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Physics I
95.141
LECTURE 22
11/29/10
95.141, F2010, Lecture 22
Department of Physics and Applied Physics
Administrative Notes
• Exam III
– In Class Wednesday, 9am
– Chapters 9-11
– 9 problems posted on Website in Practice Exam
Section. At least 1 of these problems will be on
EXAM III.
• If you have questions, start a discussion thread on Facebook,
that way my response is seen by everyone in the class.
– Review Session Tuesday Night, 6:00pm, OH150
95.141, F2010, Lecture 22
Department of Physics and Applied Physics
Outline
•
•
•
Oscillations
Simple Harmonic Motion
What do we know?
– Units
– Kinematic equations
– Freely falling objects
– Vectors
– Kinematics + Vectors = Vector
Kinematics
– Relative motion
– Projectile motion
– Uniform circular motion
– Newton’s Laws
– Force of Gravity/Normal Force
– Free Body Diagrams
– Problem solving
– Uniform Circular Motion
– Newton’s Law of Universal Gravitation
– Weightlessness
– Kepler’s Laws
95.141, F2010, Lecture 22
Department of Physics and Applied Physics
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Work by Constant Force
Scalar Product of Vectors
Work done by varying Force
Work-Energy Theorem
Conservative, non-conservative Forces
Potential Energy
Mechanical Energy
Conservation of Energy
Dissipative Forces
Gravitational Potential Revisited
Power
Momentum and Force
Conservation of Momentum
Collisions
Impulse
Conservation of Momentum and Energy
Elastic and Inelastic Collisions2D, 3D Collisions
Center of Mass and translational motion
Angular quantities
Vector nature of angular quantities
Constant angular acceleration
Torque
Rotational Inertia
Moments of Inertia
Angular Momentum
–
–
Vector Cross Products
Conservation of Angular Momentum
Review of Lecture 21
• Discussed cross product definition of angular
momentum and torque
L  I
  RF
  
Lrp
  
  r F
• Why would we ever use cross products instead
of simpler scalar expressions?
– 3D vectors
– Point masses not moving in uniform circle

 dL
nd
– Newton’s 2 Law for rotational motion  
dt
• Conservation of Angular Momentum
– No external torque, angular momentum conserved.
95.141, F2010, Lecture 22
Department of Physics and Applied Physics
Oscillations (Chapter 14)
• Imagine we have a spring/mass system, where
the mass is attached to the spring, and the
spring is massless.
A
vmax
A
95.141, F2010, Lecture 22
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Oscillations
• So we can say that the mass will move back and
forth (it will oscillate) with an Amplitude of
oscillation A.
• Can we describe what is going on
mathematically?
• Would like to determine equation of motion of
the mass. In order to do this, we need to know
the force acting on the block.
• Force depends on position: Hooke’s Law
Fspring  kx
95.141, F2010, Lecture 22
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Oscillations
Fspring  kx
dv
d  dx 
d x
 F  ma  m dt  m dt  dt   m dt 2
2
2
d x (t )
 kx (t )  m
2
dt
95.141, F2010, Lecture 22
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Possible Solutions
• What if x=At?
• What if x=Aebt?
95.141, F2010, Lecture 22
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d x (t )  k

x (t )
2
dt
m
2
Possible Solutions
• What if x=Acos(bt)?
• What if x=Acos(bt+Φ)?
95.141, F2010, Lecture 22
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d x (t )  k

x (t )
2
dt
m
2
Possible Solutions
• What if
x(t )  A cos(t   ) ,  
d x (t )  k

x (t )
2
dt
m
2
k
m
If we start with the mass displaced from equilibrium by a
distance A at t=0, then we can determine x(t).
95.141, F2010, Lecture 22
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What does motion vs. time look like?
x(t )  A cos(t ) ,  
k
m
• Plot x at t=0, π/2ω, π/ω, 3π/2ω, 2π/ω, 5π/2ω
x
A
t
-A
95.141, F2010, Lecture 22
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Harmonic Motion: terminology
• Displacement: Distance from equilibrium
• Amplitude of oscillation: max displacement of object from
equilibrium
• Cycle: one complete to-and-fro motion, from some initial
point back to original point.
• Period: Time it takes to complete one full cycle
• Frequency: number of cycles in one second
95.141, F2010, Lecture 22
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Simple Harmonic Motion
• A form of motion where the only force on the
object is the net restoring force, which is
proportional to the negative of the displacement.
• Such a system is often referred to as a simple
harmonic oscillator
• The simple harmonic oscillator’s motion is
described by:
x(t )  A cos(t   ) ,  
95.141, F2010, Lecture 22
Department of Physics and Applied Physics
k
m
What is Φ?
22
tt)t)))2
2cos(
2cos(
cos(
cos(


t

t

t




)

