Download Mechanics 105 chapter 12

Oscillatory motion (chapter
twelve)
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Motion of a particle on a spring
Simple harmonic motion
Energy in SHM
Simple pendulum
Physical pendulum
Damped oscillations
Forced oscillations
Particle attached to a spring
We can model oscillatory motion as a mass
attached to a spring (linear restoring force)


Fg  kx
Causes displaced mass to to be restored to
the equilibrium position.
Potential energy  Kinetic energy.
At equilibrium – large KE but force is now
zero.
Newton’s first law - keeps moving.
Particle attached to a spring
We can use Newton’s 2nd law to quantitatively describe
the motion



 F  ma  kx


d 2x
m 2  kx
dt
Acceleration proportional to
displacement.
Opposite direction.
Simple Harmonic Motion
Defining the ratio k/m2, the equation of motion
becomes (in one dimension)
d 2x
2



x
2
dt
This equation has the solution
x(t )  A cos(t   )
dx(t )
 A sin( t   )
dt
d 2 x(t )
2



A cos(t   )
2
dt
SHM
A: amplitude of the motion (maximum
displacement)
: =(k/m)½ – angular frequency of the motion
: phase – where the motion starts
A and  are set by the initial conditions,  is
fixed by the mass and spring constant
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SHM
Period of one full cycle of motion:
2
m
T
 2

k

1
f 

2 2
k
m
Maximum velocity and acceleration:
vmax
amax
k
 A 
A
m
k
2
 A  A
m
Energy in SHM
Kinetic energy:
1 2 1
1
mv  m 2 A2 sin 2 (t   )  kA2 sin 2 (t   )
2
2
2
Potential energy:
1 2 1 2
PE  kx  kA cos 2 (t   )
2
2
KE 
Total energy of the system:


1 2
1
kA sin 2 (t   )  cos 2 (t   )  kA2
2
2
Total energy is constant!
KE  PE 
Energy in SHM
Oscillation is repeated conversion of kinetic to potential
energy and back.
Using the expression for the total energy, we can find
the velocity as a function of position
1 2 1 2 1 2
KE  PE  mv  kx  kA
2
2
2
v   A2  x 2
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The simple pendulum
T
Ft  mat
Fg
d 2s
 mg sin   m 2
dt
s  L
d 2
 g sin   L 2
dt
Small angle approximation - sin
d 2
g



2
dt
L

L
The simple pendulum
d 2
g
 
2
dt
L
This equation has the same form as
that for the motion of the mass
attached to a spring. If we define
g
2
 
L
2
L
T
 2

g
we get the exact same differential equation,
and so the system will undergo the same
oscillatory motion as we saw earlier.
Note – the frequency (and period) of the
pendulum are independent of the mass!
The Physical Pendulum
An object hanging from a point other than its COM
  I
d 2
 mgd sin   I 2
dt
d 2
mgd


2
dt
I
mgd
2
 
I

d
COM
Damped Oscillations
If we add in a velocity dependent resistive force
 F  ma  kx  bv
d 2x
dx
m 2  kx  b
dt
dt
The solution to this DE when the resistive force is weak
x(t )  Ae
bt
2m
cos(t   )
(b  4mk )
This describes an underdamped oscillator
Damped Oscillations
1
0.5
0
0
-0.5
-1
20
40
60
80
100
Damped Oscillations
The frequency of oscillation is
2
k  b 
 b 
2
2
  
  0  

m  2m 
 2m 
2
In other words, some natural frequency plus a change
due to the damping
When b=2m, the system is critically damped (returns
to equilibrium)
For b>2m, the system is overdamped – also returns to
equilibrium (slower rate).
Forced Oscillations
If we try to drive an oscillator with a sinusoidally varying
force:
d 2x
dx
m 2  kx  b  F0 sin t
dt
dt
The steady-state solution is x(t )  A cos(t   )
F0
m
A
2
 2  02 2   bm 


where 0=(k/m)½ is the natural frequency of the
system.
The amplitude has a large increase near 0 - resonance
Forced Oscillations