  Dan Fullerton

Oscillations

Slide Duration:

Section 1: Introduction
What is Physics?

7m 12s

Intro
0:00
Objectives
0:11
What is Physics?
0:27
Why?
0:50
0:51
Matter
1:27
Matter
1:28
Mass
1:43
Inertial Mass
1:50
Gravitational Mass
2:13
A Spacecraft's Mass
3:03
What is the Mass of the Spacecraft?
3:05
Energy
3:37
Energy
3:38
Work
3:45
Putting Energy and Work Together
3:50
Mass-Energy Equivalence
4:15
Relationship between Mass & Energy: E = mc²
4:16
Source of Energy on Earth
4:47
The Study of Everything
5:00
Physics is the Study of Everything
5:01
Mechanics
5:29
Topics Covered
5:30
Topics Not Covered
6:07
Next Steps
6:44
Three Things You'd Like to Learn About in Physics
6:45
Math Review

1h 51s

Intro
0:00
Objectives
0:10
Vectors and Scalars
1:06
Scalars
1:07
Vectors
1:27
Vector Representations
2:00
Vector Representations
2:01
2:54
2:55
Graphical Vector Subtraction
5:36
Graphical Vector Subtraction
5:37
Vector Components
7:12
Vector Components
7:13
Angle of a Vector
8:56
tan θ
9:04
sin θ
9:25
cos θ
9:46
Vector Notation
10:10
Vector Notation 1
10:11
Vector Notation 2
12:59
Example I: Magnitude of the Horizontal & Vertical Component
16:08
Example II: Magnitude of the Plane's Eastward Velocity
17:59
Example III: Magnitude of Displacement
19:33
Example IV: Total Displacement from Starting Position
21:51
Example V: Find the Angle Theta Depicted by the Diagram
26:35
Vector Notation, cont.
27:07
Unit Vector Notation
27:08
Vector Component Notation
27:25
Vector Multiplication
28:39
Dot Product
28:40
Cross Product
28:54
Dot Product
29:03
Dot Product
29:04
Defining the Dot Product
29:26
Defining the Dot Product
29:27
Calculating the Dot Product
29:42
Unit Vector Notation
29:43
Vector Component Notation
30:58
Example VI: Calculating a Dot Product
31:45
Example VI: Part 1 - Find the Dot Product of the Following Vectors
31:46
Example VI: Part 2 - What is the Angle Between A and B?
32:20
Special Dot Products
33:52
Dot Product of Perpendicular Vectors
33:53
Dot Product of Parallel Vectors
34:03
Dot Product Properties
34:51
Commutative
34:52
Associative
35:05
Derivative of A * B
35:24
Example VII: Perpendicular Vectors
35:47
Cross Product
36:42
Cross Product of Two Vectors
36:43
Direction Using the Right-hand Rule
37:32
Cross Product of Parallel Vectors
38:04
Defining the Cross Product
38:13
Defining the Cross Product
38:14
Calculating the Cross Product Unit Vector Notation
38:41
Calculating the Cross Product Unit Vector Notation
38:42
Calculating the Cross Product Matrix Notation
39:18
Calculating the Cross Product Matrix Notation
39:19
Example VII: Find the Cross Product of the Following Vectors
42:09
Cross Product Properties
45:16
Cross Product Properties
45:17
Units
46:41
Fundamental Units
46:42
Derived units
47:13
Example IX: Dimensional Analysis
47:21
Calculus
49:05
Calculus
49:06
Differential Calculus
49:49
Differentiation & Derivative
49:50
Example X: Derivatives
51:21
Integral Calculus
53:03
Integration
53:04
Integral
53:11
Integration & Derivation are Inverse Functions
53:16
Determine the Original Function
53:37
Common Integrations
54:45
Common Integrations
54:46
Example XI: Integrals
55:17
Example XII: Calculus Applications
58:32
Section 2: Kinematics
Describing Motion I

23m 47s

Intro
0:00
Objectives
0:10
Position / Displacement
0:39
Object's Position
0:40
Position Vector
0:45
Displacement
0:56
Position & Displacement are Vectors
1:05
Position & Displacement in 1 Dimension
1:11
Example I: Distance & Displacement
1:21
Average Speed
2:14
Average Speed
2:15
Average Speed is Scalar
2:27
Average Velocity
2:39
Average Velocity
2:40
Average Velocity is a Vector
2:57
Example II: Speed vs. Velocity
3:16
Example II: Deer's Average Speed
3:17
Example II: Deer's Average Velocity
3:48
Example III: Chuck the Hungry Squirrel
4:21
Example III: Chuck's Distance Traveled
4:22
Example III: Chuck's Displacement
4:43
Example III: Chuck's Average Speed
5:25
Example III: Chuck's Average Velocity
5:39
Acceleration
6:11
Acceleration: Definition & Equation
6:12
Acceleration: Units
6:19
Relationship of Acceleration to Velocity
6:52
Example IV: Acceleration Problem
7:05
The Position Vector
7:39
The Position Vector
7:40
Average Velocity
9:35
Average Velocity
9:36
Instantaneous Velocity
11:20
Instantaneous Velocity
11:21
Instantaneous Velocity is the Derivative of Position with Respect to Time
11:35
Area Under the Velocity-time Graph
12:08
Acceleration
12:36
More on Acceleration
12:37
Average Acceleration
13:11
Velocity vs. Time Graph
13:14
Graph Transformations
13:59
Graphical Analysis of Motion
14:00
Velocity and acceleration in 2D
14:35
Velocity Vector in 2D
14:39
Acceleration Vector in 2D
15:26
Polynomial Derivatives
16:10
Polynomial Derivatives
16:11
Example V: Polynomial Kinematics
16:31
Example VI: Velocity Function
17:54
Example VI: Part A - Determine the Acceleration at t=1 Second
17:55
Example VI: Part B - Determine the Displacement between t=0 and t=5 Seconds
18:33
Example VII: Tortoise and Hare
20:14
Example VIII: d-t Graphs
22:40
Describing Motion II

36m 47s

Intro
0:00
Objectives
0:09
Special Case: Constant Acceleration
0:31
Constant Acceleration & Kinematic Equations
0:32
Deriving the Kinematic Equations
1:28
V = V₀ + at
1:39
∆x = V₀t +(1/2)at²
2:03
V² = V₀² +2a∆x
4:05
Problem Solving Steps
7:02
Step 1
7:13
Step 2
7:18
Step 3
7:27
Step 4
7:30
Step 5
7:31
Example IX: Horizontal Kinematics
7:38
Example X: Vertical Kinematics
9:45
Example XI: 2 Step Problem
11:23
Example XII: Acceleration Problem
15:01
Example XIII: Particle Diagrams
15:57
Example XIV: Particle Diagrams
17:36
18:46
Free Fall
22:56
Free Fall
22:57
Air Resistance
23:24
Air Resistance
23:25
Acceleration Due to Gravity
23:48
Acceleration Due to Gravity
23:49
Objects Falling From Rest
24:18
Objects Falling From Rest
24:19
Example XVI: Falling Objects
24:55
Objects Launched Upward
26:01
Objects Launched Upward
26:02
Example XVII: Ball Thrown Upward
27:16
Example XVIII: Height of a Jump
27:48
Example XIX: Ball Thrown Downward
31:10
Example XX: Maximum Height
32:27
Example XXI: Catch-Up Problem
33:53
Example XXII: Ranking Max Height
35:52
Projectile Motion

30m 34s

Intro
0:00
Objectives
0:07
What is a Projectile?
0:28
What is a Projectile?
0:29
Path of a Projectile
0:58
Path of a Projectile
0:59
Independence of Motion
2:45
Vertical & Horizontal Motion
2:46
Example I: Horizontal Launch
3:14
Example II: Parabolic Path
7:20
Angled Projectiles
8:01
Angled Projectiles
8:02
Example III: Human Cannonball
10:05
Example IV: Motion Graphs
14:39
Graphing Projectile Motion
19:05
Horizontal Equation
19:06
Vertical Equation
19:46
Example V: Arrow Fired from Tower
21:28
Example VI: Arrow Fired from Tower
24:10
Example VII: Launch from a Height
24:40
Example VIII: Acceleration of a Projectile
29:49
Circular & Relative Motion

30m 24s

Intro
0:00
Objectives
0:08
0:32
Degrees
0:35
0:40
1:08
Example I: Part A - Convert 90 Degrees to Radians
1:09
Example I: Part B - Convert 6 Radians to Degrees
2:08
Linear vs. Angular Displacement
2:38
Linear Displacement
2:39
Angular Displacement
2:52
Linear vs. Angular Velocity
3:18
Linear Velocity
3:19
Angular Velocity
3:25
Direction of Angular Velocity
4:36
Direction of Angular Velocity
4:37
Converting Linear to Angular Velocity
5:05
Converting Linear to Angular Velocity
5:06
Example II: Earth's Angular Velocity
6:12
Linear vs. Angular Acceleration
7:26
Linear Acceleration
7:27
Angular Acceleration
7:32
Centripetal Acceleration
8:05
Expressing Position Vector in Terms of Unit Vectors
8:06
Velocity
10:00
Centripetal Acceleration
11:14
Magnitude of Centripetal Acceleration
13:24
Example III: Angular Velocity & Centripetal Acceleration
14:02
Example IV: Moon's Orbit
15:03
Reference Frames
17:44
Reference Frames
17:45
Laws of Physics
18:00
Motion at Rest vs. Motion at a Constant Velocity
18:21
Motion is Relative
19:20
Reference Frame: Sitting in a Lawn Chair
19:21
Reference Frame: Sitting on a Train
19:56
Calculating Relative Velocities
20:19
Calculating Relative Velocities
20:20
Example: Calculating Relative Velocities
20:57
Example V: Man on a Train
23:19
Example VI: Airspeed
24:56
Example VII: 2-D Relative Motion
26:12
Example VIII: Relative Velocity w/ Direction
28:32
Section 3: Dynamics
Newton's First Law & Free Body Diagrams

23m 57s

Intro
0:00
Objectives
0:11
Newton's 1st Law of Motion
0:28
Newton's 1st Law of Motion
0:29
Force
1:16
Definition of Force
1:17
Units of Force
1:20
How Much is a Newton?
1:25
Contact Forces
1:47
Field Forces
2:32
What is a Net Force?
2:53
What is a Net Force?
2:54
What Does It Mean?
4:35
What Does It Mean?
4:36
Objects at Rest
4:52
Objects at Rest
4:53
Objects in Motion
5:12
Objects in Motion
5:13
Equilibrium
6:03
Static Equilibrium
6:04
Mechanical Equilibrium
6:22
Translational Equilibrium
6:38
Inertia
6:48
Inertia
6:49
Inertial Mass
6:58
Gravitational Mass
7:11
Example I: Inertia
7:40
Example II: Inertia
8:03
Example III: Translational Equilibrium
8:25
Example IV: Net Force
9:19
Free Body Diagrams
10:34
Free Body Diagrams Overview
10:35
Falling Elephant: Free Body Diagram
10:53
Free Body Diagram Neglecting Air Resistance
10:54
Free Body Diagram Including Air Resistance
11:22
Soda on Table
11:54
Free Body Diagram for a Glass of Soda Sitting on a Table
11:55
Free Body Diagram for Box on Ramp
13:38
Free Body Diagram for Box on Ramp
13:39
Pseudo- Free Body Diagram
15:26
Example V: Translational Equilibrium
18:35
Newton's Second & Third Laws of Motion

