Chapter 15 Wave Motion PDF

60 206 Wave Motion E3 16.1 Wave. A wave is a disturbance which propagates energy and momentum from one place to the other without the transport of matter. (1) Necessary properties of the medium for wave propagation : ID (i) Elasticity : So that particles can return to their mean position, after havi...

60 206 Wave Motion E3 16.1 Wave. A wave is a disturbance which propagates energy and momentum from one place to the other without the transport of matter. (1) Necessary properties of the medium for wave propagation : ID (i) Elasticity : So that particles can return to their mean position, after having been disturbed. (ii) Inertia : So that particles can store energy and overshoot their mean position. U (iii) Minimum friction amongst the particles of the medium. (iv) Uniform density of the medium. D YG (2) Characteristics of wave motion : (i) It is a sort of disturbance which travels through a medium. (ii) Material medium is essential for the propagation of mechanical waves. (iii) When a wave motion passes through a medium, particles of the medium only vibrate simple harmonically about their mean position. They do leave their position and move with the disturbance. U (iv) There is a continuous phase difference amongst successive particles of the medium i.e., particle 2 starts vibrating slightly later than particle 1 and so on. ST (v) The velocity of the particle during their vibration is different at different position. (vi) The velocity of wave motion through a particular medium is constant. It depends only on the nature of medium not on the frequency, wavelength or intensity. (vii) Energy is propagated along with the wave motion without any net transport of the medium. (3) Mechanical waves : The waves which require medium for their propagation are called mechanical waves. Example : Waves on string and spring, waves on water surface, sound waves, seismic waves. Wave Motion 207 (4) Non-mechanical waves : The waves which do not require medium for their propagation are called non- mechanical or electromagnetic waves. Examples : Light, heat (Infrared), radio waves, - rays, X-rays etc. 60 (5) Transverse waves : Particles of the medium execute simple harmonic motion about their mean position in a direction perpendicular to the direction of propagation of wave motion. (i) It travels in the form of crests and troughs. A E Vibrati on C B D Troug h Wave ID Cre st Normal Level Particl e E3 (ii) A crest is a portion of the medium which is raised temporarily above the normal position of rest of the particles of the medium when a transverse wave passes through it. U (iii) A trough is a portion of the medium which is depressed temporarily below the normal position of rest of the particles of the medium, when transverse wave passes through it. D YG (iv) Examples of transverse wave motion : Movement of string of a sitar or violin, movement of the membrane of a Tabla or Dholak, movement of kink on a rope, waves set-up on the surface of water. (v) Transverse waves can be transmitted through solids, they can be setup on the surface of liquids. But they can not be transmitted into liquids and gases. C U C T Crest T Transverse-wave in a rod ST Trough C Transverse wave in a string Transverse wave on water surface (6) Longitudinal waves : If the particles of a medium vibrate in the direction of wave motion the wave is called longitudinal. (i) It travels in the form of compression and rarefaction. (ii) A compression (C) is a region of the medium in which particles are compressed. (iii) A rarefaction (R) is a region of the medium in which particles are rarefied. C R C R C R Vibration of particle 208 Wave Motion (iv) Examples sound waves travel through air in the form of longitudinal waves, Vibration of air column in organ pipes are longitudinal, Vibration of air column above the surface of water in the tube of resonance apparatus are longitudinal. (v) These waves can be transmitted through solids, liquids and gases because for these waves propagation, volume elasticity is necessary. 60 (7) One dimensional wave : Energy is transferred in a single direction only. Example : Wave propagating in a stretched string. Example : Wave propagating on the surface of water. E3 (8) Two dimensional wave : Energy is transferred in a plane in two mutually perpendicular directions. (9) Three dimensional wave : Energy in transferred in space in all direction. Example : Light and sound waves propagating in space. ID 16.2 Important Terms Regarding Wave Motion. (1) Wavelength : (i) It is the length of one wave. U (ii) Wavelength is equal to the distance travelled by the wave during the time in which any one particle of the medium completes one vibration about its mean position. D YG (iii) Wavelength is the distance between any two nearest particles of the medium, vibrating in the same phase. (iv) Distance travelled by the wave in one time period is known as wavelength. (v) In transverse wave motion :   = Distance between the centres of two consecutive crests.  = Distance between the centres of two consecutive troughs.    = Distance in which one trough and one crest are contained.  U (vi) In longitudinal wave motion : C  = Distance between the centres of two consecutive compression. R R R C  C  ST  = Distance between the centres of two consecutive rarefaction. C  = Distance in which one compression and one rarefaction contained. (2) Frequency : (i) Frequency of vibration of a particle is defined as the number of vibrations completed by particle in one second. (ii) It is the number of complete wavelengths traversed by the wave in one second. (iii) Units of frequency are hertz (Hz) and per second. (3) Time period : (i) Time period of vibration of particle is defined as the time taken by the particle to complete one vibration about its mean position. (ii) It is the time taken by the wave to travel a distance equal to one wavelength. Wave Motion 209 (4) Relation between frequency and time period : Time period = 1/Frequency  T = 1/n (5) Relation between velocity, frequency and wavelength : v = n Velocity (v) of the wave in a given medium depends on the elastic and inertial property of the medium. 