Physics Sound PDF

8. Sound Can you recall? 1. What type of wave is a sound wave? 2. Can sound travel in vacuum? 3. What are reverberation and echo? 4. What is meant by pitch of sound? 8.1 Introduction:...

8. Sound Can you recall? 1. What type of wave is a sound wave? 2. Can sound travel in vacuum? 3. What are reverberation and echo? 4. What is meant by pitch of sound? 8.1 Introduction: agencies. In such types of waves energy gets We are all aware of the ripples created on the transferred from one point to another. Water surface of water when a stone is dropped in it. waves mentioned above are travelling waves. The water molecules oscillate up and down They keep travelling outward from the point around their equilibrium positions but they do where stone was dropped until they are stopped not move from one point to another along the by walls of the container or the boundary of the surface of water. The disturbance created by water body. Other type of waves are stationary dropping the stone, however travels outwards. waves about which we will learn in XIIth This type of wave is a periodic and regular standard. disturbance in a medium which does not cause 8.2 Common Properties of all Waves: any flow of material but causes the flow of The properties described below are valid for energy and momentum from one point to all types of waves, however, here they are another. There are different types of waves and described for mechanical waves. not all types require material medium to travel 1) Amplitude (A): Amplitude of a wave motion through. We know that light is a type of wave is the largest displacement of a particle of the and it can travel through vacuum. Here we will medium through which the wave is propagating, first study different types of waves, learn about from its rest position. It is measured in metre their common properties and then study sound in SI units. waves in particular. 2) Wavelength ( ): Wavelength is the distance Types of waves: between two successive particles which are (i) Mechanical waves: A wave is said to be in the same state of vibration. It is further mechanical if a material medium is essential explained below. It is measured in metre. for its propagation. Examples of these types 3) Period (T): Time required to complete one of waves are water waves, waves along a vibration by a particle of the medium is the stretched string, seismic waves, sound waves, period T of the wave. It is measured in seconds. etc. 4) Double periodicity: Waves possess double (ii) EM waves: These are generated due to periodicity. At every location the wave motion periodic vibrations in electric and magnetic repeats itself at equal intervals of time, hence fields. These waves can propagate through it is periodic in time. Similarly, at any given material media, however, material medium is instant, the form of wave repeats at equal not essential for their propagation. These will distances hence, it is periodic in space. In be studied in Chapter 13. this way wave motion is a doubly periodic (iii) Matter waves: There is always a wave phenomenon i.e periodic in time and periodic associated with any object if it is in motion. in space. Such waves are matter waves. These are 5) Frequency (n): Frequency of a wave is the studied in quantum mechanics. number of vibrations performed by a particle Travelling or progressive waves are during each second. SI unit of frequency waves in which a disturbance created at is hertz. (Hz) Frequency is a reciprocal one place travels to distant points and keeps 1 travelling unless stopped by some external of time period, i.e., n T 142 6) Velocity (v): The distance covered by a In the figure 8.1 (a), displacements of wave per unit time is called the velocity of the various particles along a sinusoidal wave wave. During the period (T), the wave covers a travelling along + ve x-axis are plotted against distance equal to the wavelength ( ) Therefore their respective distances from the source (at the magnitude of velocity of wave is given by, O) at a given instant. This plot is valid for distance transverse as well as longitudinal wave. Magnitude of velocity = time The state of oscillation of a particle is called wavelength its phase. In order to describe the phase at a v place, we need to know (a) the displacement (b) period the direction of velocity and (c) the oscillation  number (during which oscillation) of the v  T --- (8.1) particle there. 1 In Fig. 8.1 (a), particles P and Q (or E and but n (frequency) --- (8.2) C or B and D) have same displacements but the T  v  n --- (8.3) directions of the their velocities are opposite. This equation indicates that, the magnitude Particles B and F have same magnitude of of velocity of a wave in a medium is constant. displacements and same direction of velocity. Increase in frequency of a wave causes decrease Such particles are said to be in phase during in its wavelength. When a wave goes from one their respective oscillations. Also, these are medium to another medium, the frequency of successive particles with this property of having the wave does not change. In such a case speed same phase. Separation between these two and wavelength of the wave change. particles is wavelength. These two successive For mechanical waves to propagate through particles differ by '1' in their oscillation number, a medium, the medium should possess certain i.e., if particle B is at its nth oscillation, particle properties as given below: F will be at its (n +1)th oscillation as the wave is travelling along + x direction. Most convenient i) The medium should be continuous and way to understand phase is in terms of angle. elastic so that the medium regains original For a sinusoidal wave, the variation in the state after removal of deforming forces. displacement is a 'sine' function of distance ii) The medium should possess inertia. The from the source and of time as discussed below. medium must be capable of storing energy For such waves it is possible for us to assign and of transferring it in the form of waves. angles corresponding to the displacement (or iii) The frictional resistance of the medium time). must be negligible so that the oscillations At the instant the above graph is drawn, will not be damped. the disturbance (energy) has just reached the 7) Phase and phase difference: particle A. The phase angle