Understanding Waves: Key Concepts and Effects

[CHAPTER 8] G Chapter WAVES 8 WAVES 263 LEARNING OBJECTIVE At the end of this chapter the students will be able to: Recall the generation and propagation of waves. Describe the nature of the motions in transverse and longitudinal waves. Understand and use the terms wavelength, frequency and speed of wave. Understand and use the equation v = f. Understand and describe Newton's formula of speed of sound. 0 0 0 પ નવા ના Derive Laplace correction in Newton's formula of speed of sound. Derive the formula v = V + 0.61t. Explain and use the principle of superposition. Understand the terms interference and beats. Understand and describe reflection of waves. Explain the formation of a stationary wave using graphical method. Understand the terms node and anti-node. Understand and describe modes of vibration of string. Understand and describe Doppler's effect and its causes. Waves transport energy without transporting matter. The energy transportation is carried by a disturbance, which spreads out from a source. We are well familiar with different types of waves such as water waves in the ocean, or gently formed ripples on a still pond due to rain drop. When a musician plucks a guitar-string, sound waves are generated which on reaching our ear, produce the sensation of music. Wave disturbances may also come in a concentrated bundle like the shock waves from an aeroplane flying at supersonic speed. Whatever may be the nature of waves, the mechanism by which it [CHAPTER 8] WAVES 264 oscillations. The waves which propagate by the oscillation of material particles are known as mechanical waves. There is another class of waves which, instead of material particles, propagate out in space due to oscillations of electric and magnetic fields. Such waves are known as electromagnetic waves. We will undertake the study of electromagnetic waves at a later stage. Here we will consider the mechanical waves only. The waves generated in ropes, strings, coil of springs, water and air are all mechanical waves. So far we have been considering motion of individual particles but in case of mechanical waves, we study the collective motion of particles. An example will help us here. If you look at a black and white picture in a newspaper with a magnifying glass, you will discover that the picture is made up of many closely spaced dots. If you do not use the magnifier, you do not see the dots. What you see is the collective effect of dots in the form of a picture. Thus what we see as mechanical wave is actually the effect of oscillations of a very large number of particles of the medium through which the wave is passing. PROGRESSIVE WAVES Do You Know? Ultrasonic waves are particularly useful for undersea communication and detection systems. High frequency radio waves, used in inwater, whereas highly directional beams of ultrasonic waves can be made to travel many kilometres. radars travel just a fewcentimetres Drop a pebble into water. Ripples will be produced and spread out across the water. The ripples are the examples of progressive waves because they carry energy across the water surface. A wave, which transfers energy by moving away from the source of disturbance, is called a progressive or travelling wave. There are two kinds of progressive waves - transverse waves and longitudinal waves. Transverse and Longitudinal Waves Consider two persons holding opposite ends of a rope or a hosepipe. Suddenly one person gives one up and down jerk to the rope. This (a) disturbs the rope and creates a hump in it which travels along the rope towards the other person (Figure a and b). When this hump reaches the other person, it causes his hand to move up (Figure c). Thus the energy and momentum imparted to the end of the rope by the first person has reached the other end of the rope by travelling through the rope i.e., a wave has been set up on the rope in the form of a moving hump. We call this type of wave a pulse. The forward motion of the pulse from one end of (b) (c) the rope to the other is an example of progressive wave. The hand jerking the end of the rope is the source of the wave. The rope is the medium in which the wave moves. [CHAPTER 8] WAVES A large and loose spring coil (slinky spring) can be used to demonstrate the effect of the motion of the source in generating waves in a medium. It is better that the spring is laid on a smooth table with its one end fixed so that the spring does not sag under gravity. If the free end of the spring is vibrated from side to side, a pulse of wave having a displacement pattern shown in figure (a) will be generated which will move along the spring. 265 Transverse waves fixed end spring (a) Longitudinal waves Waves If the end of the spring is moved back and forth, along the direction of the spring itself as shown in figure (b), a wave with back and forth displacement will travel along the spring. Waves like those in figure (a) in which displacement of the spring is perpendicular to the direction of the waves are called transverse waves. Waves like those in figure (b) in which displacements are in the direction of propagation of the waves are called longitudinal waves. In this example the coil of spring is the medium, so in general we can say that (b) Transverse waves are those in which particles of the medium are displaced in a direction perpendicular to the direction of propagation of waves and longitudinal waves are those in which the particles of the medium have displacements along the direction of propagation of waves. Both types of waves can be set up in solids. In fluids, however, transverse waves die out very quickly and usually cannot be produced at all. That is why, sound waves in air are longitudinal in nature. PERIODIC WAVES Upto now we have considered wave in the form of a pulse which is set up by a single disturbance in a medium like the snapping of one end of a rope or a coil spring. Continuous, regular and rhythmic disturbances in a medium result from periodic vibrations of a source which cause periodic waves in that medium. A good example of a periodic vibrator is an oscillating mass-spring system (figure a). We have already studied in the previous chapter that the mass of such a system executes SHM. Transverse Periodic Waves Imagine an experiment where one end of a rope is fastened to a mass spring vibrator. As the mass vibrates up and down, we observe a transverse periodic wave travelling along the length of rope (figure b). The wave consists of crests and troughs. The crest is a pattern in which the rope is displaced above its equilibrium position, and in troughs, it has a displacement below its equilibrium position. As the source executes harmonic motion up and down with amplitude A and frequency f, ideally every point along the length of the rope executes SHM in turn, with the same amplitude and frequency. The wave travels towards right as crests and troughs in turn, replace one another, but the points on the rope simply oscillates up and down. The amplitude of the wave is the maximum value of the - -><-----> oo ooo Q ooooo <-----2-----> [CHAPTER 8] letter lambda 2 (figure b). WAVES In principle, the speed of the wave can be measured by timing the motion of a wave crest over a measured distance. But it is not always convenient to observe the motion of the crest. As discussed below, however, the speed of a periodic wave can be found indirectly from its frequency and wavelength. As a wave progresses, each point in the medium oscillates periodically with the frequency and period of the source. Figure illustrates a periodic wave moving to the right, as it might look in photographic snapshots taken every 1/4 period. Follow the progress of the crest that started out from the extreme left at t = 0. The time that this crest takes to move a distance of one wavelength is equal to the time required for a point in the medium to go through one complete oscillation. That is the crest moves one wavelength 2 in one period of oscillation T. The speed V of the crest is therefore, NN 266 (a) t=0 T (b) (c) (d) ity. λ (f) t=T V Distance moved Corresponding time interval λ T All parts of the wave pattern move with the same speed, so the speed of any one crest is just the speed of the wave. We can therefore, say that the speed V of the waves is 1 λ V ee ...... (1) but == f, where f is the frequency of the wave. It is the same as the frequency of the vibrator, generating the waves. Thus eq. (1) becomes V = fλ ...... (2) Phase Relationship between Two Points on a Wave F The profile of periodic waves generated by a source executing SHM is represented by a sine curve. Figure shows the snapshot of a periodic wave passing through a medium. In this figure, set of points are shown which are moving in unison as the periodic wave passes. The points C and C', as they move up and down, are always in the same state of vibration i.e., they always have identical displacements and velocities. Alternatively, we can say that as the wave passes, the points C and C' move in phase. We may also say that C' leads C by one time period of 2л radian. Any point at a distance x, C lags behind by phase angle. 2πx 2 [CHAPTER 8] WAVES 267 ...... are all in phase with each other. These points can be anywhere along the wave and need not correspond with only the highest and lowest points. For example, points such as P, P', P", ...... are all in phase. Each is separated from the next by a distance 2. Some of the points are exactly out of step. For example, when point C reaches its maximum upward displacement, at the same time D reaches its maximum downward displacement. At the instant that C begins to go down, D begins to move up. Points such as these are called one half period out of λ phase. Any two points separated from one another by 2, 3, 52, Longitudinal Periodic Waves ... are out of phase. ........ In the previous section we have considered the generation of transverse periodic waves. Now we will see how the longitudinal periodic waves can be generated. Consider a coil of spring as shown in figure. It is suspended by threads so that it can vibrate horizontally. Suppose an oscillating force F is applied to its end as indicated. The force will alternately stretch and < compress the spring, thereby sending a series of stretched regions. (called rarefaction) and compressions down the spring. We will see the oscillating force causes a longitudinal wave to move down the spring. This type of wave generated in springs is also called a compressional wave. Clearly in a compressional wave, the particles in the path of wave move back and forth along the line of propagation of the wave. F Displacement Notice in figure, the supporting threads would be exactly vertical if the spring were undisturbed. The disturbance passing down the spring causes displacements of the elements of the spring from their equilibrium positions. In figure, the displacements of the threads from the vertical are a direct measure of the displacements of the spring elements. It is, therefore, an easy way to graph the displacements of the spring elements from their equilibrium positions and this is done in the lower part of the figure. Q.1 Define progressive waves with its types. Ans. PROGRESSIVE WAVES (TRAVELLING WAVE) A wave, which transfers energy in moving away from the source of disturbance, is called a progressive or travelling waves (i) There are two kinds of traveling waves (i) Transverse waves. (ii) Longitudinal waves. Transverse Waves "Transverse waves are those in which particles of the medium are displaced in a direction perpendicular to the direction of propagation of waves." e.g., water waves, light waves. (ii) Longitudinal Waves (Compressional Waves) [CHAPTER 8] WAVES 268 Both types of waves can be setup in solids in liquids however, transverse waves die out very quickly and usually cannot be produced at all that is why sound waves in air one longitudinal in nature. Periodic Waves "Continuous, regular and rhythmic disturbances in a medium result from periodic vibrations of a source which cause periodic waves in that medium." e.g., Oscillating mass-spring system. Mean level Transverse Periodic Waves Crest "The portion of transverse wave above its mean position, is called crest." Trough "The potion of transverse wave below its mean position, is called trough." Wavelength Crest u Trough "The distance between any two consecutive crests or troughs, is called wave length." It is denoted by a Greek letter Lambda (2). Q.2 Show that V = λ f. Ans. The time that the crest takes to move a distance of one wave length is equal to the time required for a point in the medium to complete one oscillation i.e., crest moves one wave length 2' in one period of oscillation 'T', the speed 'V of the crest (wave) is As, Distance moved Corresponding time interval S = Vt Crest NNA (a) t=0 (b) (c) (d) Since, λ V S T V t -18 = = f T V = λ f Which is the relation between speed, frequency and wavelength. Longitudinal Periodic Waves (1) t=T [CHAPTER 8] WAVES 269 "The portion of longitudinal wave where particles of medium are very close to each other is called compression." Rarefaction "The portion of longitudinal wave where particles of medium are far apart from each other is called rarefaction." Q.3 Explain Newton's formula for the speed of sound in air. Ans. SPEED OF SOUND IN AIR (i) Newton's