60 Frequency (n) is characterised by the source which produces disturbance. Different sources may produce vibration of different frequencies. Wavelength () will differ to keep n  = v = constant 16.3 Sound Waves. E3 The energy to which the human ears are sensitive is known as sound. In general all types of waves are produced in an elastic material medium, Irrespective of whether these are heard or not are known as sound. According to their frequencies, waves are divided into three categories : ID (1) Audible or sound waves : Range 20 Hz to 20 KHz. These are generated by vibrating bodies such as vocal cords, stretched strings or membrane. (2) Infrasonic waves : Frequency lie below 20 Hz. U Example : waves produced during earth quake, ocean waves etc. 16.6 m. Note : D YG (3) Ultrasonic waves : Frequency greater than 20 KHz. Human ear cannot detect these waves, certain creatures such as mosquito, dog and bat show response to these. As velocity of sound in air is 332 m/sec so the wavelength of ultrasonics  < 1.66 cm and for infrasonics  > Supersonic speed : An object moving with a speed greater than the speed of sound is said to move with a supersonic speed.  Mach number : It is the ratio of velocity of source to the velocity of sound. U Mach Number = Velocity of source. Velocity of sound  Difference between sound and light waves : For propagation of sound wave material medium is required but no material ST (i) medium is required for light waves. (ii) Sound waves are longitudinal but light waves are transverse. (iii) Wavelength of sound waves ranges from 1.65 cm to 16.5 meter and for light it ranges from 4000 Å to 2000 Å. 16.4 Velocity of Sound (Wave motion). (1) Speed of transverse wave motion : (i) On a stretched string : v  (mass per unit length). T m T = Tension in the string; m = Linear density of string 210 Wave Motion (ii) In a solid body : v     = Modulus of rigidity;  = Density of the material. (2) Speed of longitudinal wave motion : (i) In a solid medium v  4  3 k = Bulk modulus;  = Modulus of rigidity;  = Density  (ii) In a liquid medium v   Y = Young's modulus of material of rod k  k  ID v Y E3 When the solid is in the form of long bar v  (iii) In gases 60 k 16.5 Velocity of Sound in Elastic Medium. When a sound wave travels through a medium such as air, water or steel, it will set U particles of medium into vibration as it passes through it. For this to happen the medium must possess both inertia i.e. mass density (so that kinetic energy may be stored) and elasticity (so D YG that PE may be stored). These two properties of matter determine the velocity of sound. i.e. velocity of sound is the characteristic of the medium in which wave propagate. v E  (E = Elasticity of the medium;  = Density of the medium) Important points (1) As solids are most elastic while gases least i.e. E S  E L  E G. So the velocity of sound is U maximum in solids and minimum in gases vsteel > vwater > vair ST 5000 m/s > 1500 m/s > 330 m/s As for sound vwater > vAir while for light vw < vA. Water is rarer than air for sound and denser for light. The concept of rarer and denser media for a wave is through the velocity of propagation (and not density). Lesser the velocity, denser is said to be the medium and vice-versa. (2) Newton's formula : He assumed that when sound propagates through air temperature remains constant.(i.e. the process is isothermal) vair = elasticity; P = Pressure. By calculation vair = 279 m/sec. K   P  As K = E = P ; E = Isothermal Wave Motion 211 However the experimental value of sound in air is 332 m/sec which is greater than that given by Newton's formula. (3) Laplace correction : He modified Newton's formula assuming that propagation of sound in air as adiabatic process.  = E (As k = E    = Adiabatic elasticity)  (  Air  1.41 ) 1.41 × 279= 331.3 m/s (4) Effect of density : v = (5) Effect of pressure : v  1 P v    60 v= k E3 v= P  RT. Velocity of sound is independent of the pressure   M (6) Effect of temperature : v  RT M ID of gas provided the temperature remains constant. ( P   when T = constant)  v  T in K  U When the temperature change is small then vt = v0 (1 +  t) D YG v0 = velocity of sound at 0°C, vt = velocity of sound at t°C ,  = temp-coefficient of velocity of sound. Value of  = 0.608 m /s = 0.61 (Approx.) o C Temperature coefficient of velocity of sound is defined as the change in the velocity of sound, when temperature changes by 1°C. (7) Effect of humidity : With increase in humidity, density of air decreases. So with rise in humidity velocity of sound increases. U This is why sound travels faster in humid air (rainy season) than in dry air (summer) at the same temperature. ST (8) Effect of wind velocity : Because wind drifts the medium (air) along its direction of motion therefore the velocity of sound in a particular direction is the algebraic sum of the velocity of sound and the component of wind velocity in that direction. Resultant velocity of sound along SL = v + w cos. (9) Sound of any frequency or wavelength travels through a given medium with the same velocity. (v = constant) For a given medium velocity remains constant. All other factors like phase, loudness pitch, quality etc. have practically no effect on sound velocity. A Wind (w) wcos  S (10) Relation between velocity of sound and root mean square velocity. L Sound (v) 212 Wave Motion RT vsound = M and vrms = 3 RT M so v rm s = v sound 3  or vsound = [  /3]1/2 vrms. 60 (11) There is no atmosphere on moon, therefore propagation of sound is not possible there. To do conversation on moon, the astronaut uses an instrument which can transmit and detect electromagnetic waves. 16.6 Reflection and Refraction of Waves. E3 When sound waves are incident on a boundary between two media, a part of incident waves returns back into the initial medium (reflection) while the remaining is partly absorbed and partly transmitted into the second medium (refraction) In case Incident of reflection and refraction of sound wave (1) The frequency of the wave remains unchanged that means ID i = r = t =  = constant (2) The incident ray, reflected ray, normal and refracted Reflected wave t Transmitted wave U ray all lie in the same plane. i r (3) For reflection angle of incidence (i) = Angle of reflection (r) sin i v i  sin t v t D YG (4) For refraction (5) In reflection from a denser medium or rigid support, phase changes by 180° and direction reverses if incident wave is y = A1 sin (t  kx ) then reflected wave becomes y = Ar sin ( t  kx   ) = – Ar sin ( t  kx ). (6) In reflection from a rarer medium or free end, phase does not change and direction reverses if incident wave is y = AI sin (  t  kx ) then Sourc e Detect or U reflected wave becomes y = Ar sin ( t  kx ) (7) Echo is an example of