Download - Oscillations Readings: Chapter 14
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Oscillations
Readings: Chapter 14
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Oscillation: Periodic Motion
T – period of motion
1f
T - frequency
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Oscillation: Periodic Motion: Simple Harmonic Motion
Simple Harmonic Motion – sinusoidal oscillation
2( ) cos cos 2x t A t A ftT
or 2( ) sin sin 2x t A t A ftT
The most general expression for sinusoidal motion
0 02( ) cos cos 2x t A t A ftT
0 - phase constant A - amplitude
0( ) cos 2 90 sin 2x t A ft A ft If then
00 90
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Oscillation: Periodic Motion: Simple Harmonic Motion
Example: Uniform circular motion
( ) cos cosx t A A t
x- component:
- angular frequency
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2( ) cos cos 2x t A t A ftT
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Oscillation: Periodic Motion: Simple Harmonic Motion
0 0 02( ) cos cos 2 cosx t A t A ft A tT
0 - phase constant
A - amplitude T – period of motion (units – s)
1f
T - frequency (units – hertz – Hz =1/s)
- angular frequency (units – rad/s)
12 2fT
A
A
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0 0 02( ) cos cos 2 cosx t A t A ft A tT
0 - phase constant
Phase constant specifies the initial position of the oscillator
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0( ) cosx t A t
Simple Harmonic Motion: Position, Velocity, and Acceleration
0
0 max 0
cos( )( )
sin sin
d tdx tv t A
dt dtA t v t
maxv A
0
2 20
sin( )( )
cos ( )
d tdv ta t A
dt dtA t x t
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0( ) cosx t A t
Simple Harmonic Motion: object oscillating on a spring
2 20( ) cos ( )a t A t x t
Newton’s second law:
2ma F m x kx km
F kx Hooke’s law:
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0( ) cosx t A t
Simple Harmonic Motion: Conservation of energy
Kinetic energy:
0( ) sinv t A t
2
2 2 20sin
2 2mv m
K A t
Potential energy:
2
2 20cos
2 2kx k
U A t
Total energy:
km
2 2 2 2 20 0
2 2 2 20 0
sin cos2 2
sin cos2 2
m kE K U A t A t
k kA t t A
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2
2 2 2 2 20 0sin sin
2 2 2mv m k
K A t A t
2
2 20cos
2 2kx k
U A t km
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0( ) cosx t A t
Example: Find the relation between kinetic and potential energy at x=A/3.
Then
2 22 12 2 9 18 9kx k A kA
U E
Then the kinetic energy at this point is
2
2kE K U A
The total energy is the sum of kinetic and potential energy. At x=A the kinetic energy is 0.
At x=A/3 the potential energy is
2 2 24 82 18 9 9k k k
K E U A A A E
2
2
11884
9
kxUK kA
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If force is proportional to displacement (it is not necessary the spring system)
then
0( ) cosx t A t
Simple Harmonic Motion:
Solution of this equation:
ma F kx
km
F kx
2
2
d xa
dt2
2
d xm kxdt
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Oscillations about equilibrium position
( ) ( )netF k L y w ky k L w ky
Equilibrium:
0( ) cosy t A t
k L w w
Lk
Net force:
Oscillations about equilibrium position
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The pendulum: small-angle approximation
0w g
k ml l
, sinnet t tF w w
then
If y is the arc length then
yl
, sinnet t ty
F w wl
If y<<l then
, 0sinnet t ty y
F w w w k yl l
The net force is the sum of two forces: tension and gravitational force.
Tangential component of the net force is
0( ) cosy t A t
0k mg gm ml l
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If there is a friction then there will be energy loss
2
2k
E A
The energy is determined by the amplitude of oscillations, so the energy loss means that the amplitude is decreasing