Formula for Speed of Sound in Air om When one particle of the medium is disturbed, the disturbance in the form of wave travel in all directions in the medium. The velocity of disturbance depends upon the density and the elasticity of the medium. The lighter the density of the medium, more quickly the disturbance moves from point to point and similarly greater the elasticity of the medium, more quickly disturbance will be propogated from point to point in the medium. Newton proposed the following formula for the velocity of sound through the materials which is as follows: V E Elasticity Density density. Where is the elasticity of the medium and p is the Newton assumed that when a sound wave travels through air, the temperature of the air during compression remains constant and pressure changes from P to (P+ AP) and therefore, the volume changes from V to (V - AV). According to Boyle's law PV (P + AP)(V − AV) Speed of sound in different media Speed ms1 Medium Solids at 20°C Lead Copper Aluminium Iron Glass 1320 3600 5100 5130 5500 or PV = PV - PAV + VAP – APAV The product APAV is very small and can be neglected. So, the above equation becomes: 0 -PAV + VAP VAP P AV ΔΡ Р || AV V Liquids at 20°C Methanol 1120 Water 1483 Gasses of S.T.P. Carbon dioxide 258 Oxygen 315 Air Helium 332 972 [CHAPTER 8] AV = Volume strain WAVES Stress Volume strain Elasticity Р V The above equation becomes: V E P 270 On substituting the values of atmospheric pressure and density of air at S.T.P. in above equation, we find that the speed of sound waves in air comes out to be 280 ms, whereas its experimental value is 332 ms1 Q.4 How laplace correct the speed of sound in air? Ans. LAPLACE CORRECTION FOR VELOCITY OF SOUND IN AIR The sound waves travel in the form of compressions and rarefactions. The compressions and rarefactions are so rapid, the temperature of air does not remain constant. The temperature increases due to compressions and the temperature decreases due to the rarefactions. Therefore during compression the air does not lose heat due to conduction and during rarefaction it does not gain heat. Thus the temperature throughout the medium does not remain constant. The relation between volume and pressure (PV = Constant) is not true but it is give PV = Constant Where y is constant and its value depends upon the nature of the gas where Y Molar specific heat at constant pressure Molar specific heat at constant volume If 'P' be the pressure then the change in pressure is very small which is P + AP therefore volume decreases from V to V - AV then PV (P + AP)(V - AV)" AP) V" (1 - AV) PV = (P+AP)\ Applying Binomial theorem: Р = or Р = (P+AP) AV (P + AP) 1-Y V AV V = P - y P2 + AP - YAP V AV V Where (y AP is negligible. Hence, we have Types of gas For Your Information Values of Constant Y 1.67 1.40 1.29 Monoatomic Diatomic Polyatomic For Your Information Ranges of Hearing Organisme Frequencies (Hz) [CHAPTER 8] or Therefore ΔΡ AV V = YP E V = - E V YP WAVES 271 Cat 60-70,000 Dog 15-50,000 Human 20-20,000 On substituting the value of atmospheric pressure and y then the speed of the sound is 333 m/s. This value of speed of sound is very close to the experimental value. Thus the laplace correction must therefore be correct. 0.5 What is effect of variation of pressure on the speed of sound? Ans. EFFECT OF PRESSURE ON SPEED OF SOUND As, V || Ρ E> Tidbits (a) V YP P→ increase (b) Ρ V→ decreases p→ increase To signal generator . Ραρ Sound waves cause the candle Since density is proportional to the pressure so the speed of flame to flicker. sound is not affected by the variation in pressure of the gas. Q.6 What is effect of density on the speed of sound? Ans. EFFECT OF DENSITY ON SPEED OF SOUND As. V y P Ρ At the same temperature and pressure for the gases having the same value of y, the velocity is inversely proportional to the square root of their density. i.e.. V 1 Note: Speed of sound in hydrogen is four times its speed in oxygen as density of the oxygen is sixteen times that of the hydrogen. Q.7 What is the effect of temperature on the speed of sound in air? Ans. EFFECT OF TEMPERATURE ON SPEED OF SOUND When a gas is heated at constant pressure, its volume is increased and hence, its density is decreased. [CHAPTER 8] Let, Vo WAVES 272 Speed of sound at 0°C Po Density of gas at 0°C Vt = Speed of sound at t°C Pt Density of gas at t°C YP then, Vo ...... (1) Do You Know? Po Wave crests y P and V1 (2) Pt Dividing equation (2) by (1) V1 Vo Vt 51 15 Vt Vo Vt Vo || y P/pt Vy P/Po SK SK Sk Po Po y P Slower than the speed of sound. Shock wave (3) afecity.l If Vo is the volume of a gas at temperature 0°C and Vt is volume at t°C, then by using volume expansion. V1 = Vo (1+ẞt) Faster than the speed of sound. What happens when a jet plane like Concorde flies faster than the speed of sound? A conical surface of concentrated Where ẞ is the coefficient of volume expansion of the gas. For sound energy sweeps over the 1 all gases, its value is about 273 Hence, As P Volume = Vo(1+273) || V Mass Density ground as a supersonic plane passes overhead. It is known as sonic boom. V1- Vox V, t Vt-Vo=ẞVot Vt=Vo+BVot = Vo(1+ẞt) m V Pt Elad 1 Pt |||| || P m Po Po ( 1+ 273 1+273) [CHAPTER 8] Vt রার রার রার রার V Vt || t WAVES P. (1+273 1 + t 273 273+t T To 273 P1 (4) 273 Where T and To are the absolute temperature. Corresponding to 5°C and 0°C respectively. Thus the speed of sound is directly proportional to the square root of the absolute temperature. Now, using Binomial theorem and neglecting high power, we have, eq. (4) as: As, Vt Vo V1 Vo (1 1+ = 1+ 1 t 273 1/2 273 tancitym 273 V1 = (1 + 3/16 Vo V1 = Vo = 332 m/s Putting the value in the 2nd factor, 332 Vo+ 546 t V1 = Vo+0.61 t This shows that one degree Celsius rise in temperature produces approximately 0.61 m/s (61 cm/s) increase in the speed of sound. Q.8 State the principle of superposition. [CHAPTER 8] WAVES 274 Ans. PRINCIPLE OF SUPERPOSITION So far, we have considered single waves. What happens when two waves encounter each other in the same medium? Suppose two waves approach each other on a coil of spring, one travelling towards the right and the other travelling towards left. Figure shows that you would see happening on the spring. The waves pass through each other without being modified. After the encounter, each wave shape looks just as it did before and is travelling along just as it was before. This phenomenon of passing through each other unchanged can be observed with all types of waves. You can easily see that it is true for surface ripples. But what is going on during the time when the two waves. overlap? Figure (c) shows that the displacements they produce just add up. At each instant, the spring's displacement at any point in the overlap region is just the sum of the displacements that would be caused by each of the two waves separately. Thus, if a particle of a medium is simultaneously acted upon by n waves such that its displacement due to each of the individual n waves be y1, y2, ......, Yn, then the resultant displacement of the particle, under the simultaneous action of these n waves is the algebraic sum of all the displacements i.e., Y = y1 + y2 + ... + Yn y2+.. This is called principle of superposition. Again, if two waves which cross each other have opposite phase, their resultant displacement will be Y У1 - У2 Particularly if y1 = y2 then result displacement Y = 0. Principle of superposition leads to many interesting phenomena with waves. Q.9 For Your Information Y Wave 1 2 Wave 2 Y=Y1+Y2 Super posed Resultant wave Superposition of two waves of the same frequency which are exactly in phase. Y1 Wave 1 Wave 2 Y2 Super posed of wave 1 and 2 Resultant wave Y = 0 Superposition of two waves of the same frequency which are exactly out of phase. (i) Two waves having same frequency and travelling in the same direction (Interference). (ii) Two waves of slightly different frequencies and travelling in the same direction (Beats). (iii) Two waves of equal frequency travelling in opposite direction (Stationary waves). State and explain the phenomenon of interference of sound. Ans. INTERFERENCE "Superposition of two waves having the same frequency and traveling in the same direction results in a phenomenon, called interference." There are two types: [CHAPTER 8] Audio generator S1 P.A. Power amplifier S2 A WAVES S1 S2 275 " (i) Microphone Fig. (a) PP Constructive Interference 口] CRO com Fig. (b) "If two waves arrive at a point in phase i.e., compression of one wave falls on compression on other wave and rarefaction of one wave falls on the rarefaction of other wave, then resultant sound is loudest." Whenever path difference is an integral multiple of wavelength the two waves are added up. This effect is called constructive interference. or Condition for constructive interference can be written as where (ii) AS = nλ n = ±1,± 2, ± 3, ..... Destructive Interference "If two waves reach a point out of phase i.e., compression of one wave falls on the rarefaction of other wave and rarefaction of one wave falls on the compression of other wave, then resultant sound will be minimum." or At points where displacements of two waves cancel each other's effect, the path difference is an odd integral multiple of half the wavelength. This effect is called destructive interference. Condition for destructive interference is where AS = λ 22 +1) 2 (2n+1) n = 0,±1,±2, ...... AAA Fig. (c) Constructive interference Large displacement is displayed on the CRO screen Fig. (d) Destructive interference Zero displacement is displayed on the CRO screen Explanation An experimental set up to observe interference effect in sound waves as shown in figure. Two loud speakers S. and S. act as two sources of harmonic sound waves of a fixed frequency [CHAPTER 8] WAVES 276 screen. The microphone is placed at various points, turn by turn, infront of the loud speakers as shown in the figure. At points P1, P3 and P5, we find that compressions met with compressions and rarefactions. So, the displacement of the two waves are added up at these points and large resultant displacement is produced. At points P2 and P4, compressions met with rarefactions so they cancel each other effect. The resultant displacement becomes zero. Now we have to find the path difference between the waves at point P1 is AS AS = S2P1 - S1P1 — 3호 1 = λ But AS = ηλ For constructive interference. Where n = 0, 1, 2, 3,...... For destructive interference enicity.com AS S2P2 - S1P2 AS = 42-352 = 12 So, AS n+ Where n = 0, 1, 2, 3,..... Q.10 What are beats? How they are produced? Show that the number of beats is equal to the difference between the frequencies of the tuning forks. Ans. BEATS other. Beat is the combined effect of two sound waves having frequencies slightly different from each Consider two tuning forks each having frequency 32 cps. Slightly load (with wax or ring) one tuning fork so that frequency decreases a little. Let the frequency becomes 30 cps. The two tuning forks are sounded together and held at equal distance from the ear. Let at t = 0, the two forks are in phase. i.e., their right prongs moves towards right producing compressions at the same time and louder sound is heard by the listener. With passage of time, the tuning fork B (30 cps) begins to fall behind 'A'. [CHAPTER 8] 1 4 WAVES At t = sec, 'A' has completed 8 vib and 'B' has completed 7 vib. The prongs are now out of phase and no sound is heard due to destructive interference. =1 B A V = 32 t = 0 Ear D v = 30 (a) A 3 Ear t = 1/4 sec At t = sec, 'A' has completed 16 vib and 'B' completes 15 vib. Att The prongs of forks become in phase and louder sound is heard. At t 4 sec, 'A' has completed 24 vib while 'B' completes 225 vib. The prongs are now in opposite phase and once again no sound is heard. At t = 1 sec, both forks have completed 32 vib and 30 vib. The prongs become in phase and max or louder sound is heard. It is observed that in 1 sec, the sound falls in intensity twice. The sudden fall of sound in intensity is known as beat. Thus two beats are produced/sec which is equal to the difference in frequencies of two forks (32-30=2). Definition of Beats The periodic alteration of sound between maximum and minimum. loudness as many times a second as the difference in frequencies is called phenomenon of beats. No of beats/sec = Difference in frequencies ±n = fA-fB Graphical Explanation of Beats The displacements of the particles of the medium due to two waves are plotted separately as function of time. The resultant A B A B A D (b) Ear t = 1/2 sec D (c) t = 3/4 sec Ear D (d) t = 1 sec Ear D (e) A (a) displacement of any particle will be the sum of the displacement due to (b) (b) NAMAN each of the two waves. The resultant wave which is produced in shown in figure (c). It is seen that amplitude of resultant waves changes with time. The change in amplitude gives rise to production of beat. Uses of Phenomenon of Beats The phenomenon of beats is used to find out: (i) Unknown frequency (ii) To tune a musical instrument Q.11 Write a note on reflection of waves. Ans. REFLECTION OF WAVES 1 S S 4 Resultant Formation of Beats 277 VVELL In an extensivemedium, a wave travels in all direction from its source with a velocity depending upon the properties of the medium. However, when the wave comes across the boundary of two media, a part of it is reflected back. The