reflection. d ST If there is a sound reflector at a distance d from the source then time interval between original sound and its echo at the site of source will be t  2d v 16.7 Reflection of Mechanical Waves Medium Longitudinal wave Transverse wave Reflection from rigid end/denser Compression as rarefaction and vice-versa Crest as crest and Trough as trough Change in direction Reversed Phase change Time change Path change  T 2  2 Wave Motion 213 medium Reflection from free end/rarer medium Compression as compression and rarefaction as rarefaction Crest as trough and trough as crest No change Zero Zero 60 16.8 Progressive Wave. Zero (1) These waves propagate in the forward direction of medium with a finite velocity. E3 (2) Energy and momentum are transmitted in the direction of propagation of waves without actual transmission of matter. (3) In progressive waves, equal changes in pressure and density occurs at all points of medium. (4) Various forms of progressive wave function. 2  x) t x (iii) y = A sin 2     T   2  (vt – x) D YG (iv) y = A sin ID (ii) y = A sin (  t – where y = displacement A = amplitude  = angular frequency n = frequency k = propagation constant T = time period  = wave length v = wave velocity t = instantaneous time x = position of particle from origin U (i) y = A sin ( t – kx) x  (v) y = A sin   t   v  Important points (a) If the sign between t and x terms is negative the wave is propagating along positive Xaxis and if the sign is positive then the wave moves in negative X-axis direction. U (b) The coefficient of sin or cos functions i.e. Argument of sin or cos function i.e. (  t - kx) = Phase. ST (c) The coefficient of t gives angular frequency  = 2 n  2 = vk. T (d) The coefficient of x gives propagation constant or wave number k  2    v. (e) The ratio of coefficient of t to that of x gives wave or phase velocity. i.e. v =  k. (f) When a given wave passes from one medium to another its frequency does not change. (g) From v = n  v   n = constant  v 1 1.  v 2 2 (5) Some terms related to progressive waves 214 Wave Motion (i) Wave number ( n ) : The number of waves present in unit length is defined as the wave 1 number ( n ) =.  Dimension = [L–1]. (ii) Propagation constant (k) : k = k  v  Phase difference between particles  = Distance between them x Angular velocity 2 and k =  2  Wave velocity  60 Unit = meter–1 ; E3 (iii) Wave velocity (v) : The velocity with which the crests and troughs or compression and    rarefaction travel in a medium, is defined as wave velocity v = = n  . 2 T k (iv) Phase and phase difference : Phase of the wave is given by the argument of sine or 2 cosine in the equation of wave. It is represented by  (x , t)  (vt  x ). ID  (v) At a given position (for fixed value of x) phase changes with time (t). U d 2v 2 2 2  d =    Time difference..dt  Phase difference = T dt T  T (vi) At a given time (for fixed value of t) phase changes with position (x). D YG d 2 2 2   d   Path difference  dx  Phase difference =   dx   Time difference = T  × Path difference Sample problems based on Progressive wave Problem 1. The speed of a wave in a certain medium is 960 m/sec. If 3600 waves pass over a certain point of the medium in 1 minute, the wavelength is U (a) 2 meters Solution : (d) v = 960 m/s ; n = (c) 8 meters (d) 16 meters v 960 3600 Hz. So     16 meters. 60 n 60 A simple harmonic progressive wave is represented by the equation y = 8 sin 2  (0.1x – 2t) where x and y are in cm and t is in seconds. At any instant the Phase difference between two particles separated by 2.0 cm in the x-direction is ST Problem 2. (b) 4 meters (a) 18° (b) 36° (c) 54° (d) 72°  x   t x  2 t  given by comparing with standard equation y = a sin 2     Solution : (d) y = 8 sin 2   T   10    10 cm So Phase Difference = Problem 3. 2  × path difference = 2 2  2   180  = 72° 10 5 The frequency of sound wave is n and its velocity is v if the frequency is increased to 4n the velocity of the wave will be (a) v (b) 2v (c) 4v (d) v/4 Wave Motion 215 Solution : (a) Wave velocity does not depends on the frequency. It depends upon the Elasticity and inertia of the medium. Problem 4. The displacement of a particle is given by x = 3 sin (5  t) + 4 cos (5  t ) The amplitude of particle is [MP PMT 1999] (b) 4 (c) 5 (d) 7 60 (a) 3 Solution : (c) Standard equation : x = a sin  t + b cos  t   x  a 2  b 2 sin  t  tan 1 b / a  x  3 sin 5 t   4 cos(5 t)  x  9  16 sin 5  t  tan 1 4 /3  x  5 sin( 5 t  tan 1 (4 / 3)) The equation of a transverse wave travelling on a rope is given by y  10 sin  (0.01 x  2.00 t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about [MP PET 1999] (a) 63 cm/sec (b) 75 cm/s ID Problem 5. E3 Given equation (c) 100 cm/sec (d) 121 cm/sec U Solution : (a) Standard eq. of travelling wave y = A sin (k x  t) By comparing with the given equation y = 10 sin (0.01  x  2 t ) D YG A = 10 cm,  = 2  Maximum particle velocity = A  = 2  × 10 = 63 cm/sec Problem 6. In a wave motion y = a sin (k x   t), y can represents (a) Electric Field (b) magnetic field (c) Displacement (d) Pressure Solution : (a,b,c,d) Problem 7. Find the ratio of the speed of sound in nitrogen gas to that of helium gas, at 300 k is 1 2 (b) 2 3 (c) 3 5 (d) 4 5 U (a) Solution : (c) v   RT M ST  N 2 M He vN   v He  He M N 2 Problem 8. 7/5 4 3. =. 5 5 / 3 28 The displacement x (in metres) of a particle performing simple harmonic motion is related   to time t (in seconds) as x = 0.05 cos  4 t  . The frequency of the motion will be 4  (a) 0.5 Hz (b) 1.0 Hz (c) 1.5 Hz Solution : (d) From the given equation, coefficient of t =   4   n=  4   2 Hz 2 2 (d) 2.0 Hz 216 Wave Motion Problem 9.   A wave is represented by the equation Y = 7 sin  7 t  0. 