Tiêu đề 15. Simple Harmonic Motion Med Ans.pdf ✅

 (c)y = a sin(2nt + ). Its phase at time t = 2t +  2. (d) y = a sin(t + ) = a sin 2 t T    +       y = 0.5 sin 2 t 0.4 2     +     y = 0.5 sin 5 t 2     +     = 0.5 cos 5t 3. (c) y = a sin(t – ) = a cos t 2     −  −     Another equation is given y = cos(t – ) So, there exists a phase difference of 2  = 90o 4. (c) From given equation  = 2 T  = 0.5  T = 4 sec Time taken from mean position to the maximum displacement = 1 4 T = 1 sec. 5. (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 sin (t + 2  ) = a cos t x = a sin (t + 2  ) = a cos t  a 2 = a cos t  t = 3   2 T  .t = 3   t = T 6 6. (d) Equation of motion is y = 5 sin 2t 6  . For y = 2.5 cm 2.5 = 5sin 2t 6   2t 6  = 6   t = 1 2 sec and phase = 2t 6  = 6  . 7. (c) 2 2 max A A A y 2   =    = =  −  A2 – y 2 = 2 A 4  y 2= 2 3A 4  y = 3A 2 8. (a) max = a = a x 2 T   a = max x T 2   a = 3 5 1.00 x 10 x (1x 10 ) 2 −  = 1.59 mm 9. (c) 2 2 2 2  =  − = − (a y ) 2 60 20 = 113 mm /s. 10. (c) It is given max = 100 cm/sec. a = 10 cm. `  max = a   = 100 10 = 10 rad/sec Hence 2 2 2 2  =  −  = − a y 50 10 (10) y  y = 5 3 cm 11. (c) At centre max = a = a. 2 T  = 0.2 x 2 0.01  = 40 12. (b) max = a = a. 2 T  = 3 x 2 6  =  cm / s 13. (d) max = a = a.2 2 a T T   = 14. (c) 2 2 2 2  =  −  =  − a y 10 a (4) and 8 = 2 2  − a (5) On solving  = 2   = 2 T  = 2  T =  sec 15. (d) F = – kx 16. (c) Acceleration = 2a at extreme position is maximum. 17. (d)a2 when it is at one extreme point. 18. (c) The velocity of simple pendulum executing SHM is maximum at mean position and minimum at extremes. 19. (d) The total energy of a body executing SHM at any position remains conserved.
20. (a) We know that during SHM, the restoring force is proportional to the displacement from equilibrium position. Hence restoring force is maximum when the displacement is maximum at its extreme position. 21. (b)  = 2 rad/s–1 , y = 20 mm, a = 60 mm v = 2 2  a – y = ( ) ( ) 2 2 2 60 – 20 = 113 mm/s–1 22. (b)Given equation is x = a cos (t– ) So, velocity dx dt = a cos (t – ) At (t – ) = 90o, the velocity will be maximum, therefore the maximum velocity is v = a. 23. (a) 24. (d) 25. (a)  =  = 2 T  T = 2. 26. (a) 27. (a)Given F = – K x ——(a) For SHM F = –K1 x ——(b) From eqn. (a) and (b) K = K1 x  F = – K1 x x = –K1 x. Here K = K1 x  K  x As x increases K increases. Ans. (a) 28. (a) a = 2 2 d x dt = – k m xor a  x 29. (d) A particle returns back to its original position in one time period. 30. (c)The total distance moved by particle in one time period is four times the amplitude. 31. (b)Position where we see the particle once in a time period that is only extreme position. twice through every other potision 32. (d) In one time period average displacement is zero hence average velocity zero hence average acceleration is zero. 33. (b)X = A + B sin t x – A = B sin t Hence, Amplitude = B = B 34. (c) dx dt = 100 cos10t = 50 or cost = 1 2 sin10t = 3 2   x = 10 3 = 17.32 cm. 35. (b) 36. (d) x = x0 sin2 t = 0 x (1– cos2 t) 2  So A = 0 x 2 Time period = 1 2  = 2 2   = / 37. (d) x = A sin t + B sin (t + /2) Ar = 2 2 A B 2ABcos 2  + + Ar = 2 2 A B+ 38. (d) y = 10 1 3 sin3 t cos3 t 2 2      +      = 10 sin (3t + 3  ) thus amplitude is 10 m or 1000 cm 39. (c)x = A cos t, y = A sin t or x 2 + y2 = A2 Thus the motion of the particle is on a circle. 40. (a) 41. (b) 42. (c) 43. (a) 44. (b) In S.H.M., velocity of particle at displacement y from mean position is  =  2 2 a – y 45. (c)Amplitude a = 2 2 4 4 + = 4 2
