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Harmonic Motion P221, November 22 nd, 2013
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Review of Simple Harmonic Motion System at rest Displace mass, stretches spring Restoring force is proportional to displacement 0x F
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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
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Sines & Cosines Restoring force is linear AND in opposite direction to displacement
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Sines & Cosines Restoring force is linear AND in opposite direction to displacement Combination of sines & cosines can solve this
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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
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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
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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
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Simple Pendulum ( coordinate) Rotational Oscillation ( ) 1-D Spring and Block (x) I wire
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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
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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
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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
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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
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Clicker Question: Discussion System 1: k is a spring constant that is independent of mass System 1 System 2 m m
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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
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Physical Pendulum RCMRCM Mg X CM
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Physical Pendulum RCMRCM Mg arc-length R CM XCMXCM RCMRCM X CM
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Physical Pendulum RCMRCM Mg arc-length R CM XCMXCM RCMRCM X CM
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Physical Pendulum X CM RCMRCM Mg arc-length R CM XCMXCM RCMRCM
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Physical Pendulum For small X CM RCMRCM Mg arc-length R CM XCMXCM RCMRCM
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Physical Pendulum For small X CM RCMRCM Mg arc-length R CM XCMXCM RCMRCM
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Physical Pendulum For small X CM RCMRCM Mg arc-length R CM XCMXCM RCMRCM
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Physical Pendulum For small X CM RCMRCM Mg arc-length R CM XCMXCM RCMRCM
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Physical Pendulum For small X CM RCMRCM Mg arc-length R CM XCMXCM RCMRCM
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The Simple Pendulum IS a Physical Pendulum The general case CM pivot R CM The simple case L
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A Specific Case: Stick Pendulum M pivot R CM CM
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A Specific Case: Stick Pendulum M pivot R CM CM
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A Specific Case: Stick Pendulum M pivot R CM CM
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A Specific Case: Stick Pendulum M pivot R CM CM
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A Specific Case: Stick Pendulum M pivot R CM CM Same period
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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
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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
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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
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m Case 2 Prelude Answer Remember period is inversely proportional to rotational frequency therefore
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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 +=
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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 +=
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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
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Case 2 m m m Case 1 Clicker: Detailed Answer
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Case 2 m m m Case 1 Clicker: Detailed Answer
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Case 2 m m m Case 1 Clicker: Detailed Answer
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Case 2 m m m Case 1 Clicker: Detailed Answer
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Case 2 m m m Case 1 Clicker: Detailed Answer
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Mechanics Lecture 21, Slide 43
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Mechanics Lecture 21, Slide 44
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Mechanics Lecture 21, Slide 45
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Mechanics Lecture 21, Slide 46
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At t 0, y 0, moving down Mechanics Lecture 21, Slide 47
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At t 0, y 0, moving down Use energy conservation to find A Mechanics Lecture 21, Slide 48
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At t 0, y 0, moving down Use energy conservation to find A Mechanics Lecture 21, Slide 49
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Mechanics Lecture 21, Slide 50
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Mechanics Lecture 21, Slide 51 Or similarly
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Mechanics Lecture 21, Slide 52
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Mechanics Lecture 21, Slide 53
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Mechanics Lecture 21, Slide 54
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Mechanics Lecture 21, Slide 55
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