8.2 A parallel plate capacitor (Fig. 8.6) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to 213 a 230 V ac supply with a (angular) frequency of 300 rad s–1. Reprint 2025-26 Physics (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates. FIGURE 8.6 8.3 What physical quantity is the same for X-rays of wavelength 10–10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500m? 8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength? 8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band? 8.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator? 8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave? 8.8 Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is n = 50.0 MHz. (a) Determine, B0,w, k, and l. (b) Find expressions for E and B. 8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hn (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation? 8.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s–1.] Reprint 2025-26

8.3 ELECTROMAGNETIC WAVES 8.3.1 Sources of electromagnetic waves How are electromagnetic waves produced? Neither stationary charges nor charges in uniform motion (steady currents) can be sources of electromagnetic waves. The former produces only electrostatic fields, while the latter produces magnetic fields that, however, do not vary with time. It is an important result of Maxwell’s theory that accelerated charges radiate electromagnetic waves. The proof of this basic result is beyond the scope of this book, but we can accept it on the basis of rough, qualitative reasoning. Consider a charge oscillating with some frequency. (An oscillating charge is an example of accelerating charge.) This produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn, is a source of oscillating electric field, and so on. The oscillating electric and magnetic fields thus regenerate each other, so to speak, as the wave propagates through the space. The frequency of the electromagnetic wave naturally equals the frequency of oscillation of the charge. The energy associated with the propagating wave comes at the expense of the energy of the source – the accelerated charge. From the preceding discussion, it might appear easy to test the prediction that light is an electromagnetic wave. We might think that all we needed to do was to set up an ac circuit in which the current oscillate at the frequency of visible light, say, yellow light. But, alas, that is not possible. The frequency of yellow light is about 6 × 1014 Hz, while the frequency that we get even with modern electronic circuits is hardly about 1011 Hz. This is why the experimental demonstration of electromagnetic 205 Reprint 2025-26 Physics wave had to come in the low frequency region (the radio wave region), as in the Hertz’s experiment (1887). Hertz’s successful experimental test of Maxwell’s theory created a sensation and sparked off other important works in this field. Two important achievements in this connection deserve mention. Seven years after Hertz, Jagdish Chandra Bose, working at Calcutta (now Kolkata), succeeded in producing and observing electromagnetic waves of much shorter 8.1 wavelength (25 mm to 5 mm). His experiment, like that of Hertz’s, was confined to the laboratory. At around the same time, Guglielmo Marconi in Italy followed Hertz’s work and succeeded in transmitting EXAMPLE electromagnetic waves over distances of many kilometres. Heinrich Rudolf Hertz Marconi’s experiment marks the beginning of the field of (1857 – 1894) German communication using electromagnetic waves. physicist who was the first to broadcast and 8.3.2 Nature of electromagnetic wavesHEINRICH receive radio waves. He It can be shown from Maxwell’s equations that electric produced electro- and magnetic fields in an electromagnetic wave are magnetic waves, sent them through space, and perpendicular to each other, and to the direction of measured their wave- propagation. It appears reasonable, say from ourRUDOLF length and speed. He discussion of the displacement current. Consider showed that the nature Fig. 8.2. The electric field inside the plates of the capacitor of their vibration, is directed perpendicular to the plates. The magnetic reflection and refraction field this gives rise to via the displacement current is was the same as that ofHERTZ along the perimeter of a circle parallel to the capacitor light and heat waves, plates. So B and E are perpendicular in this case. This establishing their identity for the first time. is a general feature. He also pioneered In Fig. 8.3, we show a typical example of a plane research on discharge of electromagnetic wave propagating along the z direction electricity through gases, (the fields are shown as a function of the z coordinate, at and discovered the(1857–1894) a given time t). The electric field Ex is along the x-axis, photoelectric effect. and varies sinusoidally with z, at a given time. The magnetic field By is along the y-axis, and again varies sinusoidally with z. The electric and magnetic fields Ex and By are perpendicular to each other, and to the direction z of propagation. We can write Ex and By as follows: Ex= E0 sin (kz–wt) [8.7(a)] By= B0 sin (kz–wt) [8.7(b)] Here k is related to the wave length FIGURE 8.3 A linearly polarised electromagnetic wave, l of the wave by the usual propagating in the z-direction with the oscillating electric field E equation along the x-direction and the oscillating magnetic field B along the y-direction. 