# Waves and sound -01-Theory

26. May 2023

### Waves and sound -01-Theory

• 1. WAVES AND SOUND WAVE MOTION A wave is a disturbance which propagates energy (and momentum) from one place to the other without the transport of matter. It is well spread over a region of space without clear cut boundaries. It cannot be said to be localized here or there. Type of waves: (a) Mechanical Waves: The waves requiring a medium means waves which are produced due to the vibration of material particles of an elastic medium e.g. vibrating string, sound wave. (b) Non-mechanical Waves: Waves which are produced due to the periodic vibration of two mutually perpendicular electric and magnetic fields are Non-mechanical waves and they do not require any material medium. Such waves known as electromagnetic waves propagate in a direction perpendicular to both electric and magnetic field. e.g. X-ray,  -ray, light waves etc. EQUATION OF A TRAVELLING WAVE Suppose, man holding a stretched string starts snapping his hand at 0  t and finishes his job at   t t .The vertical displacement y of the left end of the string is a function of time. It is zero for 0  t , has non-zero value of 0    t t and is again zero for   t t . Let us represent this function by ( ) f t . Take the left end of the string a the origin and take the X-axis along the string towards right. The function ( ) f t represents the displacement y of the particle at 0  x as a function of time ( 0, ) ( )   y x t f t The disturbance travels on the string towards right with a constant speed v . Thus, the displacement, produced at a distance x from the left end at time t was originated at the left end at the time /  t x v . But the displacement of the left end at time /  t x v is ( / )  f t x v . Hence, ( , ) ( 0, / )    y x t y x t x v ( / )   f t x v . The displacement of the particle at x at time t i.e., ( , ) y x t is generally abbreviated as y and the wave equation is written as ( / )   y f t x v . …(i) Equation (i) represents a wave traveling in the positive x-direction with a constant speed v . Such a wave is called a traveling wave or a progressive wave. The function f is arbitrary and depends on how the source moves. The time t and the position x must appear in the wave equation in the combination /  t x v only. For example, ( / ) ( / ) sin ,      t x v T t x v y A y Ae T etc. are valid wave equations. They represent waves traveling in positive x-direction with constant speed. The equation 2 2 2 2 ( ) sin   x v t y A L does not represent a wave traveling in x-direction with a constant speed. If a wave travels in negative x-direction with speed v, its general equation may be written as ( / )   y f t x v …(ii) The wave traveling in positive x-direction (equation (i)) can also be written as
• 2. PH – Waves & Sound - 2         vt x y f v or, ( )   y g x vt , …(iii) where g is some other function having the following meaning. If we put 0  t in equation (iii), we get the displacement of various particles at 0  t i.e., ( , 0) ( )   y x t g x Thus, ( ) g x represents the shape of the string at 0  t . If the displacement of the different particles at 0  t is represented by the function ( ) g x , the displacement of the particle at x at time t will be ( )   y g x vt . Similarly, if the wave is traveling along the negative x-direction and the displacement of different particles at 0  t is ( ) g x , the displacement of the particle at x at time t will be ( )   y g x vt …(iv) Thus, the function f in equation (i) and (ii) represents the displacement of the point 0  x as time passes and g in (iii) and (iv) represents the displacement at 0  t of different particles. The travelling wave moving with constant speed v towards positive x direction must satisfy the following wave function condition 2 2 2 2 2  d y d y v dt dx …(v) Illustration 1. A wave is propagating on a long stretched string along its length taken as the positive x-axis. The wave equation is given as 2 0           t x T y y e where 0 4  y mm, 1.0  T s and 4   cm. (a) Find the velocity of the wave. (b) Find the function ( ) f t giving the displacement of the particle at 0  x . (c) Find the function ( ) g x giving the shape of the string at 0  t . (d) Plot the shape ( ) g x of the string at 0  t . (e) Plot the shape of the string at 5  t s. Solution (a) The wave equation may be written as 2 2 1 / 0           x t T T y y e . Comparing with the general equation ( / )   y f t x v , we see that 4 cm 4 1.0 s     v T m/s. (b) putting 0  x in the given equation, 2 ( / ) 0 ( )   t T f t y e …(i) (c) putting 0  t in the given equation 2 ( / ) 0 ( )    x g x y e …(ii) (d) x=0
• 3. PH – Waves & Sound - 3 (e) x=0 x=20 cm Illustration 2. A traveling wave pulse is given by 2 10 5 ( 2 )    y x t . In which direction and with what velocity is the pulse propagating? What is the amplitude of pulse ? Solution: A wave pulse is a disturbance localized only in a small part of space at a given instant [as shown in figure] and its shape does not change during propagation. Though a pulse can be represented by exponential or trigonometrical functions also, it is usually expressed by the form 2 ( )     a y b x t y t=0 v x A= a b       Comparing the above with the given pulse we find that 2 ( ) ( 2 )     f x t x t i.e., the pulse is travelling along negative x-axis with velocity 2 m/s. Further amplitude is the maximum value of wave function which will be when 2 ( 2 ) 0   x t So, max 10 2 5    A y . EQUATION OF A SIMPLE HARMONIC PLANE WAVE In case of harmonic wave the displacement of successive particles of the medium is given by a sine wave or cosine function of position. The displacement y for different values of x at 0  t is given by sin  y A kx …(vi) where A and k are constants. Suppose this disturbance is propagating along positive x-direction then sin ( )   y A k x vt …(vii) Since the waveform represented by equation (vi) is based on sine function, it would repeat itself at regular distances. The first repetition would take place when 2 2 or     kx x k This distance after which the repetition takes place is called the wavelength and denoted by  . Hence 2 2 or       k k This constant k is called propagation constant or wave number. Now equation (vii) turns into 2 sin ( )     y A x vt …(viii)
