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Simple Harmonic Motion
• Simple harmonic motion (SHM) refers to acertain kind of oscillatory, or wave-like motionthat describes the behavior of many physicalphenomena:
– a pendulum – a bob attached to a spring
– low amplitude waves in air (sound), water, the ground
– the electromagnetic field of laser light
– vibration of a plucked guitar string – the electric current of most AC power supplies
– …
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SHM Position, Velocity, and Acceleration
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Simple Harmonic Motion
Periodic Motion: any motion of system
which repeats itself at regular, equal
intervals of time.
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Simple Harmonic Motion
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Simple Harmonic Motion
• Equilibrium: the position at which no net force acts onthe particle.
• Displacement: The distance of the particle from itsequilibrium position. Usually denoted as x(t) with x=0 asthe equilibrium position.
• Amplitude: the maximum value of the displacement without regard to sign. Denoted as xmax or A.
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The period and frequency of a
wave• the period T of a wave is the amount of time it takes to
go through 1 cycle
• the frequency f is the number of cycles per second
– the unit of a cycle-per-second is commonly referred to as
a hertz (Hz), after Heinrich Hertz (1847-1894), whodiscovered radio waves.
• frequency and period are related as follows:
• Since a cycle is 2p radians, the relationship betweenfrequency and angular frequency is:
T
f 1
f p 2
T
t
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• http://www.physics.uoguelph.ca/tutorials/s
hm/Q.shm.html
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Its position x as a function of time t is:
where A is the amplitude of motion : the distance from thecentre of motion to either extreme
T is the period of motion: the time for one complete cycle
of the motion.
Here is a ball moving back and forth with simple
harmonic motion (SHM):
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Restoring Force
Simple harmonic motion is the motion executed by a particle of mass m subject to
a force that is proportional to the displacement of the particle but opposite insign.
How does the restoring force act with respect to the displacement from the
equilibrium position?
F is proportional to -x
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Springs and SHM
• Attach an object of mass m to the end of a spring, pull it out to adistance A, and let it go from rest. The object will then undergo simpleharmonic motion:
• What is the angular frequency in this case?
– Use Newton’s 2nd law, together with Hooke’s law, and the
above description of the acceleration to find:
)sin()( t At v
)cos()( t At x
)cos()( 2t At a
m
k
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Springs and Simple Harmonic Motion
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Equations of Motion
Conservation of Energy allows a calculation of the velocityof the object at any position in its motion…
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Conservation of Energy For A Spring in Horizontal MotionE = Kinetic + Elastic Potential
E = ½ mv2 + ½ kx2 = Constant
• At maximum displacement, velocity is zero and all energy iselastic potential,
so total energy is equal to
½ kA2
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• To find the velocity @ any displacement… doconservation of Energy… @ some point at max
displacement• ½ mv2 + ½ kx2 = 1/2kA2
• Solving for v
• This is wonderful but what about time??!
v=dx/dt and separate variables!
22 x Am
k
dt
dx
ct m
k A
dt
dx sin
2 2k
v A x m
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2 2k v A x
m
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SHM Solution...
• Drawing of A cos( t )
• A = amplitude of oscillation
p p 2p
T = 2 p/
A
A2p
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SHM Solution...
• Drawing of A cos( t + )
p p 2p2p
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SHM Solution...
• Drawing of A cos( t - p/2 )
A
= p/2
p p 2p
= A sin( t)!
2p
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1989M3. A 2-kilogram block is dropped from a height of 0.45 m above an
uncompressed spring, as shown above. The spring has an elastic
constant of 200 N per meter and negligible mass. The block strikes the
end of the spring and sticks to it.
a. Determine the speed of the block at the instant it hits the end of thespring. 3 m/s
b. Determine the period of the simple harmonic motion that ensues. 0.63s
c. Determine the distance that the spring is compressed at the instant the
speed of the block is maximum. 0.098 m
d. Determine the maximum compression of the spring. 0.41 me. Determine the amplitude of the simple harmonic motion. 0.31 m
1989.M3 3.0 m/s, 0.63 s,
0.098 m, 0.41 m, 0.31 m
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SHM So Far
• The most general solution is x = A cos( t + )
where A = amplitude
= angular frequency
= phase
• For a mass on a spring
– The frequency does not depend on the amplitude!!! – We will see that this is true of all simple harmonic
motion!
m
k
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Pendulums
• When we were discussing the energy in asimple harmonic system, we talked about the
‘springiness’ of the system as storing the
potential energy
• But when we talk about a regular pendulum
there is nothing ‘springy’ – so where is the
potential energy stored?
