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Oscillations and Simple Harmonic Motion:
Mechanics C
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Oscillatory Motion
Oscillatory Motion is repetitive back and forth motion about an equilibrium position
Oscillatory Motion is periodic.Swinging motion and vibrations are forms of
Oscillatory Motion.
Objects that undergo Oscillatory Motion are called Oscillators.
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Simple Harmonic Motion
The time to complete one full cycle of
oscillation is a Period.
T 1f
f 1T
The amount of oscillations per second is called frequency and is measured in Hertz.
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What is the oscillation period for the broadcast of a 100MHz FM radio station?
Heinrich Hertz produced the first artificial radio waves back
in 1887!
T 1f
11108Hz
110 8s 10ns
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Simple Harmonic Motion
The most basic of all types of oscillation is depicted on
the bottom sinusoidal graph. Motion that follows
this pattern is called simple harmonic motion or SHM.
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Simple Harmonic Motion
An objects maximum displacement from its equilibrium position is
called the Amplitude (A) of the motion.
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What shape will a velocity-time graph have for SHM?
Everywhere the slope (first derivative) of the position graph is zero, the velocity
graph crosses through zero.
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2cos tx t AT
We need a position function
to describe the motion above.
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Mathematical Models of SHM
2cos tx t AT
cos 2x t A ft
cosx t A t
1Tf
2T
x(t) to symbolize position as a function of
time
A=xmax=xmin
When t=T, cos(2π)=cos(0)
x(t)=A
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Mathematical Models of SHM
sinv t A t
cosx t A t
d x tv t
dt
In this context we will call omega Angular
Frequency
What is the physical meaning of the product (Aω)?
maxv AThe maximum speed of an oscillation!
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Recall: Hooke’s LawHere is what we want to do: DERIVE AN EXPRESSION THAT DEFINES THE DISPLACEMENT FROM EQUILIBRIUM OF THE SPRING IN TERMS OF TIME.
0)(2
2
2
xmk
dtxd
dtxdmkx
dtxdamakx
maFkxF Netspring
WHAT DOES THIS MEAN? THE SECOND DERIVATIVE OF A FUNCTIONTHAT IS ADDED TO A CONSTANT TIMES ITSELF IS EQUAL TO ZERO.What kind of function will ALWAYS do this?
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Example:
An airtrack glider is attached to a spring, pulled 20cm to the right, and
released at t=0s. It makes 15 oscillations in 10 seconds.
What is the period of oscillation?15
10sec11.5oscilationsf HzT
1 1 0.671.5
T sf Hz
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Example:
An airtrack glider is attached to a spring, pulled 20cm to the right, and
released at t=0s. It makes 15 oscillations in 10 seconds.
What is the object’s maximum speed?
max2Av AT
max
0.2 21.88 /
0.67m
v m ss
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Example:
An airtrack glider is attached to a spring, pulled 20cm to the right, and
released at t=0s. It makes 15 oscillations in 10 seconds.
What are the position and velocity at t=0.8s?
cos 0.2 cos 0.8 0.0625x t A t m s m
sin 0.2 sin 0.8 1.79 /v t A t m s m s
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Example:
A mass oscillating in SHM starts at x=A and has period T. At what time, as
a fraction of T, does the object first pass through 0.5A?
2cos
( ) 0.5
tx t AT
x t A
20.5 cos tA AT
1cos 0.52T t
2 3T t
6
Tt
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Model of SHM
When collecting and modeling data of SHM your mathematical model had a value as shown below:
x(t) Acos t
x(t) Acos t C What if your clock didn’t start at x=A or x=-A?
This value represents our initial conditions. We call it the phase angle:
x(t) Acos t
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SHM and Circular Motion
Uniform circular motion projected onto one dimension is simple harmonic motion.
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SHM and Circular Motion
x(t) Acos
ddt
t
x(t) Acos t
Start with the x-component of position of the particle in UCM
End with the same result as the spring in SHM!
Notice it started at angle zero
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Initial conditions:
t 0
We will not always start our clocks at one amplitude.
x(t) Acos t 0
vx (t) Asin t 0
vx (t) vmax sin t 0
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The Phase Constant:
t 0
Phi is called the phase of the oscillation
Phi naught is called the phase constant or phase shift. This
value specifies the initial conditions.