reflected wave has the same wavelength and frequency but its phase may [CHAPTER 8] WAVES 278 Now we will discuss two most common cases of reflection at the boundary. These cases will be explained with the help of waves travelling in slinky spring. (A slinky spring is a loose spring which has small initial length but a relatively large extended length). One end of the slinky spring is tied to a rigid support on a smooth horizontal table. When a sharp jerk is given up to the free end of the slinky spring towards the side A, a displacement or a crest will travel from free end to the boundary (Figure a). It will exert a force on bound end towards the side A. Since this end is rigidly bound and acts as a denser medium, It will exert a reaction force on the spring in opposite direction. This force will produce displacement towards B and a trough will travel backwards along the spring (Figure b). A eeeeee (a) B From the above discussion it can be concluded that whenever a transverse wave, travelling in a rarer medium, encounters a denser medium, it bounces back such that the direction of its displacement is reversed. An incident crest on reflection becomes a trough. A Slinky Light string (a) leeeeeeeeeeee A A (b) This experiment is repeated with a little variation by attaching one end of a light string to a slinky spring and the other end to the rigid support as shown in figure. If now the spring is given a sharp jerk towards A, a crest travels along the spring as shown in figure. When this crest reaches the spring-string boundary, it exert a force on the string towards the side it does not oppose the motion of the spring. The end of the spring, therefore, continues its displacement towards A. The spring behaves as if it has been plucked up. In other words a spring crest is again created at the boundary of the spring-string system, which travels backwards along the spring. From this it can be concluded that when a transverse wave travelling in a denser medium, is reflected from the boundary of a rarer medium, the direction of its displacement remains the same. An incident crest is reflected as a crest. We are already familiar with the fact that the direction of displacement is reversed when there is change of 180° in the phase of vibration. So, the above conclusion can be written as follows: (i) (c) If a transverse wave travelling in a rarer medium is incident on a denser medium, it is reflected such that it undergoes a phase change of 180°. (ii) If a transverse wave travelling in a denser medium is incident on a rarer medium, it is reflected without any change in phase. Q.12 What are stationary waves? How they one produced? Define node and anti-node. Ans. STATIONARY WAVES Now let us consider the superposition of two waves moving along a string in opposite directions. Figure (a, b) shows the profile of two such waves at instants = 0, T/4, 3/4 T and T, where T is the time period of the wave. We are interested in finding out the displacements of the points 1, 2, 3, 4, 5, 6 and 7 at these instants as the waves superpose. From the figure (a, b), it is obvious that the points 1, 2, 3, etc., [CHAPTER 8] t=0 t = T/4 WAVES t = T/2 t=3T/4 t=T 279 (b) AAMMA 2 3 4 5 6 2 6 2 1 3 5 7 1 3 4 6 5 7 7 6 3 5 2 7 (a) 2 3 4 5 6 t=0 t = T/4 t = T/2 t = 3T/4 (c)% 3 5 7 1 3 1 3 5 5 7 7 1 3 3 5 5 7 1 1 3 3 5 5 7 7 1 1 ... t = T/4 t = 3T/4 6 2 4 6 4 24 6 t = T/2 t=0 1.. ง t=T always 3 5 7 at rest 3 t=T1 always 6 oscillating are distant 2/4 apart, λ being the wavelength of the waves. We can determine the resultant displacement of these points by applying the principle of superposition. Figure (c) shows the resultant displacement of the points 1, 3, 5 and 7 at the instants t = 0, T/4, T/2, 3T/4 and T. It can be seen that the resultant displacement of these points is always zero. These points of the medium are known as nodes. Figure (c) shows that the distance between two consecutive nodes is 2/2. Figure (d) shows the resultant displacement of the points 2, 4 and 6 at the instant t = 0, T/4, T/2, 3T/4 and T. The figure shows that these points are moving with an amplitude which is the sum of the amplitudes of the component waves. These points are known as antinodes. They are situated midway between the nodes and are also 2/2 apart. The distance between a node and the next antinode is 2/4. Such a pattern of nodes and antinodes is known as a stationary or standing wave. Energy in a wave moves because of the motion of the particles of the medium. The nodes always remain at rest, so energy cannot flow past these points. Hence energy remains "standing" in the medium between nodes, although it alternates between potential and kinetic forms. When the antinodes are all at their extreme displacements, the energy stored is wholly potential and when they are simultaneously passing through their equilibrium positions, the energy is wholly kinetic. An easy way to generate a stationary wave is to superpose a wave travelling down a string with its reflection travelling in opposite direction as explained in the next section. Q.13 Explain the stationary waves in a stretched string. Also calculate the frequencies. Ans. STATIONARY WAVES IN A STRETCHED STRING Consider a string of length '7' which is kept stretched by clamping its ends so that the tension in the string is F. If the string is plucked at its middle point, two transverse waves will originate from this point. One of them will move towards the left end of the string and the other towards the right end. When these waves reach the two clamped ends, they are reflected back, thus giving rise to stationary waves. The string will vibrate with such a frequency fi, so that nodes are formed at two fixed ends (clamped ends) and anti-nodes between them. Thus the string vibrates in one loop as shown in Fig. If 21 is the wavelength of this mode of vibration (1st mode of vibration) so, [CHAPTER 8] As 7 λι V |||| λι 20 WAVES = 21 λ f (i) fi |||| V λι k A NK. Pluck 不 280 f2 22 2>2 ||_|| N A N N 2 f1 fi |||| 21 1 21 F F/m V F .. V f1 21 talecity.co 1/4 1 If the string is plucked from 4 of its length, then again stationary waves will be set up, but now the string vibrate in two loops with f2. If λ2 is wave length in this case then, As, l λ + 2 2 (iii) The speed 'V' of the waves in the string depends upon the tension F of the string and mass per unit length of the string. It is given by (ii) Putting value of 21. K =21 λ1 = 21 不 Putting value of 22 f2 = 2 2) V 21 f2 2f1 → 2 2 =쫄+쓸 This shows that when string vibrates in two loops, its frequency is doubled and wave length becomes half than it vibrates in one loop. [CHAPTER 8] WAVES Similarly if the string is plucked from th of its length, it vibrates in three loops as shown in figure. As 23 23 23 1 2 ++즐 1 1 23 f3 = 3 |||| 21 23 22 alm > 3 3 V 23 1/6 281 N KAL A A N4 N -—-— 1 = 1/2 + 1/2 + 2/2- ++쓸 | Do You Know? Putting value of 23 V f3 21/3 V f3 = 3 21 f3 3f It means that when string vibrates in three loops, its frequency leexity. is three times the frequency when it vibrates in one loop. Thus we can generalize that if the string is made to vibrate in 'n' loop, then and wave length is; fn 2n nf A standing-wave pattern is formed when the length of the string is an integral multiple of half wavelength; otherwise no standing wave is formed. For Your Information where n = 1, 2, 3, ...... It is clear that as the string vibrates in more and more loops, its frequency goes on increasing and the wave length gets shorter. However the product of frequency and wave length is always equal to V, (speed of the wave). The above discussion clearly establishes that the stationary waves have a discrete set of frequencies f1, 2f1, 3f1, ........., n fj which is known as harmonic series. The fundamental frequency 'fi' is called first harmonic (over tone), 'f2' is called second harmonic (over tone), and so on. Note: The frequency of a string on a musical instrument can be changed either by varying the tension or by changing the length. For example the tension in guitar and violin strings is In an organ pipe, the primary Idriving mechanism is wavering. [CHAPTER 8] WAVES 282 Q.14 Describe the stationary waves in air column. Ans. STATIONARY WAVES IN AIR COLUMNS Stationary waves can be set in other media also, such as air column. A common example of vibrating air column is in the organ pipe. The relationship between the incident wave and the reflected wave depends on whether the reflecting end of the pipe is open or closed. If the reflecting end is open, the air molecules have complete freedom of motion and this behaves as an antinode. If the reflecting end is closed, then it behaves as a node because the movement of the molecules is restricted. The modes of vibration of an air column in a pipe open at both ends are shown in figure. In figure, the longitudinal waves set up in the pipe have been represented by transverse curved lines indicating the varying amplitude of vibration of the air particles at points along the axis of the pipe. However, it must be kept in mind that air vibrations are longitudinal along the length of the pipe. The wavelength 2' of nth harmonic and its frequency 'f' of any harmonic is given by (a) (b) V XXX (c) Fig. Stationary longitudinal waves in a pipe open at both ends. 21 λn n n V fn nv 21 ..... (1) 1, 2, 3, 4,... eine where 'v' is the speed of sound in air and '7' is the length of the pipe. The equation (10 can also be written as fn nfi (2) If a pipe is closed at one end and open at the other, the closed end is a node. The modes of vibration in this case are shown in figure. In case of fundamental note, the distance between a node and antinode is one fourth of the wavelength, Hence, Since V Hence, |||||| λι or λι = 41 4 V V f1 λι 4/ It can be proved that in a pipe closed at one end, only odd harmonics are generated, which are given by the equation. where fn nv 41 n = 1, 3, 5, ...... ...... (3) (a) 3v लोक 1 = 32, f2 = 3/1 1= = (b) X 1=52, f3 5v 41 (c) Fig. Stationary longitudinal waves This shows that the pipe, which is open at both ends, is richer in a pipe closed at one end. Only in harmonics. odd harmonics are present. [CHAPTER 8] In 1st mode 7 1 λι 221 4 λι 22 2 WAVES 283 As, V > V fi λι Putting value of '21', In 2nd mode fi 1 V 21 >2 λ2 +222 +λ2 422 4 4 taleemcity.com f2 |||||| = λε V 1 Multiply and divided by (2). In 3rd mode f2 = 2 As, f3 Putting value of 23 .. 23 +223 +223 +23 4 62.3 4 323 3+ 2~ ~lm >12 21 1 3 V 23 V f3 21/3 V f3 = 3 21 [CHAPTER 8] Case II WAVES When pipe is open at one end and closed at other end. 284 In 1st mode λι f1 ||_|| = 41 V λι Putting value of 21 V fi 41 In 2nd mode λε λε || 4 2 22+222 1 4 322 22 f |||||| 4 41 Fm 3 22 Putting value of 22 V f3 41/3 f3 f3 alemcity.c We can generalize, .. Interesting Information Echolocation allows dolphins to detect small differences in the shape, size and thickness of objects. where n = 1, 3, 5, ...... Note: The pipe which is open at both ends is richer in harmonics. Q.15 What is Doppler effect? Describe expression for apparently changed frequency when the observer is at rest while the source is in motion. (OR) What is Doppler effect? Describe the expression for apparently changed frequency when the observer is in motion while source is at rest. Ans. DOPPLER EFFECT Introduction This effect was observed by Johann Doppler while he was observing the frequency of light emitted from distant stars In some cases the frequency of light emitted from a star was found to be [CHAPTER 8] Definition WAVES 285 "The apparent change in the pitch OR frequency of a source of sound, when there is a relative motion between the source of sound and the observer, is called Doppler effect." For example, when an observer is standing on a railway platform, the pitch of the whistle of a approaching train is heard to be higher. But when the same train moves away, the pitch of the whistle becomes lower. Explanation Suppose 'V' is the velocity of the sound in a medium and a source emits a sound of frequency 'f' and wave length '2'. If both the source and the observer are stationary, then the waves received by the observer in one second are; Case-A f >2 V λ When observer 'O' is moving towards a stationary source 'S' Co If an observer ('O') moves towards, a source 'S' with velocity Uo, the relative velocity of the waves and the observer is increased to (V+ Uo), then number of waves received in one second apparent frequency increases Putting the value fA fA V >12 || λ Or V + Uo V λ V/f 1 fA S Observer Fig. An observer moving with velocity u。 