corresponding to this particle A can be taken as 0 At this instant, particle B has completed quarter oscillation and reached its positive maximum (sin 1). The phase angle of this particle Fig. 8.1 (a): Displacement as a function of B is c/2 = 90 at this instant. Similarly, phase distance along the wave. angles of particles C and E are c (180 ) and 2 c (360 ) respectively. Particle F has completed one oscillation and is at its positive maximum during its second oscillation. Hence its phase  c 5 c angle is 2   c Fig. 8.1 (b): Displacement as a function of. 2 2 time. 143  B and F are the successive particles in 8.3 Transverse Waves and Longitudinal the same state (same displacement and same Waves: direction of velocity) during their respective Progressive waves can be of two types, oscillations. Separation between these two is transverse and longitudinal waves. wavelength ( ). Phase angle between these Transverse waves : A wave in which two differs by 2 c. Hence wavelength is better particles of the medium vibrate in a direction understood as the separation between two perpendicular to the direction of propagation of particles with phase difference of 2 c. wave is called transverse wave. Water waves As noted above, waves possess double are transverse waves, as water molecules periodicity. This means the displacements of vibrate perpendicular to the surface of water particles are periodic in space (as shown in Fig.while the wave propagates along the surface. 8.1 (a)) as well as periodic time. Figure 8.1 Characteristics of transverse waves. (b) shows the displacement of one particular 1) All particles of the medium in the path of the particle as a function of time. wave vibrate in a direction perpendicular to the direction of propagation of the wave Activity : with same period and amplitude. 2) When transverse wave passes through (1) Using axes of displacement and a medium, the medium is divided distance, sketch two waves A and B into alternate the crests i.e., regions of such that A has twice the wavelength positive displacements and troughs i.e., regions of negative displacements. and half the amplitude of B. 3) A crest and an adjacent trough form one (2) Determine the wavelength and cycle of a transverse wave. The distance amplitude of each of the two waves P measured along the wave between any two and Q shown in figure below. consecutive points in the same phase (crest or trough) is called the wavelength of the wave. 4) Crests and troughs advance in the medium and are responsible for transfer of energy. 5) Transverse waves can travel through solids Characteristics of progressive wave and on surfaces of liquids only. They can 1) All vibrating particles of the medium have not travel through liquids and gases. EM same amplitude, period and frequency. waves are transverse waves but they do 2) State of oscillation i.e., phase changes from not require material medium for particle to particle. propagation. Example 8.1: The speed of sound in air is 330 6) When transverse waves advance through a m/s and that in glass is 4500 m/s. What is the medium there is no change in the pressure ratio of the wavelength of sound of a given and density at any point of medium, frequency in the two media? however shape changes periodically. Solution: vair = n air 7) If vibrations of all the particles along the path of a wave are constrained to be in vglass = n glass a single plane, then the wave is called air v air 330 polarised wave. Transverse wave can be  =   7.33  10 2 glass v glass 4500 polarised.  0.0733  7.33  102 8) Medium conveying a transverse wave must possess elasticity of shape. 144 Longitudinal waves : A wave in which requirement of the function is that it should particles of the medium vibrate in a direction describe the motion of the particle of the medium parallel to the direction of propagation of wave at that point. A sinusoidal progressive wave can is called longitudinal wave. Sound waves are be described by a sinusoidal function. Let us longitudinal waves. assume that the progressive wave is transverse Characteristics of longitudinal waves: and, therefore, the position of the particle of the 1) All the particles of medium along the path medium is described by a fixed value of x. The of the wave vibrate in a direction parallel to the direction of propagation of wave displacement from the equilibrium position can with same period and amplitude. be described by y. Such a sinusoidal wave can 2) When longitudinal wave passes through be written as follows: a medium, the medium is divided into y (x,t) = a sin (kx - t + ) --- (8.4) regions of alternate compressions and Hence a, k, and are constants. rarefactions. Compression is the region Let us see the justification for writing this where the particles of medium are crowded equation. At a particular instant say t = to, (high pressure zone), while rarefaction is y (x, t0) = a sin (kx - t0 + ) the region where the particles of medium = a sin (kx + constant ) are more widely separated, i.e. the medium gets rarefied (low pressure zone). Thus the shape of the wave at t = t0, as a 3) A compression and adjacent rarefaction function of x is a sine wave. form one cycle of longitudinal wave. The Also, at a fixed location x = x0, distance measured along the wave between y (x0,t) = a sin (kx0- t + ) any two consecutive points having the = a sin (constant - t) same phase is the wavelength of wave. Hence the displacement y, at x = x0 varies 4) For propagation of longitudinal waves, as a sine function. the medium should possess the property This means that the particles of the medium, of elasticity of volume. Thus longitudinal through which the wave travels, execute simple waves can travel through solids. liquids and gases. Longitudinal wave can not harmonic motion around their equilibrium travel through vacuum or free space. position. In addition x must increase in the 5) The compression and rarefaction advance positive direction as time t increases, so as to in the medium and are responsible for keep (kx- t + ) a constant. Thus the Eq. (8.4) transfer of energy. represents a wave travelling along the positive 6) When longitudinal wave advances through x axis. A wave represented by a medium there are periodic variations y(x, t) = a sin (kx + t + ) --- (8.5) in pressure and density along the path of is a wave travelling in the direction