. A force F applied at the free end stretches the spring. Figure 13.21 (b) shows the same spring with both ends free and attached to a mass m at either end. Each end of the spring in Fig. 13.21(b) is stretched by the same force F. Fig. 13.21 (a) What is the maximum extension of the spring in the two cases ? (b) If the mass in Fig. (a) and the two masses in Fig. (b) are released, what is the period of oscillation in each case ? Reprint 2025-26 OSCILLATIONS 277 13.14 The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed ? 13.15 The acceleration due to gravity on the surface of moon is 1.7 m s–2. What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s ? (g on the surface of earth is 9.8 m s–2) 13.16 A simple pendulum of length l and having a bob of mass M is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period ? 13.17 A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρl. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period hρ T = 2π ρ1 g where ρ is the density of cork. (Ignore damping due to viscosity of the liquid). 13.18 One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. A small pressure difference is maintained between the two columns. Show that, when the suction pump is removed, the column of mercury in the U-tube executes simple harmonic motion. Reprint 2025-26 CHAPTER FOURTEEN WAVES 14.1 INTRODUCTION In the previous Chapter, we studied the motion of objects oscillating in isolation. What happens in a system, which is a collection of such objects? A material medium provides

1.00 nF for two values of R: (i) R = 100 W and (ii) R = 200 W. For the source applied vm = 1 100 V. w0 for this case is = 1.00×106 LC FIGURE 7.14 Variation of im with w for tworad/s. cases: (i) R = 100 W, (ii) R = 200 W, We see that the current amplitude is L = 1.00 mH. maximum at the resonant frequency. Since im = vm / R at resonance, the current amplitude for case (i) is twice to that for case (ii). Resonant circuits have a variety of applications, for example, in the tuning mechanism of a radio or a TV set. The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna acts as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum. It is important to note that resonance phenomenon is exhibited by a circuit only if both L and C are present in the circuit. Only then do the voltages across L and C cancel each other (both being out of phase) and the current amplitude is vm/R, the total source voltage appearing across R. This means that we cannot have resonance in a RL or RC circuit. 189 Reprint 2025-26 Physics Example 7.6 A resistor of 200 W and a capacitor of 15.0 mF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. Solution Given R = 200 Ω, C = 15.0 µF = 15.0 × 10 −6 F V = 220 V, ν = 50Hz (a) In order to calculate the current, we need the impedance of the circuit. It is Z = R 2 + X C2 = R 2 + (2πνC )− 2 = (200 Ω )2 + (2 × 3.14 × 50 × 15.0 × 10 −6 F )−2 = (200 Ω )2 + (212.3 Ω )2 = 291.67 Ω Therefore, the current in the circuit is V 220 V I = = = 0.755 A Z 291.5 Ω (b) Since the current is the same throughout the circuit, we have V R = I R = (0.755 A)(200 Ω=) 151V VC = I X C = (0.755 A)(212.3 Ω=) 160.3 V The algebraic sum of the two voltages, VR and VC is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learnt in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem: V R +C = V R2 + VC2 7.6 = 220 V Thus, if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor EXAMPLE is equal to the voltage of the source. 7.7 POWER IN AC CIRCUIT: THE POWER FACTOR We have seen that a voltage v = vm sinwt applied to a series RLC circuit drives a current in the circuit given by i = im sin(wt + f) where v m − 1  X C − X L  i m = Z and φ = tan  R  190 Therefore, the instantaneous power p supplied by the source is Reprint 2025-26 Alternating Current p = v i = (v m sin ωt ) × [ i m sin(ωt + φ)] v m i m = [ cosφ− cos(2ωt + φ)] (7.29) 2 The average power over a cycle is given by the average of the two terms in R.H.S. of Eq. (7.29). It is only the second term which is time-dependent. Its average is zero (the positive half of the cosine cancels the negative half). Therefore, v m i m v m i m P = cos φ = cos φ 2 2 2 = V I cos φ [7.30(a)] This can also be written as, P = I 2 Z cos φ [7.30(b)] So, the average power dissipated depends not only on the voltage and current but also on the cosine of the phase angle φ between them. The quantity cosφ is called the power factor. Let us discuss the following cases: CaseCaseCaseCaseCase (i)(i)(i)(i)(i) Resistive circuit: If the circuit contains only pure R, it is called resistive. In that case φ = 0, cos φ = 1. There is maximum power dissipation. CaseCaseCaseCaseCase (ii)(ii)(ii)(ii)(ii) Purely inductive or capacitive circuit: If the circuit contains only an inductor or capacitor, we know that the phase difference between voltage and current is