towards a stationary source hears a frequency f, that is greater than the source frequency. V + Uo Hence, V fA > 1 V > f V + Uo ) f V This means that when an observer 'O' is moving towards a stationary source S, the frequency of sound increases. Case-B When observer 'O' moves away from a stationary source S [CHAPTER 8] WAVES When observer 'O' moves away from stationary source 'S' with velocity 'Uo,' then relative velocity of sound waves and the observer is, (V Uo), hence, number of waves received by the observer, per-second are reduced and is given by Observer fB V - Uo λ 286 Putting value of λ fB |||| V >14 f V-Uo V/f fB V-Uo) f V - Uo As, < 1 1 V .. fB < f S Ug = 0 Fig. An observer moving with velocity u, away from stationary source hears a frequency fg that is smaller than the source frequency. city.c This means that when observer 'O' moves away from stationary source 'S,' the apparent frequency decreases. Case-C When source 'S' is moving towards the stationary observer 'O' If the source 'S' is moving towards the observer with velocity 'Us' then in one second, the waves are compressed, (Wave length decreases), by an amount known as Doppler shift represented by A2. As, V Δλ = fλ Us f ν The compression of waves is due to the fact that same number of waves are contained in a shorter space depending upon the velocity of the source. λ Αλ. The wave length for observer 'C' is then, λ = 2-A2. Putting the values of λ and A2. .. V Us λc ff Яс V-Us f The modified frequency for observer 'C' is fc V λε S (U2 =D) (uo = 0) Us 42 K λ=2-A2 入 [CHAPTER 8] Putting value of 2c WAVES 287 V fc V-Us f V fc V V-Us ) f > 1 As, V-Us fc > f .. Observer C S Observer Dus Fig. A source moving with velocity u, towards a stationary observer C and away from stationary This means that the observed frequency increases when the observer D. Observer C hears an source is moving towards the observer. Case-D When source 'S' is moving away from stationary observer 'O' If the source is moving away from the observer 'O' with velocity 'Us,' then in one second the waves are stretched (wave length increase) by an amount Αλ. increased and observer D hears a decreased frequency. تر •min S (Ug = 0) (u2 = 0) Us Us Αλ f S The stretched of the waves is due to the fact that same number of wave are contained in larger (longer) space, depending upon the velocity of the source. Αλ The wave length for observer 'D' is, λp = 2 + A2. Putting values of 2 and A2. .. V U λD ff 51- + 2D V+Us f The modified frequency for observer 'D' will be (uo = 0) 20=2+42 Do You Know? V fp 2D Putting value of 2p V .. fp V+Us f fD -(V+U.) f V As, < 1 V+Us fD < f acoustic coupling gel backscattered transmitted signal signal blood flow blood vessel The Doppler effect can be used to monitor blood flow through major arteries. Ultrasound waves of frequencies 5 MHz to 10 MHz are directed towards the artery and a receiver detects the back scattered signal. The apparent frequency depends on the This means that the observed frequency decreases, when the velocity of flow of the blood. source is moving away from the observer [CHAPTER 8] WAVES Ans. APPLICATIONS OF DOPPLER EFFECT Doppler effect is applied in working of radar system. Radar uses radio waves to find the elevation and speed of an aeroplane. Radar is a device which transmits and receives radio waves. The radio waves transmitted from radar are reflected back from aeroplane and are received by radar. If the aeroplane is moving towards the radar then the wavelength of reflected wave is shorter. If the plane is moving away from the radar then the wavelength of reflected wave is longer as shown in figure. Plane approaching Plane receding 288 detect the motion of an aeroplane. Stationery star The difference of wavelength of transmitted and reflected A frequency shift is used in a radar to waves is used to determine the speed of aeroplane. Term SONAR (is acronym) stands for sound navigation and ranging. Sonar is the name of the technique used for detecting the presence of objects under water by acoustical echo. In Sonar "Doppler detection" depends upon relative speed of the target and the detector. It employs. The Doppler effect in which an apparent change in frequency occurs when source and observer are in relative motion with respect to one another. In military it is used for detection and location of submarine antisubmarine weapons and depth measurement of sea. Astronomers use Doppler effect to calculate the speeds of distant stars and galaxies. By comparing the line spectrum of light from the star with light from a laboratory source, the Doppler shift of stars light can be measured. Then speed of star can be calculated. Doppler effect is used to determine whether a particular star or galaxy is approaching the earth or moving away from the earth. Light from the star is measured with the help of spectrometer. It has been found that stars moving towards the earth show a blue shift. Thus is because the emitted waves by the star have shorter wavelength than of the star had been at rest. So the spectrum is shifted towards shorter wavelength i.e., to the blue end of the spectrum. It has been found that stars moving away from the earth show a red shift. This is because the emitted waves by the star have longer wavelength than of the star had been at rest. So the spectrum is shifted towards longer wavelength i.e., to the red end of the spectrum. Astronomers have discovered that all the distant galaxies are moving away from us. They have also measured their speed by measuring their red shift. Another important application of the Doppler shift using electromagnetic waves is radar speed trap. Microwaves are emitted from a transmitter in the form of short bursts. Each burst is reflected back by any moving car in the path of microwaves. The reflected microwaves are received back as Doppler's shift. By measuring Doppler shift the speed of the car can be calculated by computer programme (a) Star receding (b) (c) Star approaching ми Do You Know? [CHAPTER 8] WAVES SOLVED EXAMPLES 289 EXAMPLE 8.1 Data Find the temperature at which the velocity of sound in air is two times its velocity at 10°C. Velocity of sound Vt = 2 V10 T10 = 10+273 = 283 K To Find Temperature = T = ? SOLUTION Ve T Using V10 T10 Since V1 = 2 V10 2 V10 V10 T 283 T = 2 283 or taemcity.com Square both sides T = 4 283 T 1132 K = 1132-273 859°C Result Temperature = T = 859°C or 1132 K EXAMPLE 8.2 A tuning fork A produces 4 beats per second with another tuning fork B. It is found that by loading B with some wax the beat frequency increases to 6 beats per second. If the frequency of A is 320 Hz, determine the frequency of B when loaded. Data Number of beats After loading B, Number of beats = n = 4 || = n = 6 [CHAPTER 8] Frequency of tuning fork B = fß SOLUTION Using fA - fB = ±n 320-fp = ±4 || WAVES ? when loaded 290 fB = 320 +4 fB = 320-4 fB = 324 Hz fB = 316 Hz By loading B, its frequency will decrease. Thus if 324 Hz is original frequency the beat frequency will reduce. But in this case beat frequency increases. After loading B, fB = 316 Hz fB 316-2 fB = 314 Hz Result Frequency of fork B = 314 Hz city.com EXAMPLE 8.3 A steel wire hangs vertically from a fixed point, supporting a weight of 80 N at its lower end. The diameter of the wire is 0.50 mm and its length from the fixed point to the weight is 1.5 m. Calculate the fundamental frequency emitted by the wire when it is plucked? Data Weight of wire ale Tension in wire Diameter of wire |||| W = 80 N = F d = 80 N 0.50 mm = 0.50 x 10-3 m Length of wire = 1 = 1.5 m To Find Fundamental frequency = = fi = ? SOLUTION Using fi Where m || F 21 m Mass per unit length [CHAPTER 8] WAVES 291 Since Volume = Area x Length V = π r2 1 Put in (1) .. m = - ρπιι m -ρπ 2 f1 fi 3 .. Mass per unit length is 10 m f1 f1 = 7.8 x 103 (3.14) = 1.53 x 10 kg/m 103 1 2 (1.5) -13 113 (.50 x 10-3)2 x 1 4 80 1.53 x 10 80 x 105 1.53 80000 1.53 52287.58 f1 = 76.22 Hz Result 76.22 Hz Fundamental frequency = eemcity.com EXAMPLE 8.4 A pipe has a length of 1 m. determine the frequencies of the fundamental and the first two harmonics Data (a) If the pipe is open at both ends. (b) If the pipe is closed at one end. (Speed of sound in air = 340 ms1) 1 = 1 m V 340 ms To Find When the pipe is open at both ends (a) Next 1st harmonics Fundamental frequency = f1 = ? f2 = ? [CHAPTER 8] Fundamental frequency = f1 = ? Next 3rd harmonics f3 = ? Next 5th harmonics fs = ? SOLUTION (a) When pipe is open at both end, Using n V fn 21 WAVES 292 1, 2, 3, 4, ...... n V Here f1 21 alemcity.com So, f1 f1 f2 f2 f3 f3 = 340 2x1 170 Hz 2f1 = 340 Hz = 3f1 510 Hz When pipe is closed at one end and is open at the other Using, fn 1, 3, 5, 7, ...... Here, > (b) Result (a) and, fi fi f3 f3 फार फार क क फु = 41 340 4 x 1 85 Hz 3f1 = 3f1 fs fs = = = 3 x 85 = 255 Hz 5f1 = 5 x 85 5f1 = 425 Hz When the pipe is open at both ends Fundamental frequency = fi = 170 Hz WAVES 293 Frequency of source fo Speed of sound V To Find Apparent frequency = f' Apparent frequency when train moves away from station with same speed f" = ? SOLUTION When train is moving towards listener, then by using || V f V – us Speed of train = Us = 90 km/h 90 × 1000 m/3600s = 25 m/s = 1000 Hz 340 m/s [CHAPTER 8] When the pipe is closed at one end Fundamental frequency = f1 = (b) Next 3rd harmonics Next 5th harmonics f3 = 85 Hz 255 Hz fs = 425 Hz EXAMPLE 8.5 A train is approaching a station at 90kmh1 sounding a whistle of frequency 1000 Hz. What will be the apparent frequency of the whistle as heard by a listener sitting on the platform? What will be the apparent frequency heard by the same listener if the train moves away from the station with the same speed? Data (Speed of sound = 340msTM1) te meity.com Putting values. 340 f' × 1000 340-25 1079.4 Hz. When train is moving away from listener then by using. Putting values. f" f" || (^^) 340 1000 [CHAPTER 8] Result WAVES Apparent frequency = f' 1079.4 Hz Apparent frequency when train moves away from station with same speed f" 931.5 Hz taleemcity.com 294

[CHAPTER 8] 
G Chapter 
WAVES 
8 
WAVES 
263 
LEARNING OBJECTIVE 
At the end of this chapter the students will be able to: 
Recall the generation and propagation of waves. 
Describe the nature of the motions in transverse and longitudinal waves. 
Understand and use the terms wavelength, frequency and speed of wave. 
Understand and use the equation v = f. 
Understand and describe Newton's formula of speed of sound. 
0 0 0 પ 
નવા 
ના 
Derive Laplace correction in Newton's formula of speed of sound. 
Derive the formula v = V + 0.61t. 
Explain and use the principle of superposition. 
Understand the terms interference and beats. 
Understand and describe reflection of waves. 
Explain the formation of a stationary wave using graphical method. 
Understand the terms node and anti-node. 
Understand and describe modes of vibration of string. 
Understand and describe Doppler's effect and its causes. 
Waves transport energy without transporting matter. The energy transportation is carried by a disturbance, which spreads out from a source. We are well familiar with different types of waves such as water waves in the ocean, or gently formed ripples on a still pond due to rain drop. When a musician plucks a guitar-string, sound waves are generated which on reaching our ear, produce the sensation of music. Wave disturbances may also come in a concentrated bundle like the shock waves from an aeroplane flying at supersonic speed. Whatever may be the nature of waves, the mechanism by which it 
[CHAPTER 8] 
WAVES 
264 
oscillations. The waves which propagate by the oscillation of material particles are known as mechanical 
waves. 
There is another class of waves which, instead of material particles, propagate out in space due to oscillations of electric and magnetic fields. Such waves are known as electromagnetic waves. We will undertake the study of electromagnetic waves at a later stage. Here we will consider the mechanical waves only. The waves generated in ropes, strings, coil of springs, water and air are all mechanical 
waves. 
So far we have been considering motion of individual particles but in case of mechanical waves, we study the collective motion of particles. An example will help us here. If you look at a black and white picture in a newspaper with a magnifying glass, you will discover that the picture is made up of many closely spaced dots. If you do not use the magnifier, you do not see the dots. What you see is the collective effect of dots in the form of a picture. Thus what we see as mechanical wave is actually the effect of oscillations of a very large number of particles of the medium through which the wave is passing. 
PROGRESSIVE WAVES 
Do You Know? 
Ultrasonic waves are particularly useful for undersea communication 
and detection systems. High frequency radio waves, used in inwater, whereas highly directional beams of ultrasonic waves can be made to travel many kilometres. 
radars travel just a fewcentimetres 
Drop a pebble into water. Ripples will be produced and spread out across the water. The ripples are the examples of progressive waves because they carry energy across the water surface. A wave, which transfers energy by moving away from the source of disturbance, is called a progressive or travelling wave. There are two kinds of progressive waves - transverse waves and longitudinal waves. 
Transverse and Longitudinal Waves 
Consider two persons holding opposite ends of a rope or a hosepipe. Suddenly one person 
gives one up and down jerk to the rope. This (a) disturbs the rope and creates a hump in it which travels along the rope towards the other person (Figure a and b). 
When this hump reaches the other person, it causes his hand to move up (Figure c). Thus the energy and momentum imparted to the end of the rope by the first person has reached the other end of the rope by travelling through the rope i.e., a wave has been set up on the rope in the form of a moving hump. We call this type of wave a pulse. The forward motion of the pulse from one end of 
(b) 
(c) 
the rope to the other is an example of progressive wave. The hand jerking the end of the rope is the source of the wave. The rope is the medium in which the wave moves. 
