12 X1 T07 02 Simple Harmonic Motion
- 2. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM)
- 3. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM)
- 4. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM)
- 5. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM)
- 6. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM) (constant needs to be negative)
- 7. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM) (constant needs to be negative)
- 8. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM) (constant needs to be negative)
- 9. Simple Harmonic Motion A particle that moves back and forward in such a way that its acceleration at any instant is directly proportional to its distance from a fixed point, is said to undergo Simple Harmonic Motion (SHM) (constant needs to be negative)
- 10. when x = a , v = 0 ( a = amplitude)
- 11. when x = a , v = 0 ( a = amplitude)
- 12. when x = a , v = 0 ( a = amplitude)
- 13. NOTE: when x = a , v = 0 ( a = amplitude)
- 14. NOTE: when x = a , v = 0 ( a = amplitude)
- 15. NOTE: when x = a , v = 0 ( a = amplitude)
- 16. NOTE: Particle travels back and forward between x = - a and x = a when x = a , v = 0 ( a = amplitude)
- 24. In general;
- 25. In general;
- 26. In general;
- 27. In general;
- 28. In general;
- 29. In general;
- 30. In general; the particle has;
- 31. In general; the particle has;
- 32. In general; the particle has; (time for one oscillation)
- 33. In general; the particle has; (time for one oscillation) (number of oscillations per time period)
- 34. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t .
- 35. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion.
- 36. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion.
- 37. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion.
- 38. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion.
- 39. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion.
- 40. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion.
- 41. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion. b) Find the interval in which the particle moves and determine the greatest speed.
- 42. e.g. ( i ) A particle, P, moves on the x axis according to the law x = 4sin3 t . a) Show that P is moving in SHM and state the period of motion. b) Find the interval in which the particle moves and determine the greatest speed.
- 43. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 44. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 45. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 46. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 47. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 48. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 49. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 50. ( ii ) A particle moves so that its acceleration is given by Initially the particle is 3cm to the right of O and traveling with a velocity of 6cm/s. Find the interval in which the particle moves and determine the greatest acceleration.
- 51. NOTE:
- 52. NOTE: At amplitude; v = 0 a is maximum
- 53. NOTE: At amplitude; v = 0 a is maximum At centre; v is maximum a = 0
- 54. Exercise 3D; 1, 6, 7, 10, 12, 14ab, 15ab, 18, 19, 22, 24, 25 (start with trig, prove SHM or are told) Exercise 3F; 1, 4, 5b, 6b, 8, 9a, 10a, 13, 14 a, b( ii,iv ), 16, 18, 19 (start with ) NOTE: At amplitude; v = 0 a is maximum At centre; v is maximum a = 0