Harmonic Motion
P221, November 22nd, 2013
Review of Simple Harmonic Motion
• System at rest
• Displace mass,
stretches spring
• Restoring force is proportional
to displacement
F
0
x
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
F
v
t
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
Has a physical
• Angular interpretation
frequency is ALWAYS
(and independent of amplitude)
• “Coordinate” could be x or q or anything else
Simple Pendulum (q coordinate)
1-D Spring and Block (x)
Rotational Oscillation (q)
wire
q
I

q
• Force is proportional to
k could be mg
• Torque is proportional to
angular displacement
or torsional
strength
positional displacement
Simple Pendulum (q coordinate)
1-D Spring and Block (x)
Rotational Oscillation (q)
wire
q
I

q
Simple Pendulum (q coordinate)
1-D Spring and Block (x)
Rotational Oscillation (q)
wire
q
I

q
• Torque is proportional to
angular displacement
• Force is proportional to
positional displacement
Simple Pendulum (q coordinate)
1-D Spring and Block (x)
Rotational Oscillation (q)
wire
q
I

q
• Torque is proportional to
angular displacement
• Force is proportional to
positional displacement
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
m
System 1
q
m
System 2
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
m
System 1
q
m
System 2
Clicker Question: Discussion
• System 1: k is a spring
constant that is
independent of mass
m
System 1
q
m
System 2
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
m
System 1
q
m
System 2
Physical Pendulum
q
RC
M
XCM
Mg
Physical Pendulum
q
RC
M
XCM
Mg
q
RC
M
XC
M
arc-length
q
= RCM
Physical Pendulum
 = I
q
RC
M
XCM
Mg
q
RC
M
XC
M
arc-length
q
= RCM
Physical Pendulum
 = I
q
RC
 MgX CM
M
XCM
Mg
q
RC
M
XC
M
arc-length
q
= RCM
Physical Pendulum
 = I
For
small q
q
RC
 MgX CM
M
XCM
 MgRCM q
Mg
q
RC
M
XC
M
arc-length
q
= RCM
Physical Pendulum
 = I
For
small q
q
RC
 MgX CM
 MgRCM q
M
d 2q
I 2
dt
XCM
Mg
q
RC
M
XC
M
arc-length
q
= RCM
Physical Pendulum
 = I
For
small q
q
RC
 MgX CM
M
d 2q
 MgRCM q I 2
dt
MgRCM
d 2q
=
q
2
dt
I
XCM
Mg
q
RC
M
XC
M
arc-length
q
= RCM
Physical Pendulum
 = I
For
small q
q
RC
 MgX CM
M
d 2q
 MgRCM q I 2
dt
MgRCM
d 2q
=
q
2
dt
I
XCM
Mg
q
d 2q
2
=


q
2
dt
RC
M
XC
M
arc-length
q
= RCM
Physical Pendulum
 = I
For
small q
q
RC
 MgX CM
M
d 2q
 MgRCM q I 2
dt
MgRCM
d 2q
=
q
2
dt
I
d 2q
2
=


q
2
dt
MgRCM
=
I
XCM
Mg
q
RC
M
XC
M
arc-length
q
= RCM
The Simple Pendulum IS a Physical
Pendulum
The general case
The simple case
pivot
RCM
q
L
q
CM
MgRCM
=
I
g
MgL
=
=
2
L
ML
A Specific Case: Stick Pendulum
pivot
RCM
q
CM
M
A Specific Case: Stick Pendulum
MgRCM
=
I
pivot
RCM
q
CM
M
A Specific Case: Stick Pendulum
MgRCM
=
I
pivot
RCM
q
CM
M
L
2
1
2
ML
3
A Specific Case: Stick Pendulum
MgRCM
=
I
pivot
RCM
q
CM
M
L
2
1
2
ML
3
=
g
2
L
3
A Specific Case: Stick Pendulum
MgRCM
=
I
pivot
L
2
=
1
2
ML
3
RCM
q
CM
M
L
2
L
3
Same period
g
2
L
3
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
Case 1
Case 2
to the center of the same
stick. Which pendulum has
the longer period?
m
A) Case 1
B) 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?
Case 1
Case 2
• Which as the longer period?
A) Case 1
B) Case 2
C) Same
1
L
2
L
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?
Case 1
Case 2
• Which as the longer period?
A) Case 1
B) Case 2
C) Same
1
L
2
L
m
Prelude Answer
Case 1
Case 2
=
L
g
2
L
3
=
1
L
2
g
1
L
2
m
therefore
• Remember period is inversely proportional to
rotational frequency 
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) T1
B) T2
C) In Between
+
T2
=
m
m
m
T1
T1 > T2
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) T1
B) T2
C) In Between
+
T2
=
m
m
m
T1
T1 > T2
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!
T2
m
m
m
T1
T1 > T2
Clicker: Detailed Answer
Case 2
Case 1
m
m
m
MgRCM
=
I
Clicker: Detailed Answer
Case 2
Case 1
m
m
m
L
mg
2
MgRCM
=
I
L
2mg
2
Clicker: Detailed Answer
Case 2
Case 1
m
m
m
L
mg
2
1 2
mL
3
MgRCM
=
I
L
2mg
2
2
1 2
7
L
mL  m  = mL2
3
 2  12
Clicker: Detailed Answer
Case 2
Case 1
m
m
m
=
g
2
L
3
g
=
7
L
12
Clicker: Detailed Answer
Case 2
Case 1
m
m
m
=
g
2
L
3
g
=
7
L
12
y
ky
mg
Mechanics Lecture 21, Slide 43
y
ky
mg
mg = ky
Mechanics Lecture 21, Slide 44
mg
k=
y
y
ky
mg
mg = ky
k
=
m
Mechanics Lecture 21, Slide 45
mg
k=
y
Mechanics Lecture 21, Slide 46
At t = 0, y = 0, moving down
y(t ) =  Asin t 
v(t ) =  A cost 
a(t ) =  2 A sin t 
Mechanics Lecture 21, Slide 47
At t = 0, y = 0, moving down
y(t ) =  Asin t 
v(t ) =  A cost 
a(t ) =  2 A sin t 
Use energy conservation to find A
1 2
1 2
mvmax = kA
2
2
Mechanics Lecture 21, Slide 48
At t = 0, y = 0, moving down
y(t ) =  Asin t 
v(t ) =  A cost 
a(t ) =  2 A sin t 
Use energy conservation to find A
1 2
1 2
mvmax = kA
2
2
A = vmax
v(t ) =  A cost 
Mechanics Lecture 21, Slide 49
m
k
Mechanics Lecture 21, Slide 50
a(t ) =  2 A sin t 
amax =  2 A
Or similarly
amax
Mechanics Lecture 21, Slide 51
Fmax kymax kA
=
=
=
m
m
m
Mechanics Lecture 21, Slide 52
F (t ) = ky(t ) = kA sin (t )
Mechanics Lecture 21, Slide 53
Mechanics Lecture 21, Slide 54
1 2
U = ky
2
y(t ) =  A sin t 
Mechanics Lecture 21, Slide 55