π/2. Therefore, cos φ = 0, and no power is dissipated even though a current is flowing in the circuit. This current is sometimes referred to as wattless current. CaseCaseCaseCaseCase (iii)(iii)(iii)(iii)(iii) LCR series circuit: In an LCR series circuit, power dissipated is given by Eq. (7.30) where φ = tan–1 (Xc – XL )/ R. So, φ may be non-zero in a RL or RC or RCL circuit. Even in such cases, power is dissipated only in the resistor. CaseCaseCaseCaseCase (iv)(iv)(iv)(iv)(iv) Power dissipated at resonance in LCR circuit: At resonance Xc – XL= 0, and φ = 0. Therefore, cosφ = 1 and P = I2Z = I2 R. That is, maximum power is dissipated in a circuit (through R) at resonance. ExampleExampleExampleExampleExample 7.77.77.77.77.7 (a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain. SolutionSolutionSolutionSolutionSolution (a) We know that P = I V cosφ where cosφ is the power factor. To supply a given power at a given voltage, if cosφ is small, we have to increase current accordingly. But this will lead to large power loss (I2R) in transmission. EEEE (b)Suppose in a circuit, current I lags the voltage by an angle φ. Then power factor cosφ =R/Z. EXAMPLEXAMPLEXAMPLEXAMPLEXAMPLE We can improve the power factor (tending to 1) by making Z tend to 7.77.77.77.77.7 us understand, with the help of a phasor diagram (Fig. 7.15) R. Let 191 Reprint 2025-26 Physics FIGURE 7.15 how this can be achieved. Let us resolve I into two components. Ip along the applied voltage V and Iq perpendicular to the applied voltage. Iq as you have learnt in Section 7.7, is called the wattless component since corresponding to this component of current, there is no power loss. IP is known as the power component because it is in phase with the voltage and corresponds to power loss in the circuit. It’s clear from this analysis that if we want to improve power factor, 7.7 we must completely neutralize the lagging wattless current Iq by an equal leading wattless current I¢q. This can be done by connecting a capacitor of appropriate value in parallel so that Iq and I¢q cancel EXAMPLE each other and P is effectively Ip V. Example 7.8 A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 W, L = 25.48 mH, and C = 796 mF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor. Solution (a) To find the impedance of the circuit, we first calculate XL and XC. XL = 2 pnL = 2 × 3.14 × 50 × 25.48 × 10–3 W = 8 W 1 X C = 2 π νC 1 = 6 = 4 Ω 2 × 3.14 × 50 × 796 × 10− Therefore, Z = R 2 + ( X L − X C )2 = 3 2 + (8 − 4)2 = 5 W X C − X L 7.8 (b) Phase difference, f = tan–1 R −1  4 − 8  EXAMPLE = tan  3 = −53. 1°192 Reprint 2025-26 Alternating Current Since f is negative, the current in the circuit lags the voltage across the source. (c) The power dissipated in the circuit is P = I 2 R im 1  283  = Now, I = 2 2  5 = 40A EXAMPLE Therefore, P = (40 A )2 × 3 Ω= 4800 W (d) Power factor = cos cos –53.1 0.6 7.8 Example 7.9 Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition. Solution (a) The frequency at which the resonance occurs is 1 1 ω0 = = LC 25.48 × 10 −3 × 796 × 10 −6 = 222.1rad/s ω0 221.1 νr = = Hz = 35.4Hz 2 π 2 × 3.14 (b) The impedance Z at resonant condition is equal to the resistance: Z = R = 3 Ω The rms current at resonance is V V  283  1 = = = = 66.7 A Z R  2  3 The power dissipated at resonance is P = I 2 × R = (66.7)2 × 3 = 13.35 kW EXAMPLE You can see that in the present case, power dissipated at resonance is more than the power dissipated in Example 7.8. 7.9 Example 7.10 At an airport, a person is made to walk through the doorway of a metal detector, for security reasons. If she/he is carrying anything made of metal, the metal detector emits a sound. On what principle does this detector work? Solution The metal detector works on the principle of resonance in ac circuits. When you walk through a metal detector, you are, in fact, walking through a coil of many turns. The coil is connected to a capacitor tuned so that the circuit is in resonance. When you walk through with metal in your pocket, the impedance of the EXAMPLE circuit changes – resulting in significant change in current in the circuit. This change in current is detected and the electronic circuitry causes a sound to be emitted as an alarm. 