of the wave and also with time. negative x axis. 7) Longitudinal waves can not be polarised, as the direction of vibration of particles Symbols in Eq. (8.4): and direction of propagation of wave are y (x, t) is the displacement as a function of same or parallel. position (x) and time (t) 8.4 Mathematical Expression of a Wave: a is the amplitude of the wave. Let us describe a progressive wave is the angular frequency of the wave mathematically. Since it is a progressive wave, k is the angular wave number we require a function of both the position x and (kx0- t + ) is the argument of the time t. This function will describe the shape sinusoidal wave and is the phase of the particle of the wave at any instant of time. Another at x at time t. 145 8.5 The Speed of Travelling Waves Table 8.1: Speed of Sound in Gas, Liquids, Speed of a mechanical wave depends and Solids upon the elastic properties and density of the medium. The same medium can support both Medium Speed (m/s) transverse and longitudinal waves which have Gases different speeds. Air [0 C] 331 8.5.1 The speed of transverse waves Air [20 C] 343 The speed of a wave is determined by Helium 965 the restoring force produced in the medium Hydrogen 1284 when it is disturbed. The speed also depends Liquids on inertial properties like mass density of the Water (0 C) 1402 medium. The waves produced on a string are Water (20 C) 1482 transverse waves. In this case the restoring Seawater 1522 force is provided by the tension T in the string. Solids The inertial property i.e. the linear mass density Vulcanised Rubber 54 m, can be determined from the mass of string M Copper 3560 and its length L as m = M/L. The formula for Steel 5941 speed of transverse wave on stretched string is Granite 6000 given by Aluminium 6420 T v --- (8.6) 8.5.3 Newton’s formula for velocity of sound: m Propagation of longitudinal waves was The derivation of the formula is beyond the studied by Newton. Sound waves travel scope of this book. through a medium in the form of compressions The important point here is that the speed of a and rarefactions. The density of medium is transverse wave depends only on the properties greater at the compression while being smaller of the string, T and m. It does not depend on in the rarefaction. Hence the velocity of sound wavelength or frequency of the wave. depends on elasticity and density of the medium. 8.5.2 The speed of longitudinal waves Newton formulated the relation as In case of longitudinal waves, the particles E of the medium oscillate forward and backward v  --- (8.7)  along the direction of wave propagation. This where E is the proper modulus of elasticity of causes compression and rarefaction which medium and is the density of medium. travel in the medium as the medium possess elastic property. Newton assumed that, during propagation Speed of sound in liquids and solids is of sound, there is no change in the average higher than that in gases. The speed of sound temperature of the medium. Hence sound as a longitudinal wave in an ideal gas is given wave propagation in air is an isothermal by Newton’s formula as discussed below. process (temperature remaining constant ) and Speed of sound in different media is given in isothermal elasticity should be considered. table below. The volume elasticity of air determined under isothermal change is called isothermal bulk Always remember: modulus and is equal to the atmospheric When a sound wave goes from one pressure ‘P’. Hence Newtons formula for speed medium to another its velocity changes of sound in air is given by along with its wavelength. Its frequency, P which is decided by the source remains v   --- (8.8) constant. 146 As atmospheric pressure is given by P=hdg process but is a rapid process. If frequency is and at NTP, 256 Hz, the air is compressed and rarefied 256 h = 0.76 m of Hg times in a second. Such process must be a rapid process. Heat is produced during compression d = 13600 kg/m3-density of mercury and is lost during rarefaction. This heat does not = 1.293 kg/m3- density of air get sufficient time for dissipation. Due to this and g = 9.8 m/s2 the total heat content remains the same. Such 0.76  13600  9.8 a process is called an adiabatic process and v hence, adiabatic elasticity must be adiabatic 1.293 and not isothermal elasticity, as was assumed v = 279.9 m/s at NTP. by Newton. This is the value of velocity of sound according to Newton’s formula. But the Always remember: experimental value of velocity of sound at 00C In isothermal process temperature as determined earlier by a number of scientists remains constant while in adiabatic is 332 m/s. The difference between predicted process there is neither transfer of heat value by Newton’s formula and experimental nor of mass. value is large and it is not due to experimental The adiabatic modulus of elasticity of air error. The Experimental value is 16% greater is given by, than the value given by the formula. Newton could not give satisfactory explanation of this E= P --- (8.9) discrepancy. It was resolved by French physicist where P is the pressure of the medium (air) Pierre Simon Laplace (1749-1827). and is ratio of specific heat of air at constant Example 8.2: Suppose you are listening to an pressure (Cp) to the specific heat of air at out-door live concert sitting at a distance of 150 constant volume (Cv) called as the adiabatic m from the speakers. Your friend is listening ratio to the live broadcast of the concert in another Cp i.e., = --- (8.10) country and the radio signal has to travel 3000 Cv km to reach him. Who will hear the music first and what will be the time difference between For air the ratio of Cp / Cv is 1.41 the two? Velocity of light =3×108 m/s and that i.e. = 1.41 of sound is 330m/s. Newton's formula for speed of sound in air as Solution: Time taken by sound to reach you modified by Laplace to give 150 P s 0.4546 v  --- (8.11) 330  Time taken by the broadcasted sound (which is done by EM waves having velocity =3×108m/s) Accoriding to this formula velocity of sound at 3 NTP is 3000 km 3  10    102 s 5 3  10 km / s 3  10 5 1.41 0.76  13600  9.8 v 1.293 ∴ your friend will hear the sound first. The time difference will be = 332.3 m/s = 0.4546 - 0.01 This value is in close agreement with = 0.4446 s. the experimental value. As seen above, the velocity of sound depends