• 4. PH – Waves & Sound - 4 At 2 0 sin     t y A x …(ix) O C y x y=Asin B D /2  x 2  x Relation Between Wavelength and Velocity of Propagation: Time taken for one complete cycle of wave to pass any point is the time period (T). This is also the time taken by the disturbance in propagating a distance  .     v f T where f = frequency (Hz) 2 2     f T = circular frequency (rad/s) Different Forms of Simple Harmonic Wave Equation: sin( )      y A t kx sin 2 2                    t x A T sin 2                  x A f t v where  = phase angle. INTERFERENCE OF WAVES GOING IN SAME DIRECTION Suppose two identical sources send sinusoidal waves of same angular frequency  propagates in positive x- direction. Also, the wave velocity and hence, the wave number k is same for the two waves. One source may be started a little later than the other or the two sources may be situated at different points. The two waves arriving at a point then differ in phase. Let the amplitudes of the two waves be 1 A and 2 A and the two waves differ in phase by an angle  . Their equations may be written as 1 1 sin( )   y A kx t and 2 2 sin( )     y A kx t . According to the principle of superposition, the resultant wave is represented by 1 2 1 2 sin( ) sin( )          y y y A kx t A kx t 1 2 2 sin( ) sin( )cos cos( )sin           A kx t A kx t A kx t 1 2 2 sin( )( cos ) cos( )( sin )          kx t A A kx t A . We can evaluate it using the method described to combine two simple harmonic motions. If we write 1 2 cos cos     A A A …(x) and 2 sin sin    A A , …(xi) we get [sin( )cos cos( )sin ]         y A kx t kx t sin( )      A kx t .
• 5. PH – Waves & Sound - 5 Thus, the resultant is indeed a sine wave to amplitude A with a phase difference  with the first wave. By (x) and (xi), 2 2 2 2 2 cos sin     A A A 2 2 1 2 2 ( cos ) ( sin )      A A A 2 2 1 2 1 2 2 cos     A A A A or, 2 2 1 2 1 2 2 cos     A A A A A ….(A) Also, 2 1 2 sin sin tan cos cos         A A A A A . …(B)   A A1 A2 These relations may be remembered by using a geometrical model. We draw a vector of length 1 A to represent 1 1 sin( )   y A kx t and another vector of length 2 A at an angle  with the first one to represent 2 2 sin( )     y A kx t . The resultant of the two vectors then represents the resultant wave sin( )     y A kx t . Constructive and Destructive Interference: We see from equation (A) that the resultant amplitude A is maximum when cos 1    , or 2    n and is minimum when cos 1    , or (2 1)     n , where n is an integer. In the first case, the amplitude is 1 2  A A and in the second case, it is 1 2 | |  A A . The two cases are called constructive and destructive interferences respectively. The conditions may be written as, constructive interference : 2    n destrcutive interference : (2 1)     n . Illustration 3. Two traveling waves of equal amplitudes and equal frequencies move in opposite directions along a string. They interfere to produce a standing wave having the equation cos sin   y A kx t in which 1.0  A mm, 1.57  k cm-1 and 1 78.5    s . (a) Find the velocity of the component traveling waves. (b) Find the node closes to the origin in the region 0  x . (c) Find the antinode closes to the origin in the region 0  x . (d) Find the amplitude of the particle at 2.33  x cm. Solution: (a) The standing wave is formed by the superposition of the waves 1 sin( ) 2    A y t kx and 2 sin( ) 2    A y t kx . The wave velocity (magnitude) of either of the waves is 1 1 78.5 50 1.57       s v k cm cm/s. (b) For a node, cos 0  kx The smallest positive value of x satisfying this relation is given by 2   kx or, -1 3.14 1 2 2 1.57cm      x k cm. (c) For an antinode, | cos | 1  kx . The smallest positive x satisfying this relation is given by   kx or, 2    x k cm.
• 6. PH – Waves & Sound - 6 (d) The amplitude of vibration of the particle at x is given by | cos | A kx . For the given point, -1 7 (1.57 cm )(2.33 cm) 6 6       kx . Thus, the amplitude will be 3 (1.0mm)| cos( / 6)| mm 0.86mm 2     . Illustration 4. A transverse harmonic wave of amplitude 0.01 m is generated at one end (x = 0) of a long horizontal string by a tuning fork of frequency 500 Hz. At a given instant of time the displacement of the particle at x = 0.1 m is – 0.005 m and that of the particle at x = 0.2 m is + 0.005 m. Calculate the wavelength and the wave velocity. Obtain the equation of the wave assuming that the wave is traveling along the + x direction and that the end x = 0 is at the equilibrium position at t = 0. Solution: Since the wave is traveling along + x direction and the displacement of the end x = 0 is at time t = 0, the general equation of this wave is 2 ( , ) sin ( )            y x t A t x …(i) where A = 0.01 m when 0.1m  x , y = - 0.005 m  1 2 0.005 0.01sin ( )             t x where 1 0.1  x m or, 1 2 1 sin ( ) 2             t x  phase 1 1 2 7 sin ( ) 6         t x …(ii) when 0.2  x m y = + 0.005. Therefore, we have 1 2 0.005 0.01sin ( )             t x where 2 0.2  x m  2 1 2 2 ( ) 6         t x …(iii) From eqs. (ii) and (iii) 1 2        Now, 2      x Thus, 1 2 2 2 ( ) (0.1 0.2)           x x or 0.2   m Now, frequency n of the wave = frequency of the tuning fork = 500 Hz. Hence, wave velocity 1 500 0.2 100        n ms Substituting for A ,  and v in Equation. (i) we get ( , ) 0.01sin{10 (100 )    y x t t x This is the equation of the wave where y and x are in metres and t in seconds.