[CHAPTER 8] 
WAVES 
A large and loose spring coil (slinky spring) can be used to demonstrate the effect of the motion of the source in generating waves in a medium. It is better that the spring is laid on a smooth table with its one end fixed so that the spring does not sag under gravity. 
If the free end of the spring is vibrated from side to side, a pulse of wave having a displacement pattern shown in figure (a) will be generated which will move along the spring. 
265 
Transverse waves 
fixed 
end 
spring 
(a) 
Longitudinal waves 
Waves 
If the end of the spring is moved back and forth, along the direction of the spring itself as shown in figure (b), a wave with back and forth displacement will travel along the spring. Waves like those in figure (a) in which displacement of the spring is perpendicular to the direction of the waves are called transverse waves. Waves like those in figure (b) in which displacements are in the direction of propagation of the waves are called longitudinal waves. In this example the coil of spring is the medium, so in general we can say that 
(b) 
Transverse waves are those in which particles of the medium are displaced in a direction perpendicular to the direction of propagation of waves and longitudinal waves are those in which the particles of the medium have displacements along the direction of propagation of waves. 
Both types of waves can be set up in solids. In fluids, however, transverse waves die out very quickly and usually cannot be produced at all. That is why, sound waves in air are longitudinal in nature. 
PERIODIC WAVES 
Upto now we have considered wave in the form of a pulse which is set up by a single disturbance in a medium like the snapping of one end of a rope or a coil spring. Continuous, regular and rhythmic disturbances in a medium result from periodic vibrations of a source which cause periodic waves in that medium. A good example of a periodic vibrator is an oscillating mass-spring system (figure a). We have already studied in the previous chapter that the mass of such a system executes SHM. 
Transverse Periodic Waves 
Imagine an experiment where one end of a rope is fastened to a mass spring vibrator. As the mass vibrates up and down, we observe a transverse periodic wave travelling along the length of rope (figure b). The wave consists of crests and troughs. The crest is a pattern in which the rope is displaced above its equilibrium position, and in troughs, it has a displacement below its equilibrium position. 
As the source executes harmonic motion up and down with amplitude A and frequency f, ideally every point along the length of the rope executes SHM in turn, with the same amplitude and frequency. The wave travels towards right as crests and troughs in turn, replace one another, but the points on the rope simply oscillates up and down. The amplitude of the wave is the maximum value of the 
- 
-><-----> 
oo ooo Q 
ooooo 
<-----2-----> 
[CHAPTER 8] 
letter lambda 2 (figure b). 
WAVES 
In principle, the speed of the wave can be measured by timing the motion of a wave crest over a measured distance. But it is not always convenient to observe the motion of the crest. As discussed below, however, the speed of a periodic wave can be found indirectly from its frequency and wavelength. 
As a wave progresses, each point in the medium oscillates periodically with the frequency and period of the source. Figure illustrates a periodic wave moving to the right, as it might look in photographic snapshots taken every 1/4 period. Follow the progress of the crest that started out from the extreme left at t = 0. The time that this crest takes to move a distance of one wavelength is equal to the time required for a point in the medium to go through one complete oscillation. That is the crest moves one wavelength 2 in one period of oscillation T. The speed V of the crest is therefore, 
NN 
266 
(a) 
t=0 
T 
(b) 
(c) 
(d) 
ity. 
λ 
(f) 
t=T 
V 
Distance moved Corresponding time interval 
λ 
T 
All parts of the wave pattern move with the same speed, so the speed of any one crest is just the speed of the wave. We can therefore, say that the speed V of the waves is 
1 
λ 
V 
ee 
...... (1) 
but == f, where f is the frequency of the wave. It is the same as the frequency of the vibrator, generating 
the waves. Thus eq. (1) becomes 
V = fλ 
...... (2) 
Phase Relationship between Two Points on a Wave 
F 
The profile of periodic waves generated by a source executing SHM is represented by a sine curve. Figure shows the snapshot of a periodic wave passing through a medium. In this figure, set of points are shown which are moving in unison as the periodic wave passes. The points C and C', as they move up and down, are always in the same state of vibration i.e., they always have identical displacements and velocities. Alternatively, we can say that as the wave passes, the points C and C' move in phase. We may also say that C' leads C by one time period of 2л radian. Any point at a distance x, C lags behind by phase angle. 
2πx 
2 
[CHAPTER 8] 
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267 
...... are all in phase with each other. These points can be anywhere along the wave and need not correspond with only the highest and lowest points. For example, points such as P, P', P", ...... are all in phase. Each is separated from the next by a distance 2. 
Some of the points are exactly out of step. For example, when point C reaches its maximum upward displacement, at the same time D reaches its maximum downward displacement. At the instant that C begins to go down, D begins to move up. Points such as these are called one half period out of 
λ 
phase. Any two points separated from one another by 2, 3, 52, 
Longitudinal Periodic Waves 
... are out of phase. 
........ 
In the previous section we have considered the generation of transverse periodic waves. Now we will see how the longitudinal periodic waves can be generated. 
Consider a coil of spring as shown in figure. It is suspended by threads so that it can vibrate horizontally. Suppose an oscillating force F is applied to its end as indicated. The force will alternately stretch and < compress the spring, thereby sending a series of stretched regions. (called rarefaction) and compressions down the spring. We will see the oscillating force causes a longitudinal wave to move down the spring. This type of wave generated in springs is also called a compressional wave. Clearly in a compressional wave, the particles in the path of wave move back and forth along the line of propagation of the wave. 
F 
Displacement 
Notice in figure, the supporting threads would be exactly vertical if the spring were undisturbed. The disturbance passing down the spring causes displacements of the elements of the spring from their equilibrium positions. In figure, the displacements of the threads from the vertical are a direct measure of the displacements of the spring elements. It is, therefore, an easy way to graph the displacements of the spring elements from their equilibrium positions and this is done in the lower part of the figure. 
Q.1 Define progressive waves with its types. 
Ans. PROGRESSIVE WAVES (TRAVELLING WAVE) 
A wave, which transfers energy in moving away from the source of disturbance, is called a progressive or travelling waves 
(i) 
There are two kinds of traveling waves 
(i) 
Transverse waves. 
(ii) Longitudinal waves. 
Transverse Waves 
"Transverse waves are those in which particles of the medium are displaced in a direction perpendicular to the direction of propagation of waves." 
e.g., water waves, light waves. 
(ii) 
Longitudinal Waves (Compressional Waves) 
[CHAPTER 8] 
WAVES 
268 
Both types of waves can be setup in solids in liquids however, transverse waves die out very quickly and usually cannot be produced at all that is why sound waves in air one longitudinal in nature. 
Periodic Waves 
"Continuous, regular and rhythmic disturbances in a medium result from periodic vibrations of a source which cause periodic waves in that medium." 
e.g., Oscillating mass-spring system. 
Mean level 
Transverse Periodic Waves 
Crest 
"The portion of transverse wave above its mean position, is called crest." 
Trough 
"The potion of transverse wave below its mean position, is called trough." 
Wavelength 
Crest 
u 
Trough 
"The distance between any two consecutive crests or troughs, is called wave length." 
It is denoted by a Greek letter Lambda (2). 
Q.2 Show that V = λ f. 
Ans. The time that the crest takes to move a distance of one 
wave length is equal to the time required for a point in the medium to complete one oscillation i.e., crest moves one wave length 2' in one period of oscillation 'T', the speed 'V of the crest (wave) is 
As, 
Distance moved 
Corresponding time interval 
S = Vt 
Crest 
NNA 
(a) 
t=0 
(b) 
(c) 
(d) 
Since, 
λ 
V 
S 
T 
V 
t 
-18 
= = f 
T 
V = λ f 
Which is the relation between speed, frequency and wavelength. 
Longitudinal Periodic Waves 
(1) 
t=T 
[CHAPTER 8] 
WAVES 
269 
"The portion of longitudinal wave where particles of medium are very close to each other is called compression." 
Rarefaction 
"The portion of longitudinal wave where particles of medium are far apart from each other is 
called rarefaction." 
Q.3 
Explain Newton's formula for the speed of sound in air. 
Ans. SPEED OF SOUND IN AIR 
(i) 
Newton's Formula for Speed of Sound in Air 
om 
When one particle of the medium is disturbed, the disturbance in the form of wave travel in all directions in the medium. The velocity of disturbance depends upon the density and the elasticity of the medium. The lighter the density of the medium, more quickly the disturbance moves from point to point and similarly greater the elasticity of the medium, more quickly disturbance will be propogated from point to point in the medium. Newton proposed the following formula for the velocity of sound through 
the materials which is as follows: 
V 
E 
Elasticity Density 
density. 
Where is the elasticity of the medium and p is the 
Newton assumed that when a sound wave travels through air, the temperature of the air during compression remains constant and pressure changes from P to (P+ AP) and therefore, the volume changes from V to (V - AV). According to Boyle's law 
PV 
(P + AP)(V − AV) 
Speed of sound in different media 
Speed ms1 
Medium 
Solids at 20°C 
Lead 
Copper 
Aluminium 
Iron 
Glass 
1320 
3600 
5100 
5130 
5500 
or 
PV = PV - PAV + VAP – APAV 
The product APAV is very small and can be neglected. So, the above equation becomes: 
0 
-PAV + VAP 
VAP 
P 
AV 
ΔΡ 
Р 
|| 
AV 
V 
Liquids at 20°C 
Methanol 
1120 
Water 
1483 
Gasses of S.T.P. 
Carbon dioxide 
258 
Oxygen 
315 
Air 
Helium 
332 
972 
[CHAPTER 8] 
AV 
= 
Volume strain 
WAVES 
Stress 
Volume strain 
Elasticity 
Р 
V 
The above equation becomes: 
V 
E 
P 
270 
On substituting the values of atmospheric pressure and density of air at S.T.P. in above equation, we find that the speed of sound waves in air comes out to be 280 ms, whereas its experimental value is 332 ms1 
Q.4 How laplace correct the speed of sound in air? 
Ans. LAPLACE CORRECTION FOR VELOCITY OF SOUND IN AIR 
The sound waves travel in the form of compressions and rarefactions. The compressions and rarefactions are so rapid, the temperature of air does not remain constant. The temperature increases due to compressions and the temperature decreases due to the rarefactions. Therefore during compression the air does not lose heat due to conduction and during rarefaction it does not gain heat. Thus the temperature throughout the medium does not remain constant. The relation between volume and pressure (PV = Constant) is not true but it is give 
PV = Constant 
Where y is constant and its value depends upon the nature of the gas where 
Y 
Molar specific heat at constant pressure 
Molar specific heat at constant volume 
If 'P' be the pressure then the change in pressure is very small which is P + AP therefore volume decreases from V to V - AV then 
PV 
(P + AP)(V - AV)" 
AP) V" (1 - AV) 
PV = (P+AP)\ 
Applying Binomial theorem: 
Р 
= 
or 
Р 
= (P+AP) 
AV 
(P + AP) 1-Y V 
AV 
V 
= P - y P2 + AP - YAP V 
AV 
V 
Where (y AP is negligible. Hence, we have 
Types of gas 
For Your Information 
Values of Constant 
Y 
1.67 
1.40 
1.29 
Monoatomic 
Diatomic 
Polyatomic 
For Your Information 
Ranges of Hearing 
Organisme 
Frequencies (Hz) 
[CHAPTER 8] 
or 
Therefore 
ΔΡ 
AV 
V 
= 
YP 
E 
V 
= 
- E 
V 
YP 
WAVES 
271 
Cat 
60-70,000 
Dog 
15-50,000 
Human 
20-20,000 
On substituting the value of atmospheric pressure and y then the speed of the sound is 333 m/s. This value of speed of sound is very close to the experimental value. Thus the laplace correction must therefore be correct. 
0.5 
What is effect of variation of pressure on the speed of sound? 
Ans. EFFECT OF PRESSURE ON SPEED OF SOUND 
As, 
V 
|| 
Ρ 
E> 
Tidbits 
(a) 
V 
YP 
P→ increase 
(b) 
Ρ 
V→ decreases 
p→ increase 
To signal generator 
. Ραρ 
Sound waves cause the candle 
Since density is proportional to the pressure so the speed of flame to flicker. 
sound is not affected by the variation in pressure of the gas. 
