Energy Methods

1
Energy Methods
Energy Methods can also be used to find the
equation of motion
2
Energy Methods - Aside
Note you get the same answer if you assume
Actually the above is not really right (despite
what the book says!). It might seem as if you are
just shifting the potential energy datums for the
spring and gravity, but actually you are ALSO
saying the gravitational potential energy is zero
throughout the problem this clearly cannot be
since the height of the mass varies. So why does
it give you the right answer?
Well it is the derivative of TV that gives you
the equation of motion.
(work done - (change in PE))
i.e. it is (rate of work done being by the
forces) we are using in our derivation of the
equation of motion.
These always cancel, so it is ok to only use the
dynamic part of the spring potential energy, and
leave out the gravitational potential energy for
this purpose
3
Energy methods especially useful when masses
translate and rotate and system gets more
complex, since terms add as scalars
The oscillation will be about this equilibrium
position, and we measure x and q from the
equilibrium position and take only the dynamic
portion of the spring potential energy and leave
out the gravitational potential energy
(check you get the same result if you include the
static portion of the spring PE and the
gravitational PE)
4
Prob 8/27
3kg blob of putty dropped 2m onto initially
stationary 28kg block. Block supported by four
springs.
k for each spring is 800 N/m
Find x(t), where x is measured from initial
position of block
After impact, have a 31kg mass bouncing on four
parallel springs
So
Does the system oscillate about the initial
position of block?
No!!
The added mass of the putty changes the
equilibrium position by an amount DdST
DdST is the additional static deflection that
would occur if we added the putty gently without
causing any vibrations
From static equilibrium
5
Prob 8/27
The general solution for motion about the static
equilibrium is
(1)
This is the solution when x is measured from the
static equilibrium position
But here we are measuring x from the original
equilibrium position of the block before we added
the putty, so we will need to add on 0.0092m at
the end after we solve for x(t) given by (1).
Need initial velocity of system immediately after
impact
By conservation of momentum during the collision
By conservation of energy before the collision
So
?
6
Prob 8/27
This is the initial velocity for the oscillation,
i.e.
What is
for the oscillation?!!
is the solution for oscillation about the
equilibrium position
Remember
And the NEW equilibrium position is 0.0092 BELOW
the position where the putty hits the block (when
the oscillation starts), so
So
and
So, measuring x from the original equilibrium
position (as in the picture)
Is this the same answer as in the book?
7
Prob 8/27
Is this the same answer as in the book?
Answer in book
(1)
Want to re-write answer in book in the form
(2)
Or
Comparing (1) and (2)
So
So answer IS same as in the book
8
Damped Vibrations
Introduce dashpots for damping. Recall

• Force depends on VELOCITY
• Push quickly big resistance
• Let go it does not spring back, so
• (mechanical) energy is not stored,
• it is lost to heat etc.

9
Damped Vibrations Eqn of Motion
x again measured from the equilibrium position,
so no need to show gravity force.
Eqn of Motion
Or
Previously, introduced
Now introduce
zeta - the damping ratio
Eqn of Motion can then be re-written as
10
Damped Vibrations - Solution
(1)
Soln of Char Eqn
Two solns, as before, so general soln is sum of
the two
(Note, the undamped soln
Falls out if )
11
Damped Vibrations Possible Cases
where
(aside an effective negative damping ratio can
occur in self-excited oscillations e.g. Tacoma
Narrows bridge or aircraft flutter. We will not
deal with that here.)
Case 1 Overdamped
Both roots
real negative ? sum of decaying exponentials
(non-oscillatory)
Case 2 Critically damped
This is not the general solution since
constitutes only 1 arbitrary constant
Turns out
is also a solution in this specific case, so
general soln is
12
Overdamped Critically Damped Cases
is the critically damped solution
This is also exponentially decaying
(eventually!), non-oscillatory motion
A critically damped system with an initial
displacement, velocity or both, returns to
equilibrium faster than an overdamped system
A critically damped system can, depending on
initial conditions, pass through equilibrium
point (see t1 in fig) and the slope can change
sign ONCE (t2). A SECOND change in sign of the
slope constitutes oscillatory behaviour. So a
critically damped system is on the cusp of
oscillatory behaviour.
13
Case 3 Underdamped Case
where
Both roots
complex, so
After some manipulation (see p. 607 textbook),
this can be re-written as
()
where
Expressions for X and y can be found as before
in terms of the initial conditions (see example
shortly)
() represents oscillatory motion which is
decaying exponentially
The (damped) natural frequency is which is
different from
14
Case 3 Underdamped Case
15
Case 3 Underdamped Case
()
Experimental determination of z Apply initial
disturbance and measure response
Successive values at t1 and t2 (separated by td).
From ()
, so
But
So
Logarithmic Decrement is defined as
Thus can measure Logarithmic Decrement and
determine z from
16
Prob 8/44
Weight, W 3400 lb
100 lb force applied ? 3 displacement.
Let go. Rises, then falls to max of 0.5 below
the unloaded equilibrium posn on 1st rebound.
Treat as 1D problem, meq 0.5 truck mass. Find z
for the rear end and c for each absorber.
Behaviour
One spring and damper on each side
Static deflection gives k
Springs in parallel
17
Prob 8/44
Response oscillatory ? underdamped, zlt1
i.e.
But have 2 dashpots here in parallel so
By definition
c damping coefficient in one damper
So
18
Prob 8/41
Released from rest from initial position x0.
Find overshoot displacement x1.
k 108 N/m, c 18 Ns/m, m 3 kg
First
Next
? Underdamped (as expected)
Damped frequency
General Solution
Use initial conditions to find
Differentiate x(t)
19
So
Or
So
Want x1.
From before
So
From theory (or obviously)
So we want x(t) when
So