2 π k = (8.8) 206 λ Reprint 2025-26 Electromagnetic Waves and ω is the angular frequency. k is the magnitude of the wave vector (or propagation vector) k and its direction describes the direction of propagation of the wave. The speed of propagation of the wave is (ω/k). Using Eqs. [8.7(a) and (b)] for Ex and By and Maxwell’s equations, one finds that ω = ck, where, c = 1/ µ0ε0 [8.9(a)] The relation ω = ck is the standard one for waves (see for example, Section 14.4 of class XI Physics textbook). This relation is often written in terms of frequency, ν (=ω/2π) and wavelength, λ (=2π/k) as  2π  2 πν = c  λ  or νλ = c [8.9(b)] It is also seen from Maxwell’s equations that the magnitude of the electric and the magnetic fields in an electromagnetic wave are related as B0 = (E0/c) (8.10) We here make remarks on some features of electromagnetic waves. They are self-sustaining oscillations of electric and magnetic fields in free space, or vacuum. They differ from all the other waves we have studied so far, in respect that no material medium is involved in the vibrations of the electric and magnetic fields. But what if a material medium is actually there? We know that light, an electromagnetic wave, does propagate through glass, for example. We have seen earlier that the total electric and magnetic fields inside a medium are described in terms of a permittivity ε and a magnetic permeability µ (these describe the factors by which the total fields differ from the external fields). These replace ε0 and µ0 in the description to electric and magnetic fields in Maxwell’s equations with the result that in a material medium of permittivity ε and magnetic permeability µ, the velocity of light becomes, 1 v = µε (8.11) Thus, the velocity of light depends on electric and magnetic properties of the medium. We shall see in the next chapter that the refractive index of one medium with respect to the other is equal to the ratio of velocities of light in the two media. The velocity of electromagnetic waves in free space or vacuum is an important fundamental constant. It has been shown by experiments on electromagnetic waves of different wavelengths that this velocity is the same (independent of wavelength) to within a few metres per second, out of a value of 3×108 m/s. The constancy of the velocity of em waves in vacuum is so strongly supported by experiments and the actual value is so well known now that this is used to define a standard of length. The great technological importance of electromagnetic waves stems from their capability to carry energy from one place to another. The radio and TV signals from broadcasting stations carry energy. Light carries energy from the sun to the earth, thus making life possible on the earth. 207 Reprint 2025-26 Physics Example 8.1 A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, E = 6.3 ˆj V/m. What is B at this point? Solution Using Eq. (8.10), the magnitude of B is E B = c 6.3 V/m –8 = 8 = 2.1 × 10 T 3 × 10 m/s 8.1 To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction. Since, (+ ˆj ) × (+ ˆk ) = ˆi , B is along the z-direction. EXAMPLE Thus, B = 2.1 × 10–8 ˆk T Example 8.2 The magnetic field in a plane electromagnetic wave is given by By = (2 × 10–7) T sin (0.5×103x+1.5×1011t). (a) What is the wavelength and frequency of the wave? (b) Write an expression for the electric field. Solution (a) Comparing the given equation with   x t   By=B0 Sin 2p +  λ T  spectrum  2π We get, λ = 3 m = 1.26 cm, 0.5 × 10 1 11 and = ν= 1.5 × 10 /2 π = 23.9 GHz T ( ) 8.2 (b) E0 = B0c = 2×10–7 T × 3 × 108 m/s = 6 × 101 V/mElectromagnetic http://www.fnal.gov/pub/inquiring/more/light http://imagine.gsfc.nasa.gov/docs/science/ The electric field component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along the z-axis is obtained as EXAMPLE Ez = 60 sin (0.5 × 103x + 1.5 × 1011 t) V/m 8.4 ELECTROMAGNETIC SPECTRUM At the time Maxwell predicted the existence of electromagnetic waves, the only familiar electromagnetic waves were the visible light waves. The existence of ultraviolet and infrared waves was barely established. By the end of the nineteenth century, X-rays and gamma rays had also been discovered. We now know that, electromagnetic waves include visible light waves, X-rays, gamma rays, radio waves, microwaves, ultraviolet and infrared waves. The classification of em waves according to frequency is the electromagnetic spectrum (Fig. 8.4). There is no sharp division between one kind of wave and the next. The classification is based roughly on how the waves are produced and/or detected. We briefly describe these different types of electromagnetic waves, in 208 order of decreasing wavelengths. Reprint 2025-26 Electromagnetic Waves FIGURE 8.4 The electromagnetic spectrum, with common names for various part of it. The various regions do not have sharply defined boundaries. 8.4.1 Radio waves Radio waves are produced by the accelerated motion of charges in conducting wires. They are used in radio and television communication systems. They are generally in the frequency range from 500 kHz to about 1000 MHz. The AM (amplitude modulated) band is from 530 kHz to 1710 kHz. Higher frequencies upto 54 MHz are used for short wave bands. TV waves range from 54 MHz to 890 MHz. The FM (frequency modulated) radio band extends from 88 MHz to 108 MHz. Cellular phones use radio waves to transmit voice communication in the ultrahigh frequency (UHF) band. How these waves are transmitted and received is described in Chapter 15. 8.4.2 Microwaves Microwaves (short-wavelength radio waves), with frequencies in the gigahertz (GHz) range, are produced by special vacuum tubes (called klystrons, magnetrons and Gunn diodes). Due to their short wavelengths, they are suitable for the radar systems used in aircraft navigation. Radar also provides the basis for the speed guns used to time fast balls, tennis- serves, and automobiles. Microwave ovens are an interesting domestic application of these waves. In such ovens, the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves is transferred efficiently to the kinetic energy of 209the molecules. This raises the temperature of any food containing water. Reprint 2025-26 Physics 8.4.3 Infrared waves Infrared waves are produced by hot bodies and molecules. This band lies adjacent to the low-frequency or long-wave length end of the visible spectrum. Infrared waves are sometimes referred to as heat waves. This is because water molecules present in most materials readily absorb infrared waves (many other molecules, for example, CO2, NH3, also absorb infrared waves). After absorption, their thermal motion increases, that is, they heat up and heat their surroundings. Infrared lamps are used in physical therapy. Infrared radiation also plays an important role in maintaining the earth’s warmth or average temperature through the greenhouse effect. Incoming visible light (which passes relatively easily through the atmosphere) is absorbed by the earth’s surface and re- radiated as infrared (longer wavelength) radiations. This radiation is trapped by greenhouse gases such as carbon dioxide and water vapour. Infrared detectors are used in Earth satellites, both for military purposes and to observe growth of crops. Electronic devices (for example semiconductor light emitting diodes) also emit infrared and are widely used in the remote switches of household electronic systems such as TV sets, video recorders and hi-fi systems. 