I. Harmonic Motion Kinematics II. Simple Harmonic ... - Musowls

Advanced Placement Physics C Unit 7- Simple Harmonic Oscillators
I. Harmonic Motion Kinematics
A. Functions of time-analytical and graphical
1. position
2. velocity
3. acceleration
B. Functions of position
1. velocity
2. acceleration
II. Simple Harmonic Motion Differential Equation-
Find the natural period of …
A. Mass on a spring
B. Plane pendulum
C. Torsion Pendulum
D. Physical Pendulum
1. parallel-axis theorem
2. center of mass
E. Other Systems
III. Energy Considerations in Oscillating Systems
A. Potential Energy
B. Kinetic Energy
C. Total Energy
IV. Finding Natural Period from Energy Equations
Before taking your next AP Physics C unit test you should be familiar with all areas of
the above outline. You can enhance your knowledge of this material by studying the
following notes, doing assigned homework and reading all of chapter 15 from text. The
test may include multiple-choice conceptual questions, free response problems that are
numerical or analytical and perhaps some writing.

Lesson 2-09 Harmonic Kinematics
Read Chapter 15:1-2
Vibration Rate
How fast an object or system oscillates or vibrates can be described in any of three ways.
How long does it take to complete one cycle of vibration? This is a question of time
and is measured in seconds. The length of time is called “period”. We wish to use
the letter “t” for time but this is a special time so we use Tau. Period → τ
How many times does the system move back and forth in one second? This is the
inverse of the last question. The previous measurement was in seconds per cycle
while this question deals with how many cycles in one
second. When you count cycles rather than seconds you
measure frequency. Frequency is measured in cycles per
second or sec -1 or Hertz. The symbol for frequency is “ƒ”.
ƒ = 1 / τ
Since there are 2π radians in a circle you could relate the cycle to the number of
radians rather than cycles completed per second. This is also a frequency except that
it uses radians/sec rather than cycles/ second.
The official name is “angular frequency”
since radians are associated with angles. The ω = 2πƒ = 2π /τ
symbol for angular frequency is lower case
omega, ω. The relations are shown in the above box. These are more like
conversions rather than actual relationships.
Example #1
A child on a swing goes back and forth once every 1.40 seconds. Find the period,
frequency and angular frequency.
τ = 1.40 seconds ƒ = 0.71 Hz ω = 4.49 radians / second
Example #2
A fly flaps its wings 200 times per second. Find period, frequency & angular frequency.
ƒ = 200 Hz τ = 0.005 seconds ω = 1260 radians / second
Kinematics of Oscillation
Consider an object of mass M that is attached M
to a spring as shown in the figure to the right.
If the mass is pulled from the equilibrium
position a distance of x = A and released it will
move back and forth between x = +A and x = -A. −A x = 0 +A
We can show the motion of the object in terms of
position, velocity and acceleration for a specific time or specific position. The equations
depend upon where the object is located when the clock starts. Suppose that the object is
released at x = +A at t=0 seconds. On the other hand the object could be released and the
clock not started so that t = 0 seconds when M is in the center moving to the right. How
would starting the clock at different times change the equations?
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Starting t = 0 seconds when M is at the
far right of its motion.
The graph in this case would be as
written and graphed below.
x = A cos( ω t)
The above graph is for two complete
cycles. Recall that the slope of a
position graph shows the velocity. You
can see that M is at rest at ± A because
that is where the slope is zero. The
graph is steepest where the graph is
crossing the line for x = 0. This means
that the mass has the fastest speed at is
passes through the center. How would
we write the equation and show the
graph for velocity?
v = dx/dt = −Aωsin (ω t)
A comparison of the graphs shows that
the maximum velocity happens when x
= 0 and that maximum displacement
occurs when v = 0. What would the
acceleration equation and graph look
like?
The more that the spring is stretched or
compressed the more acceleration is
experienced by the mass. You would
expect position and acceleration graphs
to be similar. There is one small
drawback. If the mass is pulled to the
right the spring is pulling to the left. If
the mass is on the left of center the
acceleration is to the right. The negative
sign in the following equation accounts
for the opposite direction of position and
acceleration.
a = dv/dt = d 2 x/dt 2 = −Aω 2 cos (ω t)
Notice how when x = 0 then a =0 also.
Maximum values for acceleration and
position take place at the same time.
Shown below are the overlapped graphs
of v and x (upper) and a and x (lower).