• 7. PH – Waves & Sound - 7 LONGITUDINAL AND TRANSVERSE WAVE In a longitudinal wave, the particles of the medium carrying the mechanical wave move back and forth along the direction of propagation. Sound in air is a longitudinal wave. In a transverse wave, the particles of the medium oscillate in the direction perpendicular to the direction of propagation, for example the waves in a taut string. Energy of a Plane Progressive Wave: Consider a plane wave propagating with a velocity v in x-direction across an area S. An element of material medium (density = 3 / kg m ) will have a mass ( )  Sdx . The displacement of a particle from its equilibrium position is given by the wave equation sin( )    y A t kx Total energy of this element is   2 2 max 1 1 . ( ) ( ) 2 2     dE dm v Sdx A 2 2 2 (2 )    Sdx f A  Energy density 2 2 2 3 2 (J/m ) ( )     dE f A Sdx energy per unit length 2 2 2 2    f A S  Power transmitted 2 2 2 2    f A S (Watt = J/s) Intensity of the Wave (I): Intensity of the wave is defined as the power crossing per unit area 2 2 2 2    f A v …. Watt/m2 For wave propagation through a taut string,    S , the linear density in kg/m  energy per unit length 2 2 2 2    f A Wave Speed: The speed of any mechanical wave, transverse or longitudinal, depends on both an inertial property of the medium (to store kinetic energy) and an elastic property of the medium (to store potential energy). TRANSVERSE WAVE IN A STRETCHED STRING Consider a transverse pulse produced in a taut string of linear mass density  . Consider a small segment of the pulse, of length  , forming an arc of a circle of radius R. A force equal in magnitude to the tension T pulls tangentially on this segment at each end. O R T l  Let us set an observer at the centre of the pulse which moves along with the pulse towards rights. For the observer any small length dl of the string as shown will appear to move backward with a velocity v. Now the small mass of the string is in a circular path of radius R moving with speed v. Therefore, the required centripetal force is provided by the only force acting, (neglecting gravity) is the component of tension along the radius.
• 8. PH – Waves & Sound - 8 The net restoring force on the element is 2 sin( ) (2 )        F T T T R The mass of the segment is    m The acceleration of this element toward the centre of the circle is 2  v a R , where v is the velocity of the pulse. Using second law of motion, 2 ( )            v T R R or,   T v Laws of Transverse Vibrations of A String: Sonometer: The fundamental frequency of vibration of a stretched string fixed at both ends is given by 1 . 2     T From this equation, one can immediately write the following statements known as “Laws of transverse vibrations of a string” (a) Law of length – The fundamental frequency of vibration of a string (fixed at both ends) is inversely proportional to the length of the string provided its tension and its mass per unit length remain the same. 1/   L if T and  are constants. (b) Law tension – The fundamental frequency of a string is proportional to the square root of its tension provided its length and the mass per unit length remain the same.   T if L and  are constants. (c) Law of mass – The fundamental frequency of a string is inversely proportional to the square root of the linear mass density, i.e., mass per unit length provided the length and the tension remain the same. 1    if L and T are constants. Illustration 5. A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced at the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope? Solution: As the rope is heavy, its tension will be different at different points. The tension at the free end will be (2 kg)g and that at the upper end it will be (8 kg)g. 2 kg 6 kg We have,    v or,    F v or,    F v . The frequency of the wave pulse will be the same everywhere on the rope as it depends only
• 9. PH – Waves & Sound - 9 on the frequency of the source. The mass per unit length is also the same throughout the rope as it is uniform. Thus, by (i)  F is constant. Hence, 1 (2 ) (8 ) 0.06   kg g kg g m , where 1  is the wavelength at the top of the rope. This gives 1 0.12   m. Illustration 6. The vibrations of a string fixed at both ends are described by the equation -1 -1 (5.00mm)sin(1.57cm ) ]sin[(314s ) ]  y x t . (a) What is the maximum displacement of particle at 5.66 cm  x ? (b) What are the wavelength and the wave speeds of the two transverse waves that combine to give the above vibration? (c) What is the velocity of the particle at 5.66  x cm at time 2.0s?  t (d) if the length of the string is 10.0 cm, locate the nodes and the antinodes. How may loops are formed in the vibration? Solution: (a) The amplitude of the vibration of the particle at position x is -1 |(5.00 mm)sin[(1.57 cm ) ] |  A x For 5.66  x cm, (5.00 mm)sin 5.66 2          A (5.00mm)sin 2.5 3           (5.00 mm)cos 2.50 3    mm. (b) From the given equation, the wave number 1.57  k cm-1 and the angular frequency 314   s-1 . Thus, the wavelength is -1 2 2 3.14 4.00 1.57cm       k cm and the frequency is -1 -1 314s 50s 2 2 3.14       v The wave speed is -1 (50s )(4.00cm) 2.00      v m/s (c) The velocity of the particle at position x at time t is given by -1 (5.00mm)sin[(1.57cm ) ]      y x t -1 -1 [314s cos(314s ) ] t = ( 157 cm/s) sin (1.57 cm-1 ) x cos(314 s-1 )t. Putting 5.66  x and 2.00  t s, the velocity of this particle at the given instant is 5 (157cm/ s)sin cos(200 ) 2 3          