on the properties of 8.5.4 Laplace’s correction the medium. According to Laplace, the generation of compression and rarefaction is not a slow 147 8.5.5 Factors affecting speed of sound: Hence for gaseous medium obeying ideal As sound waves travel through atmosphere gas equation change in pressure has no effect (open air), some factors related to air affect the on velocity of sound unless there is change in speed of sound. temperature. Example 8.3: Consider a closed box of rigid Do you know ? walls so that the density of the air inside it Cp, the specific heat of gas at constant is constant. On heating, the pressure of this pressure, is defined as the quantity of heat enclosed air is increased from P0 to P. It is now required to raise the temperature of unit mass observed that sound travels 1.5 times faster of gas through 1 K when pressure remains 0 than at pressure P0 calculate P/P0. constant. Solution: Cv, the specific heat of gas at constant  Po volume, is defined as the quantity of heat vP   required to raise the temperature of unit mass of gas through 10 K when volume remains  Po vPo  constant.  When pressure is kept constant the v P  1.5 v P o volume of the gas increases with increase in temperature. Thus additional heat is required P  Po  1.5 to increase the volume of gas against the   external pressure. Therefore heat required P P  2.25 o to raise the temperature of unit mass of gas   through 10 K when pressure is kept constant P  2.25 Po is greater than the heat required when volume is kept constant. i.e. Cp > Cv. (b) Effect of temperature on speed of sound Suppose vo and v are the speeds of sound at a) Effect of pressure on velocity of sound T0 and T in kelvin respectively. Let 0 and be According to Laplace’s formula velocity the densities of gas at these two temperatures. of sound in air is The velocity of sound at temperature T0 and T P can be written by using Eq. (8.13), v    RT0 v0  M --- M is molar mass, n = 1 If M is the mass and V is volume of air then M  RT  v V M v RT  PV   v  --- (8.12) v0 RT0 M At constant temperature PV = constant v T   --- (8.14) according to Boyle’s law. Also M and are v0 T0 constant, hence v = constant. This equation shows that speed of sound Therefore at constant temperature, a in air is directly proportional to the square root change in pressure has no effect on velocity of of absolute temperature. Thus, speed of sound sound in air. This can be seen in another way. in air increases with increase in temperature. For gaseous medium, PV = nRT, n being the Taking T = 273 K and writing T= (273 + t) K o number of moles. where t is the temperature in degree celsius.  nRT --- (8.13) The ratio of velocity of sound in air at t C to 0 v  M that at 0 C is given by, 0 148  v 273  t m < d   = 0.81 kg/m3(at 0 C) and v0 273 m = 1.29 kg/m3(at 0 C)) v t d   1... vm > vd. v0 273 Thus, the speed of sound in moist air is v 1 greater than speed of sound in dry air. i.e speed   1   t where   v0 273 increases with increase in the moistness of air. 1 8.6 Principle of Superposition of Waves: or, v  v 0 1   t  2 Waves don’t display any repulsion towards As is very small,we can write each other. Therefore two wave patterns can  1  overlap in the same region of the space without v v0 1   t   2  affecting each other. When two waves overlap their displacements add vectorially. This  1 1  additive rule is referred to as the principle of v  v0 1   t  2 273  superposition of waves. When two or more waves travelling  t  through a medium arrive at a point of medium v  v0 1   simultaneously, each wave produces its own  546  displacement at that point independent of v0 the others. Hence the resultant displacement v  v0  t 546 at that point is equal to the vector sum of But v0 =332 m/s at 00C the displacements due to all the waves. The phenomenon of superposition will be discussed 332  v  v0  t in detail in XIIth standard. 546 8.7 Echo, reverberation and acoustics:  v v 0   0.61 t , --- (8.15) Sound waves obey the same laws of i.e., for 1 C rise in temperature velocity reflection as those of light. increases by 0.61 m/s. Hence for small 8.7.1 Echo: variations in temperature (< 50 C), the speed An echo is the repetition of the original of sound changes linearly with temperature. sound because of reflection from some rigid (c) Effect of humidity on speed of sound surface at a distance from the source of sound. Humidity (moisture) in air depends upon If we shout in a hilly region, we are likely to the presence of water vapour in it. Let m and d hear echo. be the densities of moist and dry air respectively. Why can’t we hear an echo at every place? If vm and vd are the speeds of sound in moist air At 220C, the velocity of sound in air is 344 m/s. and dry air then using Eq. (8.11). Our brain retains sound for 0.1 second. Thus for P us to hear a distinct echo, the sound should take vm  m more than 0.1s after starting from the source (i.e., from us) to get reflected and come back P to us. and v d  d distance = speed × time v d = 344 × 0.1  m  --- (8.16) vd m = 34.4 m. Moist air is always less dense than dry air, To be able to hear a distinct echo, the i.e., reflecting surface should be at a minimum 149 distance of half of the above distance i.e 17.2 8.7.3 Acoustics: m. As velocity depends on the temperature of The branch of physics which deals with air, this distance will change with temperature. the study of production, transmission and Example 8.4: A man shouts loudly close to a reception of sound is called acoustics. This is high wall. He hears an echo. If the man is at useful during the construction of theaters and 40 m from the wall, how long after the shout auditorium. While designing an auditorium, will the echo be heard ? (speed of sound in air proper care for the absorption and reflection of = 330 m/s) sound should be taken. Otherwise audience will solution: The distance travelled by the sound not be able to hear the sound clearly. wave For proper acoustics in an auditorium the = 2 × distance from man to wall. following conditions must be satisfied. = 2 × 40 1) The sound should be heard sufficiently = 80 m. loudly at all the points in the auditorium. distance... Time taken to travel the distance = The surface behind the speaker should be speed parabolic with the speaker at its focus; so 80 m that the distribution of sound is uniform = in the auditorium. Reflection of sound 330 m / s is helpful in maintaining good loudness 0.24 s... The man will hear the echo 0.24 s after he through the entire auditorium. shouts. 