Be sure that given either a graph or an
equation of position, velocity or
acceleration and the corresponding
constants that you can write the other
two equations or draw the other two
graphs.
Example #1
An object oscillates as shown in the
figure below of position in meters and
time in seconds. Answer the questions
according to the graph.
1. What is the maximum displacement?
2. How long does it take to complete
one cycle?
3. What is the angular frequency?
4. What are the units of slope of this
graph?
5. What does slope represent?
6. At what times in the graph is the
object at rest?
7. At what times after t = 0 does the
object have maximum speed?
8. What is the equation of the graph?
9. What would the equation of the
graph of velocity vs. time look like?
10. What does the slope of the velocity
graph have for units? What does the
slope of this graph tell us?
11. What does the equation of the
acceleration vs. time graph look like?
Kinematics in Terms of Position
So the previous page gives equations for
position, velocity and acceleration when
the time is known. You tell me when
you want to know what is happening and
I can use those equations to determine
the kinematics. What if I tell you where
the object is? Can you work through the
equations to find speed and acceleration?
Acceleration vs. position
Recall the equations from the last page.
x = A cos( ω t) and a = −Aω 2 cos (ω t).
Divide the acceleration equation by the
position equation. The “Acos (ω t)” in
each equation will cancel out leaving an
easy result for acceleration in terms of
position.
a = −ω 2 x
The maximum acceleration takes place
when x = ± A. Consider the graph in the
last problem. What is the acceleration
when x = +4m? Without the above
equation you would have to go back into
the position equation and solve for “t”
then plug into the acceleration equation
to solve for “a”. No need for that now!
a = - (2π rad / 4sec) 2 (4 m ) = -9.9 m/s 2
Speed vs. position
Now recall the equations x = A cos( ω t)
and v = −Aωsin (ω t). We will take
advantage of a trigonometric identity
that A 2 sin 2 ( stuff) + A 2 cos 2 (stuff) = A 2
where the above bold equations will be
used.
( v / ω ) 2 + x 2 = A 2
By isolating v you get the result:
_______
v = ±ω √ A 2 – x 2
Now you can determine the speed of the
object directly when a specific position
is known. We say speed rather than
velocity because the direction of the
object is lost in the above proof. How
fast is the object of the previous example
moving when it is at x = +4m?
Homework:
Prob.Ch 15:1, 3, 5, 11, 72, 76, 80(a-e)
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Lesson 2-10 Harmonic Differential Equation
Read Chapter 15:3, 5-7 & natural period of a system
Any system moves with simple harmonic motion
can be described by a single differential equation
as shown in the box to the right. It is critical that
the negative sign appear in the formula; other wise
there will be no oscillations. From this equation we see that if the second derivative of a
function leads to a negative constant times the original function then the system will
oscillate with a natural period that is found from τ = 2π/ω. In order to find the natural
period you can take the following general steps (See bottom half of page 392):
1. Draw a free-body diagram
2. Sum the forces or torques or both making the restoring force negative.
3. Look for small angle approximations in substitution of sinθ ≅θ or S = rθ.
4. Replace the acceleration with the corresponding differential definitions of
a = d 2 x/dt 2 or α = d 2 θ/dt 2 .
5. Collect all constants and the negative sign on the side of equation opposite to the
differential equation.
6. Equate collected and grouped constants to ω 2 = 4π 2 /τ 2 2
d 2
f =
−ω
f
2
dt
.
7. Solve for ω = 2πf = 2π/τ or f or τ.
Example#1-Mass on a Spring (see figure 16-5 on page 390)
ΣFx = -kx = ma ; a = -(k/m) x ; d 2 x/dt 2 = -(k/m) x so that ω = (k/m) 1/2 and you get
equations 16-11 and 16-12.
Example#2-Mass on a String (see figure 16-10 on page 395)
τ = 2π (L/g) 1/2
Example #3- Torsion Pendulum (see figure 16-9 on page 394
τ = 2π (I/κ) 1/2
Example#4-Physical Pendulum (see figure 16-11 on page 396)
τ = 2π (I/mgh) 1/2
You should be able to derive each of the four results above but you should also be
able to write down on paper the above working equation for each system from almost
instantaneous recall. Other physical systems are demonstrated in the Sample
Problems 15:4, 5 and 6; Example 13:4 on page 338
Homework: Prob. Ch 15: 13, 26, 43, 45, 46, 49, 50, 51, 55, 84, 106c
Take two days to complete. We will do a lab tomorrow.