7.10 193 Reprint 2025-26 Physics 7.8 TRANSFORMERS For many purposes, it is necessary to change (or transform) an alternating voltage from one to another of greater or smaller value. This is done with a device called transformer using the principle of mutual induction. A transformer consists of two sets of coils, insulated from each other. They are wound on a soft-iron core, either one on top of the other as in Fig. 7.16(a) or on separate limbs of the core as in Fig. 7.16(b). One of the coils called the primary coil has Np turns. The other coil is called the secondary coil; it has Ns turns. Often the primary coil is the input coil and the secondary coil is the output coil of the transformer. FIGURE 7.16 Two arrangements for winding of primary and secondary coil in a transformer: (a) two coils on top of each other, (b) two coils on separate limbs of the core. When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which links the secondary and induces an emf in it. The value of this emf depends on the number of turns in the secondary. We consider an ideal transformer in which the primary has negligible resistance and all the flux in the core links both primary and secondary windings. Let f be the flux in each turn in the core at time t due to current in the primary when a voltage vp is applied to it. Then the induced emf or voltage es, in the secondary with Ns turns is dφ εs = − N s (7.31) d t The alternating flux f also induces an emf, called back emf in the primary. This is dφ εp = − N p (7.32) d t But ep = vp. If this were not so, the primary current would be infinite since the primary has zero resistance (as assumed). If the secondary is an open circuit or the current taken from it is small, then to a good approximation es = vs Reprint 2025-26 Alternating Current where vs is the voltage across the secondary. Therefore, Eqs. (7.31) and (7.32) can be written as dφ v s = − N s [7.31(a)] d t dφ v p = − N p [7.32(a)] d t From Eqs. [7.31 (a)] and [7.32 (a)], we have v s N s = (7.33) v p N p Note that the above relation has been obtained using three assumptions: (i) the primary resistance and current are small; (ii) the same flux links both the primary and the secondary as very little flux escapes from the core, and (iii) the secondary current is small. If the transformer is assumed to be 100% efficient (no energy losses), the power input is equal to the power output, and since p = i v, ipvp = isvs (7.34) Although some energy is always lost, this is a good approximation, since a well designed transformer may have an efficiency of more than 95%. Combining Eqs. (7.33) and (7.34), we have i p v s N s = = (7.35) i s v p N p Since i and v both oscillate with the same frequency as the ac source, Eq. (7.35) also gives the ratio of the amplitudes or rms values of corresponding quantities. Now, we can see how a transformer affects the voltage and current. We have:  N s   N p  Vs = V p and I s = I p (7.36)  N p   N s  That is, if the secondary coil has a greater number of turns than the primary (Ns > Np), the voltage is stepped up (Vs > Vp). This type of arrangement is called a step-up transformer. However, in this arrangement, there is less current in the secondary than in the primary (Np/Ns < 1 and Is < Ip). For example, if the primary coil of a transformer has 100 turns and the secondary has 200 turns, Ns/Np = 2 and Np/Ns=1/2. Thus, a 220V input at 10A will step-up to 440 V output at 5.0 A. If the secondary coil has less turns than the primary (Ns < Np), we have a step-down transformer. In this case, Vs < Vp and Is > Ip. That is, the voltage is stepped down, or reduced, and the current is increased. The equations obtained above apply to ideal transformers (without any energy losses). But in actual transformers, small energy losses do occur due to the following reasons: (i) Flux Leakage: There is always some flux leakage; that is, not all of the flux due to primary passes through the secondary due to poor 195 Reprint 2025-26 Physics design of the core or the air gaps in the core. It can be reduced by winding the primary and secondary coils one over the other. (ii) Resistance of the windings: The wire used for the windings has some resistance and so, energy is lost due to heat produced in the wire (I 2R). In high current, low voltage windings, these are minimised by using thick wire. (iii) Eddy currents: The alternating magnetic flux induces eddy currents in the iron core and causes heating. The effect is reduced by using a laminated core. (iv) Hysteresis: The magnetisation of the core is repeatedly reversed by the alternating magnetic field. The resulting expenditure of energy in the core appears as heat and is kept to a minimum by using a magnetic material which has a low hysteresis loss. The large scale transmission and distribution of electrical energy over long distances is done with the use of transformers. The voltage output of the generator is stepped-up (so that current is reduced and consequently, the I 2R loss is cut down). It is then transmitted over long distances to an area sub-station near the consumers. There the voltage is stepped down. It is further stepped down at distributing sub-stations and utility poles before a power supply of 240 V reaches our homes. SUMMARY 1. An alternating voltage v = v m sin ω t applied to a resistor R drives a v m current i = im sinwt in the resistor, i m = . The current is in phase with R the applied voltage. 