Tiêu đề 15. Simple Harmonic motion easy Ans.pdf ✅

1. (c) vmax  v = ωA 2 = ω√A2 − y 2  A 2 − y 2 = A 2 4  y 2 = 3A 2 4  y = √3A 2 2. (d) Equation of motion is 6 2 5 sin t y  = . For y = 2.5 cm 6 2 2.5 5 sin t =  6 6 2  = t  t sec 2 1 = and phase 6 6 2  = = t . 3. (c) y = a sin( ωt − α) = a cos (ωt − α − π 2 ) Another equation is given y = cos( ωt − α) So, there exists a phase difference of π 2 = 90° 4. (d) y = a sin( t +)       = +   t T a 2 sin        = + 0.4 2 2 0.5 sin   y t       = + 2 0.5 sin 5  y  t = 0.5 cos 5t 5. (c) y = a sin(2nt +) . Its phase at time t = 2nt + 6. (c) From given equation  =   0.5 2 = T  T = 4 sec Time taken from mean position to the maximum displacement 1 4 1 = T = sec. 7. (a) It is required to calculate the time from extreme position. Hence, in this case equation for displacement of particle can be written as x a t a  t   ) cos 2 = sin( + =  a t a cos 2 =  3   t =  3 . 2  t = T  6 T t = 8. (a) v 2π T max  a = vmax 2π A = 1.00×10 3×(1×10 −5) 2π = 1.59mm 9. (c) 10. (c) 11. (c) t T y a 2 = sin  t T a a =  2 sin 2  4 sin 2 2 1 sin   t = = T  4 2  t = T  8 T t = 12. (c) 13. (d) Standard equation of S.H.M. , 2 2 2 y dt d y = − is not satisfied by y = a tan  t . 14. (b) x = a cos( t +) ....(i) and = = −a sin( t +) dt dx v ....(ii) Given at t = 0 , x = 1cm and v =  and  =  Putting these values in equation (i) and (ii) we will get a 1 sin −  = and a 1 cos =  2 2 2 2 1 1 sin cos        +      + = − a a    a = 2 cm 15. (d) y = A sint t T A sin 2 =  T t A A 2 sin 2 =  12 T t = . 16. (a) The amplitude is a maximum displacement from the mean position. 17. (c) Equation of motion y = acost  2 3 1 cos cos 2  = a t  t = t = a sec 3 2 3 2 4 2 3 3 2 =  =   =  =     T t T t 18. (a) Simple harmonic waves are set up in a string fixed at the, two ends. 19. (b) 20. (c) v1 = dy1 dt = 0.1 × 100π cos (100πt + π 3 ) v2 = dy2 dt = −0.1π sin π t = 0.1π cos (πt + π 2 ) Phase difference of velocity of first particle with respect to the velocity of 2nd particle at t = 0 is Δφ = φ1 − φ2 = π 3 − π 2 = − π 6 . 21. (b) a1 a2 = 10 25 = 2 5 22. (c) t T y a 2 = sin  3 2 sin 2 t a a  =  3 2 sin 2 1 t =  sin 6 sin 3 2  = t  3 6 2  = t  sec 4 1 t = 23. (b)
24. (a) x = a sin (ωt + π 6 )and x′ = a cos ω t = a sin (ωt + π 2 ) ∴ Δφ = (ωt + π 2 ) − (ωt + π 6 ) = π 3 25. (a) Velocity of a particle executing S.H.M. is given by v = ω√a 2 − x 2 = 2π T √A2 − A2 4 = 2π T √ 3A2 4 = πA√3 T . 26. (c) v = ω√(a 2 − y 2) = 2√60 2 − 20 2 = 113mm/s. 27. (c) It is given vmax, a = 10 cm.  vmax ω = 100 10 = 10rad/sec Hence v = ω√a 2 − y 2 50 = 10√(10) 2 − y 2  y = 5√3cm 28. (c) At centre vmax = aω = a. 2π T = 0.2×2π 0.01 = 40π 29. (b) cm s T v a a / 6 2 3 2 max    =  = =  = 30. (c) a T v = a =    2 max  v 2 m / s 2 2 2 max   =   = 31. (d) T a T a v a    .2 2 max = = = 32. (c) 2 2 v = a − y  2 2 10 =  a −(4) and 2 2 8 =  a −(5) On solving  = 2  2 2 = = T    T =  sec 33. (d) From the given equation, a = 5 and  = 4  4 (5) (3) 16 2 2 2 2 v =  a − y = − = 34. (b) m s T v a a 0.15 / 2 2 (50 10 ) 