2) Echoes and reverberation must be 8.7.2 Reverberation: eliminated or reduced. Echoes can be If the reflecting surface is nearer than 15 reduced by making the reflecting surfaces m from the source of sound, the echo joins up more absorptive. Echo will be less if the with the original sound which then seems to be auditorium is full. prolonged. Sound waves get reflected multiple 3) Unnecessary focusing of sound should be times from the walls and roof of a closed avoided and there should not be any zone room which are nearer than 15 m. This causes of poor audibility or region of silence. For a single sound to be heard not just once but that purpose curved surface of the wall or continuously. This is called reverberation. It is ceiling should be avoided. this the persistence of sound after the source has 4) Echelon effect : It is due to the mixing of switched off, as a result of repeated reflection sound produced in the hall by the echoes from walls, ceilings and other surfaces. of sound produced in front of regular Reverberation characteristics are important in structure like the stairs. To avoid this, stair the design of concert halls, theatres etc. type construction must be avoided in the If the time between successive reflections hall. of a particular sound wave reaching us is small, 5) The auditorium should be sound-proof the reflected sound gets mixed up and produces when closed, so that stray sound can not a continuous sound of increased loudness which enter from outside. can’t be heard clearly. 6) For proper acoustics no sound should be Reverberation can be decreased by making produced from the inside fittings, seats, the walls and roofs rough and by using curtains etc. Instead of fans, air conditioners may in the hall to avoid reflection of sound. Chairs be used. Soft action door closers should be and wall surfaces are covered with sound used. absorbing materials. Porous cardboard sheets, Acoustics observed in nature perforated acoustic tiles, gypsum boards, thick The importance of acoustic principles goes curtains etc. at the ceilings and at the walls are far beyond human hearing. Several animals use most convenient to reduce reverberation. 150 sound for navigation. mass transit vehicle involves the study (a) Bats depends on sound rather than light of generation and propagation of sound to locate objects. So they can fly in in the motor’s wheels and supporting total darkness of caves. They emit short structures. ultrasonic pulses of frequency 30 kHz to (c) We can study properties of the Earth by 150 kHz. The resulting echoes give them measuring the reflected and refracted information about location of the obstacle. elastic waves passing through its interior. (b) Dolphins use an analogous system for It is useful for geological studies to detect underwater navigation. The frequencies local anomalies like oil deposits etc. are subsonic about 100 Hz. They can sense 8.8 Qualities of sound: an object of about the size of a wavelength Audible sound or human response to sound: i.e., 1.4 m or larger. Whenever we talk about audible sound, Medical applications of acoustics what matters is how we perceive it. This is (a) Shock waves which are high pressure high purely a subjective attribute of sound waves. amplitude waves are used to split kidney Major qualities of sound that are of our stones into smaller pieces without invasive interest are (i) Pitch, (ii) Timbre or quality and surgery. A shock wave is produced (iii) Loudness. outside the body and is then focused by a (i) Pitch: reflector or acoustic lens so that as much This aspect refers to sharpness or shrillness of its energy as possible converges on the of the sound. If the frequency of sound is stone. When the resulting stresses in the increased, what we perceive is the increase in stone exceeds its tensile strength, it breaks the pitch or we feel the sound to be sharper. into small pieces which can be removed Tone refers to the single frequency of that wave easily. while a note may contain one or more than (b) Reflection of ultrasonic waves from one tones. We use the words high pitch or high regions in the interior of body is used for tones if frequency is higher. As sharpness is a ultrasonic imaging. It is used for prenatal subjective term, sentences like “sound of double (before the birth) examination, detection frequency is doubly sharp” make no sense. of anamolous conditions like tumour etc Also, a high pitch sound need not be louder. and the study of heart valve action. Tones of guitar are sharper than that of a base (c) At very high power level, ultrasound is guitar, sound of tabla is sharper than that of selective destroyer of pathalogical tissues a dagga, (in general) female sound is sharper in treatment of arthritis and certain type of than that of a male sound and so on. cancer. For a sound amplifier (or equaliser) when Other applications of acoustics we raise the treble knob (or treble Button), high (a) SONAR is an acronym for Sound frequencies are boosted and if we raise bass Navigational Ranging. This is a technique knob, low frequencies are boosted. for locating objects underwater by (ii) Timbre (sound quality) transmitting a pulse of ultrasonic sound During telephonic conversation with a and detecting the reflected pulse. The time friend, (mostly) you are able to know who is delay between transmission of a pulse and speaking at the other end even if you are not the reception of reflected pulse indicates told about who is speaking. Quite often we say, the depth of the object. This system is “Couldn’t you recognise the voice?” The sound useful to measure motion and position of quality in this context is called timbre. Same the submerged objects like submarine. song played on a guitar, a violin, a harmonium (b) Acoustic principle has important or a piano feel significantly different and we application to environmental problems can easily identify that instrument. Quality like noise control. The design of quiet- of sound of any sound instrument (including 151 our vocal organ) depends upon the mixture of and so on. tones and overtones in the sound generated by Hence, loudness of 20 db sound is felt that instrument. Even our