23m 57s

Intro
0:00
Objectives
0:09
Newton's 2nd Law of Motion
0:36
Newton's 2nd Law of Motion
0:37
Applying Newton's 2nd Law
1:12
Step 1
1:13
Step 2
1:18
Step 3
1:27
Step 4
1:36
Example I: Block on a Surface
1:42
Example II: Concurrent Forces
2:42
Mass vs. Weight
4:09
Mass
4:10
Weight
4:28
Example III: Mass vs. Weight
4:45
Example IV: Translational Equilibrium
6:43
Example V: Translational Equilibrium
8:23
Example VI: Determining Acceleration
10:13
Example VII: Stopping a Baseball
12:38
Example VIII: Steel Beams
14:11
Example IX: Tension Between Blocks
17:03
Example X: Banked Curves
18:57
Example XI: Tension in Cords
24:03
Example XII: Graphical Interpretation
27:13
Example XIII: Force from Velocity
28:12
Newton's 3rd Law
29:16
Newton's 3rd Law
29:17
Examples - Newton's 3rd Law
30:01
Examples - Newton's 3rd Law
30:02
Action-Reaction Pairs
30:40
Girl Kicking Soccer Ball
30:41
Rocket Ship in Space
31:02
Gravity on You
31:23
Example XIV: Force of Gravity
32:11
Example XV: Sailboat
32:38
Example XVI: Hammer and Nail
33:18
Example XVII: Net Force
33:47
Friction

20m 41s

Intro
0:00
Objectives
0:06
Coefficient of Friction
0:21
Coefficient of Friction
0:22
Approximate Coefficients of Friction
0:44
Kinetic or Static?
1:21
Sled Sliding Down a Snowy Hill
1:22
Refrigerator at Rest that You Want to Move
1:32
Car with Tires Rolling Freely
1:49
Car Skidding Across Pavement
2:01
Example I: Car Sliding
2:21
Example II: Block on Incline
3:04
Calculating the Force of Friction
3:33
Calculating the Force of Friction
3:34
Example III: Finding the Frictional Force
4:02
Example IV: Box on Wood Surface
5:34
Example V: Static vs. Kinetic Friction
7:35
Example VI: Drag Force on Airplane
7:58
Example VII: Pulling a Sled
8:41
Example VIII: AP-C 2007 FR1
13:23
Example VIII: Part A
13:24
Example VIII: Part B
14:40
Example VIII: Part C
15:19
Example VIII: Part D
17:08
Example VIII: Part E
18:24
Retarding & Drag Forces

32m 10s

Intro
0:00
Objectives
0:07
Retarding Forces
0:41
Retarding Forces
0:42
The Skydiver
1:30
Drag Forces on a Free-falling Object
1:31
Velocity as a Function of Time
5:31
Velocity as a Function of Time
5:32
Velocity as a Function of Time, cont.
12:27
Acceleration
12:28
Velocity as a Function of Time, cont.
15:16
Graph: Acceleration vs. Time
16:06
Graph: Velocity vs. Time
16:40
Graph: Displacement vs. Time
17:04
Example I: AP-C 2005 FR1
17:43
Example I: Part A
17:44
Example I: Part B
19:17
Example I: Part C
20:17
Example I: Part D
21:09
Example I: Part E
22:42
Example II: AP-C 2013 FR2
24:26
Example II: Part A
24:27
Example II: Part B
25:25
Example II: Part C
26:22
Example II: Part D
27:04
Example II: Part E
30:50
Ramps & Inclines

20m 31s

Intro
0:00
Objectives
0:06
Drawing Free Body Diagrams for Ramps
0:32
Step 1: Choose the Object & Draw It as a Dot or Box
0:33
Step 2: Draw and Label all the External Forces
0:39
Step 3: Sketch a Coordinate System
0:42
Example: Object on a Ramp
0:52
Pseudo-Free Body Diagrams
2:06
Pseudo-Free Body Diagrams
2:07
Redraw Diagram with All Forces Parallel to Axes
2:18
Box on a Ramp
4:08
Free Body Diagram for Box on a Ramp
4:09
Pseudo-Free Body Diagram for Box on a Ramp
4:54
Example I: Box at Rest
6:13
Example II: Box Held By Force
6:35
Example III: Truck on a Hill
8:46
Example IV: Force Up a Ramp
9:29
Example V: Acceleration Down a Ramp
12:01
Example VI: Able of Repose
13:59
Example VII: Sledding
17:03
Atwood Machines

24m 58s

Intro
0:00
Objectives
0:07
What is an Atwood Machine?
0:25
What is an Atwood Machine?
0:26
Properties of Atwood Machines
1:03
Ideal Pulleys are Frictionless and Massless
1:04
Tension is Constant
1:14
Setup for Atwood Machines
1:26
Setup for Atwood Machines
1:27
Solving Atwood Machine Problems
1:52
Solving Atwood Machine Problems
1:53
Alternate Solution
5:24
Analyze the System as a Whole
5:25
Example I: Basic Atwood Machine
7:31
Example II: Moving Masses
9:59
Example III: Masses and Pulley on a Table
13:32
Example IV: Mass and Pulley on a Ramp
15:47
Example V: Ranking Atwood Machines
19:50
Section 4: Work, Energy, & Power
Work

37m 34s

Intro
0:00
Objectives
0:07
What is Work?
0:36
What is Work?
0:37
Units of Work
1:09
Work in One Dimension
1:31
Work in One Dimension
1:32
Examples of Work
2:19
Stuntman in a Jet Pack
2:20
A Girl Struggles to Push Her Stalled Car
2:50
A Child in a Ghost Costume Carries a Bag of Halloween Candy Across the Yard
3:24
Example I: Moving a Refrigerator
4:03
Example II: Liberating a Car
4:53
Example III: Lifting Box
5:30
Example IV: Pulling a Wagon
6:13
Example V: Ranking Work on Carts
7:13
Non-Constant Forces
12:21
Non-Constant Forces
12:22
Force vs. Displacement Graphs
13:49
Force vs. Displacement Graphs
13:50
Hooke's Law
14:41
Hooke's Law
14:42
Determining the Spring Constant
15:38
Slope of the Graph Gives the Spring Constant, k
15:39
Work Done in Compressing the Spring
16:34
Find the Work Done in Compressing the String
16:35
Example VI: Finding Spring Constant
17:21
Example VII: Calculating Spring Constant
19:48
Example VIII: Hooke's Law
20:30
Example IX: Non-Linear Spring
22:18
Work in Multiple Dimensions
23:52
Work in Multiple Dimensions
23:53
Work-Energy Theorem
25:25
Work-Energy Theorem
25:26
Example X: Work-Energy Theorem
28:35
Example XI: Work Done on Moving Carts
30:46
Example XII: Velocity from an F-d Graph
35:01
Energy & Conservative Forces

28m 4s

Intro
0:00
Objectives
0:08
Energy Transformations
0:31
Energy Transformations
0:32
Work-Energy Theorem
0:57
Kinetic Energy
1:12
Kinetic Energy: Definition
1:13
Kinetic Energy: Equation
1:55
Example I: Frog-O-Cycle
2:07
Potential Energy
2:46
Types of Potential Energy
2:47
A Potential Energy Requires an Interaction between Objects
3:29
Internal energy
3:50
Internal Energy
3:51
Types of Energy
4:37
Types of Potential & Kinetic Energy
4:38
Gravitational Potential Energy
5:42
Gravitational Potential Energy
5:43
Example II: Potential Energy
7:27
Example III: Kinetic and Potential Energy
8:16
Example IV: Pendulum
9:09
Conservative Forces
11:37
Conservative Forces Overview
11:38
Type of Conservative Forces
12:42
Types of Non-conservative Forces
13:02
Work Done by Conservative Forces
13:28
Work Done by Conservative Forces
13:29
Newton's Law of Universal Gravitation
14:18
Gravitational Force of Attraction between Any Two Objects with Mass
14:19
Gravitational Potential Energy
15:27
Gravitational Potential Energy
15:28
Elastic Potential Energy
17:36
Elastic Potential Energy
17:37
Force from Potential Energy
18:51
Force from Potential Energy
18:52
Gravitational Force from the Gravitational Potential Energy
20:46
Gravitational Force from the Gravitational Potential Energy
20:47
Hooke's Law from Potential Energy
22:04
Hooke's Law from Potential Energy
22:05
Summary
23:16
Summary
23:17
Example V: Kinetic Energy of a Mass
24:40
Example VI: Force from Potential Energy
25:48
Example VII: Work on a Spinning Disc
26:54
Conservation of Energy

54m 56s

Intro
0:00
Objectives
0:09
Conservation of Mechanical Energy
0:32
Consider a Single Conservative Force Doing Work on a Closed System
0:33
Non-Conservative Forces
1:40
Non-Conservative Forces
1:41
Work Done by a Non-conservative Force
1:47
Formula: Total Energy
1:54
Formula: Total Mechanical Energy
2:04
Example I: Falling Mass
2:15
Example II: Law of Conservation of Energy
4:07
Example III: The Pendulum
6:34
Example IV: Cart Compressing a Spring
10:12
Example V: Cart Compressing a Spring
11:12
Example V: Part A - Potential Energy Stored in the Compressed Spring
11:13
Example V: Part B - Maximum Vertical Height
12:01
Example VI: Car Skidding to a Stop
13:05
Example VII: Block on Ramp
14:22
Example VIII: Energy Transfers
16:15
Example IX: Roller Coaster
20:04
Example X: Bungee Jumper
23:32
Example X: Part A - Speed of the Jumper at a Height of 15 Meters Above the Ground
24:48
Example X: Part B - Speed of the Jumper at a Height of 30 Meters Above the Ground
26:53
Example X: Part C - How Close Does the Jumper Get to the Ground?
28:28
Example XI: AP-C 2002 FR3
30:28
Example XI: Part A
30:59
Example XI: Part B
31:54
Example XI: Part C
32:50
Example XI: Part D & E
33:52
Example XII: AP-C 2007 FR3
35:24
Example XII: Part A
35:52
Example XII: Part B
36:27
Example XII: Part C
37:48
Example XII: Part D
39:32
Example XIII: AP-C 2010 FR1
41:07
Example XIII: Part A
41:34
Example XIII: Part B
43:05
Example XIII: Part C
45:24
Example XIII: Part D
47:18
Example XIV: AP-C 2013 FR1
48:25
Example XIV: Part A
48:50
Example XIV: Part B
49:31
Example XIV: Part C
51:27
Example XIV: Part D
52:46
Example XIV: Part E
53:25
Power