2 3 max = =  =   =  −   35. (a) n 35 Hz 2 220 2 = = =    vmax = a = 220 0.30 m / s = 66 m / s 36. (d) vmax = a and 2 Amax = a  rad sec v A 2 / 2 4 max max  = = = 37. (a) m s T a v a / 0.1 25 2 2 10 2 3 max     =   =  = = − 38. (b) A y 2 =   A y 2 rad / sec 2 8  = / = = Now vmax = a = 6  2 = 12 cm /sec 39. (c) vmax = a  4 max 10 = = a v  Now, 2 2 v =  a − y  ( ) 2 2 2 2 v =  a − y  2 2 2 2  v y = a −  2 2 2 2 2 2 (10 / 4) 5 = − = 4 −  v y a = 2 3 cm 40. (b) The particles will meet at the mean position when P completes one oscillation and Q completes half an oscillation So vP vQ = aωP aωQ = TQ TP = 6 3 = 2 1 41. (b)    1 2 max max = = a a A v 42. (a) Velocity is same. So by using v = aω  A1ω1 = A2ω2 = A3ω3 43. (d) In S.H.M. at mean position velocity is maximum So v = a (maximum) 44. (b) 45. (b) 46. (c) Acceleration A y 2 =  5 0.02 0.5  = = = y A  Maximum velocity vmax = a = 0.1 5 = 0.5 47. (d) At mean position velocity is maximum i.e., vmax = a 4 4 max 16  = = = a v   2 2 v =  a − y 2 2  8 3 = 4 4 − y 192 16 (16 ) 2  = − y 2  12 = 16 − y  y = 2 cm. . 48. (a) vmax = a = 3100 = 300 49. (a) x = 3 sin 2t + 4 cos 2t. From given equation 3, 4, a1 = a2 = and 2   =  2 2 2 a = a1 + a 3 4 5 2 2 = + = vmax = a = 5  2 = 10
50. (c) Velocity in mean position v = a, velocity at a distance of half amplitude. 2 2 v  =  a − y 4 2 2 a =  a − a v 2 3 2 3 =  = 51. (a)       = + 4 cos  x A t and       = = − + 4 sin  A  t dt dx v For maximum speed, 1 4 sin  =      +   t  4 2    t + = or 2 4    t = −    4 t = 52. (d) F = −kx 53. (c) The stone execute S.H.M. about centre of earth with time period 2 ; g R T =  where R = Radius of earth. 54. (c) Acceleration a 2 = at extreme position is maximum. 55. (d) 2 − a when it is at one extreme point. 56. (a) Maximum acceleration 2 2 2 = a = a 4 n 0.01 4 ( ) (60) 144 / sec 2 2 2 =     =  m 57. (a) Maximum acceleration Amax 2 2 2 4 T a a    = = 0.2 0.2 1 4 (3.14 ) 2    = F m A 98.596 N 0.2 0.2 0.1 4 (3.14 ) 2 max max =    =  = 58. (a) Maximum velocity = a = 16 Maximum acceleration 24 2 =  a = m a a a 3 32 24 ( ) 16 16 2 2 =   = =   59. (d) Acceleration  – displacement, and direction of acceleration is always directed towards the equilibrium position. 60. (d) Maximum force         = = 2 2 2 4 ( ) T m a ma   0.01 0.5N / 25 4 0.5 2 2  =         =   61. (d) 2 2 2 max 0.62 / 4 a a  a = cm sec      = =   [  a =1] 62. (a) For S.H.M. F = −kx.  Force = Mass  Acceleration  – x  F = – Akx; where A and k are positive constants. 63. (a) Velocity v = a = a 2 n = 0.06  2 15 = 5.65 m / s Acceleration 2 2 2 2 2 A = a = 4 n a = 5.32 10 m / s 64. (d) m A A a a 0.61 (3.5) 7.5 2 2 2 max max =  = = =   65. (a) a m 2 10 10 − =  and  = 10 rad / sec 2 2 2 2 max A = a = 10 10 10 = 10 m / sec −  66. (a) A a 2 max = 67. (d) 2 2 2 1 E = m a  2 E  a 68. (a) P.E. 2 2 2 1 = m x It is clear P.E. will be maximum when x will be maximum i.e., at x = A 69. (d) Let x be the point where K.E. = P.E. Hence 1 2 mω2 (a 2 − x 2 ) = 1 2 mω2x 2  2x 2 = a 2  T1 = 2s 70. (a) Since maximum