• 10. PH – Waves & Sound - 10 (157cm/s) cos 1 78.5cm/s. 3      (d) the nodes occur where the amplitude is zero i.e., 1 sin(1.57cm ) 0   x . or, 1 cm 0 2          x where n is an integer. Thus, 2 cm  x n The nodes, therefore, occur at 0,2cm, 4cm, 6cm, 8cm and 10cm  x . Antinodes occur in between them i.e., at 1cm, 3cm,5cm, 7cm  x and 9cm . The string vibrates in 5 loops. LONGITUDINAL WAVE IN FLUIDS Sound wave in air is a longitudinal wave. As a sound wave passes through air, potential energy is associated with periodic compressions and expansions of small volume elements of the air. The property that determines the extent to which an element of the medium changes its volume as the pressure applied to it is increased or decreases is the bulk modulus B. /    p B V V where V V is the fractional change in volume produced by a change in pressure p . Suppose air of density  is filled inside a tube of cross-sectional area A under a pressure p. Initially the air is at rest. At 0  t , the piston at left end of the tube (as shown in the figure) is set in motion toward the right with a speed u. After a time interval t , all portions of the air at the left of section 1 are moving with speed u whereas all portions at the right of the section are still at rest. The boundary between the moving and the stationary portions travels to the right with v, the speed of the elastic wave (or sound wave). In the time interval t , the piston has moved u t and the elastic disturbance has traveled a distance v t . Air (I) pA (p+ p)A  pA , p u t  v t  The mass of air that has attained a velocity u in a time t is ( )  x A . Therefore, the momentum imparted is [ ( ) ]   v t A u and the net impulse =  .   pA t Thus, impulse = change in momentum ( ) [ ( ) ]      pA t v t A u or    p vu …(xii) Since /    p B V V           V p B V where   V Av t and    V Au t
• 11. PH – Waves & Sound - 11       V Au t u V Av t v Thus,   u p B v …(xiii) From (xii) and (xiii)   B v . Velocity of Sound in an Ideal Gas: The motion of sound wave in air is adiabatic. In the case of an ideal gas, the relation between pressure p and volume V during an adiabatic process is given by   pv constant. where  is the ratio of the heat capacity at constant pressure to that at constant volume. After differentiating, we get 1 0      dp V pV dV Since     dp B V p dV     p v (Laplace correction in contrast to Newton’s formula   P v ) Using the gas equation   p RT M where M is the molar mass. Thus,   RT v M (T = Temperature in Kelvin) SOUND WAVES From practical standpoint it is easier to measure pressure variation in a sound wave than the displacements, so it is worthwhile to develop a relation between the two. Let p be the instantaneous pressure fluctuation at any point, that is, the amount by which the pressure differs from normal atmosphere pressure. If the displacements of two neighboring points x and x + x are the same, the gas between these points is neither compressed nor rarefied, there is no volume change, and consequently p = 0. Only when y varies from one point to a neighboring point there is a change of volume and therefore of pressure. The fractional volume change / V V in an element near point x turns out to be         y x , which is the rate of change of y and x as we go from one point to the neighboring point. To see why this is so, we note that / V V is proportional to change is length of an element which has length x when no wave disturbance is present, divided by x . The change in length is the value of y at the point x + x , minus the value at the point x. If x is very small, this is approximately multiplied by the derivative of y with respect to x, thus ( , ) ( , )        y y x x t y x t x x ( , ) ( , )          V y x x t y x t y V x x …(xiv) Now from the definition of the bulk modulus B, ,    V p B V and we find
• 12. PH – Waves & Sound - 12     y p B x Now sin( ) cos( )        y A t kx p BkA t kx Maximum amount by which the pressure differs from atmospheric, that is, the maximum value of p, is called the pressure amplitude, denoted P.   P BkA DISPLACEMENT WAVE AND PRESSURE WAVE A longitudinal wave in a fluid (liquid or gas) can be described either in terms of the longitudinal displacement suffered by the particles of the medium or in terms of the excess pressure generated due to the compression or rarefaction. Let us see how the two representations are related to each other. A x x+ x  s s+ s  Consider a wave going in the x-direction in a fluid. Suppose that at a time t , the particle at the undisturbed position x suffers a displacement s in the x-direction. The wave can then be described by the equation 0 sin ( / )    s s t x v …(xv) Consider the element of the material which is contained within x and   x x (figure)in the undisturbed state. Considering a cross-sectional area A , the volume of the element in the undisturbed state is  A x and its mass is   A x . As the wave passes, the ends at x and   x x are displaced by amounts s and   s s according to equation (xv) above. The increase in volume of this element at time t is    V A s = .    s A x x 0 ( / )cos( / )     As v t x v x , where s has been obtained by differentiating equation (xv) with respect to x . The element is, therefore, under a volume strain. 0 cos ( / )         As t x v x V V vA x 0 cos ( / )      s t x v v . The corresponding stress i.e., the excess pressure developed in the element at x at time t is,         V p B V , where B is the bulk modulus of the material. Thus, 0 cos ( / )     s p B t x v v ….(xvi) Comparing with (xv), we see that the pressure amplitude 0 p and the displacement amplitude 0 s are related as 0 0 0    B p s Bks v , where k is the wave number. Also, we see from (xv) and (xvi) that the pressure wave differs in phase by / 2  from the displacement wave. The pressure maxima occur where the displacement is zero and displacement maxima occur where the pressure is at its normal level. The fact that, displacement is zero where the pressure-change is maximum and vice versa, puts the two descriptions on different footings. The human ear or an electronic detector responds to the change in pressure and not to the
• 13. PH – Waves & Sound - 13 displacement in a straight forward way. Suppose two audio speakers are driven by the same amplifier and are placed facing each other. A detector is placed midway between them. The displacement of the air particles near the detector will be zero as the two sources drive these particles in opposite directions. However, both send compression waves and rarefaction waves together. As a result, pressure increases at the detector simultaneously due to both the sources. Accordingly, the pressure amplitude will be doubled, although the displacement remains zero here. A detector detects maximum intensity in such a condition. Thus, the description in terms of pressure wave is more appropriate than the description in terms of the displacement wave as far as sound properties are concerned. Illustration 7 A sound wave of wavelength 40 cm travels in air. If the difference between the maximum and minimum pressures at a given point is 3 2 1.0 10 N/m   , find the amplitude of vibration of the particles of the medium. The bulk modulus of air is 5 1.4 10  2 N/m . Solution: The pressure amplitude is 3 2 3 2 0 1.0 10 N/m 0.5 10 N/m 2       p The displacement amplitude 0 s is given by 0 0  p Bk s or, 0 0 0 2     p p s B k B 3 2 2 5 2 0.5 10 N/m (40 10 m) 2 3.14 1.4 10 N/m          10 2.2 10   m. SOUND WAVES IN SOLIDS Sound waves can travel in solids just like they can travel in fluids. The speed of longitudinal sound waves in a solid rod can be shown to be / ,   v Y where Y is the Young’s modulus of the solid and  its density. For extended solids, the speed is a more complicated function of bulk modulus and shear modulus. Table gives the speed of sound in some common materials. Medium Speed m/s Medium Speed m/s Air (dry 0º C) 332 Copper 3810 Hydrogen 1330 Aluminum 5000 Water 1486 Steel 5200 Effect of Pressure, Temperature and Humidity on the Speed of Sound in Air: We have stated that for an ideal gas, the pressure, volume and temperature of a given mass satisfy  PV T constant. As the density of a given mass is inversely proportional to its volume, the above equation may also be written as
• 14. PH – Waves & Sound - 14   P cT , where c is a constant. The speed of sound is      P v cT …(xvii) Thus, if pressure is changed but the temperature is kept constant, the density varies proportionally and / P remains constant. The speed of sound is not affected by the change in pressure provided the temperature is kept constant. If the temperature of air is changed then the speed of sound is also changed. From equation (xvii),  v T . At STP, the temperature is 0º C or 273 K. If the speed of sound at 0º C is 0 v , its value at the room temperature T (in Kelvin) will satisfy 0 273 273 273    v T t v , where t is the temperature in ºC. This may be approximated as 1/ 2 0 1 1 273 546           v t t v or, 0 1 546         t v v . The density of water vapour is less than dry air at the same pressure. Thus, the density of moist air is less than that of dry air. As a result, the speed of sound increases with increasing humidity. INTENSITY OF SOUND WAVES As a wave travels in a medium, energy is transported from one part of the space to another part. The intensity of a sound wave is defined as the average energy crossing a unit cross-sectional area perpendicular to the direction of propagation of the wave in unit time. It may also be stated as the average power transmitted across a unit cross- sectional area perpendicular to the direction of propagation. The loudness of sound that we feel is mainly related to the intensity of sound. It also depends on the frequency to some extent. Consider again a sound wave traveling along the x-direction. Let the equations for the displacement of the particles and the excess pressure developed by the wave be given by 0 0 sin ( / ) and cos ( / )       s s t x v p p t x v …(xviii) where 0 0   B s p v . Consider a cross-section of area A perpendicular to the x-direction. The medium to the left to it exerts a force pA on the medium to the right along the X-axis. The points of application of this force move longitudinally, that is along the force, with a speed   s t . Thus, the power W , transmitted by the wave from left to right across the cross-section considered, is ( )    s W pA t . By (xviii), 0 0 cos ( / ) cos ( / )       W Ap t x v s t x v 2 2 2 0 cos ( / )     A s B t x v v .
• 15. PH – Waves & Sound - 15 The average of 2 cos ( / )   t x v over a complete cycle or over a long time is 1/2. The intensity I , which is equal to the average power transmitted across unit cross-sectional area is thus, 2 2 2 2 2 0 0 1 2 2     s B B I s v v v . 2 0 2  p v I B As 2   B v , the intensity can also be written as 2 2 0 0 2 2 2     p v I p v v We see that the intensity is proportional to the square of the pressure amplitude 0 p . Loudness: Human ear is sensitive for extremely large range of intensity. So a logarithmic rather than an arithmetic scale is convenient. Accordingly, intensity level  of a sound wave is defined by the equation 0 10log         I I decibel where 12 2 0 10 /   I W m is the reference or threshold intensity level to which any intensity I is compared. Illustration 8. The pressure amplitude in a sound wave from a radio receiver is 2 2.0 10  N/m2 and the intensity at a point is 7 5.0 10  W/m2 . If by turning the “volume” knob the pressure amplitude is increased to 2 2.5 10  N/m2 evaluate the intensity. Solution: The intensity is proportional to the square of the pressure amplitude. Thus, 2 0 0          p I I p or 2 2 7 0 0 2.5 5.0 10 2.0                    p I I p W/m2 7 7.8 10   W/m2 . Illustration 9. A source emitting sound of frequency 180 Hz is placed in front of a wall at a distance of 2 m from it. A detector is also placed in front of the wall at the same distance from it. Find the minimum distance between the source and the detector for which the detector detects a maximum of sound. Speed of sound in air = 360 m/s Solution: The situation is shown in figure. Suppose the detector is placed at a distance of x meter from the sources. The wave received from the source after reflection from the wall has traveled a distance of 2 2 1/ 2 2[(2) / 4]  x meter. The difference between the two waves is 1/ 2 2 2 2 (2) 4                     x x meter. Constructive interference will take place when , 2 ,...     The minimum distance x for a maximum corresponds to    …(i) x S D