04 x   x is in meters and t is in 3  seconds. The speed of the wave is (b) 49  m/s (a) 175 m/sec (c)  In a given equation   7  , k  0.04  v  k  (d) 0.28  m / s m/s 60 Solution : (a) Standard equation y = A sin  t  kx   0  49 7 = 175 m/sec.04  E3 Problem 10. A wave is represented by the equation y = 0.5 sin (10 t + x)m. It is a travelling wave propagating along the x direction with velocity. (a) 10 m/s (b) 20 m/s (c) 5 m/s Solution : (a) v =  / k  10 / 1  10 m / s (d) None of these Problem 11. A transverse progressive wave on a stretched string has a velocity of 10 ms–1 and a D YG U ID frequency of 100 Hz. The phase difference between two particles of the string which are 2.5 cm apart will be [MP PMT 1994] (a)  / 8 (b)  / 4 (c) 3 / 8 (d)  / 2 10 Solution : (d)   v / n   0.1 m  10 cm 100 2 2  Phase difference = × path difference  × 2.5 =  10 2 Problem 12. In a stationary wave, all particles are [MP PMT 1994] (a) At rest at the same time twice in every period of oscillation (b) At rest at the same time only once in every period of oscillation (c) Never at rest at the same time (d) Never at rest at all Solution : (a) Problem 13. The path difference between the two waves U 2 x  2x    y1  a1 sin   t     is  and y 2  a 2 cos   t          2 ST (a) (b)       2  2 (c) [MP PMT 1994] 2        2 (d) 2    2x  2x       Solution : (b) y1  a1 sin   t   ; y 2  a2 sin  t    2     2x   2x          t  Phase difference =   t        2     2  Path difference =     × Phase difference     2 2 2   x Problem 14. A plane wave is described by the equation y  3 cos   10 t  . The maximum velocity of 4 the particles of the medium due to this wave is (a) 30 (b) 3  / 2 (c) 3/4 Solution : (a) Maximum velocity = A  = 3 × 10 = 30 2 (d) 40 Wave Motion 217  Problem 15. A wave represented by the given equation y = A sin (10  x  15  t  ) where x is in meter 60 3 and t is in second. The expression represents (a) A wave travelling in the positive x-direction with a velocity of 1.5 m/sec (b) A wave travelling in the negative x-direction with a velocity of 1.5 m/sec (c) A wave travelling in the negative x-direction with a wavelength of 0.2 m (d) A wave travelling in the positive x-direction with a wavelength of 0.2 m By comparing with standard equation Y = A sin k x   t   / 3  Solution : (b, c) K = 10  ,   15   k = 1.5 m/sec;   2 = 0.2 meter. k E3 We know that : v  x Problem 16. A transverse wave is described by the equation Y  y 0 sin 2   ft   The maximum particle  (a)   y 0 (b)   4 y 0 2 ID velocity is four times the wave velocity if (c)    y 0  (d)   2y 0 A = 4 f y0 2  f  4 f y 0 D YG  U Solution : (b) Maximum particle velocity = 4 wave velocity 2 Problem 17. The equation of a wave travelling in a string can be written as y = 3 cos  (100 t – x) Its wavelength is (a) 100 cm (b) 2 cm [MP PMT 1991, 94, 97; MNR 1985] (c) 5 cm (d) None of these Solution : (b) y = A cos (  t  k x)  standard equation y = 3 cos (100  t   x ) – given equation U So K =  and   2  2 cm k Problem 18. A plane wave is represented by x = 1.2 sin (314 t + 12.56 y) where x and y are distances ST measured along in x and y direction in meter and t is time in seconds. This wave has (a) A wave length of 0.25 m and travels m + ve x-direction (b) A wavelength of 0.25 m and travels in + ve y-direction (c) A wavelength of 0.5 m and travels in – ve y-direction (d) A wavelength of 0.5 m and travels in – ve x-direction Solution : (c) From given equation k = 12.56  2 0.5 m direction = – y k Problem 19. A wave is reflected from a rigid support. The change in phase on reflection will be (a)  / 4 Solution : (c) (b)  / 2 (c)  (d) 2 218 Wave Motion  Problem 20. The equation of displacement of two waves are given as y1 = 10 sin  3  t   ; y2 = 5  3 [sin 3  t  3 cos 3 t] Then what is the ratio of their amplitudes (b) 2 : 1 Solution : (c) y2 = 5 [sin 3  t  3 cos 3  t] =5 (c) 1 : 1 (d) None of these 60 (a) 1 : 2 [AIIMS 1997]     1  3 sin  3 t   = 10 sin  3  t   3 3   So, A1 = 10 and A2 = 10 velocity of wave is   4 , k   / 16  k  32 (d) 64 cm/sec in cm/sec in – x +x 4  64 cm / sec in + x direction.  / 16 D YG v (b) [MP PET 1990] U   .x Solution : (d) y = 4 sin  4 t  16   x  8 t   if x and y are in cm, then 2 8 ID (a) 64 cm/sec in – x direction direction (c) 32 cm/sec in +x direction direction  E3 Problem 21. The equation of a wave travelling on a string is y = 4 sin Problem 22. The equation of wave is y = 2 sin  (0.5 x  200 t) where x and y are expressed in cm and t in sec. The wave velocity is (a) 100 cm/sec Solution : (d) v   k  (b) 200 cm/sec (c) 300 cm/sec (d) 400 cm/sec 200  = 400 cm/sec 0.5  U 16.9 Principle of Superposition. ST The displacement at any time due to any number of waves meeting simultaneously at a point in a medium is the vector sum of the individual displacements due each one of the waves at that point at the same time.    If y 1 , y 2, y 3 ………. are the displacements at a particular time at a particular position, due to     individual waves, then the resultant displacement. y  y1  y 2  y 3 ............. Examples (i) Radio waves from different stations having different frequencies cross the antenna. But our T.V/Radio set can pick up any desired frequency. (ii) When two pulses of equal amplitude on a string approach each other [fig. (A)], then on meeting, they superimpose to produce a resultant pulse of zero amplitude [fig (B)]. After crossing, the two pulses travel (A ) (B ) (C ) Wave Motion 219 independently as shown in [fig (C)] as if nothing had happened. Important applications of superposition principle : (a) Interference of waves (b) Stationary waves (c) Beats. 16.10 Interference of Sound Waves. E3 60 When two waves of same frequency, same wavelength, same velocity (nearly equal amplitude) moves in the same direction, Their superimposition results in the interference. Due to interference the resultant intensity of sound at that point is different from the sum of intensities due to each wave separately. This modification of intensity due to superposition of two or more waves is called interference. Let at a given point two waves arrives with phase difference  and the equation of these waves is given by    y  y 1  y 2  y = A sin  t    where A  a1 2  a 2 2  2 a1 a 2 cos  and tan  = Intensity  A 2. a2 sin  a1  a 2 cos  U and since ID y1 = a1 sin  t , y2 = a2 sin (  t   ) then by the principle of superposition So I = a1 2  a 2 2  2a1 , a 2 cos   I  I1  I 2  2 I1 I 2 cos  D YG Important points (1) Constructive interference : Intensity will be maximum when   0, 2  , 4 ,...... 2 n ; where n = 0,1,2 …….. when x = 0, , 2, ……….. n; where n = 0, 1 ……. Imax = I1 + I2 + 2 I1 I 2 =  I1  I 2  2   A1  A 2 2 U It means the intensity will be maximum at those points where path difference is an integral multiple of wavelength . These points are called points of constructive interference or ST interference maxima. (2) Destructive interference : Intensity will be minimum when    , 3 ,5...... ( 2 n  1) ; where n= 1, 2, 3 …….. when x = /2, 3/2, ……….. (2n-1); Imin = I1+I2 – 2 I1 I 2  Imin =  I1  I 2 where n = 1, 2, 3……..  