Given a potential energy graph by B3x0vy

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									Given a potential energy graph, oscillations will occur between turning points
determined by
                            E  U xt 



                                                  E




                                    xt
Even the most asymmetric well can be approximated by a parabola if the system
stays close to the bottom.
Simple Harmonic Motion            k and x need not refer to
                            1 2
                         U  kx   spring parameters, but the
                                  spring-mass system is an
                            2     important example




                                        2
     dU                     d x
F      kx  kx  ma  m 2
     dx                     dt
               2
         d x    k
             2
                x0                    Standard form of harmonic oscillator equation

          dt    m                               angular frequency
                                                                        k
   Solution:        xt   A cost                              
                                                                     2
                                         
                                                                      m
                     amplitude            phase          phase angle


                   A and        determined by initial conditions


                   vt   A sin t   

Exercise: A spring-mass with k = 25 N/m and m = 5 kg is pulled out 2 m and released.
Find its position at any later time.
                                                                      0
 x0  A cos  2            v0  A sin   0
                                                                      A2
                        xt   2 cos 5t    
The motion will have completed one full cycle when the phase has
increased by 2π. This time interval is called the period of the oscillation.

              t  T     t    2
                             2
                        T
                                   
Energy Considerations

How can the total energy be constant when both K and U depend on time?
                    1         1 2
               E  mv  kx2
                    2         2
       E  m A sin t   k  A cos t 2
           1              2   1
           2                  2
         1 2
                   
     E  kA sin 2 t  cos 2 t
         2
                                             1 2
                                          E  kA
                                              2
 Vertical Springs
     When a spring mass-system                    k
     vibrates vertically, an extra
     force, gravity is involved.
     How does this affect the
     motion?                             kx
                                                                 x      m
                                     m
   netF  mg  kx
                                         mg
     d 2x
     dt 2
           k
           xg
           m
                            ?     Rewrite:
                                              d 2x
                                                2
                                                   k
                                                   x
                                                   m
                                                           mg 
                                                               0
                                                             k 
                                             dt
                                         mg                     d 2 x d 2 x
 Define new displacement:       x  x            Note that:          2
                                           k                        2
                                                                 dt    dt
                  d 2 x     k
Then we have:
                        2
                             x  0          Gravity does not change the period.
                   dt        m
 All effects of gravity can be “ignored” if we choose to measure “X” from the new
 equilibrium point.
 Example: A 1 kg mass hangs vertically from a spring with k = 20 N/m. A .1 kg dart
 moving at 20 m/s is fired into the mass and sticks into it. Find the period and
 amplitude of the resulting harmonic motion.

The old equilibrium point is              The new equilibrium point is
       110                                        1.110
 i
x0           .50 m        P0  Pf         x0f             .55 m
        20
                          .120  1.1V
                                                      20                   1kg
                                     m
                           V  1.82
                                     s
     The period just depends on k and the mass oscillating:
                         m      1.1
                  T  2    2      1.47 s
                         k      20
 Use energy conservation with “x” referred to new equilibrium point:
                    E0  E f
     1
       1.11.82  20.05  20A2
                  2 1         2 1
                                                                         A  .43m
     2              2           2
Other Oscillators

Torsion Pendulum

              As wire twists, a “restoring
              torque is exerted.
                                                       Fixed axis rotation

                 k
           I                                  I
                   d 
                    2
                                         d  2
     k  I                                      k
                                                   0
                   dt 2                      dt 2 I

                                 t    m cost   
            I
     T  2
            k
Simple Pendulum

                   T                     L
                                   


                       mg
                                              I  mL    2

                mg sin  L  I
d 2
      g
      sin   0
 dt 2 L
                       ?        Small angle approximation:

                                 sin  
        d 
          2
             g                   T  2
                                               L
              0
        dt 2 L                                 g
 Compound Pendulum


                                                   H
       mg sin  H  I                              CM

                         About pivot
                                                
                                            
d 
 2
       mgH
   2
          sin   0
dt      I
d 
 2
       mgH                             Small angle approximation:

   2
           0                        sin  
dt      I
                                          I
                                  T  2
                                         mgH

								
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