Tiêu đề 16. Waves and Sound Easy Ans.pdf ✅

1. (d) Air is more rarer for sound to travel as compared to vacuum. 2. (c) 3. (a) 4. (a) v = nλ = 2 × 5 = 10cm/sec 5. (d) v = n  n v  = 1.29m 256 330 = = 6. (d) Time lost in covering the distance of 2 km by the sound waves 6.06 330 2000 = = = v d t sec  6 sec 7. (a) vmax = a = a 2n = 0.1 2 300 = 60 cm / sec 8. (c) Audiable range of frequency is 20Hz to 20kHz 9. (c) Phase difference =   2 path difference  40 2 1.6 =       = 50 cm = 0.5m  v = n  330 = 0.5 n  n = 660 Hz 10. (a) λ = v n ; n ≈ 50,000Hz, v = 330m/sec λ = 330 50000 m = 6.6 × 10 −5 cm ≈ 5 × 10 −5 cm 11. (a) 12. (a) λ = v n = 1.7×1000 4.2×10 6 = 4 × 10 −4m 13. (d)Since maximum audible frequency is 20,000 Hz, hence 20 20,000 340 max min = =  n v  14. (c) Velocity of sound in gas v = √ γRT M  v ∝ √ γT M  vN2 vHe = √ γN2 γHe × MHe MH2 = √ 7 5 R×4 5 3 R×28 = √3 5 15. (a) Time required for a point to move from maximum displacement to zero displacement is n T t 4 1 4 = =  Hz t n 1.47 4 0.170 1 4 1 =  = = 16. (b) Wave number is the reciprocal of wavelength and is written as  1 n = . 17. (c) 1.7 200 340 = = = n v  m 18. (b) 19. (d) v ∝ λ  λ1 λ2 = v1 v2 = 2/3 3/10 = 20 9 20. (a) The time taken by the stone to reach the lake t1 = √( 2h g ) = √( 2×500 10 ) = 10sec (Using h = ut + 1 2 g ⥂ t 2 ) Now time taken by sound from lake to the man t2 = h v = 500 340 ≈ 1.5sec  Total time 21. (b) When medium changes, velocity and wavelength changes but frequency remains constant. 22. (b) 2.06 19.6 9.8 2 2 19.6 + =  = + = v v h g h t  v = 326 .7 m / s 23. (b) v ∝ √T v2 v1 = √ T2 T1 2 = √ T2 (273+0) ⇒ T2 = 273 × 4 = 1092K = 819 oC 24. (d) Velocity of sound in steel is maximum out of the given materials water and air. In vacuum sound cannot travel, it's speed is zero. 25. (b) Distance between a compression and the nearest rarefaction is 1 . 2 = m  Hence Hz v n 180 2 360 = = =  . 26. (a) v = √ γP ρ ⇒ vO2 vH2 = √ ρH2 ρo2 = √ 1 16 = 1 4 27. (d) Speed of sound in gases is v = √ γRT M  T ∝ M (Because v, -constant). Hence TH2 TO2 = MH2 MO2  TH2 (273+100) = 2 32  TH2 = 23.2K = −249.7°C
28. (c) If the temperature changes then velocity of wave and its wavelength changes. Frequency amplitude and time period remains constant. 29. (b) 30. (d) 31. (c) Path difference     =  2  2 2 1    =    = 4m Hence v = n = 120 4 = 480 m / s 32. (a) Suppose the distance between shooter and reflecting surface is d. Hence time interval for hearing echo is v d t 2 =  350 2 8 d =  d = 1400 m . 33. (b) 3.03 330 1000 Velocity Distance Time = = = sec . Sound will be heard after 3.03 sec. So his watch is set 3sec, slower. 34. (d)   P v = ; as P changes,  also changes. Hence  P remains constant so speed remains constant. 35. (b) Speed of sound in gases is given by M v M RT v 1 =     1 2 2 1 m m v v = 36. (b) By using v T M RT v =    T K C T T T T v v o 3 300 27 600 1 2 1 2 =  = = + = = 37. (a) Velocity of sound is independent of frequency. Therefore it is same (v) for frequency n and 4n. 38. (c) v = √ γRT M  v ∝ √T i.e. if v is doubled then T becomes four times, hence T2 = 4T1 = 4(273 + 27) = 1200K = 927°C 39. (d) 60 60 3600 n = = Hz  m n v 16 60 960  = = = 40. (d) Speed do sound, doesn’t depend up on pressure and density medium. 