2   A1 ~ A 2  2 (3) All maxima are equally spaced and equally loud. Same is also true for minima. Also interference maxima and minima are alternate as for maximum x  0, ,2....etc. and for  3 5  minimum x  , ,....... etc. 2 2 2 220 Wave Motion I (4) max  I min I I 1  I2 1 ~   2 I2 2   A1  A 2 2  A1 ~ A 2 2 with I1 A1 2  I2 A2 2 (5) If I 1  I 2  I0 then Imax = 4 I o and Imin =0 Imax  ( I1  I2 )2 I I1  I 2  exceeds the sum of individual intensities (I1 + I2) by an amount 2 I1 I 2 while  I1 ~ I2  2 in minima I1  I 2 2 I1 I 2 Imin  ( I1 ~ Phase 0 Diff. Path Diff.0  / 2 lacks I1  I 2  by the same amount 3 4 5 6  I2 )2  3/2 2 5/2 3 x ID 2 I1 I2. 2 E3  maximum 2 60 2 I1 I 2 (6) In interference the intensity in Hence in interference energy is neither created nor destroyed but is redistributed. 16.11 Standing Waves or Stationary Waves. U When two sets of progressive wave trains of same type (both longitudinal or both D YG transverse) having the same amplitude and same time period/frequency/wavelength travelling with same speed along the same straight line in opposite directions superimpose, a new set of waves are formed. These are called stationary waves or standing waves. Characteristics of standing waves : (1) The disturbance confined to a particular region between the starting point and reflecting point of the wave. (2) There is no forward motion of the disturbance from one particle to the adjoining particle and so on, beyond this particular region. U (3) The total energy associated with a stationary wave is twice the energy of each of incident and reflected wave. But there is no flow or transference of energy along the stationary wave. ST (4) There are certain points in the medium in a standing wave, which are permanently at rest. These are called nodes. The distance between two consecutive nodes is  2 (5) Points of maximum amplitude is known as antinodes. The distance between two consecutive antinodes is also  / 2. The distance between a node and adjoining antinode is  / 4. (6)The medium splits up into a number of segments. Each segment is vibrating up and down as a whole. (7) All the particles in one particular segment vibrate in the same phase. Particles in two consecutive segments differ in phase by 180°. (8) All the particles except those at nodes, execute simple harmonic motion about their mean position with the same time period. Wave Motion 221 (9) The amplitude of vibration of particles varies from zero at nodes to maximum at antinodes. (10) Twice during each vibration, all the particles of the medium pass simultaneously through their mean position. 60 (11) The wavelength and time period of stationary waves are the same as for the component waves. (12) Velocity of particles while crossing mean position varies from maximum at antinodes to zero at nodes. E3 (13) In standing waves, if amplitude of component waves are not equal. Resultant amplitude at nodes will be minimum (but not zero). Therefore, some energy will pass across nodes and waves will be partially standing. 16.12 Standing Waves on a String. When a string under tension is set into vibration, transverse harmonic waves propagate 2  (vt + x) vt  x       a sin 2 D YG Reflected wave y 2  a sin  x=L x=0 U Incident wave y1 = a sin 2 ID along its length. When the length of string is fixed, reflected waves will also exist. The incident and reflected waves will superimpose to produce transverse stationary waves in a string  y1 vt  x  y2 According to superposition principle : y = y1 + y2 = 2 a cos General formula for wavelength   2vt  sin 2x  2L where n = 1,2,3, … correspond to 1st , 2nd, 3rd modes n of vibration of the string. (1) 1 T 2L m normal mode of n1  vibration v 1  v  2L A N U n1  First L N 1 2 ST This mode of vibration is called the fundamental mode and the frequency is called fundamental frequency. The sound from the note so produced is called fundamental note or first harmonic. (2) Second normal mode of vibration : n 2  v 2  v 2v   2(n1 ) L 2L N A N This is second harmonic or first over tone. (3) Third normal mode of vibration : n3  v 3 This is third harmonic or second overtone. A N L = 2  3v  3n1 2L N A N A N A N 222 Wave Motion L 2L 3 L , ,............ L n n n For first mode of vibration x = 0 , x = L For second mode of vibration L ,x  L 2 x = 0, x = [Three nodes] L 2L ,x  , x  L [Four nodes] 3 3 For third mode of vibration x = 0 , x = 2  1L L 3L 5L , ,........... 2n 2n 2n 2n E3 Position of antinodes : x  [Two nodes] 60 Position of nodes : x  0, For first mode of vibration x = L/2 L 3L , 4 4 [Two antinode] ID For second mode of vibration x = [One antinode] 16.13 Standing Wave in a Closed Organ Pipe. U Organ pipes are the musical instrument which are used for producing musical sound by blowing air into the pipe. Longitudinal stationary waves are formed on account of superimposition of incident and reflected longitudinal waves. 2vt  sin 2x D YG Equation of standing wave y = 2a cos General formula for wavelength    4L 2n  1 (1) First normal mode of vibration : n1  L 1 2 v 4L U This is called fundamental frequency. The note so produced is called fundamental note or first harmonic. (2) Second normal mode of vibration : n 2  ST A N v 2  A N 3v  3 n1 4L L N A 3 2 4 This is called third harmonic or first overtone. (3) Third normal mode of vibration : n 3  5v  5 n1 4L N N A This is called fifth harmonic or second overtone. Position of nodes : x = 0, For first mode of vibration 2L 4L 6L 2nL , ,......... (2n  1) (2n  1) (2n  1) (2n  1) x=0 For second mode of vibration x = 0, x = [One node] 2L 3 [Two nodes] N A A Wave Motion 223 x = 0, x  For third mode of vibration Position of antinode : x = 2L 4 L , 5 5 [Three nodes] L 3L 5L , ,........., L 2n  1 2n  1 2n  1 For second mode of vibration x= L ,x  L 3 [Two antinode] L 3L , ,L 5 5 [Three antinode] E3 For third mode of vibration x = [One antinode] 60 For first mode of vibration x = L 16.14 Standing Waves in Open Organ Pipes. General formula for wavelength A where n = 1,2,3 ……… ID 2L  n v (1) First normal mode of vibration : n1  n2  v 2 D YG This is called fundamental frequency and the note so produced is called fundamental note or first harmonic. (2) Second  normal mode A L v 2L U 1  N of 1 2 N N A A A L = 2 vibration v  v   2   2n1  n 2  2n1 L  2L  This is called second harmonic or first overtone. v 3v (3) Third normal mode of vibration n 3  , n3  3n1  3 2 L A N N N A A L A 3 3 2 U This is called third harmonic or second overtone. Important points ST (i) Comparison of closed and open organ pipes shows that fundamental note in open organ v  v    pipe  n1  .  