8.4.4 Visible rays It is the most familiar form of electromagnetic waves. It is the part of the spectrum that is detected by the human eye. It runs from about 4 × 1014 Hz to about 7 × 1014 Hz or a wavelength range of about 700 – 400 nm. Visible light emitted or reflected from objects around us provides us information about the world. Our eyes are sensitive to this range of wavelengths. Different animals are sensitive to different range of wavelengths. For example, snakes can detect infrared waves, and the ‘visible’ range of many insects extends well into the utraviolet. 8.4.5 Ultraviolet rays It covers wavelengths ranging from about 4 × 10–7 m (400 nm) down to 6 × 10–10m (0.6 nm). Ultraviolet (UV) radiation is produced by special lamps and very hot bodies. The sun is an important source of ultraviolet light. But fortunately, most of it is absorbed in the ozone layer in the atmosphere at an altitude of about 40 – 50 km. UV light in large quantities has harmful effects on humans. Exposure to UV radiation induces the production of more melanin, causing tanning of the skin. UV radiation is absorbed by ordinary glass. Hence, one cannot get tanned or sunburn through glass windows. Welders wear special glass goggles or face masks with glass windows to protect their eyes from large amount of UV produced by welding arcs. Due to its shorter wavelengths, UV radiations can be focussed into very narrow beams for high precision applications such as LASIK (Laser- assisted in situ keratomileusis) eye surgery. UV lamps are used to kill germs in water purifiers. Ozone layer in the atmosphere plays a protective role, and hence its depletion by chlorofluorocarbons (CFCs) gas (such as freon) is a matter 210 of international concern. Reprint 2025-26 Electromagnetic Waves 8.4.6 X-rays Beyond the UV region of the electromagnetic spectrum lies the X-ray region. We are familiar with X-rays because of its medical applications. It covers wavelengths from about 10–8 m (10 nm) down to 10–13 m (10–4 nm). One common way to generate X-rays is to bombard a metal target by high energy electrons. X-rays are used as a diagnostic tool in medicine and as a treatment for certain forms of cancer. Because X-rays damage or destroy living tissues and organisms, care must be taken to avoid unnecessary or over exposure. 8.4.7 Gamma rays They lie in the upper frequency range of the electromagnetic spectrum and have wavelengths of from about 10–10m to less than 10–14m. This high frequency radiation is produced in nuclear reactions and also emitted by radioactive nuclei. They are used in medicine to destroy cancer cells. Table 8.1 summarises different types of electromagnetic waves, their production and detections. As mentioned earlier, the demarcation between different regions is not sharp and there are overlaps. TABLE 8.1 DIFFERENT TYPES OF ELECTROMAGNETIC WAVES Type Wavelength range Production Detection Radio > 0.1 m Rapid acceleration and Receiver’s aerials decelerations of electrons in aerials Microwave 0.1m to 1 mm Klystron valve or Point contact diodes magnetron valve Infra-red 1mm to 700 nm Vibration of atoms Thermopiles and molecules Bolometer, Infrared photographic film Light 700 nm to 400 nm Electrons in atoms emit The eye light when they move from Photocells one energy level to a Photographic film lower energy level Ultraviolet 400 nm to 1nm Inner shell electrons in Photocells atoms moving from one Photographic film energy level to a lower level X-rays 1nm to 10–3 nm X-ray tubes or inner shell Photographic film electrons Geiger tubes Ionisation chamber Gamma rays <10–3 nm Radioactive decay of the -do- nucleus 211 Reprint 2025-26 Physics SUMMARY 1. Maxwell found an inconsistency in the Ampere’s law and suggested the existence of an additional current, called displacement current, to remove this inconsistency. This displacement current is due to time-varying electric field and is given by dΦΕ di = ε0 dt and acts as a source of magnetic field in exactly the same way as conduction current. 2. An accelerating charge produces electromagnetic waves. An electric charge oscillating harmonically with frequency n, produces electromagnetic waves of the same frequency n. An electric dipole is a basic source of electromagnetic waves. 3. Electromagnetic waves with wavelength of the order of a few metres were first produced and detected in the laboratory by Hertz in 1887. He thus verified a basic prediction of Maxwell’s equations. 4. Electric and magnetic fields oscillate sinusoidally in space and time in an electromagnetic wave. The oscillating electric and magnetic fields, E and B are perpendicular to each other, and to the direction of propagation of the electromagnetic wave. For a wave of frequency n, wavelength l, propagating along z-direction, we have E = Ex (t) = E0 sin (kz – w t )   z     z t   = E0 sin 2 π  λ − νt  = E 0 sin 2 π  λ − T   B = By(t) = B0 sin (kz – w t)   z     z t   = B 0 sin 2 π  λ − νt  = B 0 sin  2 π  λ − T   They are related by E0/B0 = c. 5. The speed c of electromagnetic wave in vacuum is related to m0 and e0 (the free space permeability and permittivity constants) as follows: c = 1/ µ0 ε0 . The value of c equals the speed of light obtained from optical measurements. Light is an electromagnetic wave; c is, therefore, also the speed of light. Electromagnetic waves other than light also have the same velocity c in free space. The speed of light, or of electromagnetic waves in a material medium is given by v = 1/ µε where m is the permeability of the medium and e its permittivity. 