2. What are the similarities and differences between graphs? Both show oscillatory or cyclic motion Displacement Mass on spring

3. Oscillatory/Periodic Motion Repetitive Motion with a period and frequency. Motion can be driven internally - mass on spring. Externally – pendulum or tides.

4. Guitar string EEG Beta Waves

5. SHM – special case What do you notice?

6. unit is Hertz: Period T: time required for one cycle of periodic motion (sec). Frequency: number of oscillations per unit time. Vocabulary

7. What are the period and frequency of the ECG T = 1 sec. f = 1 hz,

8. Angular Frequency -  (rad/sec)  = 2  f. remember angular speed from circular motion?  =  /t  = 2  rad/s Equilibrium (0): the spot the mass would come to rest when not disturbed – F net = 0. Displacement: (s or x): distance from equilibrium. Amplitude (x o ) – max displacement from eq.

9. Representation of Oscillatory Motion Observe the motion of a bobbing mass. Where is the: Displacement positive? Displacement negative? Displacement zero? velocity positive? Where is the velocity negative? Where is the velocity zero?

10. Simple Harmonic Motion – SHM Isochronous period. Restoring Force directly proportional to displacement. Double displacement, double force etc. Further from equilibrium, more force directed toward it. Force and displacement in opposite directions.

11. Pendulum is not SHM F net not directly opposite s for small displacement angles it approximates SHM.

12. Free Body Diagram Mass on Spring Remember Hooke’s Law? F = -kx. F is the restoring force of the spring. Complete sheet.

13. Hwk Intro

14. Simple Harmonic Motion 2 Conditions. 1. Acceleration/F net proportional to displacement. 2. Acceleration/F net directed toward equilibrium. Defining Equation for SHM a = -  2 x

15. Graphs Of SHM

16. Acceleration - displacement Since F = - kx. ma = - kx. a = -k x m a  – x Graph a (Y axis) vs. displacement on the X

17. Negative gradient = when displacement is positive, acceleration and restoring force are negative ∴ a ∝ − x experimental evidence shows k = ω 2, where ω is the angular velocity, which is be ω = 2Πf : a = − ω 2 x a = − kx

18. Sketching and interpreting graphs of SHM EXAMPLE: The displacement x vs. time t for a system undergoing SHM is shown here. Sketch in red the velocity vs. time graph. SOLUTION: At the extremes, v = 0.  At x = 0, v = v MAX. The slope determines sign of v MAX. x-black v-red t (+) ( -)(+)( -) (+) x v = 0 v = v MAX -2.00.02.0

19. Sketching and interpreting graphs of simple harmonic motion EXAMPLE: The displacement x vs. time t for a system undergoing SHM is shown here. Sketch in blue the acceleration vs. time graph. SOLUTION: Since a  -x, a is just a reflection of x.  Note: x is a sine, v is a cosine, and a is a – sine wave. x-black v-red (different scale) t x v = 0 v = v MAX -2.00.02.0 a-blue (different scale)

20. Plot of displacement, velocity, acceleration, on same graph.

21. Phase and Phase Difference Two waves in Phase

22. Phase and Phase Difference Phase shift = 

23. 4. What is the phase shift?

24. The graphs all have the same period, but velocity leads displacement by ¼ of a period 90 o. Acceleration leads velocity by ¼ of a period 90 o. Phase Difference Displacement is 90 o out of phase with velocity and 180 o out of phase with acceleration. When the phase difference is 0 o or 360 o, the systems are “in phase”.

25. EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (a) Determine the maximum velocity of the mass. SOLUTION:  When the kinetic energy is maximum, the velocity is also maximum. Thus 4.0 = (1/ 2)mv MAX 2 so that 4.0 = (1/ 2)(.125) v MAX 2  v MAX = 8.0 ms -1. x -2.00.02.0

26. EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (b) Sketch E P and determine the total energy of the system. SOLUTION:  Since E K + E P = E T = CONST, and since E P = 0 when E K = E K,MAX, it must be that E T = E K,MAX = 4.0 J.  Thus the E P graph will be the “inverted” E K graph. x -2.00.02.0 ETET EPEP EKEK

27. EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (c) Determine the spring constant k of the spring. SOLUTION: Recall E P = (1/2)kx 2.  Note that E K = 0 at x = x MAX = 2.0 cm. Thus E K + E P = E T = CONST  E T = 0 + (1/ 2)kx MAX 2 so that 4.0 = (1/ 2)k 0.020 2  k = 20000 Nm -1. x -2.00.02.0

28. EXAMPLE: The kinetic energy vs. displacement for a system undergoing SHM is shown in the graph. The system consists of a 0.125-kg mass on a spring. (d) Determine the acceleration of the mass at x = 1.0 cm. SOLUTION:  From Hooke’s law, F = -kx we get F = -20000(0.01) = -200 N.  From F = ma we get -200 = 0.125a  a = -1600 ms -2. x -2.00.02.0

29. EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (a) How do you know that the mass is undergoing SHM? SOLUTION:  In SHM, a  -x. Since F = ma, then F  -x also.  The graph shows that F  -x. Thus we have SHM. x -2.00.02.0

30. EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (b) Find the spring constant of the spring. SOLUTION: Use Hooke’s law: F = -kx.  Pick any F and any x. Use k = -F / x.  Thus k = -(-5.0 N) / 1.0 m = 5.0 Nm -1. x -2.00.02.0 F = -5.0 N x = 1.0 m

31. EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (c) Find the total energy of the system. SOLUTION: Use E T = (1/2)kx MAX 2. Then E T = (1/2)kx MAX 2 = (1/2)  5.0  2.0 2 = 10. J. x -2.00.02.0

32. EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (d) Find the maximum speed of the mass. SOLUTION: Use E T = (1/2)mv MAX 2. 10. = (1/2)  4.0  v MAX 2 v MAX = 2.2 ms -1. x -2.00.02.0

33. EXAMPLE: A 4.0-kg mass is placed on a spring’s end and displaced 2.0 m to the right. The spring force F vs. its displacement x from equilibrium is shown in the graph. (e) Find the speed of the mass when its displacement is 1.0 m. SOLUTION: Use E T = (1/2)mv 2 + (1/2)kx 2. Then 10. = (1/2)(4)v 2 + (1/2)(5)1 2 v = 1.9 ms -1. x -2.00.02.0

35. Graphical Treatment Equations of SHM

36. Displacement, x, against time x = x o cos  t start point at max ampl. ** Set Calculator in Radians.

37. Displacement against time x = x o sin  t Speed = d/t = 2  R/T but 2  so v =  r but r the displacement x so. v =  x.

38. Velocity against time v = v o cos  t Starting where? Midpoint = max velocity.

39. Equations of Graphs x = x o cos  t x = x o sin  t v = -v o sin  t v = v o cos  t a = -a o cos  t -a o sin  t Released fromReleased equilibrium. top.

40. x = x o cos  t x o = 3 cm  = 2  f. = 0.4  Hz =1.26 rad/s. t = 10.66 s x = 0.03 cos (  x 10.66) = 0.019 m You must use radians on calculator. Ex 2. A mass on a spring is oscillating with f = 0.2 Hz and x o = 3 cm. What is the displacement of the mass 10.66 s after its release from the top?

41. SHM and Circular Motion Use the relationship to derive equations.

42. If an object moving with constant speed in a circular path is observed from a distant point (in the plane of the motion), it will appear to be oscillating with SHM. The shadow of a pendulum bob moves with s.h.m. when the pendulum itself is either oscillating (through a small angle) or moving in a circle with constant speed, as shown in the diagram.

43. For any s.h.m. we can find a corresponding circular motion. When a circular motion "corresponds to" a given s.h.m., i) the radius of the circle is equal to the amplitude of the s.h.m. ii) the time period of the circular motion is equal to the time period of the s.h.m.

44. From circular motion a c = v 2 /r and v c = 2  r/T Oscillating systems have acceleration too. But  = 2  /T v c = 2  r/T v c =  r But a c = v 2 /r So a c = (  r) 2 /r but r is related to displacement x. Derive Relationship between accl &  for SHM.

45. For any displacement: a = -  ²x a o = -  ²x o The negative sign shows F net & accl direction opposite displacement. Derivation of accl in Hamper pg 76. Or use 2 nd derivative of displacement.

46. Ex 3. A pendulum swings with f = 0.5 Hz. What is the size & direction of the acceleration when the bob has displacement of 2 cm right? a = -  ²x  = 2  f=  a = -(  ) 2 (0.02 m) = -0.197 m/s 2. left.

47. Ex 4: A mass is bobbing on a spring with a period of 0.20 seconds. What is its angular acceleration at a point where its displacement is 1.5 cm?  = 2  /T.  = 31 rad/s a = -  ²x a = (31rad/s)(1.5 cm) = 1480 cm/s 2. 15 m/s 2.

48. To find the velocity of an oscillating mass or pendulum at any displacement: When the mass is at equilibrium, x = 0, and velocity is maximum: v o = ±  x o. Derivation on H pg 77.

49. At max velocity v o =  x o.  f =  1) =  rad/s v o = (2  rad/s)(0.03) v o = 0.188 m/s Ex 5. A pendulum swings with f = 1 Hz and amplitude 3 cm. At what position will be its maximum velocity &what is the velocity? v o = 0.2 m/s

50. Hwk Hamper pg 75- 77 Show equations and work, hand in virtual solar system lab. http://www.learner.org/resources/series4 2.html?pop=yes&pid=565http://www.learner.org/resources/series4 2.html?pop=yes&pid=565 Mechanical Universe w/questions

51. SHM, Hooke’s Law & k.

52. For a mass undergoing SHM on a spring, what is the relation between angular frequency , and k the spring constant? Use Hooke’s law and make substitutions to derive a relation in terms of angular frequency, k, and mass.