has double the frequency of the fundamental note in closed organ pipe  n1  4L  2L    Further in an open organ pipe all harmonics are present whereas in a closed organ pipe, only alternate harmonics of frequencies n1 ,3n1 , 5 n1 ,...... etc are present. The harmonics of frequencies 2n1, 4n1, 6n1 …… are missing. Hence musical sound produced by an open organ pipe is sweeter than that produced by a closed organ pipe. 224 Wave Motion (ii) Harmonics are the notes/sounds of frequency equal to or an integral multiple of fundamental frequency (n). Thus the first, second, third, harmonics have 0.6 frequencies n 1 , 2n 1 ,3 n 1 , ……… r 60 (iii) Overtones are the notes/sounds of frequency twice/thrice/ four times the fundamental frequency (n) eg. 2 n,3n,4 n ……. and so on. (iv) In organ pipe an antinode is not formed exactly at the open end rather it is formed a little distance away from the open end outside it. The E3 distance of antinode from the open end of the pipe is known as end correction. 16.15 Vibration of a String. General formula n p  n 1 T 2L m Fixed end ID Fundamental frequency p T 2L m N A N Fixed end Plucking U L = Length of string, T = Tension in the string m = Mass per unit length (linear density), p = mode of vibration D YG Important points (1) As a string has many natural frequencies, so when it is excited with a tuning fork, the string will be in resonance with the given body if any of its natural frequencies concides with the body. (2) (i) n  1 L if L and m are constant (iii) n  if T and m are constant (ii) n  T 1 m if T and L are constant n 1 2L T 1  m 2L ST So U (3) If M is the mass of the string of length L, m  T 1  M/L 2 M L T 1 T 1 =  2 ML 2 L r  2 Lr T  where m = r2 (r = Radius,  = Density) 16.16 Comparative Study of Stretched Strings, Open Organ Pipe and Closed Organ Pipe S. No. Parameter Stretched string (1) Fundamental frequency or 1st harmonic n1  (2) Frequency of 1st overtone or 2nd n 2  2n1 v 2l Open organ pipe n1  v 2l n 2  2n1 Closed organ pipe n1  v 4l Missing Wave Motion 225 harmonic n 3  3n1 n 3  3n1 Frequency ratio of overtones 2 : 3 : 4… 2 : 3 : 4… 3 : 5 : 7… (5) Frequency ratio of harmonics 1 : 2 : 3 : 4… 1 : 2 : 3 : 4… (6) Nature of waves (4) Transverse stationary Longitudinal stationary 1 : 3 : 5 : 7… Longitudinal stationary E3 Frequency of 2nd overtone or 3rd harmonic 60 n 3  3n1 (3) 16.17 Beats. ID When two sound waves of slightly different frequencies, travelling in a medium along the same direction, superimpose on each other, the intensity of the resultant sound at a particular position rises and falls regularly with time. This phenomenon of regular variation in intensity of sound with time at a particular position is called beats. Important points U (1) One beat : If the intensity of sound is maximum at time t = 0, one beat is said to be formed when intensity becomes maximum again after becoming minimum once in between. D YG (2) Beat period : The time interval between two successive beats (i.e. two successive maxima of sound) is called beat period. (3) Beat frequency : The number of beats produced per second is called beat frequency. (4) Persistence of hearing : The impression of sound heard by our ears persist in our mind for 1/10th of a second. If another sound is heard before 1/10 second is over, the impression of the two sound mix up and our mind cannot distinguish between the two. So for the formation of distinct beats, frequencies of two sources of sound should be nearly equal (difference of frequencies less than 10) U (5) Equation of beats : If two waves of equal amplitudes 'a' and slightly different frequencies n1 and n2 travelling in a medium in the same direction are. ST y1 = a sin  1 t  a sin 2 n1 t ; y 2  a sin  2 t  a sin 2 n 2 t    By the principle of super position : y  y 1  y 2 y = A sin  (n1  n 2 ) t where A = 2a cos  (n1  n 2 ) t = Amplitude of resultant wave. (6) Beat frequency : n = n1 ~ n2. (7) Beat period : T= 1 1  Beat frequency n1 ~ n 2 16.18 Determination of Unknown Frequency. Let n2 is the unknown frequency of tuning fork B, and this tuning fork B produce x beats per second with another tuning fork of known frequency n1. As number of beat/sec is equal to the difference in frequencies of two sources, therefore n2 = n1  x 226 Wave Motion The positive/negative sign of x can be decided in the following two ways : By loading By filing If B is loaded with wax so its frequency decreases If B is filed, its frequency increases If number of beats decreases n2 = n1 – x If number of beats Increases n2 = n1 – x If number of beats Increases n2 = n1 + x If number of beats remains unchanged n2 = n1 + x If number of beats remains unchanged n2 = n1 – x If number of beats becomes zero n2 = n1 + x If number of beats becomes zero n2 = n1 – x E3 If A is loaded with wax its frequency decreases 60 If number of beats decreases n2 = n1 + x If A is filed, its frequency increases If number of beats decreases n2 = n1 + x If number of beats increases n2 = n1+ x If number of beats Increases n2 = n1 – x If number of beats remains unchanged n2 = n1 – x If number of beats remains unchanged n2 = n1 + x If number of beats becomes zero n2 = n1 – x If no of beats becomes zero n2 = n1 + x ID If number of beats decreases n2 = n1 – x U Sample problems based on Superposition of waves Problem 23. The stationary wave produced on a string is represented by the equation y = 5 D YG  x  cos   sin (40  t) where x and y are in cm and t is in seconds. The distance between  3  consecutive nodes is [MP PMT 2000] (b)  cm (a) 5 cm (c) 3 cm (d) 40 cm Solution : (c) By comparing with standard equation of stationary wave 2x y = a cos 2x = sin x U We get   3 2  vt    = 6; Distance between two consecutive nodes =  2  3 cm Problem 24. On sounding tuning fork A with another tuning fork B of frequency 384 Hz, 6 beats are ST produced per second. After loading the prongs of A with wax and then sounding it again with B, 4 Beats are produced per second what is the frequency of the tuning fork A. (a) 388 Hz (b) 80 Hz (c) 378 Hz Solution : (c) 390 390 256 384 378 YA 378 YB (d) 390 Hz Wave Motion 227 Probable frequency of A is 390 Hz and 378 Hz and After loading the beats are decreasing from 6 to 4 so the original frequency of A will be n2 = n 1 – x = 378 Hz. Problem 25. Two sound waves of slightly different frequencies propagating in the same direction produces beats due to [MP PET 2000] (b) Diffraction (c) Polarization Solution : (a) (d) Refraction 60 (a) Interference Problem 26. Beats are produced with the help of two sound waves on amplitude 3 and 5 units. The ratio of