2. For an alternating current i = im sin wt passing through a resistor R, the average power loss P (averaged over a cycle) due to joule heating is ( 1/2 )i 2mR. To express it in the same form as the dc power (P = I 2R), a special value of current is used. It is called root mean square (rms) current and is donoted by I: i m I = = 0.707 i m 2 Similarly, the rms voltage is defined by vm V = = 0.707 v m 2 We have P = IV = I 2R 3. An ac voltage v = vm sin wt applied to a pure inductor L, drives a current in the inductor i = im sin (wt – p/2), where im = vm/XL. XL = wL is called inductive reactance. The current in the inductor lags the voltage by p/2. The average power supplied to an inductor over one complete cycle is zero. Reprint 2025-26 Alternating Current 4. An ac voltage v = vm sinwt applied to a capacitor drives a current in the capacitor: i = im sin (wt + p/2). Here, v m 1 i m = , X C = is called capacitive reactance. X C ωC The current through the capacitor is p/2 ahead of the applied voltage. As in the case of inductor, the average power supplied to a capacitor over one complete cycle is zero. 5. For a series RLC circuit driven by voltage v = vm sin wt, the current is given by i = im sin (wt + f) v m where i m = R 2 + ( X C − X L )2 −1 X C − X L and φ = tan R is called the impedance of the circuit. Z = R 2 + ( X C − X L )2 The average power loss over a complete cycle is given by P = V I cosf The term cosf is called the power factor. 6. In a purely inductive or capacitive circuit, cosf = 0 and no power is dissipated even though a current is flowing in the circuit. In such cases, current is referred to as a wattless current. 7. The phase relationship between current and voltage in an ac circuit can be shown conveniently by representing voltage and current by rotating vectors called phasors. A phasor is a vector which rotates about the origin with angular speed w. The magnitude of a phasor represents the amplitude or peak value of the quantity (voltage or current) represented by the phasor. The analysis of an ac circuit is facilitated by the use of a phasor diagram. 8. A transformer consists of an iron core on which are bound a primary coil of Np turns and a secondary coil of Ns turns. If the primary coil is connected to an ac source, the primary and secondary voltages are related by  N s  Vs =  N p  V p and the currents are related by  N p  Is = I p  N s  If the secondary coil has a greater number of turns than the primary, the voltage is stepped-up (Vs > Vp). This type of arrangement is called a step- up transformer. If the secondary coil has turns less than the primary, we have a step-down transformer. 197 Reprint 2025-26 Physics Physical quantity Symbol Dimensions Unit Remarks 2 –3 v m rms voltage V [M L T A –1] V V = , v m is the 2 amplitude of the ac voltage. i m rms current I [ A] A I = , im is the amplitude of 2 the ac current. Reactance: Inductive XL [M L2 T –3 A–2] XL = L Capacitive XC [M L2 T –3 A–2] XC = 1/C Impedance Z [M L2 T –3 A–2] Depends on elements present in the circuit. 1 Resonant wr or w0 [T –1] Hz w0  for a frequency LC series RLC circuit ω0 L 1 Quality factor Q Dimensionless Q = = for a series R ω0 C R RLC circuit. Power factor Dimensionless = cosf, f is the phase difference between voltage applied and current in the circuit. POINTS TO PONDER 1. When a value is given for ac voltage or current, it is ordinarily the rms value. The voltage across the terminals of an outlet in your room is normally 240 V. This refers to the rms value of the voltage. The amplitude of this voltage is v m = 2V = 2(240) = 340 V 2. The power rating of an element used in ac circuits refers to its average power rating. 3. The power consumed in an ac circuit is never negative. 4. Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current? It cannot be derived from the mutual attraction of two parallel wires carrying ac 198 currents, as the dc ampere is derived. An ac current changes direction Reprint 2025-26 Alternating Current with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in terms of some property that is independent of the direction of the current. Joule heating is such a property, and there is one ampere of rms value of alternating current in a circuit if the current produces the same average heating effect as one ampere of dc current would produce under the same conditions. 5. In an ac circuit, while adding voltages across different elements, one should take care of their phases properly. For example, if VR and VC are voltages across R and C, respectively in an RC circuit, then the total voltage across RC combination is V RC = VR2 + VC2 and not VR + VC since VC is p/2 out of phase of VR. 6. Though in a phasor diagram, voltage and current are represented by vectors, these quantities are not really vectors themselves. They are scalar quantities. It so happens that the amplitudes and phases of harmonically varying scalars combine mathematically in the same way as do the projections of rotating vectors of corresponding magnitudes and directions. The ‘rotating vectors’ that represent harmonically varying scalar quantities are introduced only to provide us with a simple way of adding these quantities using a rule that we already know as the law of vector addition. 