Q.6 What is effect of density on the speed of sound? 
Ans. EFFECT OF DENSITY ON SPEED OF SOUND 
As. 
V 
y P 
Ρ 
At the same temperature and pressure for the gases having the same value of y, the velocity is inversely proportional to the square root of their density. 
i.e.. 
V 
1 
Note: Speed of sound in hydrogen is four times its speed in oxygen as density of the oxygen is sixteen 
times that of the hydrogen. 
Q.7 What is the effect of temperature on the speed of sound in air? 
Ans. EFFECT OF TEMPERATURE ON SPEED OF SOUND 
When a gas is heated at constant pressure, its volume is increased and hence, its density is decreased. 
[CHAPTER 8] 
Let, Vo 
WAVES 
272 
Speed of sound at 0°C 
Po 
Density of gas at 0°C 
Vt 
= 
Speed of sound at t°C 
Pt 
Density of gas at t°C 
YP 
then, 
Vo 
...... (1) 
Do You Know? 
Po 
Wave crests 
y P 
and 
V1 
(2) 
Pt 
Dividing equation (2) by (1) 
V1 
Vo 
Vt 
51 15 
Vt 
Vo 
Vt 
Vo 
|| 
y P/pt 
Vy P/Po 
SK SK Sk 
Po 
Po 
y P 
Slower than the speed of sound. 
Shock wave 
(3) 
afecity.l 
If Vo is the volume of a gas at temperature 0°C and Vt is volume at t°C, then by using volume expansion. 
V1 = Vo (1+ẞt) 
Faster than the speed of sound. What happens when a jet plane like Concorde flies faster than the speed of sound? 
A conical surface of concentrated 
Where ẞ is the coefficient of volume expansion of the gas. For sound energy sweeps over the 
1 
all gases, its value is about 
273 
Hence, 
As 
P 
Volume 
= 
Vo(1+273) 
|| 
V 
Mass Density 
ground as a supersonic plane passes overhead. It is known as sonic boom. 
V1- Vox V, t 
Vt-Vo=ẞVot 
Vt=Vo+BVot 
= Vo(1+ẞt) 
m 
V 
Pt 
Elad 
1 
Pt 
|||| 
|| 
P 
m 
Po 
Po 
( 
1+ 
273 
1+273) 
[CHAPTER 8] 
Vt 
রার রার রার রার 
V 
Vt 
|| 
t 
WAVES 
P. (1+273 
1 + 
t 
273 
273+t 
T 
To 
273 
P1 
(4) 
273 
Where T and To are the absolute temperature. Corresponding to 5°C and 0°C respectively. Thus the speed of sound is directly proportional to the square root of the absolute temperature. 
Now, using Binomial theorem and neglecting high power, we have, eq. (4) as: 
As, 
Vt 
Vo 
V1 
Vo 
(1 
1+ 
= 1+ 
1 
t 
273 
1/2 
273 
tancitym 
273 
V1 = (1 + 3/16 
Vo 
V1 
= 
Vo = 332 m/s 
Putting the value in the 2nd factor, 
332 
Vo+ 
546 
t 
V1 = Vo+0.61 t 
This shows that one degree Celsius rise in temperature produces approximately 0.61 m/s (61 cm/s) increase in the speed of sound. 
Q.8 
State the principle of superposition. 
[CHAPTER 8] 
WAVES 
274 
Ans. PRINCIPLE OF SUPERPOSITION 
So far, we have considered single waves. What happens when two waves encounter each other in the same medium? Suppose two waves approach each other on a coil of spring, one travelling towards the right and the other travelling towards left. Figure shows that you would see happening on the spring. The waves pass through each other without being modified. After the encounter, each wave shape looks just as it did before and is travelling along just as it was before. 
This phenomenon of passing through each other unchanged can be observed with all types of waves. You can easily see that it is true for surface ripples. 
But what is going on during the time when the two waves. overlap? Figure (c) shows that the displacements they produce just add up. At each instant, the spring's displacement at any point in the overlap region is just the sum of the displacements that would be caused by each of the two waves separately. 
Thus, if a particle of a medium is simultaneously acted upon by n waves such that its displacement due to each of the individual n waves be y1, y2, ......, Yn, then the resultant displacement of the particle, under the simultaneous action of these n waves is the algebraic sum of all the displacements i.e., 
Y = y1 + y2 + ... 
+ Yn 
y2+.. 
This is called principle of superposition. 
Again, if two waves which cross each other have opposite phase, their resultant displacement will be 
Y 
У1 - У2 
Particularly if y1 = y2 then result displacement Y = 0. Principle of superposition leads to many interesting phenomena with waves. 
Q.9 
For Your Information 
Y 
Wave 1 
2 
Wave 2 
Y=Y1+Y2 
Super posed 
Resultant wave 
Superposition of two waves of the same frequency which are exactly in phase. 
Y1 
Wave 1 
Wave 2 
Y2 
Super posed of wave 1 and 2 
Resultant wave 
Y = 0 Superposition of two waves of the same frequency which are exactly out of phase. 
(i) Two waves having same frequency and travelling in the same direction (Interference). 
(ii) Two waves of slightly different frequencies and travelling in the same direction (Beats). 
(iii) Two waves of equal frequency travelling in opposite direction (Stationary waves). 
State and explain the phenomenon of interference of sound. 
Ans. INTERFERENCE 
"Superposition of two waves having the same frequency and traveling in the same direction results in a phenomenon, called interference." There are two types: 
[CHAPTER 8] 
Audio generator 
S1 
P.A. 
Power amplifier 
S2 
A 
WAVES 
S1 
S2 
275 
" 
(i) 
Microphone 
Fig. (a) 
PP 
Constructive Interference 
口] CRO 
com 
Fig. (b) 
"If two waves arrive at a point in phase i.e., compression of one wave falls on compression on other wave and rarefaction of one wave falls on the rarefaction of other wave, then resultant sound is loudest." 
Whenever path difference is an integral multiple of wavelength the two waves are added up. This effect is called constructive interference. 
or 
Condition for constructive interference can be written as 
where 
(ii) 
AS = nλ 
n = ±1,± 2, ± 3, ..... 
Destructive Interference 
"If two waves reach a point out of phase i.e., compression of one wave falls on the rarefaction of other wave and rarefaction of one wave falls on the compression of other wave, then resultant sound will be minimum." 
or 
At points where displacements of two waves cancel each other's effect, the path difference is an odd integral multiple of half the wavelength. This effect is called destructive interference. 
Condition for destructive interference is 
where 
AS 
= 
λ 
22 
+1) 2 
(2n+1) 
n = 0,±1,±2, ...... 
AAA 
Fig. (c) 
Constructive interference Large displacement is displayed on the CRO screen 
Fig. (d) 
Destructive interference Zero displacement is displayed on the CRO screen 
Explanation 
An experimental set up to observe interference effect in sound waves as shown in figure. 
Two loud speakers S. and S. act as two sources of harmonic sound waves of a fixed frequency 
[CHAPTER 8] 
WAVES 
276 
screen. The microphone is placed at various points, turn by turn, infront of the loud speakers as shown in the figure. 
At points P1, P3 and P5, we find that compressions met with compressions and rarefactions. So, the displacement of the two waves are added up at these points and large resultant displacement is produced. At points P2 and P4, compressions met with rarefactions so they cancel each other effect. The resultant displacement becomes zero. Now we have to find the path difference between the waves at point P1 is 
AS 
AS 
= 
S2P1 - S1P1 
— 
3호 1 
= λ 
But 
AS 
= 
ηλ 
For constructive interference. 
Where n = 0, 1, 2, 3,...... 
For destructive interference 
enicity.com 
AS 
S2P2 - S1P2 
AS 
= 
42-352 = 12 
So, 
AS 
n+ 
Where n = 0, 1, 2, 3,..... 
Q.10 What are beats? How they are produced? Show that the number of beats is equal 
to the difference between the frequencies of the tuning forks. 
Ans. BEATS 
other. 
Beat is the combined effect of two sound waves having frequencies slightly different from each 
Consider two tuning forks each having frequency 32 cps. Slightly load (with wax or ring) one tuning fork so that frequency decreases a little. Let the frequency becomes 30 cps. The two tuning forks are sounded together and held at equal distance from the ear. Let at t = 0, the two forks are in phase. i.e., their right prongs moves towards right producing compressions at the same time and louder sound is heard by the listener. 
With passage of time, the tuning fork B (30 cps) begins to fall behind 'A'. 
[CHAPTER 8] 
1 
4 
WAVES 
At t = sec, 'A' has completed 8 vib and 'B' has completed 7 
vib. The prongs are now out of phase and no sound is heard due to destructive interference. 
=1 
B 
A 
V = 32 t = 0 
Ear 
D 
v = 30 
(a) 
A 
3 
Ear 
t = 1/4 sec 
At t = sec, 'A' has completed 16 vib and 'B' completes 15 vib. 
Att 
The prongs of forks become in phase and louder sound is heard. At t 
4 
sec, 'A' has completed 24 vib while 'B' completes 225 vib. The prongs 
are now in opposite phase and once again no sound is heard. 
At t = 1 sec, both forks have completed 32 vib and 30 vib. The prongs become in phase and max or louder sound is heard. 
It is observed that in 1 sec, the sound falls in intensity twice. The sudden fall of sound in intensity is known as beat. Thus two beats are produced/sec which is equal to the difference in frequencies of two forks (32-30=2). 
Definition of Beats 
The periodic alteration of sound between maximum and minimum. loudness as many times a second as the difference in frequencies is called phenomenon of beats. 
No of beats/sec = Difference in frequencies 
±n = fA-fB 
Graphical Explanation of Beats 
The displacements of the particles of the medium due to two waves are plotted separately as function of time. The resultant 
A 
B 
A 
B 
A 
D 
(b) 
Ear 
t = 1/2 sec 
D 
(c) 
t = 3/4 sec 
Ear 
D 
(d) 
t = 1 sec 
Ear 
D 
(e) 
A 
(a) 
displacement of any particle will be the sum of the displacement due to (b) 
(b) NAMAN 
each of the two waves. The resultant wave which is produced in shown in figure (c). It is seen that amplitude of resultant waves changes with time. The change in amplitude gives rise to production of beat. 
Uses of Phenomenon of Beats 
The phenomenon of beats is used to find out: 
(i) 
Unknown frequency 
(ii) To tune a musical instrument 
Q.11 Write a note on reflection of waves. 
Ans. REFLECTION OF WAVES 
1 
S 
S 
4 
Resultant 
Formation of Beats 
277 
VVELL 
In an extensivemedium, a wave travels in all direction from its source with a velocity depending upon the properties of the medium. However, when the wave comes across the boundary of two media, a part of it is reflected back. The reflected wave has the same wavelength and frequency but its phase may 
[CHAPTER 8] 
WAVES 
278 
Now we will discuss two most common cases of reflection at the boundary. These cases will be explained with the help of waves travelling in slinky spring. (A slinky spring is a loose spring which has small initial length but a relatively large extended length). 
One end of the slinky spring is tied to a rigid support on a smooth horizontal table. When a sharp jerk is given up to the free end of the slinky spring towards the side A, a displacement or a crest will travel from free end to the boundary (Figure a). It will exert a force on bound end towards the side A. Since this end is rigidly bound and acts as a denser medium, It will exert a reaction force on the spring in opposite direction. This force will produce displacement towards B and a trough will travel backwards along the spring (Figure b). 
A 
eeeeee 
(a) 
B 
From the above discussion it can be concluded that whenever a transverse wave, travelling in a rarer medium, encounters a denser medium, it bounces back such that the direction of its displacement is reversed. An incident crest on reflection becomes a trough. 