16m 44s

Intro
0:00
Objectives
0:06
Defining Power
0:20
Definition of Power
0:21
Units of Power
0:27
Average Power
0:43
Instantaneous Power
1:03
Instantaneous Power
1:04
Example I: Horizontal Box
2:07
Example II: Accelerating Truck
4:48
Example III: Motors Delivering Power
6:00
Example IV: Power Up a Ramp
7:00
Example V: Power from Position Function
8:51
Example VI: Motorcycle Stopping
10:48
Example VII: AP-C 2003 FR1
11:52
Example VII: Part A
11:53
Example VII: Part B
12:50
Example VII: Part C
14:36
Example VII: Part D
15:52
Section 5: Momentum
Momentum & Impulse

13m 9s

Intro
0:00
Objectives
0:07
Momentum
0:39
Definition of Momentum
0:40
Total Momentum
1:00
Formula for Momentum
1:05
Units of Momentum
1:11
Example I: Changing Momentum
1:18
Impulse
2:27
Impulse
2:28
Example II: Impulse
2:41
Relationship Between Force and ∆p (Impulse)
3:36
Relationship Between Force and ∆p (Impulse)
3:37
Example III: Force from Momentum
4:37
Impulse-Momentum Theorem
5:14
Impulse-Momentum Theorem
5:15
Example IV: Impulse-Momentum
6:26
Example V: Water Gun & Horizontal Force
7:56
Impulse from F-t Graphs
8:53
Impulse from F-t Graphs
8:54
Example VI: Non-constant Forces
9:16
Example VII: F-t Graph
10:01
Example VIII: Impulse from Force
11:19
Conservation of Linear Momentum

46m 30s

Intro
0:00
Objectives
0:08
Conservation of Linear Momentum
0:28
In an Isolated System
0:29
In Any Closed System
0:37
Direct Outcome of Newton's 3rd Law of Motion
0:47
Collisions and Explosions
1:07
Collisions and Explosions
1:08
The Law of Conservation of Linear Momentum
1:25
Solving Momentum Problems
1:35
Solving Momentum Problems
1:36
Types of Collisions
2:08
Elastic Collision
2:09
Inelastic Collision
2:34
Example I: Traffic Collision
3:00
Example II: Collision of Two Moving Objects
6:55
Example III: Recoil Velocity
9:47
Example IV: Atomic Collision
12:12
Example V: Collision in Multiple Dimensions
18:11
Example VI: AP-C 2001 FR1
25:16
Example VI: Part A
25:33
Example VI: Part B
26:44
Example VI: Part C
28:17
Example VI: Part D
28:58
Example VII: AP-C 2002 FR1
30:10
Example VII: Part A
30:20
Example VII: Part B
32:14
Example VII: Part C
34:25
Example VII: Part D
36:17
Example VIII: AP-C 2014 FR1
38:55
Example VIII: Part A
39:28
Example VIII: Part B
41:00
Example VIII: Part C
42:57
Example VIII: Part D
44:20
Center of Mass

28m 26s

Intro
0:00
Objectives
0:07
Center of Mass
0:45
Center of Mass
0:46
Finding Center of Mass by Inspection
1:25
For Uniform Density Objects
1:26
For Objects with Multiple Parts
1:36
For Irregular Objects
1:44
Example I: Center of Mass by Inspection
2:06
Calculating Center of Mass for Systems of Particles
2:25
Calculating Center of Mass for Systems of Particles
2:26
Example II: Center of Mass (1D)
3:15
Example III: Center of Mass of Continuous System
4:29
Example IV: Center of Mass (2D)
6:00
Finding Center of Mass by Integration
7:38
Finding Center of Mass by Integration
7:39
Example V: Center of Mass of a Uniform Rod
8:10
Example VI: Center of Mass of a Non-Uniform Rod
11:40
Center of Mass Relationships
14:44
Center of Mass Relationships
14:45
Center of Gravity
17:36
Center of Gravity
17:37
Uniform Gravitational Field vs. Non-uniform Gravitational Field
17:53
Example VII: AP-C 2004 FR1
18:26
Example VII: Part A
18:45
Example VII: Part B
19:38
Example VII: Part C
21:03
Example VII: Part D
22:04
Example VII: Part E
24:52
Section 6: Uniform Circular Motion
Uniform Circular Motion

21m 36s

Intro
0:00
Objectives
0:08
Uniform Circular Motion
0:42
Distance Around the Circle for Objects Traveling in a Circular Path at Constant Speed
0:51
Average Speed for Objects Traveling in a Circular Path at Constant Speed
1:15
Frequency
1:42
Definition of Frequency
1:43
Symbol of Frequency
1:46
Units of Frequency
1:49
Period
2:04
Period
2:05
Frequency and Period
2:19
Frequency and Period
2:20
Example I: Race Car
2:32
Example II: Toy Train
3:22
Example III: Round-A-Bout
4:07
Example III: Part A - Period of the Motion
4:08
Example III: Part B- Frequency of the Motion
4:43
Example III: Part C- Speed at Which Alan Revolves
4:58
Uniform Circular Motion
5:28
Is an Object Undergoing Uniform Circular Motion Accelerating?
5:29
Direction of Centripetal Acceleration
6:21
Direction of Centripetal Acceleration
6:22
Magnitude of Centripetal Acceleration
8:23
Magnitude of Centripetal Acceleration
8:24
Example IV: Car on a Track
8:39
Centripetal Force
10:14
Centripetal Force
10:15
Calculating Centripetal Force
11:47
Calculating Centripetal Force
11:48
Example V: Acceleration
12:41
Example VI: Direction of Centripetal Acceleration
13:44
Example VII: Loss of Centripetal Force
14:03
Example VIII: Bucket in Horizontal Circle
14:44
Example IX: Bucket in Vertical Circle
15:24
Example X: Demon Drop
17:38
Example X: Question 1
18:02
Example X: Question 2
18:25
Example X: Question 3
19:22
Example X: Question 4
20:13
Section 7: Rotational Motion
Rotational Kinematics

32m 52s

Intro
0:00
Objectives
0:07
0:35
Once Around a Circle: In Degrees
0:36
Once Around a Circle: In Radians
0:48
0:51
1:08
Example I: Convert 90° to Radians
1:09
Example I: Convert 6 Radians to Degree
1:23
Linear vs. Angular Displacement
1:43
Linear Displacement
1:44
Angular Displacement
1:51
Linear vs. Angular Velocity
2:04
Linear Velocity
2:05
Angular Velocity
2:10
Direction of Angular Velocity
2:28
Direction of Angular Velocity
2:29
Converting Linear to Angular Velocity
2:58
Converting Linear to Angular Velocity
2:59
Example II: Angular Velocity of Earth
3:51
Linear vs. Angular Acceleration
4:35
Linear Acceleration
4:36
Angular Acceleration
4:42
Example III: Angular Acceleration
5:09
Kinematic Variable Parallels
6:30
Kinematic Variable Parallels: Translational & Angular
6:31
Variable Translations
7:00
Variable Translations: Translational & Angular
7:01
Kinematic Equation Parallels
7:38
Kinematic Equation Parallels: Translational & Rotational
7:39
Example IV: Deriving Centripetal Acceleration
8:29
Example V: Angular Velocity
13:24
Example V: Part A
13:25
Example V: Part B
14:15
Example VI: Wheel in Motion
14:39
Example VII: AP-C 2003 FR3
16:23
Example VII: Part A
16:38
Example VII: Part B
17:34
Example VII: Part C
24:02
Example VIII: AP-C 2014 FR2
25:35
Example VIII: Part A
25:47
Example VIII: Part B
26:28
Example VIII: Part C
27:48
Example VIII: Part D
28:26
Example VIII: Part E
29:16
Moment of Inertia

24m

Intro
0:00
Objectives
0:07
Types of Inertia
0:34
Inertial Mass
0:35
Moment of Inertia
0:44
Kinetic Energy of a Rotating Disc
1:25
Kinetic Energy of a Rotating Disc
1:26
Calculating Moment of Inertia (I)
5:32
Calculating Moment of Inertia (I)
5:33
Moment of Inertia for Common Objects
5:49
Moment of Inertia for Common Objects
5:50
Example I: Point Masses
6:46
Example II: Uniform Rod
9:09
Example III: Solid Cylinder
13:07
Parallel Axis Theorem (PAT)
17:33
Parallel Axis Theorem (PAT)
17:34
Example IV: Calculating I Using the Parallel Axis Theorem
18:39
Example V: Hollow Sphere
20:18
Example VI: Long Thin Rod
20:55
Example VII: Ranking Moment of Inertia
21:50
Example VIII: Adjusting Moment of Inertia
22:39
Torque

26m 9s

Intro
0:00
Objectives
0:06
Torque
0:18
Definition of Torque
0:19
Torque & Rotation
0:26
Lever Arm ( r )
0:30
Example: Wrench
0:39
Direction of the Torque Vector
1:45
Direction of the Torque Vector
1:46
Finding Direction Using the Right-hand Rule
1:53
Newton's 2nd Law: Translational vs. Rotational
2:20
Newton's 2nd Law: Translational vs. Rotational
2:21
Equilibrium
3:17
Static Equilibrium
3:18
Dynamic Equilibrium
3:30
Example I: See-Saw Problem
3:46
Example II: Beam Problem
7:12
Example III: Pulley with Mass
10:34
Example IV: Net Torque
13:46
Example V: Ranking Torque
15:29
Example VI: Ranking Angular Acceleration
16:25
Example VII: Café Sign
17:19
Example VIII: AP-C 2008 FR2
19:44
Example VIII: Part A
20:12
Example VIII: Part B
21:08
Example VIII: Part C
22:36
Example VIII: Part D
24:37
Rotational Dynamics

56m 58s

Intro
0:00
Objectives
0:08
Conservation of Energy
0:48
Translational Kinetic Energy
0:49
Rotational Kinetic Energy
0:54
Total Kinetic Energy
1:03
Example I: Disc Rolling Down an Incline
1:10
Rotational Dynamics
4:25
Rotational Dynamics
4:26
Example II: Strings with Massive Pulleys
4:37
Example III: Rolling without Slipping
9:13
Example IV: Rolling with Slipping
13:45
Example V: Amusement Park Swing
22:49
Example VI: AP-C 2002 FR2
26:27
Example VI: Part A
26:48
Example VI: Part B
27:30
Example VI: Part C
29:51
Example VI: Part D
30:50
Example VII: AP-C 2006 FR3
31:39
Example VII: Part A
31:49
Example VII: Part B
36:20
Example VII: Part C
37:14
Example VII: Part D
38:48
Example VIII: AP-C 2010 FR2
39:40
Example VIII: Part A
39:46
Example VIII: Part B
40:44
Example VIII: Part C
44:31
Example VIII: Part D
46:44
Example IX: AP-C 2013 FR3
48:27
Example IX: Part A
48:47
Example IX: Part B
50:33
Example IX: Part C
53:28
Example IX: Part D
54:15
Example IX: Part E
56:20
Angular Momentum

