# 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



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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