Title 16. Waves and Sound Hard Ans.pdf ✅

17. (a) n v v v n s ' . − =  .500 330 30 330 ' − n =  n' = 550 Hz 18. (c) In the given condition source and listener are at the same position i.e. (car) for given condition n v v v v n car car ' . − + = .n 330 20 330 20 − + = = 140 vibration/sec 19. (a) This question is same as that of previous one so n Hz v v v v n car car ' . = 720 − + = 20. (a) When source is moving towards the stationary listener. . 857 .14 350 50 350 ' 1000  = −  = − = n n n v v v n s When source is moving away from the stationary observer s v v v n + ' ' = = 857 350 50 350  + = 750 Hz 21. (b) n v v v v n s L         − + ' =  n v v v v n             − + = 10 10 '  n f 9 11 ' = = 1.22 f. 22. (b) Height of transmitting antenna m d h 1280 2 6.4 10 (128 10 ) 2 Re 6 2 3 2 =    = = 23. (c) m Å E hc 0.8 10 0.8 14.4 10 1.6 10 6.6 10 3 10 10 3 19 34 8 =  =       = = − − −  . This wavelength belongs to X-ray region. 24. (d) A = ( ) 1 2 1 2 2 2 2 A1 + A + 2A A cos  −  Amin = A1 - A2 When 1 - 2 = , 3....... and Amax = A1 + A2 when 1 - 2 = 0, 2,....... Thus A ≥ A1 - A2 A ≤ A1 + A2 . 25. (a) P =  2 1 2A2v speed          T will not change as both T and μ are constant,  will also not change as it is property of the source that is causing the wave motion. Hence to make power half, amplitude becomes A0/√2. 26. (a) As speed          = T is same for all thus wave with maximum wavelength will have minimum angular frequency (by v = n). Also as 1 = 3 thus 1 = 3. 27. (a) Slope at any point on y-x curve in wave motion represents ratio of particle speed to wave speed. slope B > slope A hence RA > RB. 28. (c) For fully destructive interference all waves must have phase difference of 1200 with respect to each other. Thus phase constants are (450 + 1200 ) and (450 + 2400 ). 29. (d) At t = 4 sec. source turns direction of motion and starts moving towards detector. Fort = 0 to t = 4f = f0 V V s V + as Vs decreases in this duration, f increases . For t = 4 to t = 8 f = f0 V V s V + again f increase as Vs increases. Thus for whole duration f increases. 30. (a) I 0 2 2 0 '2 0 P P I I' = S L vs(+) vL = 0 S L vL = 0 vs(+) S L vs(+) vL = 0
P0' = P I I' P0 ' = 10 p 31. (b) P0 = B.k.S0= B        2 S0  P0   1 Thus pressure amplitude is highest for minimum wavelength, other parameters B and S0 being same for all. 32. (b) Along the joining line, components of velocities are towards each other. Hence effect of motions of both is to increase apparent frequency. Thus apparent frequency is f = 500 o o 340 10cos37 340 20cos53 − + = 530 Hz 33. (a) f = f0         + sound ob V V 1 sound ob 0 V V f f  (straights line) when sound ob V V = 0; 0 f f =1. and as sound ob V V → 1  0 f f → 2 34. (c) B = 10 log 0 I I If P is power of sound source then  = 10 log ( ) 0 2 I P/4r where Io = 10-12w/m2 = 10 log 0 2 4 r .I P  = 10[logP-log(4I0.r2 )] = 10[logP-log4I0-2logr] = a- b log r where a = 10 log P - 10 log I0 = constant b = 20  constant 35. (a) n 2  = L where n = 1,2,3............ using V = v  v = 2L nV , n = 1,2,3........... 36. (a) Path difference = Path B - Path A = 7d-3d =4d For being out of phase  x = 4d = 2 3 ; 2   ;................ For minimum d, 4d = 2   d = 8  37. (c) For pipe A, second resonant frequency is third harmonic thus f = 4L A 3V For pipe B, second resonant frequency is second harmonic thus f = 2L B 2V Equating, 4L A 3V = 2L B 2V LB = 3 4 LA = 3 4 .(1.5) = 2m. 38. (d) 2 4 r P  = I for an isotropic point sound source. P = I.4pr2 = (0.008w/m2 )(4..102 ) = 10 watt. 39. (c) If intensity is doubled the sound level is increased by 3.0 dB. i.e it becomes (B0 + 3) decibels. Hence when intensity is increased four times, level becomes (B0 + 6) decibels. 40. (b) If f0 is frequency of source then in t time, it sends N = f0 t waves. These are now contained in distance (V + Uw)t. Thus ' = ( ) f t V U t 0 w  +  ( ) ( ) 0 w 0 w V V U f V U  + = + 41. (d)