33m 2s

Intro
0:00
Objectives
0:09
Linear Momentum
0:44
Definition of Linear Momentum
0:45
Total Angular Momentum
0:52
p = mv
0:59
Angular Momentum
1:08
Definition of Angular Momentum
1:09
Total Angular Momentum
1:21
A Mass with Velocity v Moving at Some Position r
1:29
Calculating Angular Momentum
1:44
Calculating Angular Momentum
1:45
Spin Angular Momentum
4:17
Spin Angular Momentum
4:18
Example I: Object in Circular Orbit
4:51
Example II: Angular Momentum of a Point Particle
6:34
Angular Momentum and Net Torque
9:03
Angular Momentum and Net Torque
9:04
Conservation of Angular Momentum
11:53
Conservation of Angular Momentum
11:54
Example III: Ice Skater Problem
12:20
Example IV: Combining Spinning Discs
13:52
Example V: Catching While Rotating
15:13
Example VI: Changes in Angular Momentum
16:47
Example VII: AP-C 2005 FR3
17:37
Example VII: Part A
18:12
Example VII: Part B
18:32
Example VII: Part C
19:53
Example VII: Part D
21:52
Example VIII: AP-C 2014 FR3
24:23
Example VIII: Part A
24:31
Example VIII: Part B
25:33
Example VIII: Part C
26:58
Example VIII: Part D
28:24
Example VIII: Part E
30:42
Section 8: Oscillations
Oscillations

1h 1m 12s

Intro
0:00
Objectives
0:08
Simple Harmonic Motion
0:45
Simple Harmonic Motion
0:46
Circular Motion vs. Simple Harmonic Motion (SHM)
1:39
Circular Motion vs. Simple Harmonic Motion (SHM)
1:40
Position, Velocity, & Acceleration
4:55
Position
4:56
Velocity
5:12
Acceleration
5:49
Frequency and Period
6:37
Frequency
6:42
Period
6:49
Angular Frequency
7:05
Angular Frequency
7:06
Example I: Oscillating System
7:37
Example I: Determine the Object's Angular Frequency
7:38
Example I: What is the Object's Position at Time t = 10s?
8:16
Example I: At What Time is the Object at x = 0.1m?
9:10
Mass on a Spring
10:17
Mass on a Spring
10:18
Example II: Analysis of Spring-Block System
11:34
Example III: Spring-Block ranking
12:53
General Form of Simple Harmonic Motion
14:41
General Form of Simple Harmonic Motion
14:42
Graphing Simple Harmonic Motion (SHM)
15:22
Graphing Simple Harmonic Motion (SHM)
15:23
Energy of Simple Harmonic Motion (SHM)
15:49
Energy of Simple Harmonic Motion (SHM)
15:50
Horizontal Spring Oscillator
19:24
Horizontal Spring Oscillator
19:25
Vertical Spring Oscillator
20:58
Vertical Spring Oscillator
20:59
Springs in Series
23:30
Springs in Series
23:31
Springs in Parallel
26:08
Springs in Parallel
26:09
The Pendulum
26:59
The Pendulum
27:00
Energy and the Simple Pendulum
27:46
Energy and the Simple Pendulum
27:47
Frequency and Period of a Pendulum
30:16
Frequency and Period of a Pendulum
30:17
Example IV: Deriving Period of a Simple Pendulum
31:42
Example V: Deriving Period of a Physical Pendulum
35:20
Example VI: Summary of Spring-Block System
38:16
Example VII: Harmonic Oscillator Analysis
44:14
Example VII: Spring Constant
44:24
Example VII: Total Energy
44:45
Example VII: Speed at the Equilibrium Position
45:05
Example VII: Speed at x = 0.30 Meters
45:37
Example VII: Speed at x = -0.40 Meter
46:46
Example VII: Acceleration at the Equilibrium Position
47:21
Example VII: Magnitude of Acceleration at x = 0.50 Meters
47:35
Example VII: Net Force at the Equilibrium Position
48:04
Example VII: Net Force at x = 0.25 Meter
48:20
Example VII: Where does Kinetic Energy = Potential Energy?
48:33
Example VIII: Ranking Spring Systems
49:35
Example IX: Vertical Spring Block Oscillator
51:45
Example X: Ranking Period of Pendulum
53:50
Example XI: AP-C 2009 FR2
54:50
Example XI: Part A
54:58
Example XI: Part B
57:57
Example XI: Part C
59:11
Example XII: AP-C 2010 FR3
1:00:18
Example XII: Part A
1:00:49
Example XII: Part B
1:02:47
Example XII: Part C
1:04:30
Example XII: Part D
1:05:53
Example XII: Part E
1:08:13
Section 9: Gravity & Orbits
Gravity & Orbits

34m 59s

Intro
0:00
Objectives
0:07
Newton's Law of Universal Gravitation
0:45
Newton's Law of Universal Gravitation
0:46
Example I: Gravitational Force Between Earth and Sun
2:24
Example II: Two Satellites
3:39
Gravitational Field Strength
4:23
Gravitational Field Strength
4:24
Example III: Weight on Another Planet
6:22
Example IV: Gravitational Field of a Hollow Shell
7:31
Example V: Gravitational Field Inside a Solid Sphere
8:33
Velocity in Circular Orbit
12:05
Velocity in Circular Orbit
12:06
Period and Frequency for Circular Orbits
13:56
Period and Frequency for Circular Orbits
13:57
Mechanical Energy for Circular Orbits
16:11
Mechanical Energy for Circular Orbits
16:12
Escape Velocity
17:48
Escape Velocity
17:49
Kepler's 1st Law of Planetary Motion
19:41
Keller's 1st Law of Planetary Motion
19:42
Kepler's 2nd Law of Planetary Motion
20:05
Keller's 2nd Law of Planetary Motion
20:06
Kepler's 3rd Law of Planetary Motion
20:57
Ratio of the Squares of the Periods of Two Planets
20:58
Ratio of the Squares of the Periods to the Cubes of Their Semi-major Axes
21:41
Total Mechanical Energy for an Elliptical Orbit
21:57
Total Mechanical Energy for an Elliptical Orbit
21:58
Velocity and Radius for an Elliptical Orbit
22:35
Velocity and Radius for an Elliptical Orbit
22:36
Example VI: Rocket Launched Vertically
24:26
Example VII: AP-C 2007 FR2
28:16
Example VII: Part A
28:35
Example VII: Part B
29:51
Example VII: Part C
31:14
Example VII: Part D
32:23
Example VII: Part E
33:16
Section 10: Sample AP Exam
1998 AP Practice Exam: Multiple Choice

28m 11s

Intro
0:00
Problem 1
0:30
Problem 2
0:51
Problem 3
1:25
Problem 4
2:00
Problem 5
3:05
Problem 6
4:19
Problem 7
4:48
Problem 8
5:18
Problem 9
5:38
Problem 10
6:26
Problem 11
7:21
Problem 12
8:08
Problem 13
8:35
Problem 14
9:20
Problem 15
10:09
Problem 16
10:25
Problem 17
11:30
Problem 18
12:27
Problem 19
13:00
Problem 20
14:40
Problem 21
15:44
Problem 22
16:42
Problem 23
17:35
Problem 24
17:54
Problem 25
18:32
Problem 26
19:08
Problem 27
20:56
Problem 28
22:19
Problem 29
22:36
Problem 30
23:18
Problem 31
24:06
Problem 32
24:40
1998 AP Practice Exam: Free Response Questions (FRQ)

28m 11s

Intro
0:00
Question 1
0:15
Part A: I
0:16
Part A: II
0:46
Part A: III
1:13
Part B
1:40
Part C
2:49
Part D: I
4:46
Part D: II
5:15
Question 2
5:46
Part A: I
6:13
Part A: II
7:05
Part B: I
7:48
Part B: II
8:42
Part B: III
9:03
Part B: IV
9:26
Part B: V
11:32
Question 3
13:30
Part A: I
13:50
Part A: II
14:16
Part A: III
14:38
Part A: IV
14:56
Part A: V
15:36
Part B
16:11
Part C
17:00
Part D: I
19:56
Part D: II
21:08
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• ## Related Books 1 answer Last reply by: Professor Dan FullertonSun Jan 7, 2018 4:35 PMPost by James Glass on January 7, 2018Hello, At 1:39, why does angular velocity = (k/m)^(1/2). Thanks 1 answer Last reply by: Professor Dan FullertonSun Mar 19, 2017 4:50 PMPost by Zen Chawaldit on March 18, 2017for the ap 2013 frq 3, question b, why is gravitational potential energy not included? the skier goes up the ramp, so there must be change in height. 1 answer Last reply by: Professor Dan FullertonThu Sep 1, 2016 5:41 AMPost by El Einstein on September 1, 2016Hello Professor Fullerton, I am confused on example VII (minute 45). Why did you set U=K? Isn't U=0 when at the equilibrium. And why did you set it equal to 5J? Can you explain this part in steps for me please. 1 answer Last reply by: Professor Dan FullertonSat Jan 9, 2016 12:44 PMPost by Shehryar Khursheed on January 9, 2016On example IV when youre deriving the SHM of a pendulum, can you please explain why there is a negative on the net torque: -mglsin(theta)? You said that as the angle gets smaller, the angular acceleration gets larger. That seems counter-intuitive to me. Thanks! 1 answer Last reply by: Professor Dan FullertonMon May 11, 2015 9:14 AMPost by Joshua Bowen on May 10, 2015Im trying to understand more about wave motion when it comes the the Sinusoidal Wave function, what is the difference between these functions:x = Acos(wt) (and the derivatives of the function which equal the velocity, and acceleration)and y = Asin(kx-wt) 1 answerLast reply by: Micheal BinghamFri Apr 24, 2015 10:20 AMPost by Micheal Bingham on April 24, 2015@ About 31:29, why is the slope of the T - sqr(l) Graph, 2pi/sgr(g) ? 2 answers Last reply by: Professor Dan FullertonSun Feb 15, 2015 10:30 AMPost by Thadeus McNamara on February 14, 2015@63:47, does W ALWAYS equal change in KE? 3 answersLast reply by: Thadeus McNamaraWed Feb 18, 2015 1:03 PMPost by Thadeus McNamara on February 14, 201533:05 i still am not sure why mglsintheta is negative 1 answer Last reply by: Professor Dan FullertonSun Feb 15, 2015 10:27 AMPost by Thadeus McNamara on February 14, 2015@29:16, why is F at a maximum at the top, and why is F at a minimum at the bottom? 1 answer Last reply by: Professor Dan FullertonSun Feb 15, 2015 10:25 AMPost by Thadeus McNamara on February 14, 2015thanks for the previous answers. i have another question. @24:56, why did u write f=-k2x2 over the arrow? how is Feq equivalent to F2? You said it was because of Newton's third law but I dont know what you mean by that. 1 answer Last reply by: Professor Dan FullertonSat Feb 14, 2015 5:50 PMPost by Thadeus McNamara on February 14, 2015also during that same slide, why does w represent both angular velocity and angular frequency. what is angular frequency? 1 answer Last reply by: Professor Dan FullertonSat Feb 14, 2015 5:47 PMPost by Thadeus McNamara on February 14, 2015@4:56 is that the position, velocity, and acceleration of SHM problems or for circular problems? 1 answerLast reply by: CHARINA TECSONWed Jan 21, 2015 10:55 PMPost by CHARINA TECSON on January 21, 2015Din example 1, did you ever convert minutes to seconds?