53. ma = - k x. So a = -k x m F= - kx. a = -  ² x So  2 = -k/m

54. How can you tell this graph depicts SHM?  2 or k/m Acceleration directly  displacement, opposite direction. What is the slope? Where is the amplitude?

55. Energy in SHM

57. Pendulum is not SHM F net not directly opposite s for small displacement angles it approximates SHM.

58. Energy Conservation in Oscillatory Motion In an ideal system the total mechanical energy is conserved. A horizontal mass on a spring: Horizontal mass no PE g.

59. Determining the max KE & PE: What is PE for a stretched spring. PE = ½ k x 2 so PE max = PE = ½ k x o What is the equation for KE? KE = ½ mv 2 Determine an equation for KE max for SHM. v o =  x o, KE max = ½ m(  2 x o 2 ),

60. At any point: KE = ½ m  2 (x o 2 - x 2 ) How could you determine PE from E tot ? Subtract KE from E tot.

61. Since the total E will always equal the max KE (or PE), we can calculate the number of Joules of total E from the KE equation: E T = ½ m  2 x o 2

62. Ex 5: A 200-g pendulum bob is oscillating with Amplitude = 3 cm, and f = 0.5 Hz. How much KE will it have as it passes through the origin? KE max = ½m  2 x o 2, x o = 0.03 m  = . KE = 8.9 x 10 -4 J.

63. 6. The amplitude of an oscillating mass on a vertical spring is 8.0 cm. The spring constant is 74 N/m. Find the total energy of the oscillator and the PE and KE at a displacement = 4.8 cm. E t = 0.24 J. E k = E t – E p = 0.16 J.

64. Energy Conservation in Oscillatory Motion The total energy is constant; as the kinetic energy increases, the potential energy decreases, and vice versa.

65. Energy Conservation in Oscillatory Motion The E transforms from potential to kinetic & back, the total energy remains the same.

66. The Pendulum A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass). The angle it makes with the vertical varies with time as a sine or cosine.

67. The Pendulum The restoring force is actually proportional to sin θ, whereas the restoring force for a spring is proportional to the displacement x.

69. Units The Pendulum Damped Oscillations Driven Oscillations and Resonance

70. Mass on a Spring Since the force on a mass on a spring is proportional to the displacement, and also to the acceleration, Make substitutions to find the relationship between T and k.

71. Given that for a mass on a spring Derive an equation that relates the period of the oscillation to k, and m.

72. The Period of a Mass on a Spring Therefore, the period is How does T change as mass increases? Sketch it!

73. Mass on Spring

74. Period of pendulum

75. The Pendulum Substituting θ for sin θ allows us to treat the pendulum in a mathematically identical way to the mass on a spring. Therefore, we find that the period of a pendulum depends only on the length of the string:

76. The Pendulum For small angles, sin θ and θ are approximately equal.

77. Damped Oscillations In most physical situations, there is a force, which will tend to decrease the amplitude of the oscillation, and which is typically proportional to the speed: This causes the amplitude to decrease exponentially with time:

78. Damped Oscillations This exponential decrease is shown in the figure: The image shows a system that is lightly damped – it goes through multiple oscillations before coming to rest.

79. Damped Oscillations A critically damped system is one that relaxes back to the equilibrium position without oscillating through it. An overdamped system will also not oscillate but is damped so heavily that it takes longer to reach equilibrium.

80. Driven Oscillations and Resonance An oscillation can be driven by an oscillating driving force; the f of the driving force may or may not be the same as the natural f of the system.

81. Driven Oscillations and Resonance If the driving f is close to the natural frequency, the amplitude can become quite large, especially if the damping is small. This is called resonance.

82. Summary Period: time required for a motion to go through a complete cycle Frequency: number of oscillations per unit time Angular frequency: Simple harmonic motion occurs when the restoring force is proportional to the displacement from equilibrium.

83. Summary The amplitude is the maximum displacement from equilibrium. Position as a function of time: Velocity as a function of time:

84. Summary Acceleration as a function of time: Period of a mass on a spring: Total energy in simple harmonic motion:

85. Summary Potential energy as a function of time: Kinetic energy as a function of time: A simple pendulum with small amplitude exhibits simple harmonic motion

86. Summary Period of a simple pendulum: Period of a physical pendulum:

87. Summary An oscillating system may be driven by an external force. This force may replace energy lost to friction, or may cause the amplitude to increase greatly at resonance Resonance occurs when the driving frequency is equal to the natural frequency of the system

88. Summary Oscillations where there is a nonconservative force are called damped. Underdamped: the amplitude decreases exponentially with time: Critically damped: no oscillations; system relaxes back to equilibrium in minimum time Overdamped: also no oscillations, but slower than critical damping