Physics Beyond 2000

1. Physics Beyond 2000 Chapter 5 Simple Harmonic Motion http://library.thinkquest.org/28388/Mechanics/Motions/SHM.htm

2. Simple Harmonic Motion • It is a particular kind of oscillation. • Abbreviation is SHM. • Some terms Amplitude, period, frequency and angular frequency. http://www.physics.uoguelph.ca/tutorials/shm/Q.shm.html

3. Definition of SHM • The motion of the particle whose acceleration a is always directed towards a fixed point and is directly proportional to the distance x of the particle form that point. where ωis a constant, the angular frequency.

4. Examples of SHM The particle is at the position x = 0. It has a velocity to the right. V 0 x = 0

5. Examples of SHM The particle moves to the right with retardation a. Note that x and a are in opposite directions. a 0 x x =A A A is the maximum distance, the amplitude.

6. Examples of SHM The particle moves back to the left with acceleration a. Note that x and a are still in opposite directions. a 0 x x = 0 A A is the maximum distance, the amplitude.

7. Examples of SHM The particle moves to the left with retardation a. Note that both x and a change directions. They are still in opposite directions. a 0 x x = -A A A is the maximum distance, the amplitude.

8. Examples of SHM The particle moves to the right with acceleration a. Note that x and a are still in opposite directions. a 0 x x = 0 A A is the maximum distance, the amplitude.

9. V 0 x = 0 Examples of SHM The position x = 0 is the equilibrium position.

10. a 0 x V Examples of SHM The position x = 0 is the equilibrium position at which the net force is zero. In the oscillation, The negative sign indicates the direction of a is opposite to that of x.

11. Example 1 • Is it a SHM? • a = -16.x

12. Differential equation of SHM • acceleration a of SHM The left hand sides show the accelerations in different mathematical forms.

13. The kinematics of SHM • Displacement x of a SHM. • A solution of the differential equation is x = A.sin(ωt + ψ) t is the time, ωis the angular frequency, A is the amplitude and ψis the initial phase.

14. A t 0 2T 3T T -A The kinematics of SHM • Displacement x = A.sin(ωt + ψ) of a SHM. • The displacement x of a particle performing SHM changes sinusoidally with time t. • The period of the SHM is Displacement x time If ψ=0  x = A.sin (ωt)

15. Phase and Initial Phase • x = A.sin(ωt + ψ) ωt + ψis called the phase. • At t = 0, phase reduces to ψ. x = A.sin(ψ) ψis called the initial phase.

16. Phase • x = A.sin(ωt + ψ) • If ψ=0, x = A.sin(ωt ) • If ψ= π/2, x = A.cos(ωt ) • If ψ= π, x = -A.sin(ωt ) • etc.

17. Initial Phase ψ The value of ψis determined by the initial position of x (at t = 0). i.e. how the motion is started.

18. V a 0 x = 0 At t = 0 ψ= 0 • x = A.sin(ωt) • At t = 0, x = 0. • The motion starts at x = 0. • In the first T/4, x increases with time t and approaches the amplitude A.

19. A time t 0 2T 3T T -A ψ= 0 • x = A.sin(ωt) • At t = 0, x = 0. • The motion starts at x = 0. • In the first T/4, x increases with time t and approaches the amplitude A. Displacement x

20. a v = 0 0 x = A At t = 0 ψ= π/2 • x = A.cos(ωt) • At t = 0, x = A. • The motion starts at x = A. • In the first T/4, x decreases with time t and approaches 0.

21. Displacement x A time t 0 2T 3T T -A ψ= π/2 • x = A.cos(ωt) • At t = 0, x = A. • The motion starts at x = A. • In the first T/4, x decreases with time t and approaches 0.

22. V a 0 x = 0 At t = 0 ψ= π • x = -A.sin(ωt) • At t = 0, x = 0. • The motion starts at x = 0. • In the first T/4, x decreases with time t and approaches -A.

23. Displacement x A time t 0 2T 3T T -A ψ= π • x = -A.sin(ωt) • At t = 0, x = 0. • The motion starts at x = 0. • In the first T/4, x decreases with time t and approaches -A.

24. v=0 a 0 x = -A At t = 0 ψ= 2π/3 • x = -A.cos(ωt) • At t = 0, x = -A. • The motion starts at x = -A. • In the first T/4, x increases with time t and approaches 0.

25. Displacement x A time t 0 2T 3T T -A ψ= 2π/3 • x = -A.cos(ωt) • At t = 0, x = -A. • The motion starts at x = -A. • In the first T/4, x increases with time t and approaches 0.