### Oscillations

• Simple Harmonic Motion (SHM) is motion in which a restoring force is directly proportional to the displacement of an object.
• Nature’s response to a disturbance is often SHM.
• Angular frequency (ω) is the number of radians per second, corresponding to an angular velocity for an object traveling in uniform circular motion.
• When an object undergoes simple harmonic motion, kinetic and potential energy both vary with time, although total energy (E=K+U) remains constant.
• The period of a spring-block oscillator is proportional to the square root of mass divided by the spring constant.
• In analyzing a spring block system, if you find a new equilibrium position when the block is hanging on the spring, taking into account the effect of gravity, you can then treat the system with only the spring force to deal with, oscillating around the new equilibrium point.
• The reciprocal of the equivalent spring constant for springs in series is equal to the sum of the reciprocals of the individual spring constants.
• The equivalent spring constant for springs in parallel is equal to the sum of the individual spring constants.
• The period of a pendulum is proportional to the square root of the length of the pendulum divided by the acceleration due to gravity.
• Intro 0:00
• Objectives 0:08
• Simple Harmonic Motion 0:45
• Simple Harmonic Motion
• Circular Motion vs. Simple Harmonic Motion (SHM) 1:39
• Circular Motion vs. Simple Harmonic Motion (SHM)
• Position, Velocity, & Acceleration 4:55
• Position
• Velocity
• Acceleration
• Frequency and Period 6:37
• Frequency
• Period
• Angular Frequency 7:05
• Angular Frequency
• Example I: Oscillating System 7:37
• Example I: Determine the Object's Angular Frequency
• Example I: What is the Object's Position at Time t = 10s?
• Example I: At What Time is the Object at x = 0.1m?
• Mass on a Spring 10:17
• Mass on a Spring
• Example II: Analysis of Spring-Block System 11:34
• Example III: Spring-Block ranking 12:53
• General Form of Simple Harmonic Motion 14:41
• General Form of Simple Harmonic Motion
• Graphing Simple Harmonic Motion (SHM) 15:22
• Graphing Simple Harmonic Motion (SHM)
• Energy of Simple Harmonic Motion (SHM) 15:49
• Energy of Simple Harmonic Motion (SHM)
• Horizontal Spring Oscillator 19:24
• Horizontal Spring Oscillator
• Vertical Spring Oscillator 20:58
• Vertical Spring Oscillator
• Springs in Series 23:30
• Springs in Series
• Springs in Parallel 26:08
• Springs in Parallel
• The Pendulum 26:59
• The Pendulum
• Energy and the Simple Pendulum 27:46
• Energy and the Simple Pendulum
• Frequency and Period of a Pendulum 30:16
• Frequency and Period of a Pendulum
• Example IV: Deriving Period of a Simple Pendulum 31:42
• Example V: Deriving Period of a Physical Pendulum 35:20
• Example VI: Summary of Spring-Block System 38:16
• Example VII: Harmonic Oscillator Analysis 44:14
• Example VII: Spring Constant
• Example VII: Total Energy
• Example VII: Speed at the Equilibrium Position
• Example VII: Speed at x = 0.30 Meters
• Example VII: Speed at x = -0.40 Meter
• Example VII: Acceleration at the Equilibrium Position
• Example VII: Magnitude of Acceleration at x = 0.50 Meters
• Example VII: Net Force at the Equilibrium Position
• Example VII: Net Force at x = 0.25 Meter
• Example VII: Where does Kinetic Energy = Potential Energy?
• Example VIII: Ranking Spring Systems 49:35
• Example IX: Vertical Spring Block Oscillator 51:45
• Example X: Ranking Period of Pendulum 53:50
• Example XI: AP-C 2009 FR2 54:50
• Example XI: Part A
• Example XI: Part B
• Example XI: Part C
• Example XII: AP-C 2010 FR3 1:00:18
• Example XII: Part A
• Example XII: Part B
• Example XII: Part C
• Example XII: Part D
• Example XII: Part E

### Transcription: Oscillations

Hello, everyone, and welcome back to www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about oscillations and simple harmonic motion.0003

Our objectives include analyzing simple harmonic motion in which the displacement is expressed in the form a sin ω T or a cos ω T.0009

Recognizing simple harmonic motion when expressed in differential equations form.0018

Calculating the kinetic and potential energies of an oscillating system.0023

Analyzing problems involving horizontal and vertical masses attached to springs.0027

Finding the period of oscillations for systems involving combinations of springs and deriving the expression for the period of a pendulum,0032

both an ideal pendulum and now for the first time, a physical or real pendulum.0040

Simple harmonic motion is motion which the restoring force is directly proportional to the displacement of the object.0045

The more you displace it, the more restoring force there is trying to bring it back to its initial position.0052

The reason this is so important is, in general, nature's response to a disturbance is some sort of simple harmonic motion.0057

You can see it all over the place, the blade of grass watch it come back up, simple harmonic motion.0065

Or a snowy tree that swayed down a limb with lots of snow as the snow falls often you see the branch go back and forth.0072

All of these responses have restoring forces proportional to the displacement of the object or at least that is a good starting model in a simple harmonic motion.0079

We can even see it in the atoms of an object.0090

When they are compressed or stretched, it will vibrate back in simple harmonic motion.0092

Let us take a look at simple harmonic motion and start off with an analogy to circular motion.0098

Let us assume that we have some mass moving in a circle of radius a, it is an angular velocity ω, that at a given point in time,0101

its position vector is given by a cos θ a sin θ, where a is the radius or the magnitude, and cos θ that is the angular displacement.0113

That is our x coordinate and there is our y coordinate.0123

We are going to compare this to a system of the spring attached to a wall with the mass on the end.0126

We are going to pull the mass of displacement a from its equilibrium, our happy position, and let it go back and forth.0133

What is really amazing about this analogy is, if you we are to pull this to a and let it go,0141

you can compare its x position to what you would get is this object goes around the circle its x position.0146

As the circle goes around at any given point in time, when it is over here, the mass is going to have that same x component all the way around.0153

You get this nice analogy between what we are already familiar with circular motion and this mass moving on a spring in simple harmonic motion.0160

Let us start out by taking a look at the angular velocity there we know is the time rate of change in the angular displacement.0171

But if I rewrite that a little bit, if I separate our variables, we can write that ω dt = d θ.0180

If I integrate both sides from some T = 0 to some final time T, integral from θ = 0 to some final θ,0187

that implies what angular velocity is constant in uniform circular motion so the left hand side becomes ω T.0198

The right hand side just becomes θ, θ is given by ω T, that will come useful later.0205

If we go to the down, we want to analyze our spring block system, we can look at it from terms of Newton's law,0212

from the perspective of Newton's law F = ma and that force is restoring force – kx.0219

We also know that acceleration is the second derivative of x with respect to T².0225

This means that Md² x dt² = - kx or I could write this in a more common differential equation form d² x / dt² + k /m × x must equal 0.0233

We have a second order differential equation or the second derivative of a function + a constant × that function gives you 0.0252

Only one way I can think of to solve these sorts of things and that is the sin or cos, the only functions where their second derivative added to themselves can give you 0.0260

The general form of our solution, x is a function of time is given by a, our amplitude cos ω T.0270

Where we are going to find ω is √k/m or if we look here in our equation that piece right there, that is ω².0280

Let us take a bit further, as we look at position velocity and acceleration.0292

We started off with θ = ω T and we said that x is a position of time is a cos ω T, where there is our angular frequency ω.0299

Our velocity then is the derivative of x with respect to T which is going to be the derivative with respect to T of a cos ω T.0314

It is going to be equal to the derivative of cos is the opposite of the sin.0328

We are going to get –ω A sin ωt where our maximum velocity is going to occur, while the maximum value of the sin function is 1.0332

Our maximum velocity is going to be ωa.0345

Let us take a look at acceleration which is the derivative of velocity with respect to time or the second derivative of x with respect to time,0350

which is going to be the derivative with respect to T of – ω A sin ω T.0359

The derivative of the sin is the cos function.0371

We are going to get –ω² A cos ω T.0373

We are going to have maximum acceleration when the value of the cos function is 1 so that is going to be ω² A for maximum acceleration value.0381

We can keep going with these to find position velocity acceleration.0392

As we are doing this, oftentimes we are talking about frequency in period.0397

Probably, it is worth bringing this up again.0400

Frequency is the number of cycles or revolutions per second.0403

Its units are 1/s or hertz.0408

Period T is the time for 1 cycle or 1 complete revolution and the units are seconds.0410

We can find period when we know frequency.0417

Period is 1/ frequency or frequency is 1/period.0419

If we talk about angular frequency, angular frequency is the number of radians per second.0425

It corresponds to the angular velocity for an object traveling in uniform circular motion.0431

Note that angular velocity and angular frequency is not the same thing but they do have a strong correspondents.0435

Angular frequency ω is 2π × frequency in Hz or 2 π ÷ period.0442

You can also write that period is 2 π ÷ angular frequency.0449

Let us do an example, an oscillating system is created by releasing an object from a maximum displacement of 0.2 m.0456

The object makes 60 complete oscillations in 1 min, determine the objects angular frequency.0464

Angular frequency is 2 π × the frequency in Hz which will be 2 π × 60 Hz.0472

Pardon me, 2 π 60 complete oscillations in 1 min, that is 1 oscillation/s × 1 Hz, which is just going to be 2 π radians/s.0482

What is the objects position at time T = 10 s?0497

Position is a cos ω T which implies, since we know that ω is 2 π radians/s and our maximum displacement A is 0.2 m, the amplitude.0501

This implies then that x = 0.2 cos 2π T, which implies then if T = 10 s that x is going to be equal to 0.2 × the cos 2 π × our time of 10 s or about 0.2 m.0521

At what time is the object at x = 0.1 m?0551

We have to realize it is going to be oscillating back and forth.0555

There could be more than one answer here but let us solve for one answer.0558

That x = a cos 2π T is our function which implies then that the cos of 2 π T = x / a0561

which implies then the 2 π T = the inverse cos of x / a or T is going to be equal to the inverse cos of x/ a ÷ 2 π.0576

We substitute in our values that will be the inverse cos of 0.1/0.2 ÷ 2 π or I get 0.167 s.0594

Alright, let us take a look at a little bit more detail of a mass on a spring.0613

Here we have our mass M attached to some spring with spring constant K attach to a wall and its move to a displacement A from its equilibrium position0619

and released allowing it to oscillate back and forth between - a and a.0629

You are going to assume no thing or no loss of energy to friction.0633

If we wanted to know the period of our spring, that is going to be 2 π √ M /K, something we will derive a little bit later.0637

Or the frequency of our spring is 1 /period and which is going to be ½ π √K/M.0648

We can arrange this a little bit further, multiply both sides by 2 π to say that 2 π × the frequency = √K /M is 2 π F is what we call that angular frequency.0659