)
)
,
,
,






20
2



222cos(

t


)
,



2
x (xtxx()x(t((t)(

t


)
,


cos(t   ) ,  5 5555,5 ,,,,, 
42
22
2
Displacement
Displacement
Displacement
Displacement
Displacement
2
22
1
11
0
00
-1
-1
-1
-2
-2
-2
0
00
95.141, F2010, Lecture 22
5
10
5 time(s)
time(s)
time(s)
Department of Physics and Applied Physics
15
10
10
10
15
15
15

More terminology
• So the Φ term is known as the phase of the
oscillation. It basically shifts the x(t) plot in time.
• The term ω, which for a spring mass system, is
equal to k m , is known as the angular
frequency. x (t )  A cos(t   )
T
2

2

T
95.141, F2010, Lecture 22
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  2f
Velocity and Acceleration for SHO
• If we know x(t), we can calculate v(t) and a(t)
x (t )  A cos(t   )
95.141, F2010, Lecture 22
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Velocity and Acceleration for SHO
• If x (t )  A cos(t )
2
x(t)
1
0
-1
-2
v(t)
2
0
-2
4
a(t)
2
0
-2
-4
0
2
4
6
8
time (s)
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10
12
14
Example
• A SHO oscillates with the following properties:
– Amplitude=3m
– Period = 2s
• Give the equation of motion for the SHO
95.141, F2010, Lecture 22
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Example
• A SHO oscillates with the following properties:
– Amplitude=3m
– Period = 2s
– At t=0s, x=3m.
• Give the equation of motion for the SHO
95.141, F2010, Lecture 22
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Example
• A SHO oscillates with the following properties:
– Amplitude=3m
– Period = 2s
– At t=0s, x=1.5m
• Give the equation of motion for the SHO
95.141, F2010, Lecture 22
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Example
• A SHO oscillates with the following properties:
– Amplitude=3m
– Period = 2s
– At t=0s, v=2m/s.
• Give the equation of motion for the SHO
95.141, F2010, Lecture 22
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Energy of SHO
• The total energy of a simple harmonic oscillator
comes from the potential energy in the spring,
and the kinetic energy of the mass.
x (t )  A cos(t   )
v (t )   A sin( t   )

95.141, F2010, Lecture 22
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k
m
Example
• A spring mass system with m=4kg and k=400N/m is
displaced +0.2m from equilibrium and released.
– A) What is the equation of motion for the mass?
95.141, F2010, Lecture 22
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Example
• A spring mass system with m=4kg and k=400N/m is
displaced +0.2m from equilibrium and released.
– B) What is the total energy of the system?
95.141, F2010, Lecture 22
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Example
• A spring mass system with m=4kg and k=400N/m is
displaced +0.2m from equilibrium and released.
– C) What is the Kinetic Energy and Potential Energy of the
system at t=2s?
95.141, F2010, Lecture 22
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SHO and Circular Motion
y
• You will notice that we use the same variable for
both angular velocity and angular frequency of a
simple harmonic oscillator.

• If we imagine an object moving with uniform
circular motion (angular velocity=ω) on a flat
surface. Starting, at t=0s, at θ=0.
• We know that θ(t)=ωt
• We can write the x-position of the object as:
• And the y-position as:
95.141, F2010, Lecture 22
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A
θ
x
The pendulum
• A simple pendulum consists of a mass
(M) attached to a massless string of
length L.
• We know the motion of the mass, if
dropped from some height, resembles
simple harmonic motion: oscillates
back and forth.
• Is this really SHO? Definition of SHO
is motion resulting from a restoring
force proportional to displacement.
95.141, F2010, Lecture 22
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Simple Pendulum
L
• We can describe displacement
as:
• The restoring Force comes from
gravity, need to find component of
force of gravity along x
• Need to make an approximation
here for small θ…
95.141, F2010, Lecture 22
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θ
Δx
Simple Pendulum
• Now we have an expression for
the restoring force
F  mg sin   mg 
L
θ
x  L
mg
x
L
• From this, we can determine the
effective “spring” constant k
F 
• And we can determine the natural
frequency of the pendulum
95.141, F2010, Lecture 22
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Δx
Simple Pendulum
• If we know

g
L
L
θ
• We can determine period T
• And we can the equation of
motion for displacement in x
• …or θ
95.141, F2010, Lecture 22
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Δx