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The Simple Pendulum
• As we have already seen,the potential energy in asimple pendulum is stored inraising the bob up againstthe gravitational force
• The pendulum bob is clearlyoscillating as it moves backand forth – but is it exhibitingSHM?
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• We can see that therestoring force is:
F = -mg sinθ If we assume that the
angle θ is small, then
sinθ ~ θ and our equation
becomes
F ~ -mgθ = - mgs/L
= - (mg/L)s
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F ~ -mgθ = - mgs/L
= - (mg/L)s dividing both sides by m,
a = -(g/L)s.
Equate to a = -ω 2 x , and
ω = (g /L ) 1/2
The Physical Pendulum
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The Physical Pendulum
• Now suppose that themass is not allconcentrated in thebob?
• In this case theequations are exactlythe same, but therestoring force actsthrough the center of mass of the body (C in
the diagram) which isa distance h from thepivot point
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• Going back to our
definition of torque,we can see that
the restoring force
is producing a
torque around the
pivot point of:
sin g F L
•where L is the moment
arm of the applied
force
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The Simple Pendulum
• This doesn’t appear too promising until we make thefollowing assumption –
• that θ is small…
• If θ is small we can use the approximation thatsin θ θ
• (as long as we remember to express θ in radians)
I F L g sin
If we substitute τ = Iα, we get:
The Ph sical Pend l m
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• Making the substitution we then get:
I F L g
I
mgL
The Physical Pendulum
•So, we can reasonably say that the motion of apendulum is approximately SHM if the maximum
angular ampl i tude is smal l
which is the angular equivalent to
a = -ω 2 x
which we can then rearrange to get:
The Physical Pendulum
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• Making the substitution we then get:
I F L g
I
mgL
The Physical Pendulum
•So, we can reasonably say that the motion of apendulum is approximately SHM if the maximum
angular ampl i tude is smal l
which is the angular equivalent to
a = -ω 2 x
which we can then rearrange to get:
Th Ph i l P d l
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The Physical Pendulum
• So we go back toour previousequation for theperiod andreplace L with h to
get:
mgh
I
T p 2
Th Si l P d l
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The Simple Pendulum
where I is the moment of inertia of the pendulummgL
I T p 2
The period of a pendulum is given by:
•If all of the mass of the pendulum is concentrated in the
bob, then I = mL
2
and we get:
g
LT p 2
A A l
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An Angular
Simple Harmonic Oscillator
• The figure shows anangular version of asimple harmonic oscillator
• In this case the mass
rotates around it’s center point and twists thesuspending wire
• This is called a torsional
pendulum with torsionreferring to the twistingmotion
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Torsional Oscillator • If the disk is rotated through an angle (in either
direction) of θ , the restoring torque is given by theequation:
kx F
withComparing
mreplacedhasIandk replacedhasWhere
2givesWhich
p I T
S l P bl
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Sample ProblemA uniform bar with mass m lies symmetrically across two
rapidly rotating, fixed rollers, A and B, with distance L = 2.0cm between the bar’s center of mass and each roller. The
rollers slip against the bar with coefficient of friction µk = 0.40.
Suppose the bar is displaced horizontally by a distance x and
then released. What is the angular frequency of the resulting
horizontal motion of the bar?
ω = 14 rad/s
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Sample Problem 1
• A 1 meter stick swings about a
pivot point at one end at adistance h from its center of
mass
• What is the period of oscillation?
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A penguin dives from a uniform board that is hinged
at the left and attached to a spring at the right. The
board has length L = 2.0 m and mass m = 12 kg; the
spring constant k = 1300 N/m. When the penguindives, it leaves the board and spring oscillating with a
small amplitude.
Assume that the board is stiff enough not to bend,
and find the period T of the oscillations.
T = 0.35 s