A Slinky 
Light string 
(a) 
leeeeeeeeeeee 
A 
A 
(b) 
This experiment is repeated with a little variation by attaching one end of a light string to a slinky spring and the other end to the rigid support as shown in figure. If now the spring is given a sharp jerk towards A, a crest travels along the spring as shown in figure. When this crest reaches the spring-string boundary, it exert a force on the string towards the side it does not oppose the motion of the spring. The end of the spring, therefore, continues its displacement towards A. The spring behaves as if it has been plucked up. In other words a spring crest is again created at the boundary of the spring-string system, which travels backwards along the spring. From this it can be concluded that when a transverse wave travelling in a denser medium, is reflected from the boundary of a rarer medium, the direction of its displacement remains the same. An incident crest is reflected as a crest. We are already familiar with the fact that the direction of displacement is reversed when there is change of 180° in the phase of vibration. So, the above conclusion can be written as follows: 
(i) 
(c) 
If a transverse wave travelling in a rarer medium is incident on a denser medium, it is reflected such that it undergoes a phase change of 180°. 
(ii) If a transverse wave travelling in a denser medium is incident on a rarer medium, it is 
reflected without any change in phase. 
Q.12 What are stationary waves? How they one produced? Define node and anti-node. 
Ans. STATIONARY WAVES 
Now let us consider the superposition of two waves moving along a string in opposite directions. Figure (a, b) shows the profile of two such waves at instants = 0, T/4, 3/4 T and T, where T is the time period of the wave. We are interested in finding out the displacements of the points 1, 2, 3, 4, 5, 6 and 7 at these instants as the waves superpose. From the figure (a, b), it is obvious that the points 1, 2, 3, etc., 
[CHAPTER 8] 
t=0 
t = T/4 
WAVES 
t = T/2 
t=3T/4 
t=T 
279 
(b) 
AAMMA 
2 
3 
4 
5 
6 
2 
6 
2 
1 3 5 7 1 3 
4 6 
5 7 
7 
6 
3 
5 
2 
7 
(a) 
2 
3 
4 
5 
6 
t=0 
t = T/4 
t = T/2 
t = 3T/4 
(c)% 
3 5 7 1 3 
1 
3 
5 
5 7 
7 1 3 
3 5 5 7 1 
1 
3 3 
5 
5 7 
7 
1 
1 
... 
t = T/4 
t = 3T/4 
6 
2 4 6 
4 
24 6 
t = T/2 
t=0 1.. ง 
t=T 
always 
3 5 7 at rest 
3 
t=T1 always 
6 
oscillating 
are distant 2/4 apart, λ being the wavelength of the waves. We can determine the resultant displacement of these points by applying the principle of superposition. Figure (c) shows the resultant displacement of the points 1, 3, 5 and 7 at the instants t = 0, T/4, T/2, 3T/4 and T. It can be seen that the resultant displacement of these points is always zero. These points of the medium are known as nodes. Figure (c) shows that the distance between two consecutive nodes is 2/2. Figure (d) shows the resultant displacement of the points 2, 4 and 6 at the instant t = 0, T/4, T/2, 3T/4 and T. The figure shows that these points are moving with an amplitude which is the sum of the amplitudes of the component waves. These points are known as antinodes. They are situated midway between the nodes and are also 2/2 apart. The distance between a node and the next antinode is 2/4. Such a pattern of nodes and antinodes is known as a stationary or standing wave. 
Energy in a wave moves because of the motion of the particles of the medium. The nodes always remain at rest, so energy cannot flow past these points. Hence energy remains "standing" in the medium between nodes, although it alternates between potential and kinetic forms. When the antinodes are all at their extreme displacements, the energy stored is wholly potential and when they are simultaneously passing through their equilibrium positions, the energy is wholly kinetic. 
An easy way to generate a stationary wave is to superpose a wave travelling down a string with its reflection travelling in opposite direction as explained in the next section. 
Q.13 Explain the stationary waves in a stretched string. Also calculate the frequencies. 
Ans. STATIONARY WAVES IN A STRETCHED STRING 
Consider a string of length '7' which is kept stretched by clamping its ends so that the tension in the string is F. If the string is plucked at its middle point, two transverse waves will originate from this point. One of them will move towards the left end of the string and the other towards the right end. When these waves reach the two clamped ends, they are reflected back, thus giving rise to stationary waves. The string will vibrate with such a frequency fi, so that nodes are formed at two fixed ends (clamped ends) and anti-nodes between them. Thus the string vibrates in one loop as shown in Fig. 
If 21 is the wavelength of this mode of vibration (1st mode of vibration) so, 
[CHAPTER 8] 
As 
7 
λι 
V 
|||| 
λι 
20 
WAVES 
= 21 
λ f 
(i) 
fi 
|||| 
V 
λι 
k 
A 
NK. 
Pluck 
不 
280 
f2 
22 2>2 
||_|| 
N 
A 
N 
N 
2 
f1 
fi 
|||| 
21 
1 
21 
F 
F/m 
V 
F 
.. 
V 
f1 
21 
talecity.co 
1/4 
1 
If the string is plucked from 4 of its length, then again stationary waves will be set up, but now 
the string vibrate in two loops with f2. If λ2 is wave length in this case then, 
As, 
l 
λ + 2 2 
(iii) 
The speed 'V' of the waves in the string depends upon the tension F of the string and mass per unit length of the string. It is given by 
(ii) 
Putting value of 21. 
K 
=21 
λ1 = 21 
不 
Putting value of 22 
f2 
= 2 
2) 
V 
21 
f2 
2f1 
→ 
2 
2 
=쫄+쓸 
This shows that when string vibrates in two loops, its frequency is doubled and wave length becomes half than it vibrates in one loop. 
[CHAPTER 8] 
WAVES 
Similarly if the string is plucked from th of its length, it 
vibrates in three loops as shown in figure. 
As 
23 23 23 
1 
2 
++즐 
1 
1 
23 
f3 
= 3 
|||| 
21 
23 
22 
alm > 3 
3 
V 
23 
1/6 
281 
N 
KAL 
A 
A 
N4 
N 
-—-— 1 = 1/2 + 1/2 + 2/2- ++쓸 
| 
Do You Know? 
Putting value of 23 
V 
f3 
21/3 
V 
f3 
= 3 
21 
f3 
3f 
It means that when string vibrates in three loops, its frequency 
leexity. 
is three times the frequency when it vibrates in one loop. 
Thus we can generalize that if the string is made to vibrate in 
'n' loop, then 
and wave length is; 
fn 
2n 
nf 
A standing-wave pattern is formed when the length of the string is an integral multiple of half wavelength; otherwise no standing wave is formed. 
For Your Information 
where n = 1, 2, 3, ...... 
It is clear that as the string vibrates in more and more loops, its frequency goes on increasing and the wave length gets shorter. However the product of frequency and wave length is always equal to V, (speed of the wave). 
The above discussion clearly establishes that the stationary waves have a discrete set of frequencies f1, 2f1, 3f1, ........., n fj which is known as harmonic series. The fundamental frequency 'fi' is called first harmonic (over tone), 'f2' is called second harmonic (over tone), and so on. 
Note: The frequency of a string on a musical instrument can be 
changed either by varying the tension or by changing the length. For example the tension in guitar and violin strings is 
In an organ pipe, the primary Idriving mechanism is wavering. 
[CHAPTER 8] 
WAVES 
282 
Q.14 Describe the stationary waves in air column. 
Ans. STATIONARY WAVES IN AIR COLUMNS 
Stationary waves can be set in other media also, such as air column. A common example of vibrating air column is in the organ pipe. The relationship between the incident wave and the reflected wave depends on whether the reflecting end of the pipe is open or closed. If the reflecting end is open, the air molecules have complete freedom of motion and this behaves as an antinode. If the reflecting end is closed, then it behaves as a node because the movement of the molecules is restricted. The modes of vibration of an air column in a pipe open at both ends are shown in figure. 
In figure, the longitudinal waves set up in the pipe have been represented by transverse curved lines indicating the varying amplitude of vibration of the air particles at points along the axis of the pipe. However, it must be kept in mind that air vibrations are longitudinal along the length of the pipe. The wavelength 2' of nth harmonic and its frequency 'f' of any harmonic is given by 
(a) 
(b) 
V 
XXX 
(c) 
Fig. Stationary longitudinal waves in a pipe open at both ends. 
21 
λn 
n 
n 
V 
fn 
nv 
21 
..... (1) 
1, 2, 3, 4,... 
eine 
where 'v' is the speed of sound in air and '7' is the length of the pipe. The equation (10 can also be written as 
fn 
nfi 
(2) 
If a pipe is closed at one end and open at the other, the closed end is a node. The modes of vibration in this case are shown in figure. 
In case of fundamental note, the distance between a node and 
antinode is one fourth of the wavelength, 
Hence, 
Since 
V 
Hence, 
|||||| 
λι 
or λι = 41 
4 
V 
V 
f1 
λι 
4/ 
It can be proved that in a pipe closed at one end, only odd harmonics are generated, which are given by the equation. 
where 
fn 
nv 
41 
n 
= 1, 3, 5, ...... 
...... (3) 
(a) 
3v 
लोक 
1 = 32, f2 = 3/1 
1= 
= 
(b) 
X 
1=52, f3 
5v 
41 
(c) 
Fig. Stationary longitudinal waves 
This shows that the pipe, which is open at both ends, is richer in a pipe closed at one end. Only in harmonics. 
odd harmonics are present. 
[CHAPTER 8] 
In 1st mode 
7 
1 
λι 
221 
4 
λι 
22 2 
WAVES 
283 
As, 
V 
> 
V 
fi 
λι 
Putting value of '21', 
In 2nd mode 
fi 
1 
V 
21 
>2 
λ2 +222 +λ2 
422 4 
4 
taleemcity.com 
f2 
|||||| 
= 
λε 
V 
1 
Multiply and divided by (2). 
In 3rd mode 
f2 = 2 
As, 
f3 
Putting value of 23 
.. 
23 +223 +223 +23 4 
62.3 
4 
323 
3+ 2~ ~lm >12 
21 
1 
3 
V 
23 
V 
f3 
21/3 
V 
f3 
= 3 
21 
[CHAPTER 8] 
Case II 
WAVES 
When pipe is open at one end and closed at other end. 
284 
In 1st mode 
λι 
f1 
||_|| 
= 
41 
V 
λι 
Putting value of 21 
V 
fi 
41 
In 2nd mode 
λε 
λε 
|| 
4 
2 
22+222 
1 
4 
322 
22 
f 
|||||| 
4 
41 
Fm 
3 
22 
Putting value of 22 
V 
f3 
41/3 
f3 
f3 
alemcity.c 
We can generalize, 
.. 
Interesting Information 
Echolocation allows dolphins to detect small differences in the shape, size and thickness of objects. 
where n = 1, 3, 5, ...... 
Note: The pipe which is open at both ends is richer in harmonics. 
Q.15 What is Doppler effect? Describe expression for apparently changed frequency when the observer is at rest while the source is in motion. (OR) What is Doppler effect? Describe the expression for apparently changed frequency when the observer is in motion while source is at rest. 
Ans. DOPPLER EFFECT 
Introduction 
This effect was observed by Johann Doppler while he was observing the frequency of light 
emitted from distant stars In some cases the frequency of light emitted from a star was found to be 
[CHAPTER 8] 
Definition 
WAVES 
285 
"The apparent change in the pitch OR frequency of a source of sound, when there is a relative motion between the source of sound and the observer, is called Doppler effect." 
For example, when an observer is standing on a railway platform, the pitch of the whistle of a approaching train is heard to be higher. But when the same train moves away, the pitch of the whistle becomes lower. 
Explanation 
Suppose 'V' is the velocity of the sound in a medium and a source emits a sound of frequency 'f' and wave length '2'. If both the source and the observer are stationary, then the waves received by the observer in one second are; 
Case-A 
f 
>2 
V 
λ 
When observer 'O' is moving towards a stationary source 'S' 
Co 
If an observer ('O') moves towards, a source 'S' with velocity Uo, the relative velocity of the waves and the observer is increased to (V+ Uo), then number of waves received in one second apparent frequency increases 
Putting the value 
fA 
fA 
V 
>12 
|| 
λ 
Or 
V + Uo 
V 
λ 
V/f 
1 
fA 
S 
Observer 
Fig. An observer moving with velocity u。 towards a stationary source hears a frequency f, that is greater than the source frequency. 
V + Uo 
Hence, 
V 
fA 
> 1 
V 
> f 
V + Uo ) f 
V 
This means that when an observer 'O' is moving towards a stationary source S, the frequency of sound increases. 