Therefore, ω your angular frequency is the √k/m.0683

Let us analyze a spring block system.0692

A 5 kg block is attached to a 2000 N/m spring as shown in this place to distance of 8 cm from its equilibrium position before it released.0696

Determine the period of oscillation, the frequency, and the angular frequency for the block.0705

Let us start, it asks for period first so let us start there.0712

A period of our spring is 2π √M/K, which is going to be 2π √5kg /2000 N/m or about 0.314 s.0715

Next, it wants the frequency.0737

Frequency is 1 /period which will be 1/0.314 s or about 3.18 Hz.0739

Finally, the angular frequency for the block, ω is 2π × frequency which is 2π × 3.18 Hz or 20 radians/ s.0752

Another example, rank the following horizontal spring resting on frictionless surfaces in terms of their period from longest to shortest.0773

When we look at typically of different masses and spring constant, the way I start this is I would probably look0783

at the relationship and recognize first before looking for period.0788

The period is 2π √M/K so we could then make a table of information.0793

We will have our 4 systems, we have A, B, C, and D, and we need to know their mass, the spring constant and M/K.0801

As I start to look at these, our masses for A is 10, for B we have 7, C is 2, and D is 5.0825

Their spring constants, A is 500 N /m, B is 50, C is 2000, and D is 1000.0834

The ratio M/K it is pretty easy to calculate.0843

It is going to be 0.02, 0.14, 0.001, and 0.005.0847

We are looking for the periods from longest to shortest, since period is proportion to √M/K,0856

we are going to get the longest period where we have the greatest M /K.0863

I would rank these for period, our biggest M/KB, then A, D, C will have the shortest period.0867

As we talk about simple harmonic motion, the general form of simple harmonic motion is something where0880

we had the second derivative of function with respect to time + the constant ω² × that original function equal to 0.0885

When we do this, we find that our general solution is a cos ω T + some phase angle, some phase shift having to do0895

with where you are starting your cos function or sin function.0905

We are not going to worry a whole lot about that here in this course and note that ω here is your ω there.0910

When you go to graph this, if you have a cos function, it is going to look something like this at ω T= 0 × 0, your maximum displacement.0921

You oscillate back and forth between maximum and minimum displacement.0930

For using the sin function where you got a phase shift of 90°, you still have the same basic shape.0934

You are still oscillating for maximum A to – A.0939

Let us take a look at energy of simple harmonic motion.0949

When an object undergoes simple harmonic motion, kinetic and potential energy are both peering with time.0952

Although, the total mechanical energy, kinetic + potential remains constant.0957

If we have something like a mass spring system and we are going to compress it, the work we have to do is going to be the integral0963

from x to some 0 of f(x) dotted with dx, which is going to be the integral from x to 0 of – kx dx,0972

which will be - kx² /2 evaluated from x to 0, which is going to be ½ kx² which is the potential energy stored in the spring.0984

You have done that work on it that must be the potential energy in the spring.0996

If we wanted to look at x as a function of T, we know that a cos ω T which implies then that the potential energy in the spring as a function of time1000

is going to be ½ K × x, a cos ω T², which will be ½ K A² × √cos ω T.1013

Let us go take a look now at velocity.1034

Velocity is a function of time, is the derivative of x with respect to T, which is –ω A sin ω T.1036

Therefore, the kinetic energy of the system is going to be ½ MV² will be ½ × mass × velocity² is going to be - ω A sin ω T².1047

Or as we multiply through there, kinetic energy will be ½ M ω² a² sin ω T, square the sin ω T.1065

We know that ω² is k/m so then k if I replace ω² with K/M, our kinetic energy k is ½ k or spring constant × that amplitude² sin² ω T.1081

Let us put this all together.1100

We got our kinetic energy, we have got our spring potential energy, let us put that together to find your total energy.1101

E is kinetic + potential which is going to be our kinetic is ½ ka² sin² ω T + our potential ½ ka² cos² ω T.1110

Remember, our trigonometry sin² θ + cos² θ is equal to 1.1134

We can then factor that out to say that our total energy then ½ ka² × (sin² + cos² )is just going to be ½ KA².1142

No dependence on time because we have a constant total mechanical energy.1155

Alright, let us go back to our horizontal spring oscillator again.1163

We have mass oscillating a spring, spring constant K, force is mass × acceleration which is – kx, but a = the second derivative of x with respect to T.1168

We have md² x/dt² = - kx or d² x/ dt² = k/ mx² or + k /mx² equal 0.1184

We have mentioned before that was ω².1202

Our general solution for the horizontal spring oscillator system now looks like x(t) = A cos ω T + some phase angle.1205

Again, we are not going to have to deal with that phase angle here much, it is going to be 0 as we set up most of our systems,1217

where ω = √K/M so this works the same solution for a spring block oscillator or the period of the system is going to be 2π /ω which is 2π /√k/m we or 2π √m/k.1223

The same basic solution but now we have used this to derive the period of our spring block oscillator like we said we would a few slides ago.1248

What about a vertical spring block oscillator?1257

We have a mass, we are going to hang from the spring and we are going to let it settle to an equilibrium position1260

and then we are going to pull it down some amount A and displace it.1267

How do we deal with that where we also have this included effective gravity?1270

Let us see, we draw our free body diagram first for a mass, we will have some force of the spring which we will call ky and some force of gravity MG.1276

We are calling down the positive y direction so when I write my Newton’s second law equation.1289

The net force in the y direction that is going to be MG - ky and all of that must equal our mass × acceleration in the l direction.1294

If you are letting this hang down to the equilibrium position here, that means at this point MG - ky equilibrium must equal 01307

because at that point there is no net force and no acceleration.1317

Or we could then write that the equilibrium point on the y axis is just MG/k.1321

We have got that figured out but now we are going to take the block, we are going to pull it down or up and let it go in displacement and see what happens.1332

Our analysis says that the net force in the y direction is going to be MG - k where now our y is y equilibrium + the displacement A.1340

If we multiply through here, that means MG - ky equivalent –ka = our net force.1355

Let me just fill them in here.1367

Take a look, MG - ky at the equilibrium point is 0, we have the same thing right there net 0.1369

The net force in the y direction is just – ka, the exact same analysis we would be doing if this is a horizontal spring block oscillator without the effect of gravity.1379

All we have to do to deal with this vertical problem is find its equilibrium position once its hanging there and then treat it as if it is a horizontal spring block oscillator.1392

It just got so much more simple.1404

We will do that as sampled problem here shortly.1407

We can also have a couple of springs where they are lined up in different ways, combinations of springs.1411

If they are in series, such as this, we have k1 attached to k2, attached to a mass, we can analyze that and1417

it is nice to figure out how we could treat this as if it was just one equivalent spring.1423

Let us take a look here, force is going to be - k1x1 and by Newton’s third law that has to be equal to - k2 × the displacement x21430

which implies then the x 1 is going to be equal to k2 /k1 x2.1443

If we want to treat this as an equivalent spring, F is going to be equal to - some k equivalent × the total displacement of our block x1 + x2.1451

We can take this x1 and put it in here for x1 to find that our force is going to be equal to -k equivalent.1463

Our x1 is k2/k1 x2 + x2 not x² + x2 which implies then, since we also know our force = - k2 x2 Newton’s third law there,1474

that we could write that - k2 x2 = - k equivalent × x2.1497

We will factor that out and I'm left with k2 /k1 + 1, which implies then that k2 = equivalent k × k2 /k1 + 11507

which implies then we could write this as 1/k equivalent = 1/k1 + 1/k2.1527

You can follow that same form for any number of springs.1537

1/the equivalent spring constant is 1/ the first for spring constant + 1/ the second spring constant + 1/the third spring constant, for many springs as you might have.1540

Kind of like capacitors in series or resistors in parallel if you have done ENM.1553

You can maybe guess already what springs in parallel are going to look like.1559

They are going to look like resistors in series or capacitors in parallel.1562

But let us take a look and show that derivation, for springs in parallel now we have k1 and k2,1566

both attached to our mass M which is going to be displaced some amount x.1572

We can analyze that by looking at the force will be k1 x + k2 x, which we can say is k1 + k2 × x or k1 + k2 x is equal to some equivalent k × x.1577

Pretty easy to see, k equivalent is k1 + k2.1601

We just add up our individual spring constants to get our equivalent spring constant when they are in parallel.1604

We have spent some time on springs and we will come back to those 2.1614

We also see simple harmonic motion and pendulums.1617

If we have a pendulum here attached to a light spring, a mass on the end, and we will call this entire distance in the string length L.1623

Our pendulum swings back and forth, it is going to come up some amount H, which we will call L - L cos θ or L1 – cos θ.1632

This again is our length L and we know that when we are in our highest points here, we have our most gravitational potential energy.1648

We are down at the lowest points, when it is down there, it is kinetic energy.1655

We have talked about that energy transformation.1660

We can look at that in a little more detail though.1666

If we take a look at these different points, at their highest point here, our gravitational potential energy is MGH, its height above, its change in y position from here to here,1669

which is going to be MGL × (1 - cos θ).1682

Here all its energy is kinetic energy, back up here, we have all potential energy.1689

When we are done here, if we want to know the speed, the kinetic energy at the bottom has to equal the gravitational potential energy at the top.1697

½ MV² must equal MGL × (1 - cos θ).1705

Or rearranging V² is going to be equal to 2 GL × 1 - cos θ or V will be equal to √2 GL 1 - cos θ, when we are at that point right there.1714

As we also look at this from different perspectives, when we are at our highest point here, we have the maximum force on our mass on the string.1733

We have our maximum acceleration, we have our maximum gravitational potential energy that are kinetic energy 0 and our speed is 0.1743

Here, where the highest point of kinetic energy, the force on our mass is 0, acceleration is 0, gravitational potential energy 0,1754

but we have our maximum kinetic energy and our maximum velocity.1766

A lot of times you will see this depicted in a graph of energy showing a constant total mechanical energy as you go to higher and higher displacements.1772

B distance that is under the graph is our potential energy, the distance from your total energy to the top of a graph is your kinetic energy.1781

That total always = e total K + U.1790

But you have that shift across the different types of energy.1793

Taking a look at over here, we know our force MG is down, the force in the direction of displacement is just going to be MG sin θ,1797

where once again we have this as a restoring force leading this into a discussion of simple harmonic motion.1807

Let us look at the period and the frequency of a pendulum.1816

We have our pendulum like we have talked about previously.1820

The period of the pendulum is 2π √its length ÷ G assuming it is an ideal pendulum.1822

All the mass is at the very end, you have a light string or frequency is 1 /period which is going to be 1/2π √G /L.1829

You can even go and plot these if you wanted.1839

Period vs. Length of your pendulum and we are probably going to get some kind of roughly this shape, where T is proportional to √L.1841

If you want to linearize your graph, T and √L, to get something that is fairly linear.1850

If we did that, what happened to our slope?1862

Our slope is just going to be 2 π ÷ √G.1867

Ever want to know the acceleration due to gravity on the Moon?1874

Here is a great way to find that out.1877

Go up to the moon, take a bunch of pendulums, different lengths, measure their periods compared to their length, plot them and take the slope.1879

