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Simple Harmonic Motion
Simple Harmonic Motion
Simple harmonic motion is
the repetitive motion of an
oscillating object (simple
harmonic oscillator).
Example: A spring mass
system.
Like circular motion,
simple harmonic motion
has a regular period and
frequency.
In fact, simple harmonic
motion and circular motion
are linked both
mathematically and
graphically.
Equilibrium
The spring mass system
shown below is at
equilibrium ( x = 0 ).
Equilibrium for a spring is
its natural rest position.
This spring can either be
stretched or compressed,
displacing the mass from its
current equilibrium
position.
+F
0
To stretch the spring a
force, F , must be applied in
the +x direction.
This new unbalanced force
will displace the mass from
equilibrium.
A force is applied
As the mass is displaced the
spring generates an
increasing force of it own.
+F
0
A force is applied
−FS
0
As the mass is displaced the
spring generates an
increasing force of it own.
+F
A force is applied
−FS
0
As the mass is displaced the
spring generates an
increasing force of it own.
+F
A force is applied
As the mass is displaced the
spring generates an
increasing force of it own.
−FS
0
+F
A force is applied
As the mass is displaced the
spring generates an
increasing force of it own.
This force is known as a
restoring force, as it acts to
restore the spring to its
original equilibrium
position.
−FS
0
+F
Amplitude
When the applied force, F ,
and the restoring force, FS ,
are equal and opposite the
oscillator is stationary.
R = A = xmax
The oscillator has reached
maximum displacement, xmax .
Maximum displacement is
known as the amplitude, A .
This is equal to the radius, R ,
of a circle matching the
springs displacement.
−FS
0
x
+F
+A = xmax
The Oscillator is Released
When the applied force, F,
is removed, the restoring
force moves the mass left.
x
−FS
+F
x
0
+A
The Oscillator is Released
When the applied force, F,
is removed, the restoring
force moves the mass left.
x
−FS
x
0
+A
The Oscillator is Released
When the applied force, F,
is removed, the restoring
force moves the mass left.
x
−FS
x
0
+A
The Oscillator is Released
When the applied force, F,
is removed, the restoring
force moves the mass left.
x
−FS
x
0
+A
The Oscillator is Released
When the applied force, F,
is removed, the restoring
force moves the mass left.
x
−FS
x
0
+A
The Oscillator is Released
R = A = xmax
When the applied force, F,
is removed, the restoring
force moves the mass left.
ω
Δθ
x
The radius of the circle is
equal to the maximum
displacement (amplitude)
R = A = xmax
The radius, R = A , and
displacement, x , are two of the
sides of a right triangle.
While the mass moved to the
left the corresponding point on
the circle moved with an
angular velocity, ω , through
an angle, Δθ .
−FS
x
0
+A
Conclusion
R = A = xmax
ω
Δθ
x
An oscillator’s displacement
equals the x-component of
the radius of another object
that is in uniform circular
motion with the same
amplitude (R = A = xmax).
Cosine solves for x .
( )
x = Acos Dq
In uniform circular motion
Dq = w t
−FS
Therefore,
( )
x = Acos w t
x
0
+A
Conclusion
Period and frequency are
related to angular velocity.
R = A = xmax
ω
2p
1
T=
=
w
f
Which rearrange to
Δθ
x
2p
w=
= 2p f
T
These can be substituted
into the previous equation
( )
x = Acos w t
−FS
(
x = Acos 2p ft
x
0
+A
)
Putting all together
(
x = Acos 2p ft
)
The motion described by
this equation looks like…
x(t)
FS
x(t)
−A
0
+A
Putting all together
(
x = Acos 2p ft
)
The motion described by
this equation looks like…
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Putting all together
(
x = Acos 2p ft
ω
Δθ
x(t)
FS
x(t)
−A
0
+A
)
The motion described by
this equation looks like…
Graphing Oscillations
+
A
0
x(t)
−
A
FS
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
x(t)
+
A
0
−
A
FS
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
+
A
0
x(t)
−
A
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
+
A
0
x(t)
−
A
FS
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
+
A
0
x(t)
−
A
FS
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
+
A
0
x(t)
−
A
FS
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
+
A
0
x(t)
−
A
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
x(t)
+
A
0
−
A
FS
x(t)
−A
0
+A
π
2π
Putting all together
ω
Δθ
+
A
0
x(t)
−
A
FS
x(t)
−A
0
+A
π
2π
Analyzing Oscillation Graphs: Amplitude
Amplitude is the largest displacement and it can be easily read on
the graph.
A = 8m
Analyzing Oscillation Graphs: Period
T = 12s
Period is the time of one cycle (1 revolution). One cycle is one
complete wave form. This wave form extends from 3 seconds to 15 s.
The time of this wave form is 15 − 3 = 12 s
Analyzing Oscillation Graphs: Frequency
Frequency is the inverse of period.
1 1
f = = = 0.0833s
T 12
Analyzing Oscillation Graphs: Displacement
Displacement can be solved using the displacement equation and
substituting the values determined in the previous slides.
(
f = 0.0833s
) A = 8m
x = (8) cos ( 2p ( 0.0833) t )
the given time to determine the
x = (8) cos ( 0.523t ) Substitute
matching displacement.
x = Acos 2p ft
Analyzing Oscillation Graphs: Maximum Speed
Maximum speed can be found using the equation for speed in uniform
circular motion. Set the radius equal to amplitude.
2p r
v=
T
()
( )
2p A 2p 8
v=
=
= 4.19m s
T
12
Analyzing Oscillation Graphs: Qualitatively
When is the object moving to the right at maximum speed?
This occurs at equilibrium (x = 0) and when object is moving toward a
greater displacement value: 9 and 21 seconds
When is the object moving to the left at maximum speed?
This occurs at equilibrium (x = 0) and when object is moving toward a
lesser displacement value: 3 and 15 seconds
When is the object instantaneously at rest?
This occurs at maximum displacement: 0, 6, 12, 18, and 24 seconds.
Circular Motion
R
Δθ
x(t)
y(t)
ω
Oscillation is only concerned with
the component of the circular
motion equation that is parallel to
the oscillation.
If a spring is oscillating horizontally
( )
x(t) = Acos w t
If a spring is oscillating vertically
( )
y(t) = Asin w t
Circular motion involves both equations simultaneously.
Amplitude is equal radius R , and Δθ = ω t .
x(t) = Rcos Dq
y(t) = Rsin Dq