• 16. PH – Waves & Sound - 16 The wavelength is -1 360 m/s 2 180s     u m v Thus, by (i), 1/ 2 2 2 2 (2) 2 4          x x or, 1/ 2 2 4 1 4 2          x x or, 2 2 4 1 4 4     x x x or, 3  x . Thus, the detector should be placed at a distance of 3 m from the sources. Note that there is no abrupt phase change. SUPERPOSITION OF WAVES Two or more waves can traverse the same space independently of one another. Thus the displacement of any particle in the medium at any given time is simply the vector sum of displacements that the individual waves would give it. This process of the vector addition of the displacement of a particle is called superposition. Interference: When two waves of the same frequency, superimpose each other, there occurs redistribution of energy in the medium which causes either a minimum intensity or maximum intensity which is more than the sum of the intensities of the individual sources. This phenomenon is called interference of waves. Let the two waves be 1 1 2 2 sin( ), sin( )        y A kx t y A kx t According to the principal of superposition 1 2   y y y 1 2 sin( ) sin( )         A kx t A kx t 1 2 2 sin( ) sin( )cos cos( ) ( sin )            A kx t A kx t kx t A 1 2 2 sin( ) ( os ) cos( ) ( sin )         kx t A A c kx t A sin( )      R kx t where 1 2 cos cos     A A R and 2 sin sin    A R and 2 2 2 1 2 2 ( cos ) ( sin )      R A A A 2 2 1 2 1 2 2 cos     A A A A If 1 I and 2 I are intensities of the interfering waves and  is the phase difference, then the resultant intensity is given by 1 2 1 2 2 cos     I I I I I Now, 2 max 1 2 ( ) for 2      I I I n 2 min 1 2 ( ) for (2 1)       I I I n Illustration 10. Two coherent sound sources are at distances 1 0.2  x m and 0.27  x m from a point. Calculate the intensity of the resultant wave at the at point if the frequency of each wave is 800  f Hz and velocity of wave in the medium is 224 /  v m s . The intensity of each wave is 2 0 60 /  I W m . Solution: Path difference, 2 1 0.27 0.2 0.07      p x x m 2 2 2 (800)(0.07) 224 2                 f p p v
• 17. PH – Waves & Sound - 17 1 2 1 2 2 cos     I I I I I or 0 0 0 2 cos( / 2)     I I I I 2 0 2 2(60) 120 W/m    I . STANDING WAVES A standing wave is formed when two identical waves traveling in the opposite directions along the same line, interfere. On the path of the stationary wave, there are points where the amplitude is zero, they are known as NODES. On the other hand there are points where the amplitude is maximum, they are known as ANTINODES. The distance between two consecutive nodes or two consecutive antinodes is 2  . The distance between a node and the next antinode is 4  . Consider two waves of the same frequency, speed and amplitude, which are traveling in opposite directions along a string. Two such waves may be represented by the equations 1 sin( )    y a kx t and 2 sin( )    y a kx t Hence the resultant may be written as 1 2 sin( ) sin( )         y y y a kx t a kx t 2 sin cos   y a kx t This is the equation of a standing wave. Reflection of Waves: (a) Waves on reflection from a fixed end undergoes a phase change of 180º. Incident Wave Reflected Wave (b) While a wave reflected from a free end is reflected without a change in phase. Incident Wave Reflected Wave STATIONARY WAVES IN STRINGS N A N L (b) A N N A N L (c) N A A N A string of length L is stretched between two points. When the string is set into vibrations, a transverse progressive wave begins to travel along the string. It is reflected at the fixed end. The incident and the reflected waves interfere to produce a stationary transverse wave in which the ends are always nodes. (a) In the simplest form, the string vibrates in one loop in which the ends are the nodes and the centre is the antinode. This mode of vibration is known as the fundamental mode and the frequency of vibration is known as the fundamental frequency or first harmonic.
• 18. PH – Waves & Sound - 18 1 2   L  1 2   L If 1 f is the fundamental frequency of vibration, then the velocity of transverse waves is given as, 1 1 1 f or f / 2    v v L  1 2 f  v L …(xix) (b) The same string under the same conditions may also vibrate in two loops, such that the centre is also the node.  2 2 2 2      L L If 2 f is the frequency of vibrations, then the velocity of transverse waves is given as, 2 2 2 2 f f or f /      v v L v L = 2f1 …(xx) The frequency 2 f is known as second harmonic or first overtone. (c) The same string under the same conditions may also vibrate in three segments.  3 3 2 3 2 3      L L If 3 f is the frequency in this mode of vibration, then, 3 3 3 3 2 f f or f 3 / 2 3      v v L v L = 3f1 …(xxi) The frequency 3 f is known as the third harmonic or second overtone. Thus a stretched string in addition to the fundamental node, also vibrates with frequencies which are integral multiples of the fundamental frequencies. These frequencies are known as harmonics. The velocity of transverse wave in a stretched string is given as   T v where T = tension in the string.  = linear density or mass per unit length of string. If the string fixed at two end, vibrates in its fundamental mode, then 1 2 f f 2     T v L L  = volume of unit length × density 2 2 1 4        D r where D = diameter of the wire,  = density. Illustration 11. A sonometer wire 100 cm in length has a fundamental frequency of 330 Hz. Find (a) the velocity of propagation of transverse waves along the wire and (b) the wavelength of the resulting sound in air if velocity of sound in air is 330 m/s. Solution: (a) In case of transverse vibration of string for fundamental mode : ( / 2), . ., 2 2 1 2        L i e L m i.e., the wavelength of transverse waves propagating on string is 2 m. Now as the frequency of wire is given to be 330 Hz, so from    f velocity of transverse waves
• 19. PH – Waves & Sound - 19 along the wire will be 330 2 660     wire m/s i.e., for transverse mechanical waves propagating along the wire, 330  f Hz, 2   m and 660   m/s (b) Here vibrating wire will act as source and produce sound, i.e., longitudinal waves in air. Now as frequency does not change with change in medium so 330  f Hz and as velocity in air is given to be = 330 m/s so from    f ( / ) (330/ 330) 1      air air f m i.e., for sound (longitudinal mechanical waves) in air produced by vibrations of wire (body), 330  f Hz, 1   m and 330   m/s. STATIONARY WAVES IN AIR COLUMN Open pipe: If both ends of a pipe are open and a system of air is directed against an edge, standing longitudinal waves can be set up in the tube. The open ends are a displacement antinodes and pressure nodes L A N (a) (b) (c) 1/2 A A 2/4 2/2 3/4 3/2 (a) For fundamental mode of vibrations, 1 1 2 2      L L 1 1 1 2     v f v Lf or 1 2  v f L …(xxii) (b) For the second harmonic or first overtone, 2 2 or     L L 2 2 2     v f v Lf or 2 2 2 2    v v f f L L …(xxiii) (c) For the third harmonic or second overtone, 3 3 2 3 2 3       L L 3 3 3 2 3     v f v Lf or 2 3 2 f L   …(xxiv) From (xxii), (xxiii) and (xxiv) we get, 1 2 3 : : :.............. 1: 2 :3:..............  f f f i.e. for a cylindrical tube, open at both ends, the harmonics excitable in the tude are all integral multiples of its fundamental.