46. (a) (a) In SHM , velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme positions. Therefore, the time taken for the particle to go from 0 to A/2 will be less than the time taken the time taken to go it from A/2 to A. or T1 < T2 . From the equations of SHM we can show that T1 = T0 – A/2 = T/12 [ A/2 = A sin T1 or T1 = /6]] and T2 = TA/2 – A = T/6 . So that T1 + T2 = T0 – A = T/4 47. (d)T = 8 second., A = 1 cm, x = A sint. = 1sin 2 8  t. a = – 2x = – 2 2 8        sin 2 8        t cm/s2 At, t = 4 3 second. a = – 2 2 8        sin 3  = – 2 3 32  cm/s2 . 48. (a)Maximum speed of a particle executing SHM is umax = a = a(2n) n = umax 2 a Here, umax = 31.4 cm/s, a = 5cm Substituting, thegiven values, we have n = 31.4 2 3.14 5   = 1Hz 49. (c)Let displacement equation of particle executing SHM is y = a sin t As particle travels half of the amplitude from the equilibrium position, so y = a 2 Therefore, a 2 = a sin t or sin t = 1 2 = sin 6  or t = 6  or t = 6   or t = 2 6 T         2 as T     =     or t = T 12 Hence, the particles travels half of he amplitude from the equilibrium in T 12 sec. 50. (a) The displacement equation of particle executing SHM is x = a cos (t + ) ... (i) velocity, v = dx dt = – a isn (t + ) ... (ii) and acceleration, A = d dt  = – a2 cos (t + ) ... (iii) Fig. (i) is a plot of Eq. (i) with  = 0. Fig. (ii) shows Eq. (ii) also with  = 0. Fig. (iii) is a plot of Eq. (iii). It should be noted that in the figures the curve of  is shifted ( to the left) from the cvrve of of x by one- quarter period 1 T 4       . Similarly, the acceleration curve of A is shifted (to the left) by 1 4 T relative to the velocity is 90° (0.5) out of phase with the displacement and the acceleration is 90° (0.5) out of phase with the velocity but 180° () outo of phase with displacement. 51. (b)Acceleration of simple harmonic motion is amax = – 2A
or 2 max 1 1 2 max 2 2 (a ) (a )  =  (as A remains same) or 2 2 max 1 2 max 2 (a ) (100) 1 (a ) 10 (1000)   = =     = 1 : 102 52. (d) Velocity is the time derivative of displacement. Writing the given equation of a point performing SHM x = a sin t 6     +     ......(i) Differentiating Eq. (i), w.r.t. time, we obtain v = dx a cos t dt 6    =   +     It is given that v = a 2  , so that a a cos t 2 6     =   +     or 1 cos t 2 6    =  +     or cos cos t 3 6     =  +     or t 6 3    + =  t 6   = or t = T T 6 6 2 12   = =    Thus, at T 12 velocity of the point will be equal to half of its maximum velocity. 53. (b) 2 2 d x dt = x ........ (i) We know a = 2 2 d x dt = – 2x ........ (ii) From Eq. (i) and (ii), we have 2 =    = 2 T   or  T = 2 a  54. (b) A = vmax T = 2  = max 2 A v  = 0.01 sec. 55. (c) A2 = g  A = g/2 . 56. (a)|v| = (2 × 10–2 ) ()sint For |v| to be maximum sint = 1 t = 2  , 3 2  ,........... t = 1 2 s. 57. (d) x = x0 cos t 4     −     , v = x0 sin t 4     −     a = x02 cos t 4     −     , a = x02 cos t 4     − +      a = x02 3 cos t 4     +     . 58. (d) 59. (a) Maximum acceleration = a2 = a x 4 2n 2 = a2 = a x 4 2n 2 = 0.01 x 4 () 2 x (60)2 = 144 2m/sec2 60. (a)Maximum acceleration Amax = a2 = 2 2 2 a x 4 1 x 4 x (3.14) T 0.2 x 0.2  = Fmax = m x Amax = 2 0.1 x 4 x (3.14) 0.2 x 0.2 = 98.596 N 61. (d) Acceleration  – displacement, and direction of acceleration is always directed towards the equilibrium position. 62. (d) amax = 2a = 2 4        a = 0.62 cm / sec2 [ a = 1] 63. (a)  2 = 100  w = 102 V =  2 2 A – x = 10 2 2 (1.5) – (0.5) = 10 2 64. (d)v = dy dt = A cos t = A 2 1– sin t 