An Angular Simple Harmonic Oscillator

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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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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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• 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 Simple 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 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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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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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 Physical Pendulum

• The other difference in this case is that the
  rotational inertia will not be a simple I mL2
  but rather something more complicated which will
  depend on the shape of the body

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The Physical Pendulum

• For any physical pendulum that oscillates around
  a point O with period T, there is a simple
  pendulum of length L0 which oscillates with the
  same period
• The point P on the physical pendulum a distance
  L0 from O is called the center of oscillation

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Measuring g

• The equation for the period of a physical
  pendulum gives us a very nice and neat
  relationship between T, I and g

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• Suppose we have a uniform rod of length L which
  we allow to rotate from one end as shown
• the moment of inertia I mL2/3
• h L/2

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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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Sample Problem 1
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Sample Problem 2

• What is the distance L0 between the pivot point
  of the stick and the center of oscillation of the
  stick?

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P 17.15
Value of r for which T is minimum can be obtained
by solving
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P 17.16
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P 17.18
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P 17.11
Momentum conservation gives
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E 17.32
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P 17.21
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P 17.22
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P 17.19
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2nd Method
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• At the right we have a plot of data recorded by
  Galileo of an object (the moon Callisto) that
  moved back and forth relative to the disk of
  Jupiter

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• The circles are Galileos data points and the
  curve is a best fit to that data
• This would strongly suggest that Callisto
  exhibits SHM

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• But in fact Callisto is moving with pretty much a
  constant speed in a nearly circular orbit about
  Jupiter
• So what is it that we are seeing in the data?

• What Gallileo saw and what the data tells us
• is that SHM is the projection of uniform
• circular motion in the plane of the motion
• In other words, SHM is uniform circular motion
  viewed edge-on

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• SHM is the projection of uniform circular motion
  on a diameter of the circle in which the latter
  motion takes place

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Simple Harmonic Motion Uniform Circular Motion

• We have at the right a reference circle the
  particle at point P is moving on that circle at
  a constant angular speed ?
• The radius of our reference circle is xm
• Finally, the projection of the position of P
  onto the x axis is the point P

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Simple Harmonic Motion Uniform Circular Motion

• We can easily see that the position of the
  projection point P is given by the formula
• Which is the formula for the displacement of an
  object exhibiting SHM

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Simple Harmonic Motion Uniform Circular Motion

• Similarly, if we look at the velocity of our
  particle (and use the relationshipv ?r) we can
  see that it obeys

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Simple Harmonic Motion Uniform Circular Motion

• And finally, if we look at the acceleration of
  our particle (and use the relationship ar ?2r)
  we can see that it obeys

• So regardless of whether we look at displacement,
  velocity or acceleration, we see that the
  projection of uniform circular motion does indeed
  obey the rules of SHM

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Theory for Problem Nos17.40,17.4117.42
Lissajous figure
Curve which shows the path followed by a particle
having SHM in two perpendicular directions
simultaneously is called Lissajous figure on the
name Mr. J.A.Lissajous(1822-1880)
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Lissajous with irrational frequency ratios
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