Damped Vibrations Summary

1
Damped Vibrations Summary
Equation of Motion
Or
where
2
Solution Possible Cases
where
Case 1 Overdamped
Both roots
real negative ? sum of decaying exponentials
(non-oscillatory)
For solving problems, use above equations as-is
Case 2 Critically damped
Two roots identical real neg. Non-oscillatory
(slope changes sign at most ONCE).
For solving problems, use
Case 3 Underdamped
Oscillatory. Both roots complex.
For solving problems, use
In each case find the constants (e.g. C, A, X, y)
from initial values of x and
3
Logarithmic Decrement (Underdamped Case only)
4
Prob 8/43
Released from rest at x0 6, when t0
Find displacement at t 0.5 sec, if
(a) c 12 lb-sec/ft and (b) c 18 lb-sec/ft
Convert k first
(a)
So
Next
? critically damped
General solution for critically damped case is
Use initial conditions to find A1 and A2
Differentiate x(t)
So
Or
So
and
So
5
Prob 8/43
Released from rest at x0 6, when t0
Find displacement at t 0.5 sec, if
(b) c 18 1b-sec/ft
? overdamped
General solution for overdamped case is
where
Here
So
(1)
Use initial conditions to find C1 and C2
Differentiate x(t)
(2)
So
(1)(2) ?
6
Prob 8/43
For part (b), the solution is
So
Compare answers
Released from rest at x0 6, when t0
Part (a) critically damped
Part (b) overdamped
x(t) is reducing fastest (from its initial value
x0 ) in part (a), i.e. system approaches
equilibrium faster when critically damped than
when overdamped
7
Forced Vibrations
Free Vibrations Provide initial conditions, then
let go
Forced Vibrations (more common) Some external
force is continuously driving the system
throughout time
The driving force can be periodic (repeating),
like in (a) and (b) in the figure
or non-periodic like in (c)
Turns out that all the periodic functions in (b)
can be created by forming infinite series of
harmonic functions (fourier analysis).
8
Types of Harmonic Forcing
9
Resonance Sneak Preview
The effect of forcing on the system is a matter
of timing
Pushing a child on a swing will increase the
oscillation amplitude IF we push in time with the
natural oscillation frequency of the swing. By
doing this we are causing resonance of the system.
If there is damping in the system, the damping
force will grow as the speed of the motion grows,
and a maximum steady state amplitude will be
achieved (fortunately for our children). The more
damping in the system, the smaller the amplitude
at resonance.
But if the damping is very small or the applied
force very large, the oscillation amplitude at
resonance can grow to be VERY large.
Resonance occurs when the forcing frequency is
near the natural frequency of the system
See http//www.lightandmatter.com/html_books/3vw/c
h02/ch02.html for a nice, intuitive discussion of
resonance
10
Resonance Sneak Preview
11
Resonance Sneak Preview
Resonance can have catastrophic effects
but is critical to many technologies
12
Down to Work!
x again measured from the equilibrium position,
so no need to show gravity force.
RHS not zero any more ? A Non-homogeneous equation
Eqn of Motion
Or
Or, with
defined as before
Solution of Non-homogeneous equation has two
parts
Complementary soln solution with F0 0 same
as before (unforced)
Particular soln ANY solution to the equation
with
13
Finding a Particular Solution
Try
i.e. we guess the displacement will be at the
same frequency as the force, but perhaps out of
phase
Then
and
Insert into equation of motion
Expand the sin, cos terms
Equate terms in
and terms in
14
Finding a Particular Solution
(1)
(2)
Solve (2) for f
Or
A little manipulation of (1) gives
so
But notice that
where
here, is the displacement that F0 would cause if
applied statically
(NOT the static displacement due to gravity!)
15
The Complete Solution
The complete solution is thus
Or
Recall complementary soln takes on different
forms for z lt, or gt 1
e.g. for z lt 1 (underdamped case), the complete
solution can be written
Regardless of whether z is lt, or gt 1, the
complementary soln always dies out over time,
whereas the particular solution persists.
Thus the complementary part of the solution is
called the transient solution, and the particular
part of the solution is called the steady state
solution. (hence the subscripts tr and ss
above).
16
The Complete Solution
The complete solution thus looks something like
the following
Force
Displacement
Time
Time
The steady-state vibrations are at the forcing
frequency and persist. They are independent of
the initial conditions, and are constant.
Transient vibrations at the natural frequency
occur at the start (superimposed on the SS
vibrations). The transient vibrations depend on
the initial conditions and die off if there is
the slightest bit of energy loss.
17
The Steady State Solution
The steady state solution is usually of most
interest, so we focus on that from now on.
Here it is again
where
(often given the symbol M) is called the
amplitude ratio or magnification factor since it
tells you how big the vibration amplitude is,
compared to if the same force was applied
statically.