7. There are no power losses associated with pure capacitances and pure inductances in an ac circuit. The only element that dissipates energy in an ac circuit is the resistive element. 8. In a RLC circuit, resonance phenomenon occur when XL = XC or 1 ω0 = . For resonance to occur, the presence of both L and C LC elements in the circuit is a must. With only one of these (L or C ) elements, there is no possibility of voltage cancellation and hence, no resonance is possible. 9. The power factor in a RLC circuit is a measure of how close the circuit is to expending the maximum power. 10. In generators and motors, the roles of input and output are reversed. In a motor, electric energy is the input and mechanical energy is the output. In a generator, mechanical energy is the input and electric energy is the output. Both devices simply transform energy from one form to another. 11. A transformer (step-up) changes a low-voltage into a high-voltage. This does not violate the law of conservation of energy. The current is reduced by the same proportion. 199 Reprint 2025-26 Physics EXERCISES 7.1 A 100 W resistor is connected to a 220 V, 50 Hz ac supply. (a) What is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? 7.2 (a) The peak voltage of an ac supply is 300 V. What is the rms voltage? (b) The rms value of current in an ac circuit is 10 A. What is the peak current? 7.3 A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. 7.4 A 60 mF capacitor is connected to a 110 V, 60 Hz ac supply. Determine the rms value of the current in the circuit. 7.5 In Exercises 7.3 and 7.4, what is the net power absorbed by each circuit over a complete cycle. Explain your answer. 7.6 A charged 30 mF capacitor is connected to a 27 mH inductor. What is the angular frequency of free oscillations of the circuit? 7.7 A series LCR circuit with R = 20 W, L = 1.5 H and C = 35 mF is connected to a variable-frequency 200 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle? 7.8 Figure 7.17 shows a series LCR circuit connected to a variable frequency 230 V source. L = 5.0 H, C = 80mF, R = 40 W. FIGURE 7.17 (a) Determine the source frequency which drives the circuit in resonance. (b) Obtain the impedance of the circuit and the amplitude of current at the resonating frequency. (c) Determine the rms potential drops across the three elements of the circuit. Show that the potential drop across the LC combination is zero at the resonating frequency. Reprint 2025-26 Chapter Eight ELECTROMAGNETIC WAVES 8.1 INTRODUCTION In Chapter 4, we learnt that an electric current produces magnetic field and that two current-carrying wires exert a magnetic force on each other. Further, in Chapter 6, we have seen that a magnetic field changing with time gives rise to an electric field. Is the converse also true? Does an electric field changing with time give rise to a magnetic field? James Clerk Maxwell (1831-1879), argued that this was indeed the case – not only an electric current but also a time-varying electric field generates magnetic field. While applying the Ampere’s circuital law to find magnetic field at a point outside a capacitor connected to a time-varying current, Maxwell noticed an inconsistency in the Ampere’s circuital law. He suggested the existence of an additional current, called by him, the displacement current to remove this inconsistency. Maxwell formulated a set of equations involving electric and magnetic fields, and their sources, the charge and current densities. These equations are known as Maxwell’s equations. Together with the Lorentz force formula (Chapter 4), they mathematically express all the basic laws of electromagnetism. The most important prediction to emerge from Maxwell’s equations is the existence of electromagnetic waves, which are (coupled) time- varying electric and magnetic fields that propagate in space. The speed of the waves, according to these equations, turned out to be very close to Reprint 2025-26 Physics the speed of light( 3 ×108 m/s), obtained from optical measurements. This led to the remarkable conclusion that light is an electromagnetic wave. Maxwell’s work thus unified the domain of electricity, magnetism and light. Hertz, in 1885, experimentally demonstrated the existence of electromagnetic waves. Its technological use by Marconi and others led in due course to the revolution in communication that we are witnessing today. In this chapter, we first discuss the need