Different values of the phase constant correspond to different starting points on the circle and thus to
different initial conditions
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Phase Shifts:
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An object on a spring oscillates with a period of 0.8s and an amplitude of 10cm. At t=0s, it is 5cm to the left of
equilibrium and moving to the left. What are its position and direction of motion at t=2s?
x(t) Acos t 0
x0 5cm Acos 0 Initial conditions:
0 cos 1 x0
A
cos 1 5cm
10cm
120
23 rads
From the period we get:
2T
2
0.8s7.85rad /s
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An object on a spring oscillates with a period of 0.8s and an amplitude of 10cm. At t=0s, it is 5cm to the left of
equilibrium and moving to the left. What are its position and direction of motion at t=2s?
x(t) Acos t 0
7.85rad /s
0 23 rads
A 0.1mt 2s
x(t) 0.1cos 7.85 2 23
x(t) 0.05m
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We have modeled SHM mathematically. Now comes the physics.
Total mechanical energy is conserved for our SHM example of a spring with
constant k, mass m, and on a frictionless surface.
E K U 12
mv2 12
kx2
The particle has all potential energy at x=A and x=–A, and the particle has purely kinetic energy at x=0.
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At turning points:
E U 12
kA2
At x=0:
E k 12
mvmax2
From conservation:
12
kA2 12
mvmax2
Maximum speed as related to amplitude:
vmax km
A
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From energy considerations:
From kinematics:
Combine these:
vmax km
A
vmax A
km
f 1
2km
T 2 mk
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a 500g block on a spring is pulled a distance of 20cm and released. The subsequent oscillations are measured to
have a period of 0.8s. at what position or positions is the block’s speed 1.0m/s?
The motion is SHM and energy is conserved.
12
mv2 12
kx2 12
kA2
kx2 kA2 mv2
x A2 mk
v2
x A2 v2
2
2T
2
0.8s7.85rad /s
x 0.15m
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Dynamics of SHM
Acceleration is at a maximum when the particle is at maximum and minimum displacement from x=0.
ax dvx (t)
dt
d Asin t dt
2Acos t
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Dynamics of SHM
Acceleration is proportional to the
negative of the displacement.
ax 2Acos t
ax 2x
x Acos t
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Dynamics of SHM
As we found with energy considerations:
ax 2x
F max kx
max kx
ax km
x
According to Newton’s 2nd Law:
ax d2xdt 2
Acceleration is not constant:
d2xdt 2
km
x
This is the equation of motion for a mass on a spring. It is of a general
form called a second order differential equation.
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2nd-Order Differential Equations:
Unlike algebraic equations, their solutions are not numbers, but functions.
In SHM we are only interested in one form so we can use our solution for many objects undergoing SHM.
Solutions to these diff. eqns. are unique (there is only one). One common method of solving is guessing the
solution that the equation should have…
d2xdt 2
km
xFrom
evidence, we expect
the solution:
x Acos t 0
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2nd-Order Differential Equations:
Let’s put this possible solution into our equation and see if we guessed right!
d2xdt 2
km
x
IT WORKS. Sinusoidal oscillation of SHM is a result of Newton’s laws!
x Acos t 0
d2xdt 2 2Acos t
dxdt
Asin t
2Acos t km
Acos t
2 km
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Vertical springs oscillate differently than horizontal springs because there are 2 forces acting.
The equilibrium position gets shifted downward
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What about vertical oscillations of a spring-mass system??
Fnet kL mg 0Hanging at rest:
kL mg
L mk
g
this is the equilibrium position of the system.
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Now we let the system oscillate. At maximum:
But:
Fnet k L y mg
Fnet kL mg ky
kL mg 0So:
Fnet ky
Everything that we have learned about horizontal oscillations is equally valid for
vertical oscillations!
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You need to show how to derive the Period of a Pendulum equationT = 2∏√l/g
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The Pendulum
Fnet t mgsin ma t
d2sdt 2 gsin
Equation of motion for a pendulum
s L
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Small Angle Approximation:
d2sdt 2 gsin
When θ is about 0.1rad or less, h and
s are about the same.
sin
cos 1
tan sin 1
d2sdt 2 g
sL
Fnet tm d2s
dt 2 mgsL
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The Pendulum
Equation of motion for a pendulum
d2sdt 2
gsL
gL
(t) max cos t 0
x(t) Acos t 0
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A Pendulum Clock
What length pendulum will have a period of exactly 1s?
gL
T 2 Lg
g T2
2
L
L 9.8m/s2 1s2
2
0.248m
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Conditions for SHM
Notice that all objects that we look at are described
the same mathematically.
Any system with a linear restoring force will undergo simple
harmonic motion around the equilibrium position.
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A Physical Pendulum
d2dt 2
mglI
I mgd mglsin
when there is mass in the
entire pendulum, not just the bob.
