1. Periodic Motion - 1

2. Oscillatory/Periodic Motion Repetitive Motion Simple Harmonic Motion SHM special case of oscillatory.
Object w equilibrium position has restoring force pushing it back to equilibrium.

3. Simple Harmonic Motion
A spring exerts a restoring force directily proportional to the displacement fr. eq:
Remember Hooke’s Law
Negative sign Fnet opposite to s.

4. Pendulum is not SHM
Fnet not directly opposite s
for small displacement angles it approximates SHM.

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

6. Equilibrium (0): the spot the mass would come to rest when not disturbed – Fnet = 0.
Displacement: (x) distance from equilibrium.
Amplitude (xo) – max displacement from eq.
Angular frequency w, multiply f by 2p. Reported as cy/sec or rad/sec. 2p = 1 full cycle.

7. Angular frequency w is a measure that comes from circular motion.
It is measured in radians per second.
f, Hz is cycles per second.

8. An object makes 1 revolution per second
An object makes 1 revolution per second. How many radians does it complete in 1 second?
2p rad s-1 = 1 revolution per second.

9. 2p rad s-1 = also describes 1 non-circular cycle or oscillation.

10. To find w, from frequency f, or period T, use: w = 2pf or w = 2p T

11. Ex 1. A pendulum completes 4 swings in 0. 5 seconds
Ex 1. A pendulum completes 4 swings in 0.5 seconds. What is its angular frequency?
T = s
w = 2p/T = 50 rad/sec.

12. Free Body Diagram Mass on Spring
Complete sheet.

14. Simple Harmonic Motion 2 Conditions.
1. Acceleration/Fnet proportional to displacement.
2. Acceleration/Fnet directed toward equilibrium.
Defining Equation for SHM
a = -w2x

15. Sketch Graphs.

16. Simple Harmonic Motion On a bobbing mass:
No friction
Weight stays constant.
Tension increases with stretch.
Fnet / accl toward equilibrium position.
Fnet increases with is proportional to displacement.

17. Graphical Treatment
Equations of SHM

18. Displacement, x, against time
x = xo cos wt
start point at max ampl.
** Set Calculator in Radians.

19. Displacement against time
x = xosinwt
Speed = d/t = 2pR/T but
2p /T = w so
v = wr but r the displacement x so.
v = wx.

20. Velocity against time v = vocoswt
Starting where?
Midpoint = max velocity.

21. Equations of Graphs x = xo cos wt x = xo sin wt
v = -vo sin wt v = vo cos wt
a = -aocos wt -ao sin wt
Released from Released equilibrium.
top.

22. Ex 2. A mass on a spring is oscillating with f = 0. 2 Hz and xo = 3 cm
Ex 2. A mass on a spring is oscillating with f = 0.2 Hz and xo = 3 cm. What is the displacement of the mass s after its release from the top?
x = xo cos wt xo = 3 cm
w = 2pf. = 0.4 p Hz =1.26 rad/s.
t = s
x = 0.03 cos (1.26 x 10.66) = m
You must use radians on calculator.

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

24. 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.

25. For any s. h. m. we can find a corresponding circular motion
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.

26. Derive Relationship between accl & w for SHM.
From circular motion ac = v2/r and vc = 2pr/T Oscillating systems have acceleration too.
But w = 2p/T
vc = 2pr/T
vc = wr
But ac = v2/r
So ac = (wr)2/r
but r is related to displacement x.

27. The negative sign shows Fnet & accl direction opposite displacement.
For any displacement:
a = -w²x
ao = -w²xo
The negative sign shows Fnet & accl direction opposite displacement.
Derivation of accl in Hamper pg 76.
Or use 2nd derivative of displacement.

28. a = -w²x w = 2pf = p a = -(p)2 (0.02 m) = -0.197 m/s2. left.
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 = -w²x
w = 2pf = p
a = -(p)2 (0.02 m) = m/s2. left.

29. Ex 4: A mass is bobbing on a spring with a period of 0. 20 seconds
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?
w = 2p/T
.w = 31 rad/s
a = -w²x
a = (31rad/s)(1.5 cm) = cm/s2.
15 m/s2.

30. SHM, Hooke’s Law & k.

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

32. F= - kx.
ma = - k x.
So a = -k x m
a = -w² x
So w2 = -k/m

33. When the mass is at equilibrium, x = 0, and velocity is maximum:
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:
vo = ± wxo.
Derivation on H pg 77.

34. Ex 4. A pendulum swings with f = 1 Hz and amplitude 3 cm
Ex 4. A pendulum swings with f = 1 Hz and amplitude 3 cm. At what position will be its maximum velocity &what is the velocity?
At max velocity vo = wxo.
w = 2pf = 2p(1) = 2p rad/s
vo = (2p rad/s)(0.03)
vo = m/s
vo = 0.2 m/s

35. Mechanical Universe w/questions
Hwk Hamper pg Show equations and work, hand in virtual solar system lab.
Mechanical Universe w/questions

36. Units of Chapter 4 The Pendulum Damped Oscillations
Driven Oscillations and Resonance

37. The Period of a Mass on a Spring
Since the force on a mass on a spring is proportional to the displacement, and also to the acceleration, we find that .
Make substitutions to find the relationship between T and k.

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

40. Period of pendulum

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

42. Determining the max KE & PE:
PE = ½ k x2 for a stretched spring.
So PEmax = PE = ½ k xo
KE = ½ mv2
vmax = wxo,
KEmax = ½ m(w2xo 2),

43. At any point: KE = ½ mw2 (xo2 - x2 )
How could you determine PE from Etot?
Subtract KE from Etot.

44. 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:
ET = ½ mw2xo2

45. 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?
KEmax = ½ w2xo 2,
xo = 0.03 m
w = p.
KE = 8.9 x 10-4 J.

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

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

48. 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.

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

50. The Pendulum
However, for small angles, sin θ and θ are approximately equal.

51. 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:

53. Damped Oscillations
In most physical situations, there is a nonconservative force of some sort, 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:

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

55. Damped Oscillations
A critically damped system is one that relaxes back to the equilibrium position without oscillating and in minimum time;
an overdamped system will also not oscillate but is damped so heavily that it takes longer to reach equilibrium.

56. 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.

57. 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.

58. 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.

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

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

61. 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

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

63. 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

64. 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