maximum to minimum intensity in the beats is (b) 5 : 3 2 Solution : (d) (c) 4 : 1 (d) 16 : 1 E3 (a) 2 : 1 2  A  A2   53  Imax  =    1   5  3  = 16:1 Imin  A1  A2    Problem 27. Two tuning forks have frequencies 380 and 384 hertz respectively. When they are sounded (b) 1/4 sec (c) 1/8 sec [MP PMT/PET 1998] (d) 1/16 sec U (a) 1/2 sec ID together, they produce 4 beats. After hearing the maximum sound, how long will it take to hear the minimum sound Solution : (c) Beats period = Time interval between two minima 1 1  sec n1  n2 4 D YG T= Time interval between maximum sound and minimum sound = T/2 = 1/8 sec Problem 28. Two tuning fork A and B give 4 beats per second when sounded together. The frequency of A is 320 Hz. When some wax is added to B and it is sounded with A, 4 beats per second are again heard. The frequency of B is (a) 312 Hz (b) 316 Hz (c) 324 Hz (d) 328 Hz Solution : (c) Since there is no change in beats. Therefore the original frequency of B is U n2 = n 1 + x = 320 + 4 = 324 Problem 29. 41 forks are so arranged that each produces 5 beat/sec when sounded with its near fork. If ST the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork respectively (a) 200, 400 [MP PMT 1997] (b) 205, 410 (c) 195, 390 (d) 100, 200 Solution : (a) Let the frequency of first tuning fork = n and that of last = 2n n, n + 5, n + 10, n + 15 …… 2n this forms A.P. Formula of A.P l = a + (N – 1) r Common difference where l = Last term, a = First term, N = Number of term, r = 2n = n + (41 – 1) 5 2n = n + 200 n = 200 and 2n = 400 228 Wave Motion Problem 30. In stationary waves, antinodes are the points where there is [MP PMT 1996] (a) Minimum displacement and minimum pressure change (b) Minimum displacement and maximum pressure change (c) Maximum displacement and maximum pressure change 60 (d) Maximum displacement and minimum pressure change Solution : (d) At Antinodes displacement is maximum but pressure change is minimum. Problem 31. The equation y = 0.15 sin 5x cos 300 t, describes a stationary wave. The wavelength of the (a) Zero meter (b) 1.256 meter Solution : (b) By comparing with standard equation  E3 stationary wave is (c) 2.512 meter 2x  5x    (d) 0.628 meter 2   = 1.256 meter 5 x  ID Problem 32. The equation of a stationary wave is y = 0.8 cos  [MP PMT 1995]  sin 200  t where x is in cm. and t is in  20  sec. The separation between consecutive nodes will be (b) 10 cm (c) 40 cm U (a) 20 cm 2x  sin 2vt  D YG Solution : (a) Standard equation y = A cos By comparing this equation with given equation. Distance Between two nodes = (d) 30 cm 2x x    = 40 cm  20  = 20 cm. 2 Problem 33. Which of the property makes difference between progressive and stationary waves (a) Amplitude (b) Frequency (c) Propagation of energy (d) Phase of the wave U Solution : (c) In stationary waves there is no transfer of energy. Problem 34. If amplitude of waves at distance r from a point source is A, the amplitude at a distance 2r ST will be [MP PMT 1985] (a) 2A Solution : (c) I  A2 and I  (b) A 1 r 2 (c) A/2 so r  1 ; A r r1 A 2   A2 = A1  1 r2 A1  r2 (d) A/4  1  = A   = A/2 2  Problem 35. If two waves of same frequency and same amplitude respectively on superimposition produced a resultant disturbance of the same amplitude the wave differ in phase by (a)  (b) 2 / 3 (c)  / 2 2 2 Solution : (b) A  A1  A 2  2 A1 A 2 cos  A2 = A2 + A2 + 2A2 cos  [A1 = A2 = A given] (d) zero Wave Motion 229 cos  = – 1/2    120   2 3 Problem 36. The superposition takes place between two waves of frequency f and amplitude a. The total intensity is directly proportional to (c) 2a2 (b) 2a Solution : (d) I  a1  a2 2 (d) 4a2 60 (a) a [As a1 = a2 = a] I  4a 2 Problem 37. The following equation represent progressive transverse waves E3 z1 = A cos ( t – kx) [MP PET 1993] z2 = A cos ( t + kx) z3 = A cos ( t + ky) ID z4 = A cos ( 2 t – 2ky) A stationary wave will be formed by superposing (a) z1 and z2 (b) z1 and z4 (c) z2 and z3 (d) z3 and z4 Solution : (a) The direction of wave must be opposite and frequencies will be same then by superposition, U standing wave formation takes place. Problem 38. When two sound waves with a phase difference of  / 2 and each having amplitude A and D YG frequency  are superimposed on each other, then the maximum amplitude and frequency of resultant wave is (a) A ; / 2 [MP PMT 1989] (b) 2 A ; (c) 2A; 2 a1 2  a2 2  2a1a 2 cos  = Solution : (d) Resultant Amplitude =  2 (d) A 2  A 2  2 A 2 cos  2 = 2 A; 2A and frequency remains same = . U Problem 39. There is a destructive interference between the two waves of wavelength  coming from ST two different paths at a point. To get maximum sound or constructive interference at that point, the path of one wave is to be increased by (a)  / 4 (b)  / 2 (c) 3 4 Solution : (b) Destructive interference means the path difference is 2n  1 (d)   2 If it is increased by  / 2 Then new path difference (2n – 1)  2   2 = n which is the condition of constructive interference. Problem 40. The tuning fork and sonometer wire were sounded together and produce 4 beats/second when the length of sonometer wire is 95 cm or 100 cm. The frequency of tuning fork is 230 Wave Motion (a) 156 Hz Solution : (a) Frequency n  (b) 152 Hz 1 l  As n  1 2l (c) 148 Hz (d) 160 Hz T m If n is the frequency of tuning fork. 1 1  n–4   (n + 4 ) 95 = (n – 4) 100  n = 156 Hz. 95 100 60 n+4  Problem 41. A tuning fork F1 has a frequency of 256 Hz and it is observed to produce 6 beats/second (a) 253 Hz E3 with another tuning fork F2. When F2 is loaded with wax. It still produces 6 beats/second with F1. The frequency of F2 before loading was (b) 262 Hz (c) 250 Hz (d) 259 Hz 16.19 Doppler Effect. ID Solution : (b) No of beats does not change even after loading then n2 = n1 + x = 256 + 6 = 262 Hz. Whenever there is a relative motion between a source of sound and the listener, the U apparent frequency of sound heard by the listener is different from the actual frequency of sound emitted by the source. D YG When the distance between the source and listener is decreasing the apparent frequency increases. It means the apparent frequency is more than the actual frequency of sound. The reverse is also true. General expression for apparent frequency n' = v  v m   v L n v  v m   v S  L S v v S Here n = Actual frequency; vL = Velocity of listener; vS = Velocity of source vL vm = Velocity of medium and v = Velocity of sound wave U Sign convention : All velocities along the direction S to L are taken as positive and all