With no damping (z 0) the amplitude ratio goes
to infinity when
? If the forcing frequency matches the natural
frequency of the system, then even for tiny
forces, the displacement becomes infinite!
RESONANCE!!!
18
The Steady State Solution
With damping (z gt 0) the amplitude ratio at
is
which is finite
We see that this is gt 1 for z lt 0.5 and lt
1 for z gt 0.5
i.e. we get amplification at for z
lt 0.5
For small values of z the amplitude ratio at
resonance can still be very large.
19
The Steady State Solution
How about
when ?
We see it 90o, regardless of z !
? The displacement is 90o out of phase with the
force when
Well, look at the velocity
Significant?
So the velocity is ALWAYS 90o out of phase with
the displacement (at any forcing frequency) since
sine is 90o out of phase with cosine.
So when
The velocity is IN-PHASE with the force!
? When the mass is moving to the right, the force
is to the right when the mass is moving to the
left, the force is to the left (or up/down etc)
? When we force the system at its natural
frequency, energy is being added to the system in
the most efficient way possible we are not
fighting the natural dynamics of the system
20
The Steady State Solution
We get the FULL picture by plotting the amplitude
ratio and f as a function of
Recall
This is called the frequency response.
21
The Steady State Solution
We get the FULL picture by plotting the amplitude
ratio and f as a function of
Recall
This is called the frequency response.
No amplification at high
Little amplification at low
No amplification at high
22
The Steady State Solution
Why dont we get amplification at low and high
?
Imagine the childs swing again.
Say we could somehow push with a sinusoidally
varying force instead of short duration impulse
forces
The swing has a natural frequency it wants to
swing at determined by the chain length and g
(i.e. gravity provides the spring in this case).
Imagine trying to force the swing to move at a
very low frequency
When the swing starts to come back from its the
highest point, you would actually be holding your
child up, trying to slow down his/her motion.
i.e. you would be wasting much of your energy,
fighting gravity, or fighting the spring force.
Imagine trying to force the swing to move at a
high frequency
When the swing is on its way up, you would have
to slow it down, to get it to turn around faster,
i.e. if you leave it to its natural devices, it
will take too long to get to the top of its
motion (the frequency will not be high enough).
Again, you would be wasting much of your energy,
fighting gravity (actually fighting the inertia
force in this case).
23
The Steady State Solution
Why dont we get amplification at low and high
?
In both these cases, f is not 90o, so the force
is not in phase with the velocity
The only way NOT to waste energy fighting the
spring and/or inertia forces in the system, is
to force the system at its natural frequency.
This is the most efficient way to put energy into
the system.
Then the applied force is in phase with the
velocity and in fact the only force it needs to
overcome is the damping force.
If no damping, there is nothing to fight and the
displacement ? infinity!!
24
Prob 8/51
c 2.4 lb-sec/ft
Determine the range of driving frequencies for
which Xss lt 3
Get k and m in standard units
?
What else do we need?
Want
Or
25
Prob 8/51
c 2.4 lb-sec/ft
Determine the range of driving frequencies for
which Xss lt 3
an inequality.
There are 2 solutions to the equality
Safe zones
and
Or
and
To satisfy the inequality, need
Relevant curve (z0.1)
Or
Recall
26
Prob 8/50
Redo for no damping case
when no damping
?
Note this actually has two solutions
(Use when )
(Use when )
And
Using the two possible solutions in the above way
keeps X positive for all w
27
Aside
Another way to look at it is to use
for all w
(i.e. allow X to change sign as w passes through
the natural frequency)
Since the time-varying steady state solution is
a change in sign of X as w passes through the
natural frequency is the same as a phase change
of 180o since
You can see this sudden phase change for the
undamped case in the frequency response plot
You can also see it if you examine the equation
for f around in the limit as z ? 0
In solving problems, the approach on the previous
page is probably easiest
28
Prob 8/50 (continued)
Want
for
SO
?
for
SO
?
Comparing with the damped case in Prob 8/51, the
range of unsafe frequencies is larger here
(unsurprisingly)
29
Rotor Excitation
Rotating machinery not always perfectly balanced
e.g. your cars wheels. Imperfectly machined
rotating disks etc
w
Leads to a sinusoidally-varying force
Unbalanced. More mass one side than other.
Model as an effective unbalanced mass m with an
effective eccentricity e, in a machine of mass M
30
Rotor Excitation
Let M be the mass of the machine, INCLUDING the
eccentric mass
The non-rotating part of the mass has position x,
while the rotating part of the mass has position
Equation of motion
?
Like before, but the forcing function is
And the driven mass is M (not m), so
? Note
31
Rotor Excitation
For external forcing with force F0, had
where
was the static displacement caused by F0
There isnt really a static displacement here
since only operates when
the rotor is switched on
But if we say is the static displacement
that would occur if a static force of value
was applied to the system, we
can use the same equation to solve.
i.e. use above eqn for as before, if
define