own sound quality double that of 10 db, but its intensity is 10 during morning (after we get up) and in the times that of the 10 db sound. Now, we feel 40 evening is different. It is drastically affected if db sound twice as loud as 20 db sound but its we are suffering from cold or cough. Concept intensity is 100 times as that of 20 db sound of overtones will be discussed during XIIth and 10000 times that of 10 db sound. This is the standard. power of logarithmic or exponential scale. (iii) Loudness: If we move away from a (practically) point Intensity of a wave is a measurable source, the intensity of its sound varies inversely quantity which is proportional to square of 1 the amplitude (I A2) and is measured in with square of the distance, i.e., I. r2 the (SI) unit of W/m2. Human perception of Whenever you are using earphones or jam intensity of sound is loudness. Obviously, if your mobile at your ear, the distance from the intensity is more, loudness is more. The human source is too small. Obviously, such a habit for response to intensity is not linear, i.e., a sound a long time can affect your normal hearing. of double intensity is louder but not doubly Example 8.5: When heard independently, two loud. This is also valid for brightness of light. sound waves produce sensations of 60 db and In both cases, the response is approximately 55 db respectively. How much will the sensation logarithmic. Using this property, the loudness be if those are sounded together, perfectly in (and brightness) can be measured. phase? Under ideal conditions, for a perfectly Solution: healthy human ear, the least audible intensity is I I L1  60 db  10 log10 1  1  106 or I1  106 I 0 I0 = 10-12 W/m2. Loudness of a sound of intensity I0 I0 I, measured in the unit bel is given by Similarly, I 2 105.5 I 0  I  Lbel  log10   --- (8.17) As the waves combine perfectly in phase,  I0  the vector addition of their amplitudes will be 2 2 2 2 Popular or commonly used unit for loudness is given by A  ( A1  A2 )  A1  A2  2 A1 A2 decibel. As intensity is proportional to square of the We know, 1 decimetre or 1 dm = 0.1m. amplitude. Similarly, 1 decibel or 1 db = 0.1 bel. ∴1 bel =  I  I1  I 2  2 I1 I 2 10 db. Thus, loudness in db is 10 times loudness in bel   105 I 0 101  100.5  2 101.5   I   Ldb  10 Lbel  10 log10   I  0   105 I 0 10  3.1623  2  100.75  For sound of least audible intensity I0,  24.41 105 I 0  2.441 106 I 0 I  Ldb 10 log10  0   10 log10 1  0 --- (8.18)  I0  This corresponds to threshold of hearing For sound of 10 db,  I   I  10 10 log10      101 or I  10 I 0  I0   I0  For sound of 20 db,  I   I  It is interesting to note that there is only a 20  10 log10     102 or I  100 I 0 marginal increase in the loudness.  I0   I0  152  Table 8.2: Approximate Decibel Ratings of Some Audible Sounds Source or description of noise Loudness, Ldb Effcet Immediate ear Extremely loud 160 damage Jet aeroplane, near 25 m 150 Rupture of eardrum Auto horn, within a metre, Aircraft tale 110 Strongly painful off, 60 m Diesel train, 30 m, Average factory 80 Highway traffic, 8 m 70 Uncomfortable Conversion at a restaurant 60 Conversation at home 50 Quiet urban background sound 40 Quiet rural area 30 Virtual silence Whispering of leaves, 5 m 20 Normal breathing 10 Threshold of hearing 0 8.9 Doppler Effect: Do you know ? Have you ever heard an approaching train According to the world health and noticed distinct change in the pitch of the organisation a billion young people could be sound of its whistle, when it passes away ? at risk of hearing loss due to unsafe listening Same thing similar happens when a listener practices. Among teenagers and young moves towards or away from the stationary adults aged 12-35 years (i) about 50% are source of sound. Such a phenomenon was exposed to unsafe levels of sound from use first identified in 1842 by Austrian physicist of personal audio devices and (ii) about 40% Christian Doppler (1803-1853) and is known are exposed to potentially damaging sound as Doppler effect. levels at clubs, discotheques and bars. When a source of sound and a listener are 8.9.1 Source Moving and Listener Stationary: in motion relative to each other the frequency Consider a source of sound S, moving away of sound heard by listener is not the same as the from a stationary listener L (called relative frequency emitted by the source. recede) with velocity vs. Speed of sound waves Doppler effect is the apparent change with respect to the medium is v which is always in frequency of sound due to relative motion positive. Suppose the listener uses a detector for between the source and listener. Doppler effect counting each wave crest that reaches it. is a wave phenomenon. It holds for sound Initially (at t = 0), source which is at point waves and also for EM waves. But here we S1 emits a crest when at distance d from the shall consider it for sound waves only. listener see Fig. 8.2 (a). This crest reaches the The changes is frequency can be studied under listener at time t1= d/v. Let T0 be the time period 3 different conditions: at which the waves are emitted. Thus, at t = 1) When listener is stationary but source is T0 the source moves the distance = vs To and moving. reaches the point S2. Distance of S2 from the 2) When listener is moving but source is listener is (d+vsTo). when at S2, the source emits stationary. second crest. This crest reaches the listener at 3) When listener and source both are moving.  d  v sT0  t2 =T0    --- (8.19)  v  153  Similarly at time pTo, the source emits its (p+1)th crest (where, p is an integer, p = 1,2,3,...). It reaches the listener at time  d + pv sT0  Fig. 8.2 (b): Doppler effect detected when t p+1 = pT0 +    v  the listener is moving and source is at rest Hence the listener’s detector counts p in the medium. crests in the time interval 8.9.2 Listener Approaching a Stationary  d + pv sT0  d Source with Velocity vL: t p+1 - t1 = pT0 +  -  v Consider a listener approaching with  v  Hence the period of wave as recorded by velocity vL towards a stationary source S as the listener is shown in Fig. 8.2 (b). Let the first wave be (t - t ) emitted by the source at t = 0, when the listener T = p+1 1 or was at L1 at an initial distance d from the source. p Let t1 be the instant when the listener receives  d + pv sT0 d  this (wave), his position being L2. During time  pT0 + v -  v T =  t1, the listener travels distance vLt1 towards the p stationary source. In this time, the sound wave v T travels distance  d  v L t1  with speed v. T =T0 + s 0 v d  v L t1 d  t1   t1   v  v v  vL T =T0 1 + s   v  Second wave is emitted by the source at t T0 v +v s  = the time period of the waves emitted by the T =T0    v  source. Let t2 be the instant when the listener receives second wave. During time t2 , the 1 1  v   =   distance travelled by the listener is v L t2. Thus, T T0  v + v s  the distance to be travelled by the sound to  v  reach the listener is then d v L t2.  