Slope = 2π /√G, you could then go calculate G, the acceleration due to gravity on the Moon.1887

How do we get this 2π √L /G?1897

Let us see if we can derive that.1900

Here, we have a simple pendulum, an ideal pendulum.1904

We have a mass M on the string about some point P, we have a very light string, and all of the masses here at M.1907

There is no friction just the ideal case.1914

If I look at the forces here, I have the force of gravity MG down on my mass and I am also going to define distance1918

from our point there q to our mass as the position function from point B.1927

Their length is L.1935

If we look at this from the torque perspective, torque about point P is our P crossed with F, which is going to be our P crossed with MG our force.1938

Which implies then that the magnitude of our torque about point B is going to be MGL sin θ,1955

which implies then that - MGL sin θ = our moment of inertia about point P × angular acceleration.1968

Why that negative? If you take a look as θ is getting smaller, we are going to larger values of angular acceleration.1978

So we have to take that into account there.1985

We are going to use what is known as the small angle approximation.1988

For small angles of θ, they are under about 15°.1992

The sin θ is a very close the θ itself.1995

This will allow us to simplify our analysis and write this as - MGL θ = the moment of inertia about point B × α.1999

We also know that α is the second derivative of θ.2011

We can write this as - MGL θ is equal to our moment of inertia P × the second derivative of θ with respect to T.2018

We have a second order differential equation.2030

Let us put into that standard form so we can see a little bit more clearly, d² θ / dt² + we will have MGL / the moment of inertia about point P × θ = 0.2032

By the way, if you remember our form dist is what we called ω².2052

The solution to our problem, our general solution is θ is going to be equal to a cos ω T.2059

The moment of inertia of a single mass at the end of the string, our moment of inertia about point P is going to be ML².2067

That means that ω² will be MGL /ML² IP which is going to be G /L which implies then that ω = √G /L.2076

If we want that period, period is 2π / ω which is going to be 2π ÷ √G/L or 2π √L /G.2093

We derived the formula for the period of a simple pendulum.2110

That is easy, what about real pendulums?2116

Alright, let us try that.2120

Here, we have a non ideal pendulum, a pendulum now it is made out of a rod where it has a center of mass in the middle.2122

We are going to have a pivot point P there and the distance between our pivot point and our center of mass we are going to call D.2129

We can use the same basic analysis in order to find its period.2136

Net torque = R cross F, we know θ is going to be equal to a cos ω T.2140

We have already done that analysis or ω² if we go through all of that again is MGD /our moment inertia about point P,2150

which implies then that ω is √MGD /our moment of inertia about point P.2161

The difference here is we have a different moment of inertia.2169

To find our moment of inertia about point P, all that is going to be the moment of inertia about our center of mass.2175

We can use the parallel axis theorem to shift our pivot point some distance and D away from the center of mass, that will be + the mass of our pendulum bar × D².2181

The mass of our rod rotated about its center is ML² /12 + MD².2196

If period is 2π / ω that is going to be 2π / ω.2208

Here, our √MGD/√IP, this is going to be √IP is ML² /12 + MD² /MGD,2214

which implies then that our period is going to be equal to 2π.2231

We can factor an MN out of all of that and I would have L² /12 + D² /GD or2235

if I wanted to make that a little bit prettier, we can write that as 2π × √L² + 12 D² ÷ 12 GD.2247

If you want to test this out, I lot of my students quite regularly is to take a meter stick.2264

Find its center of mass, drill a hole in it, near one of the end some distance from that center of mass measure that2270

and then put it on a hook and swing it back and forth.2278

Let it go at about 10, 20 times, measure the entire time for 20 revolutions, 20 periods, and divide by 20,2281

and see how that does compared to when you plug it into the equation and you will find it is very accurate.2287

Alright, let us go back to our spring block system for a summary here.2295

If we want to take a look at our spring block system with displacement velocity, force acceleration, potential energy in the spring2299

and kinetic energy and a bunch of different positions as we go from A-B to A-C and back again is we are oscillating some displacement x and –x.2307

Sometimes it is useful to graph all of these just because you see them come up so often.2316

Let us make a graph, we will start here by saying on the 0 point on our time axis, we will call when we have position A,2321

then we will pull the B to A to C and back again.2330

I'm just going to plot some of these lines in here very loosely so that we have a common reference frame.2333

I think we will do it there and let us do the same thing here for our couple of energy graphs.2359

If this is position A, that will be B, back to A to C to A back to B again.2384

The same thing here, A to B back to A, C to A to B again.2393

We will have graphs of let us see.2401

We will have this as our x.2406

At time 0, when its position A our displacement is 0.2409

At position B, where there are maximum displacement which we call x or typically capital A.2413

Back to A 0 to C to A, back to B.2421

Our plot would look something like that.2426

In that position A, where our 0 displacement at B our displacement is x.2433

At C, it would be –x.2438

We can take a look now at velocity which we know is the derivative of x.2441

As we look at that, we know that at A where maximum velocity but it points B and C we are always going to be at 0.2447

We can fill those points in and we know that 2 because if we take the derivative of position that will be the slope.2455

We have a 0 here, we have a 0 there, we have a 0 slope there, so we have 0 values for velocity there.2460

We are going to start at A and the maximum velocity come down to a minimum or maximum speed in the opposite direction2467

and end up with a graph that looks kind of like that.2477

How about if we take a look at the force?2482

If we look at the force, we are going to have, let us fill in our table here toward velocity.2486

We have maximum velocity at A, at B we have 0 and at C we have 0.2492

Now looking at force, when we are at A we have 0 force, so anytime we are at A we can fill 0.2500

When we are at B, we have maximum force but in the negative direction so we will have a negative there.2507

We will have a positive here, when we are at C maximum force back to the right restoring force and so on.2514

We can graph our force working something like that.2520

And our acceleration of course, should follow the same thing, the same general pattern by Newton’s second law.2527

They should tolerate along.2539

We will have 0 force and 0 acceleration when we are at point A.2540

At point B, we are going to have the negative maximum for force and acceleration going back toward the left.2544

At C, we will have our maximum positive force and acceleration.2552

And again, if we want to look at our derivative slope relationship, look at the velocity here at A.2558

The slope of that is 0 therefore, we have 0 acceleration right there as well and force correlate right along with that.2565

How about potential energy and kinetic energy?2573

As I look at potential energy, we know when we are at A that is going to be 0 so we can fill that in right now.2576

A for potential energy is 0 there, we will have a 0 at A.2584

At B and C, we are going to have our maximum values.2590

We do not have a negative because this is a scalar.2594

We will have a graph that looks something like this.2597

When we look at kinetic energy, that one with our table here, we are going to have maximums at B and C for potential energy.2608

For kinetic B and C, for that split second, it stopped.2619

Kinetic will have 0 here and we will have our maximum when we are at position A.2623

We can draw this, we will put our 0 in here to make our drawing a little bit simpler.2629

We are just going to come down and up, the complimentary graph to our potential energy function.2634

You get a feel for how the graphs of all of these would look in simple harmonic motion and how they all relay to each other.2644

Let us do an analysis of the simple harmonic later.2653

We have a 2 kg block attached to a spring, a force of 20 N stretches a spring to a displacement of ½ m, find the spring constant.2658

We can do that using F = kx or k = f /x which is 20 N /0.5 m or 40 N/m.2667

If we want the total energy, our total energy is going to be the total energy stored in our spring when we are at maximum displacement2683

or ½ kx² which is ½ × 40 N /m × maximum displacement 0.5 m² or about 5 joules, the speed at the equilibrium position.2691

To do that, the potential energy of the spring when it is at its maximum displacement must be equal to the kinetic energy.2710

When it is at 0 displacement, which is ½ MV², all of that must equal 5 joules which implies then that V must be equal to √2 × 5/2,2716

which is just going to be √5 or about 2.24 m/s.2731

If we want the speed at 0.3 m, let us do this from a total energy perspective again.2739

That is going to be the spring potential energy + the kinetic energy all has to be that constant 5 joules.2747

But that is going to be ½ kx² + ½ MV² which implies then that MV² is going to be equal to 2 × our energy total - kx²,2753

which implies that V is going to be equal to the √2 × our total energy - kx² ÷ M,2768

which implies that V = 2 × 5 joules - our spring constant 40 in our displacement 0.3 m² ÷ our mass of 2 kg.2780

The square root of that whole thing gives me 1.79 m/s.2793

Moving on, find the speed when it is at -0.4 m.2805

We can use the same formula, the same analysis just with different values, V = √2 × our total energy - kx² /M2811

which will be 2 × 5 joules - spring constant 40 N/m × displacement -0.4² ÷ our mass 2.2822

Square root of all of that is 1.34 m/s, the acceleration of the equilibrium position.2835

Remember, at equilibrium, at x = 0, our force is 0.2846

Therefore, our acceleration is 0.2851

How about the magnitude of the acceleration at 0.5 m?2856

The force is - kx which is going to be -40 N/m × 0.5 m or 20 N - 20 N.2861

The acceleration which by Newton’s second law is force/mass is -20 N / 2 kg or -10 m/s².2870

Continuing on, find the net force of the equilibrium position.2885

We have already said that at equilibrium, the acceleration is 0.2891

Therefore, the force is 0, the net force at 0.25 m though we can do that, F = kx is going to be -40 Nm × 0.25 m or -10 N.2898

And where does the kinetic energy equal the potential energy?2914

The total is 5 that means kinetic has to equal 2.5 joules and potential has to equal 2.5 joules.2918

We can solve that from our spring potential energy equation.2925

That has to equal 2.5 joules which is ½ kx² which implies that x² is going to be 2 × 2.5 joules ÷ our k 40 or 0.125 m².2931

Therefore, x is the square root of that or 0.354 m.2948

It is important to note here that is in not halfway between the equilibrium position and maximum displacement.2957

You do not have equal amounts of kinetic and potential energy when you are at half that displacement.2966

It does not work that way.2971

Another example, let us rank some spring systems here.2975

We have the spring block oscillators with combinations of springs, rank them from highest to lowest in terms of equivalent of spring constant and period of oscillation.2979

It looks like this to would be another great spot for a table.2990

We are going to make a table that has our system A, B, C, and D.2993

We will look at our mass, our equivalent spring constant, and M/k which is proportional to our period is 2π √M /K.2999

It looks like we can fill on our masses first.3023

I have 3, 6, 2, and 4, to find the equivalent spring constant this is going to be 1/20 + 1/5 and reciprocal of that will give us our equivalent or 4.3025

For B, because they are in parallel we just add them, that one was easy 25.3039

For C, we have 2 in series again so that is going to be the equivalent spring constant 1 /k equivalent is 1/15 + 1/15.3044

When we solve that, we will get 7.5 N /m.3054

The last one is in series again, just add them up 20.3058

M/k is 0.75 6/25 about 0.242, 27.5 about 0.27 and 4/20 is 1/5 or 0.2.3063

If we wanted rank these in terms of the equivalent spring constant that is going to be, we will start with B, D, C, and A from highest to lowest.3074