Case-B 
When observer 'O' moves away from a stationary source S 
[CHAPTER 8] 
WAVES 
When observer 'O' moves away from stationary source 'S' with velocity 'Uo,' then relative velocity of sound waves and the observer is, (V Uo), hence, number of waves received by the observer, per-second are reduced and is given by 
Observer 
fB 
V - Uo λ 
286 
Putting value of 
λ 
fB 
|||| 
V 
>14 
f 
V-Uo V/f 
fB 
V-Uo) f 
V - Uo 
As, 
< 1 1 
V 
.. 
fB 
< f 
S 
Ug = 0 
Fig. An observer moving with velocity u, away from stationary source hears a frequency fg that is smaller than the source frequency. 
city.c 
This means that when observer 'O' moves away from stationary source 'S,' the apparent frequency decreases. 
Case-C 
When source 'S' is moving towards the stationary observer 'O' 
If the source 'S' is moving towards the observer with velocity 'Us' then in one second, the waves are compressed, (Wave length decreases), by an amount known as Doppler shift represented by A2. 
As, 
V 
Δλ 
= fλ 
Us f 
ν 
The compression of waves is due to the fact that same number of waves are contained in a shorter space depending upon the velocity of the source. 
λ Αλ. 
The wave length for observer 'C' is then, λ = 2-A2. 
Putting the values of λ and A2. 
.. 
V Us 
λc 
ff 
Яс 
V-Us f 
The modified frequency for observer 'C' is 
fc 
V 
λε 
S 
(U2 =D) 
(uo = 0) 
Us 
42 
K 
λ=2-A2 
入 
[CHAPTER 8] 
Putting value of 2c 
WAVES 
287 
V 
fc 
V-Us f 
V 
fc 
V 
V-Us 
) f 
> 1 
As, 
V-Us 
fc 
> f 
.. 
Observer C 
S 
Observer Dus 
Fig. A source moving with velocity u, towards a stationary observer C and away from stationary 
This means that the observed frequency increases when the observer D. Observer C hears an source is moving towards the observer. 
Case-D 
When source 'S' is moving away from stationary observer 'O' 
If the source is moving away from the observer 'O' with velocity 'Us,' then in one second the waves are stretched (wave length increase) by an amount Αλ. 
increased and observer D hears a decreased frequency. 
تر 
•min 
S 
(Ug = 0) 
(u2 = 0) 
Us 
Us 
Αλ 
f 
S 
The stretched of the waves is due to the fact that same number of wave are contained in larger (longer) space, depending upon the velocity of the source. 
Αλ 
The wave length for observer 'D' is, λp = 2 + A2. Putting values of 2 and A2. 
.. 
V U 
λD 
ff 
51- 
+ 
2D 
V+Us f 
The modified frequency for observer 'D' will be 
(uo = 0) 
20=2+42 
Do You Know? 
V 
fp 
2D 
Putting value of 2p 
V 
.. 
fp 
V+Us f 
fD 
-(V+U.) 
f 
V 
As, 
< 1 
V+Us 
fD 
< f 
acoustic 
coupling 
gel 
backscattered 
transmitted signal 
signal 
blood flow 
blood vessel 
The Doppler effect can be used to monitor blood flow through major arteries. Ultrasound waves of frequencies 5 MHz to 10 MHz are directed towards the artery and a receiver detects the back scattered signal. The apparent frequency depends on the 
This means that the observed frequency decreases, when the velocity of flow of the blood. 
source is moving away from the observer 
[CHAPTER 8] 
WAVES 
Ans. APPLICATIONS OF DOPPLER EFFECT 
Doppler effect is applied in working of radar system. Radar uses radio waves to find the elevation and speed of an aeroplane. Radar is a device which transmits and receives radio waves. The radio waves transmitted from radar are reflected back from aeroplane and are received by radar. If the aeroplane is moving towards the radar then the wavelength of reflected wave is shorter. If the plane is moving away from the radar then the wavelength of reflected wave is longer as shown in figure. 
Plane approaching 
Plane receding 
288 
detect the motion of an aeroplane. 
Stationery star 
The difference of wavelength of transmitted and reflected A frequency shift is used in a radar to waves is used to determine the speed of aeroplane. Term SONAR (is acronym) stands for sound navigation and ranging. Sonar is the name of the technique used for detecting the presence of objects under water by acoustical echo. In Sonar "Doppler detection" depends upon relative speed of the target and the detector. It employs. The Doppler effect in which an apparent change in frequency occurs when source and observer are in relative motion with respect to one another. 
In military it is used for detection and location of submarine antisubmarine weapons and depth measurement of sea. Astronomers use Doppler effect to calculate the speeds of distant stars and galaxies. By comparing the line spectrum of light from the star with light from a laboratory source, the Doppler shift of stars light can be measured. Then speed of star can be calculated. 
Doppler effect is used to determine whether a particular star or galaxy is approaching the earth or moving away from the earth. Light from the star is measured with the help of spectrometer. It has been found that stars moving towards the earth show a blue shift. Thus is because the emitted waves by the star have shorter wavelength than of the star had been at rest. So the spectrum is shifted towards shorter wavelength i.e., to the blue end of the spectrum. 
It has been found that stars moving away from the earth show a red shift. This is because the emitted waves by the star have longer wavelength than of the star had been at rest. So the spectrum is shifted towards longer wavelength i.e., to the red end of the spectrum. Astronomers have discovered that all the distant galaxies are moving away from us. They have also measured their speed by measuring their red shift. Another important application of the Doppler shift using electromagnetic waves is radar speed trap. Microwaves are emitted from a transmitter in the form of short bursts. Each burst is reflected back by any moving car in the path of microwaves. The reflected microwaves are received back as Doppler's shift. By measuring Doppler shift the speed of the car can be calculated by computer 
programme 
(a) 
Star receding 
(b) 
(c) 
Star approaching 
ми 
Do You Know? 
[CHAPTER 8] 
WAVES 
SOLVED EXAMPLES 
289 
EXAMPLE 8.1 
Data 
Find the temperature at which the velocity of sound in air is two times its velocity at 10°C. 
Velocity of sound 
Vt = 2 V10 
T10 
= 
10+273 = 
283 K 
To Find 
Temperature 
= T 
= ? 
SOLUTION 
Ve 
T 
Using 
V10 
T10 
Since 
V1 
= 2 V10 
2 V10 
V10 
T 
283 
T 
= 2 
283 
or 
taemcity.com 
Square both sides 
T 
= 4 
283 
T 
1132 K 
= 1132-273 
859°C 
Result 
Temperature = T = 859°C or 1132 K 
EXAMPLE 8.2 
A tuning fork A produces 4 beats per second with another tuning fork B. It is found that by loading B with some wax the beat frequency increases to 6 beats per second. If the frequency of A is 320 Hz, determine the frequency of B when loaded. 
Data 
Number of beats 
After loading B, 
Number of beats 
= n 
= 4 
|| 
= n 
= 6 
[CHAPTER 8] 
Frequency of tuning fork B = fß 
SOLUTION 
Using 
fA - fB = ±n 
320-fp = ±4 
|| 
WAVES 
? when loaded 
290 
fB 
= 
320 +4 
fB 
= 320-4 
fB 
= 324 Hz 
fB = 316 Hz 
By loading B, its frequency will decrease. Thus if 324 Hz is original frequency the beat frequency will reduce. But in this case beat frequency increases. 
After loading B, 
fB 
= 316 Hz 
fB 
316-2 
fB 
= 314 Hz 
Result 
Frequency of fork B 
= 314 Hz 
city.com 
EXAMPLE 8.3 
A steel wire hangs vertically from a fixed point, supporting a weight of 80 N at its lower end. The diameter of the wire is 0.50 mm and its length from the fixed point to the weight is 1.5 m. Calculate the fundamental frequency emitted by the wire when it is plucked? 
Data 
Weight of wire 
ale 
Tension in wire 
Diameter of wire 
|||| 
W 
= 80 N 
= F 
d 
= 80 N 
0.50 mm 
= 
0.50 x 10-3 m 
Length of wire 
= 1 
= 
1.5 m 
To Find 
Fundamental frequency 
= 
= fi 
= ? 
SOLUTION 
Using 
fi 
Where 
m 
|| 
F 
21 
m 
Mass per unit length 
[CHAPTER 8] 
WAVES 
291 
Since 
Volume 
= Area x Length 
V 
= π r2 1 
Put in (1) 
.. 
m 
= 
- ρπιι 
m 
-ρπ 
2 
f1 
fi 
3 
.. Mass per unit length is 
10 
m 
f1 
f1 
= 7.8 x 103 (3.14) 
= 1.53 x 10 kg/m 
103 
1 
2 (1.5) 
-13 
113 
(.50 x 10-3)2 
x 1 
4 
80 
1.53 x 10 
80 x 105 
1.53 
80000 
1.53 
52287.58 
f1 
= 76.22 Hz 
Result 
76.22 Hz 
Fundamental frequency = 
eemcity.com 
EXAMPLE 8.4 
A pipe has a length of 1 m. determine the frequencies of the fundamental and the first two 
harmonics 
Data 
(a) If the pipe is open at both ends. 
(b) If the pipe is closed at one end. 
(Speed of sound in air = 340 ms1) 
1 
= 1 m 
V 
340 ms 
To Find 
When the pipe is open at both ends 
(a) 
Next 1st harmonics 
Fundamental frequency = f1 = ? 
f2 = ? 
[CHAPTER 8] 
Fundamental frequency = f1 = ? 
Next 3rd harmonics 
f3 = ? 
Next 5th harmonics 
fs = ? 
SOLUTION 
(a) 
When pipe is open at both end, 
Using 
n V 
fn 
21 
WAVES 
292 
1, 2, 3, 4, ...... 
n 
V 
Here 
f1 
21 
alemcity.com 
So, 
f1 
f1 
f2 
f2 
f3 
f3 
= 
340 
2x1 
170 Hz 
2f1 
= 340 Hz 
= 
3f1 
510 Hz 
When pipe is closed at one end and is open at the other 
Using, 
fn 
1, 3, 5, 7, ...... 
Here, 
> 
(b) 
Result 
(a) 
and, 
fi 
fi 
f3 
f3 
फार फार क क फु 
= 
41 
340 
4 x 1 
85 Hz 
3f1 
= 3f1 
fs 
fs 
= 
= 
= 3 x 85 
= 255 Hz 
5f1 
= 5 x 85 
5f1 
= 425 Hz 
When the pipe is open at both ends 
Fundamental frequency 
= 
fi 
= 
170 Hz 
WAVES 
293 
Frequency of source 
fo 
Speed of sound 
V 
To Find 
Apparent frequency = f' 
Apparent frequency when train moves away from station with same speed f" = ? 
SOLUTION 
When train is moving towards listener, then by using 
|| 
V 
f V – us 
Speed of train 
= 
Us 
= 
90 km/h 
90 × 1000 m/3600s 
= 25 m/s 
= 1000 Hz 
340 m/s 
[CHAPTER 8] 
When the pipe is closed at one end 
Fundamental frequency = f1 = 
(b) 
Next 3rd harmonics 
Next 5th harmonics 
f3 
= 
85 Hz 
255 Hz 
fs = 425 Hz 
EXAMPLE 8.5 
A train is approaching a station at 90kmh1 sounding a whistle of frequency 1000 Hz. What will be the apparent frequency of the whistle as heard by a listener sitting on the platform? What will be the apparent frequency heard by the same listener if the train moves away from the station with the same speed? 
Data 
(Speed of sound = 
340msTM1) 
te meity.com 
Putting values. 
340 
f' 
× 1000 
340-25 
1079.4 Hz. 
When train is moving away from listener then by using. 
Putting values. 
f" 
f" 
|| 
(^^) 
340 
1000 
[CHAPTER 8] 
Result 
WAVES 
Apparent frequency = f' 
1079.4 Hz 
Apparent frequency when train moves away from station with same speed 
f" 
931.5 Hz 
taleemcity.com 
294

Transcript for:Understanding Waves: Key Concepts and Effects

Transcript for:
Understanding Waves: Key Concepts and Effects