• 20. PH – Waves & Sound - 20  In the general case, 2   L n where 1,2.......  n Frequency 2    v nv L where 1,2......  n Closed pipe: If one end of a pipe is closed the reflected wave is 180º out of phase with the incoming wave. Thus the displacement of the small volume elements at the closed end must always be zero. Hence the closed end must be a displacement node and pressure antinode L A N (a) 2 4  1 2  A N (b) A 2 4  N A (c) A N 3 4  A N 3 2  N 3 2  (a) This represents the fundamental mode of vibration, 1 1 4 4      L L , if 1 f is the fundamental frequency, then the velocity of sound waves is given as, 1 1 1 4     v f v Lf or 1 4  v f L …(xxv) (b) This is the third harmonic or first overtone. 2 2 4 3 4 3       L L 2 2 2 4 3     v f v Lf or 2 2 3 , 4   v f f L …(xxvi) (c) This is the fifth harmonic or seconds overtone. 3 3 4 5 4 5      L L 3 3 3 4 5     v f v Lf or 3 5 5 4   V f f L …(xxvii) From (xxv), (xxvi) and (xxvii) we get, 1 2 3 : : :...... 1: 3:5 :........  f f f In general, 4 (2 1)    l n where n = 0, 1, 2 …… Velocity of sound = v Frequency (2 1) 4   n v L where n = 0, 1, 2 …… Illustration 12. A tube closed at one end has a vibrating diaphragm at the other end, which may be assumed to be displacement node. It is found that when the frequency of the diaphragm is 2000 Hz, a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm. When the frequency is gradually reduced, the stationary wave pattern disappears but another stationary wave pattern
• 21. PH – Waves & Sound - 21 reappears at a frequency of 1600 Hz. Calculate (i) the speed of sound in air, (ii) the distance between adjacent nodes at a frequency of 1600 Hz, (iii) the distance between the diaphragm and the closed end (iv) the next lower frequencies at which stationary wave patterns will be obtained. Solution : Since the node-to-node distance is / 2  / 2  = 0.08 or  = 0.16 m (i)   c n  1 2000 0.16 320ms    c (ii) 320 1600   or  = 0.2 m  distance between nodes = 0.2/2 = 0.1 m = 10 cm. (iii) Since there are nodes at the ends, the distance between the closed end and the memberane must be exact integrals of / 2  .  5 0.4 2 / 2 ' 0.2/ 2 ' 4       n n n When 5, ' 4   n n 0.16/ 2 0.4 40     l n m cm . (iv) For the next lower frequency  n 3, 2, 1  0.4 3 / 2   or 0.8/3   Since 320 , 1200 0.8/3     c n n Hz Again 0.4 1. / 2   or 0.4   m  320/ 0.4 800   n Hz Again 0.4 1. / 2   or =0.8  m  320 / 0.8 400   n Hz. BEATS When two interacting waves have slightly different frequencies the resultant disturbance at any point due to the superposition periodically fluctuates causing waxing and waning in the resultant intensity. The waxing and waning in the resultant intensity of two superposed waves of slightly different frequency are known as beats. Let the displacement produced at a point by one wave be 1 1 1 sin(2 )    y A f t And the displacement produced at the point produced by the other wave of equal amplitude as 2 2 2 sin(2 )     y A f t By the principle of superposition, the resultant displacement is 1 2 1 1 2 2 sin(2 ) sin(2 )          y y y A f t A f t 1 2 1 2 1 2 2 sin 2 cos2 2 2 2                                  f f f f y A t t
• 22. PH – Waves & Sound - 22 1 2 1 2 sin 2 2 2                          f f Y R t where, R = 2 A cos 2 1 2 2        f f t The time for one beat is the time between consecutive maximum or minima. First maxima would occur when 1 2 cos2 1 2           f f t Then 1 2 2 0 2          f f t  0  t For second maxima would occur when 1 2 cos2 1 2           f f t Then 1 2 2 2           f f t or, 1 2 1   t f f The time for one beat = 1 2 1 2 1 1 0           f f f f Similarly it may also be shown that time between two consecutive minima is 1 2 1  f f . Hence, frequency of beat i.e. number of beats in one second or Beat frequency = 1 2 ~ f f . Illustration 13. A string of length 25 cm and mass 2.5 g is under tension. A pipe closed at one end is 40 cm long. When the string is set vibrating in its first overtone and the air in the pipe in its fundamental frequency, 8 beats per second are heard. It is observed that decreasing the tension in the string decreased the beat frequency. If the speed of sound in air is 320 ms-1 , find the tension in the string. Solution: n (fundamental frequency of the string) = 1 2 T l m or 2 1 20 2 0.25 10    T n T The fundamental frequency of a closed pipe ' 4  c n l  320 ' 200 4 0.40    n Hz The frequency of the first overtone of the string = 2 40  n T Since there are 8 beats / ,2 ~ ' 8  s n n or 40 ~ 200 8  T