6. The spectrum of electromagnetic waves stretches, in principle, over an infinite range of wavelengths. Different regions are known by different names; g-rays, X-rays, ultraviolet rays, visible rays, infrared rays, microwaves and radio waves in order of increasing wavelength from 10–2 Å or 10–12 m to 106 m. They interact with matter via their electric and magnetic fields which set in oscillation charges present in all matter. The detailed interaction and so the mechanism of absorption, scattering, etc., depend on the wavelength of the electromagnetic wave, and the nature of the atoms and molecules 212 in the medium. Reprint 2025-26 Electromagnetic Waves POINTS TO PONDER 1. The basic difference between various types of electromagnetic waves lies in their wavelengths or frequencies since all of them travel through vacuum with the same speed. Consequently, the waves differ considerably in their mode of interaction with matter. 2. Accelerated charged particles radiate electromagnetic waves. The wavelength of the electromagnetic wave is often correlated with the characteristic size of the system that radiates. Thus, gamma radiation, having wavelength of 10–14 m to 10–15 m, typically originate from an atomic nucleus. X-rays are emitted from heavy atoms. Radio waves are produced by accelerating electrons in a circuit. A transmitting antenna can most efficiently radiate waves having a wavelength of about the same size as the antenna. Visible radiation emitted by atoms is, however, much longer in wavelength than atomic size. 3. Infrared waves, with frequencies lower than those of visible light, vibrate not only the electrons, but entire atoms or molecules of a substance. This vibration increases the internal energy and consequently, the temperature of the substance. This is why infrared waves are often called heat waves. 4. The centre of sensitivity of our eyes coincides with the centre of the wavelength distribution of the sun. It is because humans have evolved with visions most sensitive to the strongest wavelengths from the sun. EXERCISES

14.5 THE PRINCIPLE OF SUPERPOSITION the medium. If the waves arrive in a region OF WAVES simultaneously, and therefore, overlap, the net displacement y (x,t) is given by What happens when two wave pulses travelling in opposite directions cross each other y (x, t) = y1(x, t) + y2(x, t) (14.25) (Fig. 14.9)? It turns out that wave pulses If we have two or more waves moving in the continue to retain their identities after they have medium the resultant waveform is the sum of crossed. However, during the time they overlap, wave functions of individual waves. That is, if the wave pattern is different from either of the the wave functions of the moving waves are Reprint 2025-26 288 PHYSICS y1 = f1(x–vt), y2 = f2(x–vt), .......... .......... yn = fn (x–vt) then the wave function describing the disturbance in the medium is y = f1(x – vt)+ f2(x – vt)+ ...+ fn(x – vt) n = ∑ f ( x − vt ) (14.26) i i =1 The principle of superposition is basic to the phenomenon of interference. For simplicity, consider two harmonic travelling waves on a stretched string, both with the same ω (angular frequency) and k (wave number), and, therefore, the same wavelength Fig. 14.10 The resultant of two harmonic waves of λ. Their wave speed will be identical. Let us equal amplitude and wavelength further assume that their amplitudes are equal according to the principle of superposition. and they are both travelling in the positive The amplitude of the resultant wave depends on the phase difference φ, whichdirection of x-axis. The waves only differ in their is zero for (a) and π for (b)initial phase. According to Eq. (14.2), the two waves are described by the functions: φ between the constituent two waves: y1(x, t) = a sin (kx – ωt) (14.27) A(φ) = 2a cos ½φ (14.32) For φ = 0, when the waves are in phase, and y2(x, t) = a sin (kx – ωt + φ ) (14.28) y ( x , t ) = 2a sin ( kx − ωt ) (14.33) The net displacement is then, by the principle i.e., the resultant wave has amplitude 2a, theof superposition, given by largest possible value for A. For φ = π , the y (x, t) = a sin (kx – ωt) + a sin (kx – ωt + φ) waves are completely, out of phase and the (14.29) resultant wave has zero displacement   ( kx − ωt ) + ( kx − ωt + φ)  φ everywhere at all times = a  2sin   cos  y (x, t) = 0 (14.34)   2  2  Eq. (14.33) refers to the so-called constructive (14.30) interference of the two waves where the where we have used the familiar trignometric amplitudes add up in the resultant wave. Eq. identity for sin A + sin B . We then have (14.34) is the case of destructive intereference where the amplitudes subtract out in the φ  φ resultant wave. Fig. 14.10 shows these two cases y ( x , t ) = 2 a cos sin  kx − ωt +  (14.31) 2  2  of interference of waves arising from the principle of superposition.Eq. (14.31) is also a harmonic travelling wave in the positive direction of x-axis, with the same 14.6 REFLECTION OF WAVES frequency and wavelength. However, its initial So far we considered waves propagating in an φ unbounded medium. What happens if a pulse phase angle is . The significant thing is that 2 or a wave meets a boundary? If the boundary is its amplitude is a function of the phase difference rigid, the pulse or wave gets reflected. The Reprint 2025-26 WAVES 289 phenomenon of echo is an example of reflection If on the other hand, the boundary point is by a rigid boundary. If the boundary is not not rigid but completely free to move (such as in completely rigid or is an interface between two the case of a string tied to a freely moving ring different elastic media, the situation is some on a rod), the reflected pulse has the same phase what complicated. A part of the incident wave is and amplitude (assuming no energy dissipation) reflected and a part is transmitted