Simple Harmonic Motion

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Simple Harmonic Motion

• Simple harmonic motion (SHM) refers to a certain
  kind of oscillatory, or wave-like motion that
  describes the behavior of many physical
  phenomena
• 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 on the particle.
• Displacement The distance of the particle from
  its equilibrium position. Usually denoted as x(t)
  with x0 as the equilibrium position.
• Amplitude the maximum value of the displacement
  with out 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), who discovered radio waves.
• frequency and period are related as follows
• Since a cycle is 2p radians, the relationship
  between frequency and angular frequency is

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• http//www.physics.uoguelph.ca/tutorials/shm/Q.shm
  .html

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Here is a ball moving back and forth with simple
harmonic motion (SHM)
Its position x as a function of time t is
                   where A is the amplitude of
motion the distance from the centre of motion
to either extreme T is the period of motion the
time for one complete cycle of the motion.
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Restoring Force
How does the restoring force act with respect to
the displacement from the equilibrium position?
F is proportional to -x
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 in sign.
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Springs and SHM

• Attach an object of mass m to the end of a
  spring, pull it out to a distance A, and let it
  go from rest. The object will then undergo simple
  harmonic motion
• What is the angular frequency in this case?
• Use Newtons 2nd law, together with Hookes law,
  and the above description of the acceleration to
  find

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Springs and Simple Harmonic Motion
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Equations of Motion
Conservation of Energy allows a calculation of
the velocity of the object at any position in its
motion
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Conservation of Energy For A Spring in Horizontal
Motion

• E Kinetic Elastic
  Potential
• E ½ mv2 ½ kx2
  Constant
• At maximum displacement, velocity is zero and all
  energy is elastic potential,
• so total energy is equal to

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• To find the velocity _at_ any displacement do
  conservation of Energy _at_ some point at max
  displacement
• ½ mv2 ½ kx2 1/2kA2
• Solving for v
• This is wonderful but what about time??!
• vdx/dt and separate variables!
• ?

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SHM Solution...

• Drawing of A cos(?t )
• A amplitude of oscillation

T 2?/?
A
?
??
?
??
-??
A
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SHM Solution...

• Drawing of A cos(?t ?)

?
?
??
?
??
-??
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SHM Solution...

• Drawing of A cos(?t - ?/2)

? ??/2
A
?
??
?
??
-??
A sin(?t)!
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1989.M3 3.0 m/s, 0.63 s, 0.098 m, 0.41 m, 0.31 m

• 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 the spring. 3 m/s
• b. Determine the period of the simple harmonic
  motion that ensues. 0.63s
• 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 m
• e. Determine the amplitude of the simple harmonic
  motion. 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!

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Pendulums

• When we were discussing the energy in a simple
  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
  a simple pendulum is stored in raising the bob up
  against the gravitational force
• The pendulum bob is clearly oscillating as it
  moves back and forth but is it exhibiting SHM?

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• We can see that the restoring force is
• F -mg sin?
• If we assume that the
• angle ? is small, then
• sin? ?and our equation
• becomes
• F -mg? - mgs/L
• - (mg/L)s

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• F -mg? - mgs/L
• - (mg/L)sdividing both sides
  by m,
• a -(g/L)s.
• Equate to a -? 2 x, and
• ? (g/L)1/2

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The Physical Pendulum

• Now suppose that the mass is not all concentrated
  in the bob?
• In this case the equations are exactly the same,
  but the restoring force acts through the center
  of mass of the body (C in the diagram) which is a
  distance h from the pivot 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

• where L is the moment arm of the applied force

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The Simple Pendulum
If we substitute t Ia, we get

• This doesnt appear too promising until we make
  the following assumption
• that ? is small
• If ? is small we can use the approximation that
  sin ? ? ?
• (as long as we remember to express ? in radians)

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The Physical Pendulum

• Making the substitution we then get

which we can then rearrange to get
which is the angular equivalent to
a -? 2 x

• So, we can reasonably say that the motion of a
  pendulum is approximately SHM if the maximum
  angular amplitude is small

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The Physical Pendulum

• Making the substitution we then get

which we can then rearrange to get
which is the angular equivalent to
a -? 2 x

• So, we can reasonably say that the motion of a
  pendulum is approximately SHM if the maximum
  angular amplitude is small

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The Physical Pendulum

• So we go back to our previous equation for the
  period and replace L with h to get

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The Simple Pendulum
The period of a pendulum is given by

• where I is the moment of inertia of the pendulum

• If all of the mass of the pendulum is
  concentrated in the bob, then I mL2 and we
  get

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An AngularSimple Harmonic Oscillator

• The figure shows an angular version of a simple
  harmonic oscillator
• In this case the mass rotates around its center
  point and twists the suspending wire
• This is called a torsional pendulum with torsion
  referring to the twisting motion

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Torsional Oscillator

• If the disk is rotated through an angle (in
  either direction) of ?, the restoring torque is
  given by the equation

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Sample Problem
A uniform bar with mass m lies symmetrically
across two rapidly rotating, fixed rollers, A and
B, with distance L 2.0 cm between the bars
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?
? 14 rad/s
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Sample Problem 1

• A 1 meter stick swings about a pivot point at one
  end at a distance 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 penguin dives, 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