26. Angular frequency ω • Period T = time for one complete oscillation. • Frequency f = number of oscillations in one second. • Angular frequency = 2f

27. Angular frequency ω • In a SHM, x = A.sin(ωt+ψ) • After a time T, x must be the same again. • A.sin(ωt+ψ) = A.sin(ωt+ ωT +ψ) • ωT = 2π Unit of ω is rad s-1

28. Isochronous oscillation • The period T in a SHM is independent of the amplitude A. • A SHM is an isochronous oscillation.

29. Velocity in SHM • x = A.sin(ωt+ψ)

30. Velocity in SHM • x = A.sin(ωt+ψ) What is the maximum speed in this motion? vo = Aω The maximum speed occurs at the equilibrium position. i.e. when x = 0.

31. Velocity in SHM • Example 2.

32. Acceleration in SHM

33. Acceleration in SHM What is the maximum acceleration in this motion? ao = -Aω2 or Aω2 The maximum acceleration occurs at the positions with maximum displacement. i.e. when x = A or -A.

34. Displacement, Velocity and Acceleration in SHM • x = A.sin(ωt + ψ) • v = Aω.cos(ωt + ψ) • a = -Aω2.sin(ωt + ψ)

35. Aω t 0 A 2T 3T T -Aω t 0 2T 3T Aω2 T -A t 0 a = -Aω2.sin(ωt) 2T 3T T -Aω2 Displacement, Velocity and Acceleration in SHM with ψ=0 x x = A.sin(ωt) v v = Aω.cos(ωt) a

36. Acceleration and Displacementin SHM • x = A.sin(ωt + ψ) • a = -Aω2.sin(ωt + ψ)  • a = - ω2.x • This is a characteristic of a SHM.

37. x x = A.sin(ωt) A t 0 2T 3T Aω2 T -A t 0 a = -Aω2.sin(ωt) 2T 3T T -Aω2 Acceleration and Displacementin SHM • a = - ω2.x • acceleration and displacement in SHM are in antiphase.(i.e. in opposite directions.) a

38. x Aω t x = A.sin(ωt) 0 A 2T 3T T -Aω t 0 2T 3T T -A Velocity and Displacement in SHM • v leads x by π/2 or 900 (or x lags behind x by π/2 or 900 ). v v = Aω.cos(ωt)

39. Velocity and Displacement in SHM • v leads x by π/2 or 900.

40. Aω t 0 2T 3T T -Aω Aω2 t 0 a = -Aω2.sin(ωt) 2T 3T T -Aω2 Velocity and Acceleration in SHM • a leads v by π/2 or 900 (or v lags behind a by π/2 or 900 ). a v v = Aω.cos(ωt)

41. The kinematics of SHM • We may look upon SHM as a projection of a uniform circular motion on its diameter. • A green ball is performing a uniform circular motion with angular velocity . ω

42. The kinematics of SHM • A green ball is performing a uniform circular motion with angular velocity . • Its projection on the diameter is the red ball performing SHM. ω http://id.mind.net/~zona/mstm/physics/mechanics/simpleHarmonicMotion/vrmlshm.html

43. The kinematics of SHM • A green ball is performing a uniform circular motion with angular velocity . • Its projection on the diameter is the red ball performing SHM. ω

44. The kinematics of SHM • A green ball is performing a uniform circular motion with angular velocity . • Its projection on the diameter is the red ball performing SHM. ω

45. The kinematics of SHM • A green ball is performing a uniform circular motion with angular velocity . • Its projection on the diameter is the red ball performing SHM. ω

46. The kinematics of SHM • A green ball is performing a uniform circular motion with angular velocity . • Its projection on the diameter is the red ball performing SHM. ω http://www.phy.ntnu.edu.tw/java/shm/shm.html

47. Period • The period of the circular motion is the same as the period of the SHM.  Period T = time to move to and fro once. Period T = time to complete one revolution

48. A  O Displacement of SHM ω At time = 0, the position of the green ball and the red ball are as shown. Let A = radius of the circle It is also the amplitude of the SHM Let  be the starting angle of the green ball. It is the initial phase of the red ball.

49. Displacement of SHM After time = t, the green travels and angular displacement t and the red ball moves a displacement x as shown. ω A  t O The displacement of SHM is x = A.sin(t + ) t +  is the phase of the SHM. x

50. Velocity of SHM The linear speed of the green ball is A. The velocity of the red ball is the horizontal component of A . ω A +t O A  v v = A .cos(t + ) Let A  = vo v = vo.cos(t + )