Lesson 2-11 Energy Considerations
Read Chapter 15:4, 8
From the relationship between force and acceleration and by comparison to one of the
kinematics results we found a method for getting the period of oscillation of any system.
Could a similar result come from using energy and speed? Recall that the key players in
energy considerations are (1) total mechanical energy, (2) potential energy and (3) kinetic
energy.
Again look at the figure of the oscillating mass on page 2 of these notes. If we pull
the object back to x = +A and release we fix the total energy of the system to be a
specific amount. At any instant the total mechanical energy is the sum of kinetic energy
and potential energy. Since M is released from rest the initial kinetic energy is zero. The
potential energy at the instant of release is PE = ½ kx 2 or since x=+A the potential energy
is ½ kA 2 . The total energy at release is E = KE + PE or 0 + ½ kA 2 . As the mass moves
in towards the center the PE decreases while the KE increases yet the total energy will
remain fixed at ½ kA 2 . Shown below are graphs of total mechanical energy, potential
energy and kinetic energy as functions of position for the mass on page 2. The mass is
pulled back to x = +1m and released from rest. Spring constant is 10 N/m; mass is ½ Kg.
Total Mechanical Energy Spring Potential Energy Kinetic Energy
E = ½ kA 2 = 5J
No matter where M is
located the total energy is a
constant value of 5 J.
PE = ½ k x 2
The potential energy goes
from a maximum of 5J at
either end to 0 J in the
center.
KE = ½ k(A 2 – x 2 )
The kinetic energy is
merely total energy minus
potential energy or the
difference in the graphs.
Although potential energy and kinetic energy vary, they do so in such a way to keep the
total energy constant. What is also nice is the result you find algebraically if you take the
expression in the third column and set it equal to ½ mv 2 . By allowing ½ k (A 2 – x 2 ) to
be equal to ½ mv 2 and solving for the speed in terms of x you get the result shown on
page 4 and the result shown on page 5.
____ _________ _____
v = √(k/m) √ (A 2 – x 2 ) where ω = √(k/m)
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Energy vs. time
The previous text demonstrates conservation of energy in terms of position. What about
conservation of energy in terms of time? What do the equations and graphs of time look
like? We shall use the numbers from our previous example: k=10N/m, m = ½ Kg &
A=1m. As a result ω = √ ( 10 / 0.5) or 4.47 radians / second. The position graphs and
velocity graphs are shown below.
Position vs. time graph
x = 1m cos (4.47t)
__
Y1= 1cos(√(20)x)
for TI-83
Velocity vs. time graph
v = -1m (4.47/sec) sin(4.47t)
___ __
Y2= -(√(20) sin (√(20)x)
for TI-83
With the above equations and graphs in mind we can now consider graphs and equations
for PE , KE and ETOT.
Potential Energy vs. time
PE = ½ k x 2 = ½ (10){1cos(√(20)t)} 2
Y3 = 5 Y1 2 for TI-83
Note there are no negative energies.
Kinetic Energy vs. time
KE = ½ m v 2
= ½ (0.5){-1m (4.47/sec)sin(4.47t)} 2
Y4 = ¼ Y2 2 for TI-83
Again there are no negative energies.
By combining the equations for PE and KE you can conclude the total energy to be a
constant of E = 5Joules. You could also graph Y5 = Y3 + Y4 to convince yourself that the
total energy is time independent. Both of these recommendations are left as an exercise
to the diligent student.
Homework Problems Ch 15: 27 – 33, Good AP problems are 14, 35, 36.

Lesson 2-12 Energy Considerations
Read These Notes
Recall that we found the speed of a particle as a function of position in the first lesson of
this unit. When finding this expression we had not yet specified what type of system we
were considering. The expression is shown below.
By isolating v you get the result:
We now see a second method for determining the natural period of a spring system using
energy considerations.
1. You can use write an energy equation in terms of kinetic energy terms and a
position term for potential energy.
2. You then set your equation equal to ½ kA 2 .
3. Isolate or solve for speed.
4. Group all constants in front of the radical.
5. Set the constants equal to ω.
6. Use τ = 2π/ω to get the equation for the natural period.
I would not attempt this method on pendulum systems such as plane pendulums, torsion
pendulums or physical pendulums.
Some examples are shown below.
Example#1-Mass on a Spring (see figure 15-5 on page 390)
E = U + K or ½ kA 2 = ½ kx 2 + ½mv 2 and v = (k/m) 1/2 (A 2 – x 2 ) 1/2 so that ω = (k/m) 1/2
Example#2-Sample Problem 15:6 (see figure 15-12 on page 398)
E = U + K or ½ kA 2 = ½ kx 2 + ½ Iω 2 so that ½ kA 2 = ½ kx 2 + ½ (1/3 ML 2 )(v 2 /L 2 ) or
v = (k/3M) 1/2 (A 2 – x 2 ) 1/2 . You find that ω = (k/3M) 1/2
Example #3
Consider figure 15-60 and problem 106. For this case the object rolls without
slipping and there are two kinetic energy terms, translational and rotational.
E = U + K or ½ kA 2 = ½ kx 2 + ½ Iω 2 + ½ mv 2 so that ½ kA 2 = ½ kx 2 + ½ (1/2
MR 2 )(v 2 /R 2 ) + ½ Mv 2 or kA 2 – kx 2 = (3/2)Mv 2 so that v = (2k/3M) 1/2 (A 2 – x 2 ) 1/2 .
You find that ω = (2k/3M) 1/2 .
One final note about objects that roll without slipping must be made. You can use the
differential equation method discussed in a previous lesson with one twist. Summing the
torques about the contact point with the floor enables you to construct a pure rotational
equation, like a torsion pendulum about the contact point. Don’t forget that the inertia is
modified using the parallel-axis theorem.
Expect a test real soon!
_______
v = ±ω √ A 2 – x 2
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