n = n0   --- (8.20) v +v s  ∴ Sound (second wave) travels this distance d  v L t2 where n is the frequency recorded by the listener with speed v in time  v and no is the frequency emitted by the source. However, this time should be counted after T0, If source of sound is moving towards the as the second wave was emitted at t T0. listener with speed vs (called relative approach), d  v L t2  t  vT0  d the second term from Eq. (8.18) onwards, will  t 2  T0  2 v  vL v be negative (or will be subtracted). Thus, in this case, d  v L t3 2vT0  d Similarly, t3  2T0   t3  v v  vL  v  Extending this argument to (p+1)th wave, we n = n0   --- (8.21) v - v can write,  s  d  v L t p 1 pvT0  d t p 1  pT0   t p 1  v v  vL Time duration between instances of receiving Fig. 8.2 (a): Doppler effect detected when successive waves is the observed or recorded the source is moving and listener is at rest period T. in the medium. 154  Case (III) If |vL| < |vs|, n < no as now there is pvT0  d d pvT0  pT  t p 1  t1    relative recede (source recedes faster, listener v  vL v  vL v  vL approaches slowly).  v  8.9.4 Common Properties between Doppler T  T0   --- (8.22)  v  vL  Effect of Sound and Light: A) Wherever there is relative motion between 1 1  v  vL  listener (or observer) and source (of sound   T T0  v  or light waves), the recorded frequency is different than the emitted frequency.  v  vL   n  n0   --- (8.23) B) Recorded frequency is higher (than emitted  v  frequency), if there is relative approach. 8.9.3 Both Source and Listener are Moving: C) Recorded frequency is lower, if there is In general when both the source and relative recede. listener are in motion, we can write the observed D) If vL or vs are much smaller then wave frequency speed (speed of sound or light) we can use  v  vL  vr as relative velocity. In this case, using n = n0   --- (8.24) Eq. (8.24)  v vs  n v r  --- (8.26) Where the upper signs (in both numerator and n v  denominator) should be chosen during relative approach while lower signs should be chosen where n is Doppler shift or change in the during relative recede. It must be remembered recorded frequency, i.e., |n - no| and is that ‘when you are deciding the sign for any the recorded change in wavelength. one of these, the other should be considered to n  n0 v r  be at rest’. n v Illustration:  v   n  n0  1  r  --- (8.27) Consider an observer or listener and a  v  source moving with respective velocities vL and vS along the same direction. In this case, listener Once again upper sign is to be used during is approaching the source with vL (irrespective relative approach while lower sign is to be of whether source is moving or not). Thus, the used during relative recede. upper, i.e., positive sign, should be chosen E) If velocities of source and observer for numerator. However, the source is moving (listener) are not along the same line with vS away form the listener irrespective of their respective components along the listener's motion. Thus the lower sign in the line joining them should be chosen for denominator which is positive has to be chosen. longitudinal Doppler effect and the same  v + vL  mathematical treatment is applicable.  n  no   --- (8.25) 8.9.5 Major Differences between Doppler  v + vs  Effects of Sound and Light: Case (I) If |vL| = |vs|, n = no. Thus there is no A) As the speed of light is absolute, only Doppler shift as there is no relative motion, relative velocity between the observer and even if both are moving. the source matters, i.e., who is in motion is Case (II) If |vL| > |vs|, numerator will be greater, not relevant. n > no. This is because there is relative approach B) Classical and relativistic Doppler effects as the listener approaches the source faster and are different in the case of light, while in the source is receding at a slower rate. case of sound, it is only classical. 155  C) For obtaining exact Doppler shift for sound  v + vL  waves, it is absolutely important to know n' = n   who is in motion.  v  D) If wind is present, its velocity alters the  330  220  n  3600  speed of sound and hence affects the  330  Doppler shift. In this case, component of n  6000 Hz the wind velocity (vw) is chosen along the The frequency of echo detected by rocket = line joining source and observer. This is to 6000 Hz be algebraically added with the velocity of sound. Hence 'v' is to be replaced by (v Example 8.7: A bat, flying at velocity VB = 12.5 vw) in all the above expressions. Positive m/s, is followed by a car running at velocity sign to be used if v and vw are along VB = 50 m/s. Actual directions of the velocities the same direction (remember that v is of the car and the bat are as shown in the figure always positive and always from source to below, both being in the same horizontal plane listener). Negative sign is to be used if v (the plane of the figure). To detect the car, the and vw are oppositely directed. bat radiates ultrasonic waves of frequency 36 kHz. Speed of sound at surrounding Example 8.6: A rocket is moving at a speed of temperature is 350 m/s. 220 m/s towards a stationary target. It emits a wave of frequency 1200 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) The frequency of sound detected by the target and (2) The frequency of echo detected by rocket (velocity of sound= 330 m/s.) There is an ultrasonic frequency detector fitted in the car. Calculate the frequency Solution: Given, target stationary, i.e., recorded by this detector. vL = 0, vs = 220 m/s, v = 330 m/s The ultrasonic waves radiated by the bat n0 = 1200 Hz are reflected by the car. The bat detects these To find the frequency of sound detected by waves and from the detected frequency, it the target we have to used Eq. (8.25) knows about the speed of the