velocities along the direction L to S are taken as negative. If the medium is stationary vm = 0   n  ST  v  vL then n' =   v  vS Special cases : (1) Source is moving towards the listener, but the listener at rest n'  v.n v  vS (2) Source is moving away from the listener but the listener is at rest n'  v.n v  vS v  vL n v v  vL (4) Source is at rest and listener is moving towards the source n' .n v (3) Source is at rest and listener is moving away from the source n'  Wave Motion 231  v  vL (5) Source and listener are approaching each other n'    v  vS   n   v  vL (6) Source and listener moving away from each other n'    v  vS  n  60 (7) Both moves in the same direction with same velocity n' = n, i.e. there will be no Doppler effect because relative motion between source and listener is zero. (8) Source and listener moves at right angle to the direction of wave propagation. n' = n E3 It means there is no change in frequency of sound heard if there is a small displacement of source and listener at right angle to the direction of wave propagation but for a large displacement the frequency decreases because the distance between source of sound and listener increases. Important points ID (i) If the velocity of source and listener is equal to or greater than the sound velocity then Doppler effect is not seen. (ii) Doppler effect gives information regarding the change in frequency only. It does not says about intensity of sound. U (iii) Doppler effect in sound is asymmetric but in light it is symmetric. 16.20 Some Typical Features of Doppler’s Effect in Sound. D YG (1) When a source is moving in a direction making an angle  w.r.t. the listener : The apparent frequency heard by listener L at rest L (Listener at rest) nv When source is at point A is n   v  v s cos  As source moves along AB, value of  increases, cos decreases, n  goes on decreasing. At point C,   90 , cos   cos 90 o  0 , n   n. vs cos  s A o  vs C vs B vs cos U At point B, the apparent frequency of sound becomes nv n   v  v s cos  ST (2) When a source of sound approaches a high wall or a hill with a constant velocity v s , the reflected sound propagates in a direction opposite to that of direct sound. We can assume that the source and observer are approaching each other with same velocity i.e. v s  v L   v  vL n     v  vs   n  (3) When a listener moves between two distant sound sources : Let v L be the velocity of (v  v L )n listener away from S 1 and towards S 2. Apparent frequency from S 1 is n   v and apparent frequency heard from S L is n   (v  v L )n v 232 Wave Motion  Beat frequency  n   n   2nv L v  (4) When source is revolving in a circle and listener L is on one side O nv nv and n min  v  vs v  vs S r 60 v s  r so n max  (5) When listener L is moving in a circle and the source is on one side (v  v L )n (v  v L )n and n min  v v  E3 v L  r so n max  (6) There will be no change in frequency of sound heard, if the source O L r is situated at the centre of the circle along which listener is moving. (ii) When medium alone is moving. ID (7) Conditions for no Doppler effect : (i) When source (S) and listener (L) both are at rest. U (iii) When S and L move in such a way that distance between S and L remains constant. (iv) When source S and listener L, are moving in mutually perpendicular directions. D YG Sample problems based on Doppler effect Problem 42. A source of sound of frequency 90 vibration/sec is approaching a stationary observer with a speed equal to 1/10 the speed of sound. What will be the frequency heard by the observer (a) 80 vibration/sec (c) 100 vibration/sec (d) 120 vibration/sec v v 10 10  90.n  n' .n  n'  n  100 vibration/sec v v  vs 9 9 v 10 vs(+) S vL = 0 L U Solution : (c) n'  (b) 90 vibration/sec Problem 43. A source of sound of frequency 500 Hz is moving towards an observer with velocity 30 m/s. ST The speed of the sound is 330 m/s. The frequency heard by the observer will be (a) 550 Hz Solution : (a) n'  (b) 458.3 Hz v 330.n  n' .500  n' = 550 Hz v  vs 330  30 (c) 530 Hz vs(+) S (d) 545.5 Hz vL = 0 L Problem 44. A motor car blowing a horn of frequency 124 vibration/sec moves with a velocity 72 km/hr towards a tall wall. The frequency of the reflected sound heard by the driver will be (velocity of sound in air is 330 m/s) [MP PET 1997] (a) 109 vibration/sec (b) 132 vibration/sec (c) 140 vibration/sec (d) 248 vibration/sec Solution : (c) In the given condition source and listener are at the same position i.e. (car) for given condition Wave Motion 233 n'  v  vcar 330  20.n .n = 140 vibration/sec v  vcar 330  20 Problem 45. The driver of car travelling with a speed 30 meter/sec. towards a hill sounds a horn of (a) 720 Hz (b) 555.5 Hz (c) 550 Hz Solution : (a) This question is same as that of previous one so n'  60 frequency 600 Hz. If the velocity of sound in air is 330 m/s the frequency of reflected sound as heard by the driver is [MP PMT 1996] (d) 500 Hz v  vcar.n  720 Hz v  vcar E3 Problem 46. The source of sound s is moving with a velocity 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him ? The [MP PMT 1994] (a) 750 Hz (c) 1143 Hz (b) 857 Hz (d) 1333 Hz ID velocity of sound in the medium is 350 m/s Solution : (a) When source is moving towards the stationary listener. vs(+) U v 350 n'  n  1000 .n  n  857.14 v  vs 350  50 Hz D YG When source is moving away from the stationary observer n' '  S vL = 0 L 350 v =  857 = 750 v  vs 350  50 Problem 47. A source and listener are both moving towards each other with speed v/10 where v is the speed of sound. If the frequency of the note emitted by the source is f, the frequency heard by the listener would be nearly (a) 1.11 f   n  v  v  10  n'   v  v  10  U  v  vL Solution : (b) n'    v  vs (b) 1.22 f   n     n'  [MP PMT 1994] (c) f (d) 1.27 f 11 f = 1.22 f. 9 ST Problem 48. A man is watching two trains, one leaving and the other coming in with equal speed of 4 m/s. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air = 320 m/s) will be equal to (a) 6 (b) 3 (c) 0 Solution : (a) App. Frequency due to train which is coming in n1  App. Frequency due to train which is leaving n2  1   1  So number of beats n1 – n2 =   320  240  316 324  (d) 12 v.n v  vs v.n v  vs  n1 – n2 = 6 Problem 49. At what speed should a source of sound move so that observer finds the apparent frequency equal to half of the original frequency 234 Wave Motion (a) v / 2 v.n v  vs  n v .n 2 v  vs (c) v / 4 (d) v  vs = v ST U D YG U ID E3 60 Solution : (d) n'  (b) 2v