Period is 2π √M/k so we can go in the same order as M/k here which is going to be A, C, B, and final D.3088

Alright, let us do a vertical spring block oscillator system.3104

A 2 kg block attached to an un stretched spring, a spring constant k = 200 N/m as shown is released from rest.3107

Determine the period of the blocks oscillation and the maximum displacement of the block from its equilibrium while undergoing simple harmonic motion.3116

We can start with the period that is going to be 2π √M /k which is 2π × √2kg /200 N/m or 0.63 s.3125

To find the maximum displacement, we will do a little bit more work here.3145

The gravitational potential energy at the block starting point MG Δ y must equal the elastic potential energy stored in the spring and its lowest point.3151

We can solve for Δ y using the gravitational potential energy = stored potential energy in the spring3160

which implies that MG Δ y must be equal to ½ K Δ y².3167

½ kx² which implies if we solve for Δ y that is just going to be and divided Δ y at both sides and get 2 MG /k,3175

that is the entire displacement from the top and the bottom.3188

We want the displacement from its equilibrium position to its maximum displacement so that is going to be half of that.3191

Our amplitude will be Δ y/ 2 which is 2 MG /2k or just MG /K with a mass of 2kg, G of 9.8 m/s² ÷ k 200 N /m.3197

I come up with about 0.1 m for our maximum displacement from its equilibrium position while it is in that simple harmonic motion.3216

Let us rank the periods of some pendulums.3230

We have these 4 pendulums of uniform mass density from highest to lowest frequency.3232

Period is 2π √L /G, so the frequency is 1 /the period that must be 1/2 π × √G /L.3239

If we want the shortest period, if we want the shortest frequency, the smallest frequency, we want the biggest length.3254

If we want the highest frequency, we want the shortest length.3267

The shorter ones are going to go back and forth a lot more quickly than the long ones.3269

This would be D, A, B, C, because frequency is only a function of the length.3274

Alright, let us finish this off by looking at a couple of free response problems from old AP exams.3287

Let us start off with a 2009 exam free response question number 2.3292

As we look at this one, we are given a long thin rectangular bar of mass M, length L.3300

We are to determine its moments of inertia.3306

It has a non uniform mass density.3308

First A1, we are going to write the differential equation for the angle θ.3311

To do this, I'm going to draw it at its extreme just to make it easier to see what is going on.3316

We would not raise it at that height because we would lose our small angle approximation.3321

If we brought it up to the side like this, we will have some MG wherever that center of mass happens to be.3325

We will call x the distance from our pivot to our center of mass and there is a our angle θ.3331

We know net torque = Iα so we can write that our net torque we are going to have - mgx × sin θ must equal our moment of inertia IB × α.3338

Αlpha we know is the second derivative of θ so we can write this now as - MG × x sin θ = IB D² θ / DT².3359

That should cover us for writing the differential equation.3376

Now for part A2, it says apply the small angle approximation to calculate the period of the bars motion.3380

The small angle approximation says for small θ is about under about 15°, sin θ is approximately equal the θ.3387

It would not work as we have not draw on our diagram, we would not actually raise it that high.3398

If we did this with a small angle and let it go, we could then write that - mgx × θ = IB D² θ / DT² or3403

putting this into a more familiar form we can write this as DT D² θ / DT² + it looks like we will have mgx / IB × θ = 0.3417

There is a more familiar form, we have the second derivative of θ + some constant mgx / IB θ where that if you recall is our ω².3432

If we want that period, it looks like we are going to be looking for period is 2π / ω which is going to be 2π ÷ √mgx/ IB3444

which is going to be 2π × √ of the moment of inertia of our bar ÷ mgx.3462

That should cover us for part A2.3474

Let us give ourselves more room here for part B.3478

Describe the experimental procedure you would use to make the additional measurements needed to determine IB3484

and how you would use your measurements to minimize experimental error?3489

When I talked about this and we talked about finding that period of the real pendulum, I displace the bar a little bit of under 15°.3493

It will go back and forth say 10 times and measure the amount of time it takes to go back and forth 10 times.3503

Then divide that total time by 10 to get the period, that would give your T and then you could calculate your moment of inertia.3509

Once you have your period, you know period is 2π √ of the moment of inertia of your bar / mgx.3517

So then T² = 4π² IB / mgx which implies then that IB is just going to be equal to T² mgx / 4π².3527

You could use that in order to find your moment of inertia of the bar.3547

Let us take a look at part C, now suppose you are not given the location of the center of mass of the bar, how could you determine it and what equipment would you need?3553

That should be pretty easy too.3563

What I would do is take something like a razorblade, have the razor blade up vertically and put the bar on top of it and move it until you balance it.3564

Wherever you balance it, make a duct and that is going to be your location of the center of mass of the bar.3576

You could also hang it from different points and look where all the plumb bob lines across.3583

The easiest, I think probably needed to set it on some sort of edge and move it back and forth until you get to balance and that will be your center of mass point.3587

In words, something about balancing it and make sure you list any equipment you would need, pen and some sort of flat edge, pick whatever you want to use for that edge.3598

I think that covers the 2009 question, let us do one more.3609

Let us go to the 2010 exam free response 3.3618

Here we have a skier of mass M pulled up the hill by a rope and the magnitude of the acceleration is modeled by that equation there3624

where you have sin function between 0 and T.3631

After T it is not accelerating anymore, acceleration is 0, greater than T.3635

Alright derive an expression for the velocity of the skiers, a function of time.3642

Assume the skier starts from rest, for part A, if we are given the acceleration we can find the velocity as the integral from some value of T= 0 to some final time T.3647

The acceleration is a function of time which will be the integral from 0 to T of looks like acceleration is a max × sin π t/T.3662

We can pull other constant that is going to be A max × the integral to the sin π t /T.3677

I should not forget my DT here and the integral of the sin is going to be the opposite of the cos.3690

We need to have the argument in here, we are going to have our du which is π / T.3696

We need to have T /π out here to make a ratio of 1.3701

And then we can integrate that to say that V is going to be equal to - a max T/π cos of π t /T.3706

All evaluated from 0 to T which is going to be - a max T / π and we will have that cos π t / T – 1.3724

Or putting that negative through, we can rewrite this as V= A max t / π × (1 - cos π t /T) and that should work for T in the integral from 0 to T.3743

For part B, find the work done by the net force on the skier from rest until it reaches terminal speed there at T.3769

We could use the work energy theorem here.3779

Work is going to be change in kinetic energy, our net torque which is going to be our final kinetic energy - our initial kinetic energy3782

which is ½ MV final² -1/2 MV² initial.3790

V initial is 0 that part it is easy.3796

V final is just going to be V at time T which is going to be, we will have a max T /π × 1 - cos π which is going to be 1 - -1 so that is going to be 2a max capital T /π.3799

We can plug that in for our final velocity, this piece we said was 0 which implies then that the work done is going to be equal to ½ M × our final velocity²3828

which is going to be 4 a max² T² /π² or putting all of this together that will be 2 × M × maximum acceleration² × T² /π².3843

And that should cover us for part B.3866

Taking a look at part C, determine the magnitude of the force exerted by the rope on the skier at terminal speed.3872

Here I'm going to look at my free body diagram, we will draw our axis in first.3882

We have the normal force of the skier, we have the force of tension from the rope, and we have the weight of the skier down.3889

If I wanted to draw our pseudo free body diagram, breaking up that MG into components parallel3899

and perpendicular with the axis, I would have still, we have our normal force.3906

We have our force of tension from the rope.3912

We have MG sin θ and of course MG cos θ.3916

I can write my Newton’s second law expression in the x direction.3924

Net force in the x direction which we know is going to be 0 because it is moving at constant speed, at terminal speed is going to be equal to tension force - MG sin θ.3929

Therefore, force of tension in the rope must be MG sin θ.3942

That should cover us for C.3951

Part D, derive an expression for the total impulse imparted to the skier during the acceleration.3955

Impulse is the integral of force with respect to time.3962

Impulse is going to be the integral of FD T which will be the integral our forces mass × acceleration which is going to be the integral from some value T = 03966

to some final value T of M, our acceleration function is given we have a max sin π t /T DT.3979

We will pull other constants again that is going to be equal to, we can pull out M, we can pull out a max integral of sin π t /T DT.3992

We already talk about how we integrate that, we are going to π /T here so we are going to need HT T / π out here in order to integrate that.4011

That is going to be Ma max T / π and then we will have that cos π t /T.4024

Let us see, I will evaluate it from 0 to T.4045

We got to have our negative sign and their 2 the integral of the sin is the opposite of cos.4050

That is going to be equal to - M a max T/ π × cos π, when we substitute T in for t.4054

- the cos 0 which is 1, cos π is -1, -1 is going to be 2.4070

-2 we will have -2 × negative this will give us 2 ma max T /π.4079

One more piece to the puzzle, for part E, suppose the magnitude of the acceleration is instead modeled as the exponential for T greater than 0.4094

On the axis below, sketch the graphs of the force exerted by the rope on the skier for the 2 models from T = 0 to some time T greater than terminal velocity.4107

Label them F1 and F2.4116

First thing I want to do is I'm going to figure out what my functions are.4120

If we look at the original first, F1 is what we will call that our original.4123

We had the FT, tension force in the rope - MG sin θ all had the equal MA which is MA max sin π t/T.4131

Since we are going to plot the force of tension in the rope, FT must equal MG sin θ + MA max sin π t /T.4147

There is our first function, our new function says that FT - MG sin θ is now going to be equal to mass × new acceleration function a max E t / 2 T.4163

Therefore, the function we will be modeling here for the force on the rope is going to be MG sin θ + MA max e t /2 T.4186

Let us try our axis and see if we can graph these.4203

Finish this one up in style.4206

There is our y our force, here is our x our time, we have got some value T and force.4211

Let us label here MG sin θ and now for regional force we know that at time T =0,4225

we are going to have the sin of 0 so we are going to start at just MG sin θ so we can start with our first point there.4237

We are going to have our maximum value halfway to T, when we have sin π /2.4244

That will be at that point we will have MG sin θ + M a max.4252

We could label that point up here as well, that will be important.4258

MG sin θ + MA max right there.4262

If I we are to plot this, we know when we get the T, we are going to have sin π which is going to be back to 0.4269

This will be back to MG sin θ and after that, it is a constant flat line.4278

We have that part and in between we have our sin function which is going to bring us up to a maximum here.4283

It is kind of something that looks like that for initial force.4293

Our new force is MG sin θ + Ma max E to all of this.4300

At time T = 0, e⁰ is 1 so we are going to start at MG sin θ + MA max.4306

This one, we will start up there at that point and it looks like we are going to have the decaying as T gets very big, this is going to become a very large number in the denominator.4313

This piece is going to go to 0 and we are going to decay down the MG sin θ.4324

I would say that this one would look something like that, may be not quite like that.4328

Like that with some sort of exponential decay there so there would be our F2.4334

Hopefully, that gets you a great start on oscillations.4340

Thank you so much for joining us here today at www.educator.com.4343

I look forward to seeing you again soon and make it a great day everyone.4347

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