car. Calculate the  v  frequency of the reflected waves as detected by n = n0   the bat. (sin 37 = cos 53 0.6, sin 53 = cos  v  vs  37 0.8)  330  Solution: As shown in the figure below, the n  1200    330  220  components of velocities of the bat and the car, n  3600 Hz along the line joining them, are 0 1 The frequency of sound detected by the VC cos 53  50  0.6  30 m s and target = 3600 Hz. VB cos 370  12.5  0.8  10 m s 1. When echo is heard by rocket’s detector, These should be used while calculating the target is considered as source doppler shifted frequencies.... vs = 0 The frequency of sound emitted by the source (i.e. target) is n0 = 3600 Hz, and the frequency detected by rocket is n'. Now listener is approaching a the source and so we have to use. 156   v  vL  emitted frequency n0  38  103 Hz , n ? Doppler shifted frequency, n  n0  ;  v vs  Car, the source, is approaching the listener upper signs to be used during approach, lower (bat). signs during recede. Thus, v S vC cos 530 30 m/s Part I: Frequency radiated by the bat n0 = 36 0 ×103 Hz, Frequency detected by the detector in Thus, v L v B cos 37 10 m/s the car = n = ? Now bat-the listener is receding while car the In this case, bat is the source which is moving  v  vL  away from the car (receding) while the detector source is approaching  n  n0    v  vs  in the car is the listener, who is approaching the  350  10   n  38  103   source (bat). v s VB cos370 10 m/s and  350  30  v L VC cos 530 30 m/s 34  38  103  The source (bat) is receding, while the listener 33  v  vL  3  39.15 10 Hz (car) is approaching  n  n0    v  vs   39.15 kHz 3  350  30   n  36  10    350  10  Internet my friend  38  103 Hz = 38 kHz Part II: Reflected frequency, as detected by https://hyperphysics.phys-astr.gsu.edu/ the bat: Frequency reflected by the car is the hbase/hframe.html Doppler shifted frequency as detected at the car. Thus, this time, the car is the source with Ex erc ises Exercises 1. Choose the correct alternatives concerns should i) A sound carried by air from a sitar to a (A) amplify sound (B) reflect sound listener is a wave of following type. (C) transmit sound (D) absorb sound (A) Longitudinal stationary 2. Answer briefly. (B)Transverse progressive i) Wave motion is doubly periodic. (C) Transverse stationary Explain. (D) Longitudinal progressive ii) What is Doppler effect? ii) When sound waves travel from air to water, iii) Describe a transverse wave. which of these remains constant ? iv) Define a longitudinal wave. (A) Velocity (B) Frequency v) State Newton’s formula for velocity of (C) Wavelength (D) All of above sound. iii) The Laplace’s correction in the expression vi) What is the effect of pressure on velocity for velocity of sound given by Newton is of sound? needed because sound waves vii) What is the effect of humidity of air on (A) are longitudinal velocity of sound? (B) propagate isothermally viii) What do you mean by an echo? (C) propagate adiabatically ix) State any two applications of acoustics. (D) are of long wavelength x) Define amplitude and wavelength of a iv) Speed of sound is maximum in wave. (A) air (B) water xi) Draw a wave and indicate points which (C) vacuum (D) solid are (i) in phase (ii) out of phase (iii) v) The walls of the hall built for music have a phase difference of /2. 157 xii) Define the relation between velocity, what value should be the speed of sound wavelength and frequency of wave. in air? xiii) State and explain principle of b) Now, she moves to another location superposition of waves. and finds that she should now make xiv) State the expression for apparent 45 claps in 1 minute to coincide with frequency when source of sound and successive echoes. Calculate her distance listener are for the new position from the wall. i) moving towards each other [Ans: a) 340 m/s b) 255 m] ii) moving away from each other v) Sound wave A has period 0.015 s, sound xv) State the expression for apparent wave B has period 0.025. Which sound frequency when source is stationary and has greater frequency? listener is [Ans : A] 1) moving towards the source vii) At what temperature will the speed of 2) moving away from the source sound in air be 1.75 times its speed at xvi) State the expression for apparent N.T.P? frequency when listener is stationary [Ans: 836 k = 563 C] and source is. i) moving towards the listener viii) A man standing between 2 parallel eliffs ii) moving away from the listener fires a gun. He hearns two echos one xvii) Explain what is meant by phase of a after 3 seconds and other after 5 seconds. wave. The separation between the two cliffs is xviii) Define progressive wave. State any four 1360 m, what is the speed of sound? properties. [Ans:340m/s] xix) Distinguish between traverse waves and ix) If the velocity of sound in air at a given longitudinal waves. place on two different days of a given xx) Explain Newtons formula for velocity week are in the ratio of 1:1.1. Assuming of sound. What is its limitation? the temperatures on the two days to be 3. Solve the following problems. same what quantitative conclusion can i) A certain sound wave in air has a speed your draw about the condition on the 340 m/s and wavelength 1.7 m for this two days? wave, calculate [Ans: Air is moist on one day a) the frequency b) the period. and dry = 1.12 moist = 1.21 moist ] [Ans a) 200 Hz, b) 0.005s] x) A police car travels towards a stationary ii) A tuning fork of frequency 170 Hz observer at a speed of 15 m/s. The siren produces sound waves of wavelength 2 on the car emits a sound of frequency 250 m. Calculate speed of sound. Hz. Calculate the recorded frequency. [Ans: 340 m/s] The speed of sound is 340 m/s. iii) An echo-sounder in a fishing boat [Ans : 261.54 Hz] receives an echo from a shoal of fish xi) The sound emitted from the siren of an 0.45 s after it was sent. If the speed of ambulance has frequency of 1500 Hz. sound in water is 1500 m/s, how deep is The speed of sound is 340 m/s. Calculate the shoal? the difference in frequencies heard by [Ans : 337.5 m] a stationary observer it the ambulance iv) A girl stands 170 m away from a high initially. travels towards and then away wall and claps her hands at a steady rate from the observer at a speed of 30 m/s. so that each clap coincides with the echo [Ans : 1645-2676 Hz] of the one before. a) If she makes 60 claps in 1 minute, *** 158