• 23. PH – Waves & Sound - 23 Since on decreasing the tension, the beat frequency decreases, 2n is definitely greater than ' n .  40 200 8 or 27.04N    T T . Illustration 14. A tuning fork of frequency 256 Hz and an open organ pipe of slightly lower frequency are at 17ºC. When sounded together, they produce 4 beats per second. On altering the temperature of the air in the pipe, it is observed that the number of beats per second first diminishes to zero and then increases again to 4. By how much and in what direction has the temperature of the air in the pipe been altered? Solution: 17 2  c n l where l = length of the pipe  17 256 4 2   c l or 17 252 2  c l Since beats decrease first and then increase to 4, the frequency of the pipe increases. This can happen only if the temperature increases. Let t be the final temperature, in Celsius, Now 256 4 2   t c l or 260 2  t c l Dividing 17 260 252  t c c or 273 260 273 17 252    t ( )  c T or 308.7 273 35.7    t ºC  rise in temperature = 35.7 – 17 = 18.7ºC. Illustration 15. Find the fundamental frequency and the first four overtones of a 15 cm pipe (a) if the pipe is closed at one end, (b) if the pipe is open at both ends (c) How many overtones may be heard by a person of normal hearing in each of the above cases? Velocity of sound in air = 330 ms-1 . Solution: (a) 0 4  c n l where 0 n = frequency of the fundamental  0 330 550 4 0.15    n Hz The first four overtones are 0 0 0 0 3 , 5 , 7 and 9 n n n n  So, the required frequencies are 550, 1650, 2750, 3850, 4950 Hz. (b) 0 330 1100 2 2 0.15     c n l Hz. The first four overtones are 0 0 0 0 2 , 3 , 4 and 5 n n n n So, the required frequencies are 1100, 2200, 3300, 4400, 5500 Hz. (c) The frequency of the nth overtone is (2n + 1) 0 n .  (2n + 1) 0 n = 20000or(2n + 1) 550 = 20000 or n = 17.68.
• 24. PH – Waves & Sound - 24 The acceptable value is 17. The frequency of the nth overtone is (n +1) 0 n .  (n +1) 0 n = 20000or (n +1) 100 = 20000 or n = 17.18 The acceptable value is 17. Illustration 16. Two tuning fork A and B sounded together give 6 beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are 24 cm and 25 cm respectively. Calculate the frequencies of forks. Solution: Let the frequency of the first fork be 1 f and that of second be 2 f . We then have, 1 4 24   v f and 2 4 25   v f We also see that 1 2  f f  1 2 6   f f …(i) and 1 2 24 25  f f …(ii) Solving (i) and (ii), we get 1 150  f Hz and 2 144  f Hz. DOPPLER EFFECT The apparent shift in frequency of the wave motion when the source of sound or light moves with respect to the observer, is called Doppler Effect. Calculation of Apparent Frequency: Suppose v is the velocity of sound in air, s v is the velocity of the source of sound(s), 0 v is the velocity of the observe (O), and f is the frequency of the source. (i) Source moves towards stationary observer. If the source were stationary the f waves sent out in one second towards the observer O would occupy a distance v , and the wavelength would be v/f. If S moves with a velocity s v towards (O), the f waves sent out occupy a distance ( )  s v v because S has moved a distance s v towards O in 1s. So the apparent wavelength would be '    s v v f Thus, apparent frequency Velocity of sound relative to O ' Wavelenght of wave reaching O  f ' '           s v v f f v v
• 25. PH – Waves & Sound - 25 (ii) Source moves away from stationary observer. Now, apparent wavelength '    s v v f  Apparent frequency ' / '   f v n or '         s v f f v v (iii) Observer moves towards stationary source. Velocity of sound relative to O ' Wavelenght of wave reaching O  f here, velocity of sound relative to O = v + 0 v and wavelength of waves reaching O = v/f  0 0 ' /           v v v v f f v f v (iv) Observer moves away from the stationary source. 0 0 ' /           v v v v f f v f v (v) Source and observer both moves toward each other. 0 0 '             s s v v v v f f v v v v f (vi) Both moves away from each other. 0 '          s v v f f v v (vii) Source moves towards observer but observer moves away from source 0 '          s v v f f v v (viii) Source moves away from observer but observer moves towards source. 0 '          s v v f f v v . Illustration 17. A siren emitting a sound of frequency 2000 Hz moves away from you towards a cliff at a speed of 8 m/s. (a) What is the frequency of the sound you hear coming directly from the siren. (b) What is the frequency of sound you hear reflected off the cliff. Speed of sound in air is 330 m/s. Solution: (a) The frequency of Sound heard directly. 1 0         s v f f v v 8  s V m/s.  1 330 2000 330 8          f 1 330 2000 338   f .
• 26. PH – Waves & Sound - 26 (b) The frequency of the reflected sound is given by 2 0         s v f f v v  2 330 2000 330 8          f 2 330 2000 322   f = 2050 Hz.