value of t 2 cos is 1. o Ko K = K t = 2 max cos Also Kmax = PEmax = Ko 71. (a) F = −kx  dW = Fdx = −kxdx So ∫ dW W 0 = ∫ −kxdx x 0  W = U = − 1 2 kx 2 72. (c) Suppose at displacement y from mean position potential energy = kinetic energy  1 2 m(a 2 − y 2 )ω2 = 1 2 mω2y 2  a 2 = 2y 2  y = a √2 73. (c) Total energy in SHM E = 1 2 mω2a 2 ; (where a = amplitude) Potential energy U = 1 2 mω2 (a 2 − y 2 ) = E − 1 2 mω2y 2 When y = a 2  U = E − 1 2 mω2 ( a 2 4 ) = E − E 4 = 3E 4
74. (b) Potential energy (U) Total energy (E) = 1 2 mω2y 2 1 2 mω2a 2 = y 2 a 2 So 2.5 E = ( a 2 ) 2 a 2  E = 10J 75. (d) Kinetic energy T = 1 2 mω2 (a 2 − x 2 ) and potential energy, V = 1 2 mω2x 2 ∴ T V = a 2−x 2 x 2 76. (c) 2 2 2 2 max 2 1 2 1 m a m y U U   =  2 2 4 1 a y =  2 a y = 77. (c) Kinetic energy K = 1 2 mω2 (a 2 − y 2 ) = 1 2 × 10 × ( 2π 2 ) 2 [10 2 − 5 2 ] = 375 π 2ergs 78. (b) 4 2 1 2 1 2 1 2 2 2 2 2 2 2 =       = = = a a a y m a m y E U   79. (a) 80. (a) The time period of potential energy and kinetic energy is half that of SHM. 81. (b) If at any instant displacement is y then it is given that U =  E 2 1        =  2 2 2 2 2 1 2 1 2 1 m y m a  cm a y 4.2 2 6 2 = = = 82. (b) So a = 6cm,  = 100 rad /sec K m a 1 (100 ) (6 10 ) 18 J 2 1 2 1 2 2 2 2 2 max = =     = −  83. (c) In S.H.M., frequency of K.E. and P.E. = 2  (Frequency of oscillating particle) 84. (b) Total energy 2 2 1 U = Ka 85. (d) 2 2 2 2 2 2 2 1 2 1 a y m a m y E U = =    16 4 9 3 80 2 2 =       = a a U  U = 45 J 86. (c) 87. (c) Maximum potential energy position is y = a and maximum kinetic energy position is y = 0 88. (c) Mg = Kl  U Kl mgl 2 1 2 1 2 max = = 89. (b) 4 2 1 2 1 2 1 2 2 2 2 2 2 2 2 =       = = = a a a y m a m y E U    4 E U = 90. (b) In S.H.M., at mean position i.e. at x = 0 kinetic energy will be maximum and pE will be minimum. Total energy is always constant. 91. (a) In SHM for a complete cycle average value of kinetic energy and potential energy are equal i.e. = 2 2 4 1 = m a 92. (c) Total energy 2 2 2 1 = m a = constant 93. (c) Kinetic energy at mean position, . 2 1 2 Kmax = mv max m K v max max 2  = 0.32 2 16 = = 100 = 10 m / s. 94. (a) E = 1 2 ma 2ω 2 = 1 2 ma 2 ( 4π 2 T2 ) E ∝ a 2 T2 95. (b) K E = 1 2 mω2(a 2−y 2) 1 2 mω2a2 = a 2−y 2 a2 = 1 − y 2 a2 So, ( 3E 4 ) E = 1 − y 2 a2 ⇒ y 2 a2 = 1 − 3 4 = 1 4 ⇒ y = a 2 . 96. (c) Kinetic energy K = 1 2 mv 2 = 1 2 ma 2ω2 cos2 ω t = 1 2 mω2a 2 (1 + cos 2 ωt) hence kinetic energy varies periodically with double the frequency of S.H.M. i.e. 2ω. 97. (c) 2 2 2 1 E = m a  2 2 a a E E  =   2 2 4 3 a a E E       =        a  = a 4 3   E E 16 9  = 98. (d) In simple harmonic motion, energy changes from kinetic to potential and potential to kinetic but the sum of two always remains constant. 99. (b) Body collides elastically with walls of room. So, there will be no loss in its energy and it will remain colliding with walls of room, so it’s motion will be periodic. There is no change in energy of the body, hence there is no acceleration, so it’s motion is not SHM.