Understanding driven oscillations....

1. Physics 430: Lecture 13
Driven Oscillations and
Resonance
Dale E. Gary
NJIT Physics Department

2. October 15, 2009
5.5 Driven Damped Oscillations
 Last time we solved the homogeneous equation for damped oscillations:
 We now wish to consider the case when there is another forcing that
depends on time, i.e. driven oscillations:
 Again writing 2b = b/m, and now writing f(t) = F(t)/m, we have the driven
damped oscillator equation of motion:
 The left side of this equation is linear in derivatives of x, in which case it is
always possible to define a differential operator
where you can think of the operator D as shorthand for the entire right
side.
 Our equation can then be written simply as Dx = f. However, the concept of
differential operators is much more than a simple shorthand notation. It is a
powerful mathematical tool with applications throughout physics.
.
0


 kx
x
b
x
m 


).
(t
F
kx
x
b
x
m 

 


).
(
2 2
o t
f
x
x
x 

 
b 


,
2 2
o
2
2

b 


dt
d
dt
d
D

3. October 15, 2009
Linear Differential Operators
 Let’s look at some operations with the operator that we
have defined.
 Because it is a linear operator, we have
where a is a constant.
 Likewise, for a and b constant, we can write
for any two functions x1(t) and x2(t). Any operator that satisfies this equation
is called a linear operator.
 Back to our original, now very simple appearing equation: Dx = f. We can
write the homogeneous equation (the damped oscillator one we already
solved) as Dx = 0. Let’s call the solution to this the homogeneous solution
xh =C1er1
t + C2er2
t. Let’s also call the solution to the inhomogeneous equation
the particular solution xp.
 Then the sum of these two solutions is also a solution to the inhomogeneous
equation, because:
.
)
(
and
)
( 2
1
2
1 Dx
Dx
x
x
D
aDx
ax
D 



,
2 2
o
2
2

b 


dt
d
dt
d
D
2
1
2
1 )
( bDx
aDx
bx
ax
D 


.
0
)
( f
f
Dx
Dx
x
x
D p
h
p
h 






4. October 15, 2009
Sinusoidal Driving Force
 The general solution to Dx = f, with the operator that
we have defined, is therefore
where the homogeneous solution xh already carries the required two
arbitrary constants.
 Again, the driving function f(t) = F(t)/m is so far an arbitrary forcing term.
Let’s look at the special (although quite common) case that the forcing term
is sinusoidal, i.e. f(t) = focos(t)., where fo is the amplitude of the driving
force.
 The equation of motion then becomes:
 Note that the two omegas in this equation are different—  is the driving
frequency, and o is the “resonant frequency” of the equivalent undamped
oscillator without forcing.
 As before, we can consider this the real part of a complex solution of an
equation for z = x + iy,
,
2 2
o
2
2

b 


dt
d
dt
d
D
,
2
1
2
1 p
t
r
t
r
p
h x
e
C
e
C
x
x 



).
cos(
2 2
o t
f
x
x
x o 

b 

 


.
2 2
o
t
i
oe
f
z
z
z 

b 

 



5. October 15, 2009
Sinusoidal Driving Force-2
 With the equation in this form, we seek a solution of the form
which after substitution into
yields
 We can now solve for the constant C, and rewrite the constant as a complex
number
 Now we have to find expressions for A and d. To do that, first note that
CC*=A2, where C* denotes the complex conjugate of C. Thus
 Also, d is the arctangent of the imaginary/real part of
(arctan(y/x)),
,
2 2
o
t
i
oe
f
z
z
z 

b 

 


,
t
i
Ce
z 

  .
2 2
o
2 t
i
o
t
i
e
f
Ce
i 



b
 



.
2
2
2
o
o d

b


i
Ae
i
f
C 




 
.
4 2
2
2
2
2
o
2
o
2

b

 


f
A
.
2
arctan 2
2
o 

b
d


2b
2
2
o 
 
d

b

 i
2
2
2
o 


6. October 15, 2009
Sinusoidal Driving Force-3
 With these substitutions, the solution looks very familiar:
and the real part is the solution we want:
 Remember, this is the particular solution, so to this we have to add the
general solution to the homogeneous equation:
 You may think that this equation has four arbitrary constants, but in fact
both A and d are already determined, so only C1 and C2 are arbitrary.
 There are a couple of amazing things about this solution. The first is that
the initial conditions set the value of C1 and C2, but these both die out
exponentially with time. Thus, the behavior at long times is just set by the
driving force—initial conditions do not matter in the long term!
 Consider the case of weak damping (b < o). We saw that the homogeneous
term becomes an exponentially damped sinusoidal oscillation
where we distinguish between the A and d of the transient part (defined by
the initial conditions) by adding the subscript tr.
 
,
)
( d
 
 t
i
Ae
t
z
 .
cos
)
( d
 
 t
A
t
x
 .
cos
)
( 2
1
2
1 d
 




 t
A
e
C
e
C
x
x
t
x t
r
t
r
p
h
   .
cos
cos
)
( 1 d

d

b





 
t
A
t
e
A
x
x
t
x tr
t
tr
p
h
Applet

7. October 15, 2009
5.6 Resonance
 The long term solution for the driven oscillator is
for t >> 1/b, where the amplitude, as we have seen, is
 This amplitude expression is interesting, because it says that as you drive a
system at frequency , its amplitude depends on both how far off you are
from the resonant frequency o, and also on how big the damping is.
 If there is no damping at all, and you drive the system at the resonant
frequency, then both terms
 In this case, the amplitude goes to infinity. The way to think about this is
that the driving force pumps energy into the oscillator (like pushing a child
on a swing), and if there is no dissipation, there is no loss of energy and
the energy grows to become infinite.
 If you tune the driver frequency (variable ) for a given oscillator (fixed
o), what is the value of  for which A2 is maximum?
 ,
cos
)
( d
 
 t
A
t
x
 
.
4 2
2
2
2
2
o
2
o
2

b

 


f
A
  .
0
4
and
0 2
2
2
2
2
o 

 
b


o

A2
tuning a
radio
2 2
2 o 2 .
   b
  

8. October 15, 2009
Keeping Track of Frequencies
 We have met a number of frequencies in this discussion, so it might be
helpful to list them in one place to help keep them straight.
 To find out the maximum amplitude of a particular driven oscillator, just let
 ~ o in
i.e.
 From this you can see that the amplitude goes as b1. It also turns out that
the width of the resonance increases with b as FWHM ~ 2b. (Prob. 5.41)
2 2
1 o
2 2
2 o
/ natural frequency of undamped oscillator
frequency of damped oscillator
frequency of driving force
2 value of at which response is max
o k m

  b

  b 
 
  

  
 
.
4 2
2
2
2
2
o
2
o
2

b

 


f
A
.
2 o
o
max

b
f
A 
o

A2 b = 0.1o
b = 0.2o
b = 0.3o

9. October 15, 2009
Quality Factor and Phase at
Resonance
 There is a quantitative measure of how sharp the resonance is, which you
may have heard of before—the quality factor, or Q.
 It is just the ratio of the resonance peak frequency o to the peak width 2b.
 In the case of a pendulum, which you might think is a good resonator
(keeps good time), the Q might be about 100. For a quartz clock, on the
other hand, Q may reach 10000.
 An alternative way of thinking about the Q of a resonator
is to consider it as
 What about the phase? Remember our expression for d:
.
2
/
2
/
1
period
decay time
b



b

 o
o
Q 


.
2
o
b


Q
o

A2 high Q
low Q
.
2
arctan 2
2
o 

b
d


o

d
high Q
low Q
/2

0