for displacement current and its consequences. Then we present a descriptive account of electromagnetic waves. James Clerk Maxwell The broad spectrum of electromagnetic waves, (1831 – 1879) Born in stretching from g rays (wavelength ~10–12 m) to long Edinburgh, Scotland, radio waves (wavelength ~106 m) is described. was among the greatest physicists of the nineteenth century. He 8.2 DISPLACEMENT CURRENT derived the thermal We have seen in Chapter 4 that an electrical current velocity distribution of produces a magnetic field around it. Maxwell showed molecules in a gas and that for logical consistency, a changing electric field must was among the first to obtain reliable also produce a magnetic field. This effect is of great estimates of molecular importance because it explains the existence of radio parameters from waves, gamma rays and visible light, as well as all other measurable quantities forms of electromagnetic waves. like viscosity, etc. To see how a changing electric field gives rise to Maxwell’s greatest a magnetic field, let us consider the process of acheivement was the unification of the laws of charging of a capacitor and apply Ampere’s circuital electricity and law given by (Chapter 4) magnetism (discovered by Coulomb, Oersted, “B.dl = m0 i (t) (8.1)(1831–1879) Ampere and Faraday) to find magnetic field at a point outside the capacitor. into a consistent set of Figure 8.1(a) shows a parallel plate capacitor C which equations now called Maxwell’s equations. is a part of circuit through which a time-dependent From these he arrived at current i (t) flows . Let us find the magnetic field at a the most important point such as P, in a region outside the parallel plateMAXWELL conclusion that light is capacitor. For this, we consider a plane circular loop of an electromagnetic radius r whose plane is perpendicular to the direction wave. Interestingly, of the current-carrying wire, and which is centred Maxwell did not agree symmetrically with respect to the wire [Fig. 8.1(a)]. FromCLERK with the idea (strongly suggested by the symmetry, the magnetic field is directed along the Faraday’s laws of circumference of the circular loop and is the same in electrolysis) that magnitude at all points on the loop so that if B is theJAMES electricity was magnitude of the field, the left side of Eq. (8.1) is B (2p r). particulate in nature. So we have B (2pr) = m0i (t) (8 .2) 202 Reprint 2025-26 Electromagnetic Waves Now, consider a different surface, which has the same boundary. This is a pot like surface [Fig. 8.1(b)] which nowhere touches the current, but has its bottom between the capacitor plates; its mouth is the circular loop mentioned above. Another such surface is shaped like a tiffin box (without the lid) [Fig. 8.1(c)]. On applying Ampere’s circuital law to such surfaces with the same perimeter, we find that the left hand side of Eq. (8.1) has not changed but the right hand side is zero and not m0i, since no current passes through the surface of Fig. 8.1(b) and (c). So we have a contradiction; calculated one way, there is a magnetic field at a point P; calculated another way, the magnetic field at P is zero. Since the contradiction arises from our use of Ampere’s circuital law, this law must be missing something. The missing term must be such that one gets the same magnetic field at point P, no matter what surface is used. We can actually guess the missing term by looking carefully at Fig. 8.1(c). Is there anything passing through the surface S between the plates of the capacitor? Yes, of course, the electric field! If the plates of the capacitor have an area A, and a total charge Q, the magnitude of the electric field E between the plates is (Q/A)/e0 (see Eq. 2.41). The field is perpendicular to the surface S of Fig. 8.1(c). It has the same magnitude over the area A of the capacitor plates, and vanishes outside it. So what is the electric flux FE through the surface S ? Using Gauss’s law, it is 1 Q Q ΦE = E A = A = (8.3) ε0 A ε0 Now if the charge Q on the capacitor plates changes with time, there is a current i = (dQ/dt), so that using Eq. (8.3), we have dΦE d  Q  1 d Q = FIGURE 8.1 A d t d t  ε0 = ε0 d t parallel plate capacitor C, as part of This implies that for consistency, a circuit through which a time  dΦE  dependent current i (t) flows, (a) a loop of ε0  d t  = i (8.4) radius r, to determine This is the missing term in Ampere’s circuital law. If we generalise magnetic field at a this law by adding to the total current carried by conductors through point P on the loop; the surface, another term which is e0 times the rate of change of electric (b) a pot-shaped surface passingflux through the same surface, the total has the same value of current i through the interior for all surfaces. If this is done, there is no contradiction in the value of B between