1. Oscillations Simple Harmonic Motion

2. Simple Harmonic Motion
Simple harmonic motion is the repetitive motion of an oscillating object (simple harmonic oscillator).
Example: A spring mass system.
Like circular motion, simple harmonic motion has a regular period and frequency.
In fact, simple harmonic motion and circular motion are linked both mathematically and graphically.

3. Equilibrium
The spring mass system shown below is at equilibrium ( x = 0 ).
Equilibrium for a spring is its natural rest position.
This spring can either be stretched or compressed, displacing the mass from its current equilibrium position.
To stretch the spring a force, F , must be applied in the +x direction.
This new unbalanced force will displace the mass from equilibrium. +F

4. A force is applied
As the mass is displaced the spring generates an increasing force of it own. +F

5. A force is applied +F −FS
As the mass is displaced the spring generates an increasing force of it own. +F
−FS

6. A force is applied −FS +F
As the mass is displaced the spring generates an increasing force of it own. +F
−FS

7. A force is applied −FS +F
As the mass is displaced the spring generates an increasing force of it own. +F
−FS

8. A force is applied +F −FS
As the mass is displaced the spring generates an increasing force of it own.
This force is known as a restoring force, as it acts to restore the spring to its original equilibrium position. +F
−FS

9. Amplitude
When the applied force, F , and the restoring force, FS , are equal and opposite the oscillator is stationary.
The oscillator has reached maximum displacement, xmax .
Maximum displacement is known as the amplitude, A .
This is equal to the radius, R , of a circle matching the springs displacement.
+A = xmax +F
−FS

10. The Oscillator is Released
When the applied force, F, is removed, the restoring force moves the mass left. x +A
−FS +F

11. The Oscillator is Released
When the applied force, F, is removed, the restoring force moves the mass left. x +A
−FS

12. The Oscillator is Released
When the applied force, F, is removed, the restoring force moves the mass left. x
−FS +A

13. The Oscillator is Released
When the applied force, F, is removed, the restoring force moves the mass left. x
−FS +A

14. The Oscillator is Released
When the applied force, F, is removed, the restoring force moves the mass left. x
−FS +A

15. The Oscillator is Released
When the applied force, F, is removed, the restoring force moves the mass left.
The radius of the circle is equal to the maximum displacement (amplitude)
R = A = xmax
The radius, R = A , and displacement, x , are two of the sides of a right triangle.
While the mass moved to the left the corresponding point on the circle moved with an angular velocity, ω , through an angle, Δθ .
R = A = xmax Δθ ω x +A
−FS

16. Conclusion ω R = A = xmax Δθ x −FS
An oscillator’s displacement equals the x-component of the radius of another object that is in uniform circular motion with the same amplitude (R = A = xmax).
Cosine solves for x .
In uniform circular motion
Therefore, +A
−FS x
R = A = xmax Δθ ω

17. Conclusion ω R = A = xmax Δθ x −FS
Period and frequency are related to angular velocity.
Which rearrange to
These can be substituted into the previous equation +A
−FS x
R = A = xmax Δθ ω

18. Putting all together FS x(t)
The motion described by this equation looks like…
x(t) −A +A FS

19. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… ω Δθ
x(t) −A +A FS
x(t)

20. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… ω Δθ
x(t) −A +A FS
x(t)

21. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

22. Putting all together ω Δθ x(t) x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A
x(t)

23. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

24. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

25. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

26. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

27. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

28. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

29. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

30. Putting all together ω Δθ x(t) x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A
x(t)

31. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

32. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

33. Putting all together ω Δθ x(t) FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS
x(t)

34. Putting all together ω Δθ FS x(t)
The motion described by this equation looks like… Δθ
x(t) −A +A FS

35. Graphing Oscillationsπ 2π
x(t) −A +A FS

36. Putting all together +A −A π 2π ω Δθ
x(t) −A +A FS
x(t)

37. Putting all together +A −A π 2π ω Δθ
x(t) −A +A
x(t)

38. Putting all together +A −A π 2π ω Δθ
x(t) −A +A FS
x(t)

39. Putting all together +A −A π 2π ω Δθ
x(t) −A +A FS
x(t)

40. Putting all together +A −A π 2π ω Δθ
x(t) −A +A FS
x(t)

41. Putting all together +A −A π 2π ω Δθ
x(t) −A +A
x(t)

42. Putting all together +A −A π 2π ω Δθ
x(t) −A +A FS
x(t)

43. Putting all together +A −A π 2π ω Δθ
x(t) −A +A FS

44. Analyzing Oscillation Graphs: Amplitude
Amplitude is the largest displacement and it can be easily read on the graph.

45. Analyzing Oscillation Graphs: Period
Period is the time of one cycle (1 revolution). One cycle is one complete wave form. This wave form extends from 3 seconds to 15 s. The time of this wave form is 15 − 3 = 12 s

46. Analyzing Oscillation Graphs: Frequency
Frequency is the inverse of period.

47. Analyzing Oscillation Graphs: Displacement
Displacement can be solved using the displacement equation and substituting the values determined in the previous slides.
Substitute the given time to determine the matching displacement.

48. Analyzing Oscillation Graphs: Maximum Speed
Maximum speed can be found using the equation for speed in uniform circular motion. Set the radius equal to amplitude.

49. Analyzing Oscillation Graphs: Qualitatively
When is the object moving to the right at maximum speed?
This occurs at equilibrium (x = 0) and when object is moving toward a greater displacement value: 9 and 21 seconds
When is the object moving to the left at maximum speed?
This occurs at equilibrium (x = 0) and when object is moving toward a lesser displacement value: 3 and 15 seconds
When is the object instantaneously at rest?
This occurs at maximum displacement: 0, 6, 12, 18, and 24 seconds.

50. Circular Motion R ω y(t) Δθ x(t)
Oscillation is only concerned with the component of the circular motion equation that is parallel to the oscillation.
If a spring is oscillating horizontally
If a spring is oscillating vertically
x(t)
y(t) Δθ ω R
Circular motion involves both equations simultaneously.
Amplitude is equal radius R , and Δθ = ω t .