41. (d) If d is the distance between man and reflecting surface of sound then for hearing echo d v t d 170 m 2 340 1 2 =  =   = 42. (c) n = 54 60 Hz, λ = 10mv = nλ = 9m/s. 43. (a) M RT v  =  M v 1  . Since M is minimum for H2 so sound velocity is maximum in H2. 44. (d) 2d =vt , where v = velocity of sound = 332 m / s t = Persistence of hearing = sec 10 1 .  16.5 2 10 1 332 2 =  =  = v t d m 45. (c) Since solid has both the properties (rigidity and elasticity) 46. (b) If d is the distance between man and reflecting surface of sound then for hearing echo 247 .5 2 330 1.5 2 =  d = v t  d = m 47. (d) Speed of sound v  T and it is independent of pressure. 48. (b) Frequency of wave is n Hz 2 60 3600  =  25.3 30 760 = = = n v  m. 49. (a) Speed of sound d P v  =  1 2 2 1 d d v v = ( P - constant) 50. (d) 384 352 = = n v  ; during 1 vibration of fork sound will travel m 384 352 during 36 vibration of fork sound will travel 36 33 384 352  = m 51. (c) At given temperature and pressure  1 v   2 : 1 1 4 1 2 2 1 = = =   v v 52. (c) v ∝ √T√ T2 T1 = v2 v1 ⇒ T2 = T1 ( v2 v1 ) 2 ⇒ T2 = 273 × 4 = 1092K O d
53. (c) n̄ = 1 λ = 1 6000×10−10 = 1.66 × 10 6m−1 54. (b) v ∝ 1 √M ⇒ vH2 vO2 = √ MO2 MH2 = √ 32 2 ⇒ vH2 vO2 = 4 1 55. (a) The minimum distance between compression and rarefaction of the wire 4  l =  Wave length  = 4l Now by 90 4 1 360 =  v = n  n = sec . −1 56. (a) vsound ∝ 1 √ρ ⇒ v1 v2 = √ ρ2 ρ1 = √ 4 1 = 2  v2 = v1 2 = vs 2 57. (a) Suppose the distance between two fixed points is d then t = d v also v ∝ √T ⇒ t1 t2 = v2 v1 = √ T2 T1  2 t2 = √ 303 283  t2= 1.9 sec. 58. (a) The density of moist air (i.e. air mixed with water vapours) is less than the density of dry air Hence from v = √ γP ρ  vmoistair > vdryair 59. (d) Frequency of sound does not change with medium, because it is characteristics of source. 60. (c) Since M RT v  = i.e., v  T 61. (a) Frequency of waves remains same, i.e. 60 Hz and wavelength 3 60 10 330  = = n v  = 5.5 mm. 62. (c) Path difference 2 2 3 6         =  =  = 63. (d) Interference, diffraction and reflection occurs in both transverse and longitudinal waves. Polarisation occurs only in transverse waves. 64. (c) Water waves are transverse as well as longitudinal in nature. 65. (c) 66. (a) In transverse waves medium particles vibrate perpendicular to the direction of propagation of wave. 67. (d) 68. (a) Wave on a plucked string is stationary wave. Light waves are EM waves. Water waves are transverse as well as longitudinal. 69. (b) 70. (b) Transverse wave can propagate in solids but not in liquids and gases. 71. (b) Because sound waves in gases are longitudinal. 72. (d) 73. (c) Since distance between two consecutive crests is , so 2 . 2      =  = 74. (d) Comparing given equation with standard equation of progressive wave. The velocity of wave cm s k x t v 400 / 0.5 200 (Co - efficient of ) (Co - efficient of ) = = =    75. (c) Comparing with y = a cos(t + kx −) , We get k 0.02 100 cm 2 = =  =   Also, it is given that phase difference between particles . 