into the as the incident pulse. The net maximum second medium. If a wave is incident obliquely displacement at the boundary is then twice the on the boundary between two different media amplitude of each pulse. An example of non- rigid the transmitted wave is called the refracted boundary is the open end of an organ pipe. wave. The incident and refracted waves obey To summarise, a travelling wave or pulse Snell’s law of refraction, and the incident and suffers a phase change of π on reflection at a reflected waves obey the usual laws of rigid boundary and no phase change on reflection. reflection at an open boundary. To put this Fig. 14.11 shows a pulse travelling along a mathematically, let the incident travelling stretched string and being reflected by the wave be boundary. Assuming there is no absorption of y 2 ( x , t ) = a sin ( kx − ωt )energy by the boundary, the reflected wave has the same shape as the incident pulse but it At a rigid boundary, the reflected wave is given suffers a phase change of π or 1800 on reflection. by This is because the boundary is rigid and the yr(x, t) = a sin (kx – ωt + π). disturbance must have zero displacement at all = – a sin (kx – ωt) (14.35) times at the boundary. By the principle of At an open boundary, the reflected wave is given superposition, this is possible only if the reflected by and incident waves differ by a phase of π, so that yr(x, t) = a sin (kx – ωt + 0). the resultant displacement is zero. This = a sin (kx – ωt) (14.36) reasoning is based on boundary condition on a Clearly, at the rigid boundary, y = y 2 + y r = 0rigid wall. We can arrive at the same conclusion dynamically also. As the pulse arrives at the wall, at all times. it exerts a force on the wall. By Newton’s Third 14.6.1 Standing Waves and Normal Modes Law, the wall exerts an equal and opposite force We considered above reflection at one boundary. on the string generating a reflected pulse that But there are familiar situations (a string fixed differs by a phase of π. at either end or an air column in a pipe with either end closed) in which reflection takes place at two or more boundaries. In a string, for example, a wave travelling in one direction will get reflected at one end, which in turn will travel and get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called standing waves or stationary waves. To see this mathematically, consider a wave travelling along the positive direction of x-axis and a reflected wave of the same amplitude and wavelength in the negative direction of x-axis. From Eqs. (14.2) and (14.4), with φ = 0, we get: y1(x, t) = a sin (kx – ωt) y2(x, t) = a sin (kx + ωt) The resultant wave on the string is, according to the principle of superposition: Fig. 14.11 Reflection of a pulse meeting a rigid boundary. y (x, t) = y1(x, t) + y2(x, t) Reprint 2025-26 290 PHYSICS = a [sin (kx – ωt) + sin (kx + ωt)] nodes; the points at which the amplitude is the largest are called antinodes. Fig. 14.12 showsUsing the familiar trignometric identity a stationary wave pattern resulting fromSin (A+B) + Sin (A–B) = 2 sin A cosB we get, superposition of two travelling waves in y (x, t) = 2a sin kx cos ωt (14.37) opposite directions. Note the important difference in the wave The most significant feature of stationary pattern described by Eq. (14.37) from that waves is that the boundary conditions constrain described by Eq. (14.2) or Eq. (14.4). The terms the possible wavelengths or frequencies of kx and ωt appear separately, not in the vibration of the system. The system cannot combination kx - ωt. The amplitude of this wave oscillate with any arbitrary frequency (contrast is 2a sin kx. Thus, in this wave pattern, the this with a harmonic travelling wave), but is amplitude varies from point-to-point, but each characterised by a set of natural frequencies or element of the string oscillates with the same normal modes of oscillation. Let us determine angular frequency ω or time period. There is no these normal modes for a stretched string fixed phase difference between oscillations of different at both ends. elements of the wave. The string as a whole First, from Eq. (14.37), the positions of nodes vibrates in phase with differing amplitudes at (where the amplitude is zero) are given by different points. The wave pattern is neither sin kx = 0 . moving to the right nor to the left. Hence, they which implies are called standing or stationary waves. The kx = nπ; n = 0, 1, 2, 3, ... amplitude is fixed at a given location but, as Since, k = 2π/λ , we get remarked earlier, it is different at different locations. The points at which the amplitude is nλ zero (i.e., where there is no motion at all) are x = ; n = 0, 1, 2, 3, ... (14.38) 2 Fig. 14.12 Stationary waves arising from superposition of two harmonic waves travelling in opposite directions. Note that the positions of zero displacement (nodes) remain fixed at all times. Reprint 2025-26 WAVES 291 Clearly, the distance between any two speed of wave determined by the properties of λ the medium. The n = 2 frequency is called the successive nodes is . In the same way, the second harmonic; n = 3 is the third harmonic 2 and so on. We can label the various harmonics bypositions of antinodes (where the amplitude is the symbol νn ( n = 1, 2, ...).the largest) are given by the largest value of sin Fig. 14.13 shows the first six harmonics of akx : sin k x = 1 stretched string fixed at either end. A string need not vibrate in one of these modes only.which implies Generally, the vibration of a string will be a kx = (n + ½) π ; n = 0, 1, 2, 3, ... superposition of different modes; some modes With k = 2π/λ, we get may be more strongly excited and some less. Musical instruments like sitar or violin are λ based on this principle. Where the string is x = (n + ½) ; n = 0, 1, 2, 3, ... (14.39) 2 plucked or bowed, determines which modes are Again the distance between any two consecutive more prominent than others. Let us next consider normal modes of λ antinodes is . Eq. (14.38) can be applied to oscillation of an air column