Harmonic Motion P221, November 22 nd, 2013

Review of Simple Harmonic Motion System at rest Displace mass, stretches spring Restoring force is proportional to displacement 0x F

More Review No external forces  energy conserved Kinetic is converted to potential, vice versa Velocity at ends is 0 “turning point” Fastest at center Frequency is constant t F v

Sines & Cosines Restoring force is linear AND in opposite direction to displacement

Sines & Cosines Restoring force is linear AND in opposite direction to displacement Combination of sines & cosines can solve this

Sines & Cosines Restoring force is linear AND in opposite direction to displacement Combination of sines & cosines can solve this Angular frequency is ALWAYS (and independent of amplitude) “Coordinate” could be x or  or anything else

Sines & Cosines Restoring force is linear AND in opposite direction to displacement Combination of sines & cosines can solve this Angular frequency is ALWAYS (and independent of amplitude) “Coordinate” could be x or  or anything else Has a physical interpretation

Simple Pendulum (  coordinate) Rotational Oscillation (  ) Torque is proportional to angular displacement 1-D Spring and Block (x) Force is proportional to positional displacement  I wire    could be mg or torsional strength

Simple Pendulum (  coordinate) Rotational Oscillation (  ) 1-D Spring and Block (x)  I wire  

Simple Pendulum (  coordinate) Rotational Oscillation (  ) Torque is proportional to angular displacement 1-D Spring and Block (x) Force is proportional to positional displacement  I wire  

Simple Pendulum (  coordinate) Rotational Oscillation (  ) Torque is proportional to angular displacement 1-D Spring and Block (x) Force is proportional to positional displacement  I wire  

Clicker Question Systems 1 & 2 are oscillating at their own frequencies. We then double the masses. Do the frequencies change? A) Both change B) Neither change C) Only system 1 changes D) Only system 2 changes  System 1 System 2 m m

Clicker Question Systems 1 & 2 are oscillating at their own frequencies. We then double the masses. Do the frequencies change? A) Both change B) Neither change C) Only system 1 changes D) Only system 2 changes  System 1 System 2 m m

Clicker Question: Discussion System 1: k is a spring constant that is independent of mass  System 1 System 2 m m

Clicker Question: Discussion System 1: k is a spring constant that is independent of mass System 2: both restorative force and moment of inertia are proportional to mass  System 1 System 2 m m

Physical Pendulum  RCMRCM Mg X CM

Physical Pendulum  RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM X CM

Physical Pendulum  RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM X CM

Physical Pendulum  X CM RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM

Physical Pendulum For small   X CM RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM

Physical Pendulum For small   X CM RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM

Physical Pendulum For small   X CM RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM

Physical Pendulum For small   X CM RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM

Physical Pendulum For small   X CM RCMRCM Mg  arc-length  R CM  XCMXCM RCMRCM

The Simple Pendulum IS a Physical Pendulum The general case CM pivot  R CM The simple case  L

A Specific Case: Stick Pendulum M pivot  R CM CM

A Specific Case: Stick Pendulum M pivot  R CM CM

A Specific Case: Stick Pendulum M pivot  R CM CM

A Specific Case: Stick Pendulum M pivot  R CM CM

A Specific Case: Stick Pendulum M pivot  R CM CM Same period

Clicker Question In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attached to the center of the same stick. Which pendulum has the longer period? A) Case 1 B) Case 2 C) Same Case 1 m Case 2 m m

Clicker Question: Prelude In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attached to a string of length L  2 ? Which as the longer period? A) Case 1 B) Case 2 C) Same Case 1 m

Case 2 Clicker Question: Prelude In Case 1 a stick of mass m and length L is pivoted at one end and used as a pendulum. In Case 2 a point particle of mass m is attached to a string of length L  2 ? Which as the longer period? A) Case 1 B) Case 2 C) Same Case 1 m

m Case 2 Prelude Answer Remember period is inversely proportional to rotational frequency  therefore

Clicker Question: Prelude 2 We know that T 1 > T 2. Now suppose these pendula are “glued” together from the same pivot. What is the new period? A) T 1 B) T 2 C) In Between m T1T1 T2T2 T1  T2T1  T2 m m +=

Clicker Question: Prelude 2 We know that T 1 > T 2. Now suppose these pendula are “glued” together from the same pivot. What is the new period? A) T 1 B) T 2 C) In Between m T1T1 T2T2 T1  T2T1  T2 m m +=

Clicker Question: Discussion We know that T 1 > T 2 and T of the “glued” pendulum is in between. We have proven T 1 is the longest. But, let’s calculate in detail! m T1T1 T1  T2T1  T2 T2T2 m m

Case 2 m m m Case 1 Clicker: Detailed Answer

Case 2 m m m Case 1 Clicker: Detailed Answer

Case 2 m m m Case 1 Clicker: Detailed Answer

Case 2 m m m Case 1 Clicker: Detailed Answer

Case 2 m m m Case 1 Clicker: Detailed Answer

Mechanics Lecture 21, Slide 43

Mechanics Lecture 21, Slide 44

Mechanics Lecture 21, Slide 45

Mechanics Lecture 21, Slide 46

At t  0, y  0, moving down Mechanics Lecture 21, Slide 47

At t  0, y  0, moving down Use energy conservation to find A Mechanics Lecture 21, Slide 48

At t  0, y  0, moving down Use energy conservation to find A Mechanics Lecture 21, Slide 49

Mechanics Lecture 21, Slide 50

Mechanics Lecture 21, Slide 51 Or similarly

Mechanics Lecture 21, Slide 52

Mechanics Lecture 21, Slide 53

Mechanics Lecture 21, Slide 54

Mechanics Lecture 21, Slide 55