2   = Hence path difference between them 25 cm 4 100 2 2 2 4  =  =  = = =        76. (b) Phase difference between two successive crest is 2. Also, phase difference () = T 2 time interval (t)  n Hz T T 5 sec 5 1 0.2 2 2 1 =   =  =  −  77. (c) Comparing with the standard equation, ( ) 2 y = A sin vt − x   , we have v = 200 cm / sec ,  = 200 cm ;  = = 1  v n 1 sec − 78. (d) Let the phase of second particle be  . Hence phase difference between two particles is  = x    2  15 60 2 3  =       −    6 5 3 2      − =  = 79. (d) The given equation can be written as       = − 16 4 sin 4 x y t    Co - efficient of ( ) Co - efficient of ( ) ( ) x K t v  =
 v 64 cm / sec / 16 4 = =   along +x direction. 80. (c) x t v Co - efficient of Co - efficient of = 20 cm / sec 31.4 628 = = 81. (d) sin( ) 1 y = a t − kx and       = − = − + 2 2 cos( ) sin  y a t k x a  t k x Hence phase difference between these two is . 2  82. (c) I ∝ A 2 ∝ 1 d2 A ∝ 1 d 83. (c) I1 I2 = a1 2 a2 2 = ( 0.06 0.03) 2 = 4 1 84. (c) After reflection from rigid support, a wave suffers a phase change of . 85. (c) The given equation representing a wave travelling along –y direction (because ‘+’ sign is given between t term and x term). On comparing it with x = A sin ( t + ky) We get 12.56 2 = =   k  0.5 m 12.56 2 3.14 =   = 86. (c) Comparing with y = a sin( ωt − kx) a = 10 π , ω = 200π ∴ v 10 π secmax and ω = 2π T ⇒ 200π = 2π T  T = 10 −3 sec 87. (b) Comparing the given equation with y = a cos( t −kx) We get    = = 2 k   = 2cm 88. (b) Comparing the given equation with y = a sin( ωt − kx), We get a = Y0,  = 2f, k = 2π λ . Hence maximum particle velocity (vparticle0max and wave velocity (v)wave = ω k = 2πf 2π/λ = fλ ∵ (vParticleWavemax  Y0 × 2πf = 4fλ  λ = πY0 2 . 89. (b)  ( t x ) x y a vt       0.5 cos 4 2 2 2 cos  = +      = + 90. (b) x t v Co - efficient of Co - efficient of = 2 / sec 50 100 = = m . 91. (d) y = f(x 2 − vt 2 ) doesn’t follows the standard wave equation. 92. (a) 93. (b)       = −    x y a t 2 1 1 sin and       = − +    x y a t 2 cos 2 2       = − + + 2 2 2 sin      x a t So phase difference = 2   + and  =       + 2 2     94. (a) Both waves are moving opposite to each other . 95. (a) The velocity of wave m s k x t v 10 / 1 10 (Co - efficient of ) (Co - efficient of ) = = =  96. (a) x t v Co - efficient of Co - efficient of = 175 m / s 0.04 7 = =  . 97. (a) The given equation is y = 10 sin(0.01x − 2t) Hence  = coefficient of t = 2  Maximum speed of the particle vmax = a = 10  2 = 10  2  3.14 = 62.8  63 cm/s 98. (d) On comparing the given equation with standard equation y = a sin 2π λ (vt − x). It is clear that wave speed (v)wave = v and maximum particle velocity (vparticlemax = y0 ×co- efficient of t = y0 × 2πv λ ∵ (vparticlewavemax  a×2πv λ = 2v  λ = πy0 99. (a) Given y = A sin(kx − t)  dt dy v = = −A cos(kx − t) :  vmax = A 100.(a) Comparing with y = (x, t) = a sin( t −kx)    0.01 2 k = =   = 200 m. 101. (b) 102.(d) Comparing the given equation with standard equation       = −   x T t y a sin 2  T = 0.04 sec  Hz T 25 1  = = Also 3 0.04 2 2 ( ) 2 2 2 max          =      = =    a T A a =7.4  104 cm/sec2 . 103.(b) From the given equation amplitude a = 0.04m vmax = a = 3 10 = 30