the capacitor obtained anywhere using the generalised Ampere’s law. B at the point P plates with the loop shown in (a) as itsis non-zero no matter which surface is used for calculating it. B at a rim; (c) a tiffin- point P outside the plates [Fig. 8.1(a)] is the same as at a point M just shaped surface with inside, as it should be. The current carried by conductors due to flow of the circular loop as charges is called conduction current. The current, given by Eq. (8.4), is a its rim and a flat circular bottom S new term, and is due to changing electric field (or electric displacement, between the capacitor an old term still used sometimes). It is, therefore, called displacement plates. The arrows show uniform electriccurrent or Maxwell’s displacement current. Figure 8.2 shows the electric field between the and magnetic fields inside the parallel plate capacitor discussed above. capacitor plates. The generalisation made by Maxwell then is the following. The source of a magnetic field is not just the conduction electric current due to flowing 203 Reprint 2025-26 Physics charges, but also the time rate of change of electric field. More precisely, the total current i is the sum of the conduction current denoted by ic, and the displacement current denoted by id (= ✒0 (d✂E/dt)). So we have dΦE i = i c + i d = i c + ε0 (8.5) d t In explicit terms, this means that outside the capacitor plates, we have only conduction current ic = i, and no displacement current, i.e., id = 0. On the other hand, inside the capacitor, there is no conduction current, i.e., ic = 0, and there is only displacement current, so that id = i. The generalised (and correct) Ampere’s circuital law has the same form as Eq. (8.1), with one difference: “the total current passing through any surface of which the closed loop is the perimeter” is the sum of the conduction current and the displacement current. The generalised law is dΦE B idl = µ0 i c + µ0 ε0 (8.6) Ñ∫ dt and is known as Ampere-Maxwell law. In all respects, the displacement current has the same physical effects as the conduction current. In some cases, for example, steady electric fields in a conducting wire, the displacement current may be zero since the electric field E does not change with time. In other FIGURE 8.2 (a) The cases, for example, the charging capacitor above, both conduction electric and magnetic and displacement currents may be present in different regions of fields E and B between space. In most of the cases, they both may be present in the same the capacitor plates, at region of space, as there exist no perfectly conducting or perfectly the point M. (b) A cross insulating medium. Most interestingly, there may be large regions sectional view of Fig. (a). of space where there is no conduction current, but there is only a displacement current due to time-varying electric fields. In such a region, we expect a magnetic field, though there is no (conduction) current source nearby! The prediction of such a displacement current can be verified experimentally. For example, a magnetic field (say at point M) between the plates of the capacitor in Fig. 8.2(a) can be measured and is seen to be the same as that just outside (at P). The displacement current has (literally) far reaching consequences. One thing we immediately notice is that the laws of electricity and magnetism are now more symmetrical*. Faraday’s law of induction states that there is an induced emf equal to the rate of change of magnetic flux. Now, since the emf between two points 1 and 2 is the work done per unit charge in taking it from 1 to 2, the existence of an emf implies the existence of an electric field. So, we can rephrase Faraday’s law of electromagnetic induction by saying that a magnetic field, changing with time, gives rise to an electric field. Then, the fact that an electric field changing with time gives rise to a magnetic field, is the symmetrical counterpart, and is * They are still not perfectly symmetrical; there are no known sources of magnetic field (magnetic monopoles) analogous to electric charges which are sources of 204 electric field. Reprint 2025-26 Electromagnetic Waves a consequence of the displacement current being a source of a magnetic field. Thus, time- dependent electric and magnetic fields give rise to each other! Faraday’s law of electromagnetic induction and Ampere-Maxwell law give a quantitative expression of this statement, with the current being the total current, as in Eq. (8.5). One very important consequence of this symmetry is the existence of electromagnetic waves, which we discuss qualitatively in the next section. MAXWELL’S EQUATIONS IN VACUUM 1. “E.dA = Q/✒0 (Gauss’s Law for electricity) 2. “B.dA = 0 (Gauss’s Law for magnetism) –dΦB 3. “E.dl == (Faraday’s Law) d t d ΦE + µ0 ε0 (Ampere – Maxwell Law) 4. “B.dl = µ0 i c d t