harmonic motion p221, november 22 nd, 2013. review of simple harmonic motion system at rest displace...
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Harmonic Motion
P221, November 22nd, 2013
Review of Simple Harmonic Motion
• System at rest
• Displace mass,stretches spring
• Restoring force is proportionalto displacement
0 x
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 q 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 q or anything else
Has a physical interpretation
Simple Pendulum (q coordinate)
Rotational Oscillation (q)
• Torque is proportional to angular displacement
1-D Spring and Block (x)
• Force is proportional to positional displacement
q I wire
k could be mg or torsional
strength
Simple Pendulum (q coordinate)
Rotational Oscillation (q) 1-D Spring and Block (x)
q I wire
Simple Pendulum (q coordinate)
Rotational Oscillation (q)
• Torque is proportional to angular displacement
1-D Spring and Block (x)
• Force is proportional to positional displacement
q I wire
Simple Pendulum (q coordinate)
Rotational Oscillation (q)
• Torque is proportional to angular displacement
1-D Spring and Block (x)
• Force is proportional to positional displacement
q I wire
Clicker Question
• Systems 1 & 2 are oscillating at their own frequencies. We then double the masses. Do the frequencies change?
A) Both changeB) Neither changeC) Only system 1 changesD) Only system 2 changes
q
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 changeB) Neither changeC) Only system 1 changesD) Only system 2 changes
q
System 1
System 2
m
m
Clicker Question: Discussion
• System 1: k is a spring constant that is independent of mass
q
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
q
System 1
System 2
m
m
Physical Pendulum
q RC
M
Mg
XCM
Physical Pendulum
q RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
XCM
Physical Pendulum
Iq RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
XCM
Physical Pendulum
Iq
CMMgXXCM
RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
Physical Pendulum
Forsmall q
Iq
CMMgXXCM
RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
CMMgR
Physical Pendulum
Forsmall q
I
2
2
dt
dI
q
CMMgXXCM
RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
CMMgR
Physical Pendulum
Forsmall q
I
2
2
dt
dI
q
CMMgXXCM
RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
CMMgR
I
MgR
dt
d CM2
2
Physical Pendulum
Forsmall q
I
2
2
dt
dI
q
CMMgXXCM
RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
CMMgR
I
MgR
dt
d CM2
2
22
2
dt
d
Physical Pendulum
Forsmall q
I
2
2
dt
dI
q
CMMgXXCM
RC
M
Mg
q
arc-length = RCM q
XCM
RC
M
I
MgRCM
CMMgR
I
MgR
dt
d CM2
2
22
2
dt
d
The Simple Pendulum IS a Physical Pendulum
The general case
CM
pivot
q
RCM
I
MgRCM
The simple case
qL
2ML
MgL
L
g
A Specific Case: Stick Pendulum
M
pivot
q
RCM
CM
A Specific Case: Stick Pendulum
M
pivot
q
RCM
CM
I
MgRCM
A Specific Case: Stick Pendulum
2
L
2
3
1ML
M
pivot
q
RCM
CM
I
MgRCM
A Specific Case: Stick Pendulum
L
g
32
2
L
2
3
1ML
M
pivot
q
RCM
CM
I
MgRCM
A Specific Case: Stick Pendulum
L
g
32
2
L
2
3
1ML
M
pivot
q
RCM
CM
I
MgRCM
Same period
L3
2
L
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 attachedto the center of the samestick. Which pendulum hasthe longer period?
A) Case 1B) Case 2C) Same
Case 1
m
Case 2
m
m
Case 2
1
2L
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 1B) Case 2C) Same
L
Case 1
m
Case 2
1
2L
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 1B) Case 2C) Same
L
Case 1
m
L
Case 1
L
g
32
m
Case 2
1
2L L
g
21
Prelude Answer
• Remember period is inversely proportional to rotational frequency w
therefore
Clicker Question: Prelude 2
• We know that T1 > T2. Now suppose these pendula are “glued” together from the same pivot. What is the new period?
A) T1B) T2C) In Between
mT1
T2
T1 > T2
mm
+ =
Clicker Question: Prelude 2
• We know that T1 > T2. Now suppose these pendula are “glued” together from the same pivot. What is the new period?
A) T1B) T2C) In Between
mT1
T2
T1 > T2
mm
+ =
Clicker Question: Discussion
• We know that T1 > T2 and T of the “glued” pendulum is in between. We have proven T1 is the longest. But, let’s calculate in detail!
mT1
T1 > T2
T2
mm
Case 2m
m
mCase 1
I
MgRCM
Clicker: Detailed Answer
2
Lmg
Case 2m
m
mCase 1
I
MgRCM
Clicker: Detailed Answer
22
Lmg
2
Lmg
Case 2m
m
mCase 1
I
MgRCM
Clicker: Detailed Answer
22
Lmg
2
3
1mL 2
22
12
7
23
1mL
LmmL
Case 2m
m
mCase 1
Clicker: Detailed Answer
L
g
32
L
g
127
Case 2m
m
mCase 1
Clicker: Detailed Answer
L
g
32
L
g
127
Mechanics Lecture 21, Slide 43
y yk
mg
Mechanics Lecture 21, Slide 44
y yk
mg
ykmg y
mgk
Mechanics Lecture 21, Slide 45
y yk
mg
ykmg y
mgk
m
k
Mechanics Lecture 21, Slide 46
At t = 0, y = 0, moving down
Mechanics Lecture 21, Slide 47
tAty sin)(
tAtv cos)( tAta sin)( 2
At t = 0, y = 0, moving down
Use energy conservation to find A
Mechanics Lecture 21, Slide 48
tAty sin)(
tAtv cos)( tAta sin)( 2
22max 2
1
2
1kAmv
At t = 0, y = 0, moving down
Use energy conservation to find A
Mechanics Lecture 21, Slide 49
tAty sin)(
tAtv cos)( tAta sin)( 2
22max 2
1
2
1kAmv
k
mvA max
tAtv cos)(
Mechanics Lecture 21, Slide 50
Mechanics Lecture 21, Slide 51
tAta sin)( 2
Aa 2max
Or similarly
m
kA
m
yk
m
Fa
maxmax
max
Mechanics Lecture 21, Slide 52
Mechanics Lecture 21, Slide 53
)(sin)()( tkAtkytF
Mechanics Lecture 21, Slide 54
Mechanics Lecture 21, Slide 55
tAty sin)(
2
2
1kyU
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