Small Angle Approx.
mgl
I
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Damped Oscillations
All real oscillators are damped oscillators. these are any that slow
down and eventually stop.a model of drag force for
slow objects:
Fdrag bv
b is the damping constant (sort of like a coefficient of friction).
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Damped Oscillations
F Fs Fdrag kx bv ma
kx b dxdt
m d2xdt 2 0
Another 2nd-order diff eq.
Solution to 2nd-order diff eq:
x(t) Ae bt / 2m cos t 0
km
b2
4m2
02
b2
4m2
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Damped Oscillations
x(t) Ae bt / 2m cos t 0
A slowly changing line that provides a border to
a rapid oscillation is called the envelope of
the oscillations.
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Driven Oscillations
Not all oscillating objects are disturbed from rest then allowed to move undisturbed.
Some objects may be subjected to a periodic external force.
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DrivenOscillations
All objects have a natural frequency at which they tend to vibrate when disturbed.
Objects may be exposed to a periodic force with a particular driving frequency.
If the driven frequency matches
the natural frequency of an
object, RESONANCE occurs
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SpringBest example of simple harmonic oscillator.
T = 2m/k
m
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Simple PendulumActs as simple harmonic oscillator only when angle of swing is small.
T = 2L/g
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Conical PendulumNot really a simple harmonic oscillator, but equation is similar to simple pendulum.
T = 2L(cos )/g
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The small angle approximation for a simple pendulum
mg
mgcos
mgsinglT
TLgLg
ifLgmLLmg
IFr
pendulum
2
2,
0)(
sin,sin)()(sin
sin2
A simple pendulum is one where a mass is located at the end of string. The string’s length represents the radius of a circle and has negligible mass.
Once again, using our sine function model we can derive using circular motion equations the formula for the period of a pendulum.
If the angle is small, the “radian” value for theta and the sine of the theta in degrees will be equal.
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Torsional PendulumTwists back and forth through equilibrium position.
T = 2I/
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Physical PendulumAnything that doesn’t fall into any of the other categories of pendulums.
T = 2I/
= Mgd
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K1 + U1 = K2 + U2◦K = 1/2mv2
◦U = mgh1/2mv1
2+mgh1 =1/2mv22+ mgh2
v12 + 2gh1 = v2
2+ 2gh2
Energy Conservation in Pendulums
h
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The Physical Pendulum
mgdIT
TImgdImgd
ifImgd
LdIdmg
IFr
pendulumphysical
2
2,
0)(
sin,2,sin
sin
A physical pendulum is an oscillating body that rotates according to the location of its center of mass rather than a simple pendulum where all the mass is located at the end of a light string.
It is important to understand that “d” is the lever arm distance or the distance from the COM position to the point of rotation. It is also the same “d” in the Parallel Axes theorem.
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ExampleA spring is hanging from the ceiling. You know that
if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:
kkkxFs )3)((330Spring Constant
Angular frequency 5
1102
mk
mk
110 N/m
4.7 rad/s
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ExampleA spring is hanging from the ceiling. You know
that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:
Amplitude
Frequency and Period
Stated in the question as 1.5 m
7.4222
7.42
22
T
f
Tf
0.75 Hz
1.34 s
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ExampleA spring is hanging from the ceiling. You know
that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:
2
22
)5.1)(110(21
21
21
U
kAkxU sTotal Energy
Maximum velocity )7.4)(5.1(Av
123.75 J
7.05 m/s
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Position of mass at maximum velocity
Maximum acceleration of the mass
Position of mass at maximum acceleration
At the equilibrium position
)5.1()7.4( 22Aa 33.135 m/s/s
At maximum amplitude, 1.5 m
A spring is hanging from the ceiling. You know that if you elongate the spring by 3.0 meters, it will take 330 N of force to hold it at that position: The spring is then hung and a 5.0-kg mass is attached. The system is allowed to reach equilibrium; then displaced an additional 1.5 meters and released. Calculate the:
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a) Find the length of a simple seconds pendulum.
b) What assumption have you made in this calculation?
A “seconds pendulum” beats seconds; that is, it takes 1 s for half a cycle.
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Problem #2
A thin uniform rod of mass 0.112 kg and length 0.096 m is suspended by a wire through its center and perpendicular to its length. The wire is twisted and the rod set to oscillating. The period is found to be 2.14 s. When a flat body in the shape of an equilateral triangle is suspended similarly through its center of mass, the period is 5.83. Find the rotational inertia of the triangle.
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Problem #3
A uniform disk is pivoted at its rim. Find the period for small oscillations and the length of the equivalent simple pendulum.
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THE
END