with one end closed 2 the case of a stretched string of length L fixed at both ends. Taking one end to be at x = 0, the boundary conditions are that x = 0 and x = L are positions of nodes. The x = 0 condition is already satisfied. The x = L node condition requires that the length L is related to λ by λ L = n ; n = 1, 2, 3, ... (14.40) 2 Thus, the possible wavelengths of stationary waves are constrained by the relation 2L λ = ; n = 1, 2, 3, … (14.41) n with corresponding frequencies nv v = , for n = 1, 2, 3, (14.42) 2L We have thus obtained the natural frequencies - the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic. For the stretched string fixed at either end v it is given by v = , corresponding 2 L Fig. 14.13 The first six harmonics of vibrations of a stretched to n = 1 of Eq. (14.42). Here v is the string fixed at both ends. Reprint 2025-26 292 PHYSICS and the other open. A glass tube partially filled modes of this system is more complex. This with water illustrates this system. The end in problem involves wave propagation in two contact with water is a node, while the open end dimensions. However, the underlying physics is is an antinode. At the node the pressure the same. changes are the largest, while the displacement is minimum (zero). At the open end - the u Example 14.5 A pipe, 30.0 cm long, is open antinode, it is just the other way - least pressure at both ends. Which harmonic mode of the change and maximum amplitude of pipe resonates a 1.1 kHz source? Will displacement. Taking the end in contact with resonance with the same source be water to be x = 0, the node condition (Eq. 14.38) observed if one end of the pipe is closed ? is already satisfied. If the other end x = L is an Take the speed of sound in air as antinode, Eq. (14.39) gives 330 m s–1.  1  λ n +L = , for n = 0, 1, 2, 3, …  2  2 Answer The first harmonic frequency is given by The possible wavelengths are then restricted by v v the relation : ν1 = λ1 = 2 L (open pipe) where L is the length of the pipe. The frequency 2 L λ = , for n = 0, 1, 2, 3,... (14.43) of its nth harmonic is: ( n + 1 / 2 ) nv νn = 2L , for n = 1, 2, 3, ... (open pipe) The normal modes – the natural frequencies – of the system are First few modes of an open pipe are shown in Fig. 14.15.  1  v For L = 30.0 cm, v = 330 m s–1, ; n = 0, 1, 2, 3, ... (14.44) ν =  n + 2  2 L n 330 (m s − 1 ) νn = = 550 n s–1 The fundamental frequency corresponds to n = 0, 0.6 (m) v Clearly, a source of frequency 1.1 kHz will and is given by . The higher frequencies resonate at v2, i.e. the second harmonic. 4 L are odd harmonics, i.e., odd multiples of the v v fundamental frequency : 3 , 5 , etc. 4 L 4 L Fig. 14.14 shows the first six odd harmonics of air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode. It is then easily seen that an open air column at both ends generates all harmonics (See Fig. 14.15). The systems above, strings and air columns, can also undergo forced oscillations (Chapter 13). If the external frequency is close to one of the natural frequencies, the system shows resonance. Normal modes of a circular membrane rigidly clamped to the circumference as in a tabla are determined by the boundary condition that no Fundamental point on the circumference of the membrane or third fifth vibrates. Estimation of the frequencies of normal first harmonic harmonic harmonic Reprint 2025-26 WAVES 293 Fig. 14.15 Standing waves in an open pipe, first four harmonics are depicted. while tuning their instruments with each other. They go on tuning until their sensitive ears do seventh ninth eleventh not detect any beats. harmonic harmonic harmonic To see this mathematically, let us consider two harmonic sound waves of nearly equal Fig. 14.14 Normal modes of an air column open at angular frequency ω1 and ω2 and fix the location one end and closed at the other end. Only to be x = 0 for convenience. Eq. (14.2) with a the odd harmonics are seen to be possible. suitable choice of phase (φ = π/2 for each) and, assuming equal amplitudes, gives Now if one end of the pipe is closed (Fig. 14.15), it follows from Eq. (14.15) that the fundamental s1 = a cos ω1t and s2 = a cos ω2t (14.45) frequency is Here we have replaced the symbol y by s, v v since we are referring to longitudinal not transverse displacement. Let ω1 be the (slightly) ν1 = λ1 = 4 L (pipe closed at one end) greater of the two frequencies. The resultant and only the odd numbered harmonics are displacement is, by the principle of present : superposition, s = s1 + s2 = a (cos ω1 t + cos ω2 t) 3v 5v ν3 = , ν5 = , and so on. Using the familiar trignometric identity for 4 L 4 L cos A + cosB, we get For L = 30 cm and v = 330 m s–1, the (ω1 - ω2 ) t (ω1 + ω2 ) t fundamental frequency of the pipe closed at one = 2 a cos cos (14.46) end is 275 Hz and the source frequency 2 2 corresponds to its fourth harmonic. Since this which may be written as : harmonic is not a possible mode, no resonance s = [2 a cos ωb t ] cos ωat (14.47) will be observed with the source, the moment If |ω1 – ω2| <<ω1, ω2, ωa >> ωb, th one end is closed. ⊳ where 14.7 BEATS (ω1 − ω2 ) (ω1 + ω2 ) ωb = and ωa = ‘Beats’ is an interesting phenomenon arising 2 2 from interference of waves. When two harmonic Now if we assume |ω1 – ω2| <<ω1, which means sound waves of close (but not equal) frequencies ωa >> ωb, we can interpret Eq. (14.47) as follows. are heard at the same time, we hear a sound of The resultant wave is oscillating with the average similar frequency (the average of two close angular frequency ωa; however its amplitude is frequencies), but we hear something else also. not constant in time, unlike a pure harmonic We hear audibly distinct waxing and waning of wave. The amplitude is the largest when the the intensity of the sound, with a frequency term cos ωb t takes its limit +1 or –1. In other equal to the difference in the two close words, the intensity of the resultant wave waxes frequencies. Artists use this phenomenon often and wanes with a frequency which is 2ωb = ω1 – Reprint 2025-26 294 PHYSICS ω2. Since ω = 2πν, the beat frequency νbeat, is given by νbeat = ν1 – ν2 (14.48) Fig. 14.16 illustrates the phenomenon of beats for two harmonic waves of frequencies 11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency of 2 Hz. Musical Pillars Temples often have some pillars portraying human figures playing musical instru- ments, but seldom do these pillars themselves produce music. At the Nellaiappar temple in Tamil Nadu, gentle taps on a cluster of pillars carved out of a single piece of rock produce the basic notes of Indian classical music, viz. Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa. Vibrations of these pillars depend on elasticity of the stone used, Fig. 14.16 Superposition of two harmonic waves, one its density and shape. of frequency 11 Hz (a), and the other of Musical pillars are categorised into three frequency 9Hz (b), giving rise to beats of frequency 2 Hz, as shown in (c). types: The first is called the Shruti Pillar, as it can produce the basic notes — the “swaras”. The second type is the Gana u Example 14.6 Two sitar strings A and B Thoongal, which generates the basic tunes playing the note ‘Dha’ are slightly out of that make up the “ragas”. The third variety tune and produce beats of frequency 5 Hz. is the Laya Thoongal pillars that produce The tension of the string B is slightly “taal” (beats) when tapped. The pillars at the increased and the beat frequency is found Nellaiappar temple are a combination of the to decrease to 3 Hz. What is the original Shruti and Laya types. frequency of B if the frequency of A is Archaeologists date the Nelliappar 427 Hz ? temple to the 7th century and claim it was built by successive rulers of the Pandyan Answer Increase in the tension of a string dynasty. increases its frequency. If the original frequency The musical pillars of Nelliappar and of B (νB) were greater than that of A (νA ), further several other temples in southern India like increase in νB should have resulted in an those at Hampi (picture), Kanyakumari, and increase in the beat frequency. But the beat Thiruvananthapuram are unique to the frequency is found to decrease. This shows that country and have no parallel in any other νB < νA. Since νA – νB = 5 Hz, and νA = 427 Hz, we part of the world. get νB = 422 Hz. ⊳ Reprint 2025-26 WAVES 295 SUMMARY 1. Mechanical waves can exist in material media and are governed by Newton’s Laws. 2. Transverse waves are waves in which the particles of the medium oscillate perpendicular to the direction of wave propagation. 3. Longitudinal waves are waves in which the particles of the medium oscillate along the direction of wave propagation. 4. Progressive wave is a wave that moves from one point of medium to another. 5. The displacement in a sinusoidal wave propagating in the positive x direction is given by y (x, t) = a sin (kx – ωt + φ) where a is the amplitude of the wave, k is the angular wave number, ω is the angular frequency, (kx – ωt + φ) is the phase, and φ is the phase constant or phase angle. 6. Wavelength λ of a progressive wave is the distance between two consecutive points of the same phase at a given time. In a stationary wave, it is twice the distance between two consecutive nodes or antinodes. 7. Period T of oscillation of a wave is defined as the time any element of the medium takes to move through one complete oscillation. It is related to the angular frequency ω through the relation 2π T = ω 8. Frequency v of a wave is defined as 1/T and is related to angular frequency by ω ν = 2 π ω λ 9. Speed of a progressive wave is given by v = = = λν k T 10. The speed of a transverse wave on a stretched string is set by the properties of the string. The speed on a string with tension T and linear mass density µ is T v = µ 11. Sound waves are longitudinal mechanical waves that can travel through solids, liquids, or gases. The speed v of sound wave in a fluid having bulk modulus B and density ρ is B v = ρ The speed of longitudinal waves in a metallic bar is Y v = ρ For gases, since B = γP, the speed of sound is γP v = ρ Reprint 2025-26 296 PHYSICS 12. When two or more waves traverse simultaneously in the same medium, the displacement of any element of the medium is the algebraic sum of the displacements due to each wave. This is known as the principle of superposition of waves n f i ( x − vt ) y = ∑ i =1 13. Two sinusoidal waves on the same string exhibit interference, adding or cancelling according to the principle of superposition. If the two are travelling in the same direction and have the same amplitude a and frequency but differ in phase by a phase constant φ, the result is a single wave with the same frequency ω :  1   1  y (x, t) =  2a cos 2 φ sin  kx − ωt + 2 φ If φ = 0 or an integral multiple of 2π, the waves are exactly in phase and the interference is constructive; if φ= π, they are exactly out of phase and the interference is destructive. 14. A travelling wave, at a rigid boundary or a closed end, is reflected with a phase reversal but the reflection at an open boundary takes place without any phase change. For an incident wave yi (x, t) = a sin (kx – ωt ) the reflected wave at a rigid boundary is yr (x, t) = – a sin (kx + ωt ) For reflection at an open boundary yr (x,t ) = a sin (kx + ωt) 15. The interference of two identical waves moving in opposite directions produces standing waves. For a string with fixed ends, the standing wave is given by y (x, t) = [2a sin kx ] cos ωt Standing waves are characterised by fixed locations of zero displacement called nodes and fixed locations of maximum displacements called antinodes. The separation between two consecutive nodes or antinodes is λ/2. A stretched string of length L fixed at both the ends vibrates with frequencies given by n v v = , n = 1, 2, 3, ... 2 L The set of frequencies given by the above relation are called the normal modes of oscillation of the system. The oscillation mode with lowest frequency is called the fundamental mode or the first harmonic. The second harmonic is the oscillation mode with n = 2 and so on. A pipe of length L with one end closed and other end open (such as air columns) vibrates with frequencies given by v v = ( n + ½) , n = 0, 1, 2, 3, ... 