551 Hz fork can produce beats of frequencies 1, 2, 7 with 550 Hz, 553 Hz and 558 Hz forks respectively. 553 Hz fork can produce beat frequencies 3 and 5 with 550 Hz and 558 Hz forks respectively. 558 Hz can produces 8 Hz beat with 550 Hz fork. 42. (c) It should be remembered that the oscillator reading is correct and the tuning fork frequency marked is wrong. When the oscillator reading is 514, two beats are heard. Hence the frequency of the tuning fork is 514  2 = 516 or 512. When the oscillator reading is 510, the frequency of the tuning fork is 510  6 = 516 or 504. The common value is 516. Hence the frequency is 516 Hz. 43. (a) The relative velocity of sound waves with respect to the walls is V + v. Hence the apparent frequency of the waves striking the surface of the wall is (V + v)/ . The number of positive crests striking per second is the same as frequency. In three seconds the number is [3 (V + v)]/ . 44. (c) The resultant amplitude A of two waves of amplitudes a1 and a2 and phase difference  is ( 1/ 2 1 2 2 2 2 1 (a + a + 2a a cos) . Substituting a1 = 10, a2 = 10 and  = 900 , we get A = 14.1. 45. (a) When two SHM of different amplitudes and same period are superposed, the resulting motion is in general elliptical. When the phase difference is 0 or 1800 , the motion will be a straight line. When the phase difference is 90 or 2700 , the motion will be an ellipse whose axes coincide with coordinate axes. If the amplitudes are the same and the phase difference is 900 or 2700 , the motion will be circular. 46. (c) The energy is inversely proportional to the square of distance. Hence the amplitude is inversely proportional to the distance. 47. (b) The maximum particle velocity of a SHM of amplitude y0 and frequency f is 2fY0 . The wave velocity is f. For 2 fy0 to be equal to 4f,  has to be y0 / 2 (Here  = a). 48. (a) Inside a gas, only the longitudinal mode of transmission is possible for sound waves. 49. (a) The maximum restoring force of S.H.M. is the weight of the object in the platform. If A is the amplitude, we have m 2A = mg, where  = 2 f. This solves to f = 980 2 1 a g 2 1  =  50. (b) Writing the general expression for y in terms of x as y = 2 1 (x vt) 1 + − at t = 0, y = 1/ (1 + x)2 . At t = 2 s, y = 2 1 [x v(2)] 1 + − Comparing with the given equation we get 2v = 1 and v = 0.5 m/s. 51. (c) It should be remembered that the oscillator reading is correct and the tuning fork frequency marked is wrong. When the oscillator reading is 514, two beats are heard. Hence the frequency of the tuning fork is 514  2 = 516 or 512. When the oscillator reading is 510, the frequency of the tuning fork is 510  6 = 516 or 504. The common value is 516. Hence the frequency is 516 Hz. 52. (a) The relative velocity of sound waves with respect to the walls is V + v. Hence the apparent frequency of the waves striking the surface of the wall is (V + v)/ . The number of positive crests striking per second is the same as frequency. In three seconds the number is [3 (V + v)]/ . 53. (c) The resultant amplitude A of two waves of amplitudes a1 and a2 and phase difference  is ( . Substituting a1 = 10, a2 = 10 and  = 900 , we get A = 14.1. 54. (a) When two SHM of different amplitudes and same period are superposed, the resulting motion is in general elliptical. When the phase difference is 0 or 1800 , the motion will be a straight line. When the phase difference is 90 or 2700 , the motion will be an ellipse whose axes 1/ 2 1 2 2 2 2 1 (a + a + 2a a cos)