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Oscillations

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					Oscillations
Consider a spring-mass system
undergoing oscillations or vibrations
on a frictionless horizontal surface.
        -x                           +x

                                 Fnet = 0 N


                     X= 0 (equilibrium position)
When the displacement is positive the
restoring force is negative.
                              +x

             Fnet



                    X= 0
When the displacement is negative the
restoring force is positive.
           -x

                              Fnet



                    X= 0
     Amplitude =     X

The amplitude is equal to the
maximum displacement from the
equilibrium position.
The period (T) is the
  length of time it
requires to complete
   one oscillation
  The frequency (ƒ)is
equal to the number of
  oscillations in one
 second or how often
 an oscillation occurs.
     T = 1/ƒ
The unit for period is the second.

  1 / s = 1 hertz = 1 Hz
Which represents one oscillation
         per second.
Any repetitive motion
 such as a mass on a
spring or a pendulum
 can be described as
  harmonic motion
     Simple Harmonic
      Motion (SHM)

SHM occurs if the restoring
force is directly proportional
to the displacement.
Therefore in SHM the
   restoring force:
       F = -kx
  Free-body diagram

                                    k x

       x
                                    mg
                     m
Gently place a mass on the spring
so that it can hang at rest
      Apply a light
     force to extend
        the spring
x
     k (x+ x)
 x


     mg
x               New
                 equilibrium
                 position
 x
     k (x+ x)

     mg
   The force acting on the mass is

 F = -k (x+ x) + mg   = -k (x+ mg ) + mg
                                 k

      therefore F = -k x
So a good rule of practice … for vertical
springs: find the new equilibrium point,
measure the displacements from there …
and ignore gravity.
Period of a Pendulum
  Another example of Simple
      Harmonic Motion
Period as a function of Length
Using radian measure x = L  which represents the arc
path of the pendulum or its actual displacement.



      x=L                    
   Frestore = - mg (sin 
   )
           Frestore = - kx



                                   x
                                       L mg (sin  )
                             x=0
                                Period of a Pendulum vs Length

             25




             20
Period (s)




             15




                                         Period is directly proportional to the
             10
                                         ________ Root of the Length.

              5




              0
                  0   20   40       60         80         100      120       140   160
                                          Lenght (cm)
140




120




100


                            1/2
 80             y = k (x)
Y




 60



                                                2
 40
            y = k (x)                 y = k (x)
 20




    0
        0   2           4         6         8       10   12   14
                                       X
                                         Period as a function of Square Root of the Length

           25
Period of the Pendulum




           20

                                 Period as a function of Square
                                 Root of the Length
           15
                                                                                             ^T

           10




                  5

                                                     ^ L
                  0
                         0   2              4            6           8           10          12   14
                                                  Square Root of the Length
Equating Acceleration (ax) a
Pendulum to the Centripetal
       Acceleration


 Ac = 4π2r    Then T2 = 4π2r
       T2               Ac
SHM (reference circle)
  using Newton’s 2nd Law
    Frestore = - Kx = ma

                   a = -K x
                        m
which infers that any object in SHM will
have an acceleration that is directly
proportional to its displacement.
If an object is moving in a vertical
 circle with constant speed v0 and
 radius R. Consider a light source
  shining from above that casts a
     shadow below the object
        v0

    R

             Shadow moves back and forth with properties
             equal to the horizontal component of the
             circularly moving object.
Can we show that acceleration of the
   shadow is proportional to the
          displacement.

                                Recall from unit circle that:
              At the ends          x = R cos 
         v0   vhorizontal = 0
                                   ac = v02 cos
     R
                                        R

                            ac = v02 R(cos) = v02 (x)
                                 R2           R2
                  +x             Since v and R are constant,
                                 ac is proportional to x
Again since v and R are constant, ac is
proportional to x. So if the acceleration of
the shadow is directly proportional to its
displacement, this motion is SHM.
                                   Recall from unit circle that:
                 At the ends          x = R cos 
            v0   vhorizontal = 0
                                      ac = v02 cos
       R
                                           R

                               ac = v02 R(cos) = v02 (x)
                                    R2           R2
                     +x
 If we reverse our argument, we
  will find that for every system
    undergoing SHM, you can
visualize a reference circle where
   the uniform circular motion
corresponds to the real system…
Correspondence of circle to system
Reference circle             Real System
• Radius (R)                 • Amplitude (A)
• Constant tangential        • Maximum speed (v0 )
  speed
• Period of revolution (T)   • Period of oscillation (T)

  2πA = v0T             > T = 2πA
                               v0
Using a spring-mass system, we can
  compare the energy stored in the
   spring at full extension to the
 kinetic energy as it passes through
        the equilibrium point
   ES = EK
½ kSA2 = ½ m v02      A =    m
                      v0     kS
          A =   m
          v0    kS

Substituting into our equation
 for the systems period (T):
                          The period
T = 2πA = 2π m            depends on the
     v0              kS   system
                          properties
To determine the period of a pendulum
  we can replace kS with      mg/L


Spring-mass
system
              T = 2π m
                                   kS
                                                 The period of a
Pendulum      T = 2π               L             pendulum is
                                                 independent of
                                   g             mass
  This formula we had previously derived from one of your first
  labs and then through vector analysis of centripetal acceleration
  experienced during uniform circular motion.

				
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posted:3/7/2012
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