2L The set of frequencies represented by the above relation are the normal modes of oscillation of such a system. The lowest frequency given by v/4L is the fundamental mode or the first harmonic. 16. A string of length L fixed at both ends or an air column closed at one end and open at the other end or open at both the ends, vibrates with certain frequencies called their normal modes. Each of these frequencies is a resonant frequency of the system. 17. Beats arise when two waves having slightly different frequencies, ν1 and ν2 and comparable amplitudes, are superposed. The beat frequency is νbeat = ν1 ~ ν2 Reprint 2025-26 WAVES 297 POINTS TO PONDER 1. A wave is not motion of matter as a whole in a medium. A wind is different from the sound wave in air. The former involves motion of air from one place to the other. The latter involves compressions and rarefactions of layers of air. 2. In a wave, energy and not the matter is transferred from one point to the other. 3. In a mechanical wave, energy transfer takes place because of the coupling through elastic forces between neighbouring oscillating parts of the medium. 4. Transverse waves can propagate only in medium with shear modulus of elasticity, Longitudinal waves need bulk modulus of elasticity and are therefore, possible in all media, solids, liquids and gases. 5. In a harmonic progressive wave of a given frequency, all particles have the same amplitude but different phases at a given instant of time. In a stationary wave, all particles between two nodes have the same phase at a given instant but have different amplitudes. 6. Relative to an observer at rest in a medium the speed of a mechanical wave in that medium (v) depends only on elastic and other properties (such as mass density) of the medium. It does not depend on the velocity of the source. EXERCISES 14.1 A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? 14.2 A stone dropped from the top of a tower of height 300 m splashes into the water of a pond near the base of the tower. When is the splash heard at the top given that the speed of sound in air is 340 m s–1 ? (g = 9.8 m s–2) 14.3 A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1. γP 14.4 Use the formula v = to explain why the speed of sound in air ρ (a) is independent of pressure, (b) increases with temperature, (c) increases with humidity. Reprint 2025-26 298 PHYSICS 14.5 You have learnt that a travelling wave in one dimension is represented by a function y = f (x, t) where x and t must appear in the combination x – v t or x + v t, i.e. y = f (x ± v t). Is the converse true? Examine if the following functions for y can possibly represent a travelling wave : (a) (x – vt )2 (b) log [(x + vt)/x0] (c) 1/(x + vt) 14.6 A bat emits ultrasonic sound of frequency 1000 kHz in air. If the sound meets a water surface, what is the wavelength of (a) the reflected sound, (b) the transmitted sound? Speed of sound in air is 340 m s –1 and in water 1486 m s–1. 14.7 A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in the tissue in which the speed of sound is 1.7 km s–1 ? The operating frequency of the scanner is 4.2 MHz. 14.8 A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + π/4) where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave ? If it is travelling, what are the speed and direction of its propagation ? (b) What are its amplitude and frequency ? (c) What is the initial phase at the origin ? (d) What is the least distance between two successive crests in the wave ? 14.9 For the wave described in Exercise 14.8, plot the displacement (y) versus (t) graphs for x = 0, 2 and 4 cm. What are the shapes of these graphs? In which aspects does the oscillatory motion in travelling wave differ from one point to another: amplitude, frequency or phase ? 14.10 For the travelling harmonic wave y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35) where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of (a) 4 m, (b) 0.5 m, (c) λ/2, (d) 3λ/4 14.11 The transverse displacement of a string (clamped at its both ends) is given by  2π  y(x, t) = 0.06 sin  3 x  cos (120 πt) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10–2 kg. Answer the following : (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave ? Reprint 2025-26 WAVES 299 (c) Determine the tension in the string. 14.12 (i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end? 14.13 Given below are some functions of x and t to represent the displacement (transverse or longitudinal) of an elastic wave. State which of these represent (i) a travelling wave, (ii) a stationary wave or (iii) none at all: (a) y = 2 cos (3x) sin (10t) (b) y = 2 x − vt (c) y = 3 sin (5x – 0.5t) + 4 cos (5x – 0.5t) (d) y = cos x sin t + cos 2x sin 2t 14.14 A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10–2 kg and its linear mass density is 4.0 × 10–2 kg m–1. What is (a) the speed of a transverse wave on the string, and (b) the tension in the string? 14.15 A metre-long tube open at one end, with a movable piston at the other end, shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tube length is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature of the experiment. The edge effects may be neglected. 14.16 A steel rod 100 cm long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be 2.53 kHz. What is the speed of sound in steel? 14.17 A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s–1). 14.18 Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B? 14.19 Explain why (or how): (a) in a sound wave, a displacement node is a pressure antinode and vice versa, (b) bats can ascertain distances, directions, nature, and sizes of the obstacles without any “eyes”, (c) a violin note and sitar note may have the same frequency, yet we can distinguish between the two notes, (d) solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases, and (e) the shape of a pulse gets distorted during propagation in a dispersive medium. Reprint 2025-26