Topic 2 damped oscillation

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Topic 1-2 Damped SHMUEEP1033 Oscillations and Waves
Topic 2:
Damped Oscillation

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• For ideal SHM, total energy remained constant and
displacement followed a simple sine curve for infinite time
• In practice some energy is always dissipated by a resistive or
viscous process
• Example, the amplitude of a freely swinging pendulum will
always decay with time as energy is lost
• The presence of resistance to motion means that another force
is active, which is taken as being proportional to the velocity

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frictional force acts in the direction opposite to that of the
velocity (see figure below)
where r is the constant of proportionality and has the
dimensions of force per unit of velocity and the present of
this term will always result in energy loss
Newton’s Second law becomes:

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The problem now is to find the behaviour of the displacement
x from the equation:
where the coefficients m, r and s are constant
When these coefficients are constant a solution of the form
can always be found
C has the dimension of x (i.e. length)
 has the dimension of inverse time
differential
equation

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Taking C as a constant length:
The equation of motion become:
So:
Or:

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Solving the quadratic equation in  gives:
The displacement can now be expressed as:
the sum of both these terms:

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= positive, zero or negative
depending on the relative magnitude of the two terms inside it
 three possible solutions and each solution describes a
particular kind of behaviour
: heavy damping results in a dead beat system
damping resistance term stiffness term
: balance between the two terms results in a
critically damped system
: system is lightly damped and gives
oscillatory damped simple harmonic motion

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Case 1. Heavy Damping
Let: and
If now F = C1 + C2 and G = C1  C2


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Hyperbolic sine:
Hyperbolic cosine:

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• This represents non-oscillatory behaviour
• The actual displacement will depend upon the initial or
boundary conditions
• If x = 0 at t = 0, then F = 0 and displacement x become
Case 1. Heavy Damping

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If x = 0 at t = 0
0
0)0sinh0cosh(0

 
F
GFe p


















t
m
s
m
r
Gex
qtGex
mrt
pt
2/1
2
2
2/
4
sinh
sinh

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Case 1: Heavy Damping
It will return to zero
displacement quite
slowly without
oscillating about its
equilibrium position

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Case 2: Critical Damping
 q = 0

m
r
2

The quadratic equation
in  has equal roots
In the differential equation
solution, demands that C
must be written:
C = A + Bt
A = a constant length B = a given velocity which depends
on the boundary conditions

satisfies

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Example
Mechanical oscillators which experience sudden impulses and are
required to return to zero displacement in the minimum time
Suppose: at t = 0, displacement x = 0 (i.e. A = 0) and receives an
impulse which gives it an initial velocity V
at t = 0:
Complete solution:

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At maximum displacement:

At this time the displacement is:

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Case 3: Damped Simple Harmonic Motion
when
= imaginary quantity
So the displacement is:

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The bracket has the dimensions of inverse time (i.e. of frequency)

 the behaviour of the displacement x is oscillatory with a
new frequency
 = frequency of ideal simple harmonic motion


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To compare the behaviour of the damped oscillator with
the ideal case, we have to express the solution in a form
similar to that in the ideal case, i.e.
Rewrite:
Choose:
where A and  are constant which depend on the motion at t = 0

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
SHM with new frequency:
Amplitude A is modified by e-rt/2m (which decays with time)

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Summary
solution
solution
: heavy damped
: critically damped
: damped SHM

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: heavy damped
Summary

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: critically damped
Summary

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: damped SHM
Amplitude A is modified by e-rt/2m (which decays with time)
Summary

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Methods of Describing the Damping of an Oscillator
Logarithmic Decrement, 
Relaxation Time or Modulus of Decay
Q-value of a Damped Simple Harmonic Oscillator

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Energy of an oscillator:
E  a2proportional to the square of its amplitude:
In the presence of a damping force
the amplitude decays with time as
So the energy decay will be proportional to
E 
• The larger the value of the damping force r the more rapid
the decay of the amplitude and energy
Methods of Describing the Damping of an OscillatorMethods of Describing the Damping of an Oscillator

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= measures the rate at which the amplitude dies away
Suppose in the expression:
Choose  = /2, x = A0 at t = 0:

Logarithmic Decrement

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then one period later the amplitude is given by
If the period of oscillation is  where  = 2/

- logarithmic decrement
where:

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The logarithmic decrement  is the logarithm of the ratio of
two amplitudes of oscillation which are separated by one
period

Experimentally, the value of is best found by comparing
amplitudes of oscillations which are separated by n periods.
The graph of versus n for different value of n has a
slope 

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Relaxation Time or Modulus of Decay
Another way of expressing the damping effect:
time taken for the amplitude to decay to
of its original value Ao
 at time

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Quality Factor or Q-value measures the rate at which
the energy decays from E0 to E0e-1
the decay of the amplitude:
the decay of energy is proportional to

where E0 is the energy value at t = 0

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• Time for the energy E decay to E0e-1 : t = m/r
• During this time the oscillator will have vibrated
through m/r rad
define the Quality Factor or Q-value:
Q-value = number of radians through which the
damped system oscillates as its energy
decays to

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If r is small, then Q is very large 

to a very close approximation:
which is a constant of the damped system


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Q is a constant implies that the ratio
is also a constant
is the number of cycles (or
complete oscillations) through
which the system moves in
decaying to


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If
the energy lost per cycle is:
where (the period of oscillation)

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Example
For the damped oscillator shown,
m = 250 g, k = 85 N m-1, and
r = 0.070 kg s-1.
(a) What is the period of the
motion?
(b) How long does it take for the
amplitude of the damped
oscillation to drop to half of its
initial value?
(c) How long does it take for the
mechanical energy to drop to
one half its initial value.
r

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Solution
(a) Period is approximately that of undamped oscillator:
s34.0
85
25.0
22  
k
m
T
  5.0ln2/ln 2/

mrte mrt
(b) Amplitude of oscillation: Ae-rt/2m = 0.5A
s95.4
070.0
)6931.0(25.025.0ln2





r
m
t
Example

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(c) Mechanical energy at time t is mrt
ekA /2
2
1 
s48.2
070.0
)6931.0)(25.0(2
1
ln
2
1
ln
2
1
2
1
2
1 2/2













r
m
t
m
rt
kAekA mrt
Solution
Example

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Damped SHM in an Electrical Circuit
Electrical circuit of inductance, capacitance and resistance
capable of damped simple harmonic oscillations
The sum of the voltages around the circuit
is given from Kirchhoff ’s law

which, for , gives oscillatory
behaviour at a frequency
LCR circuit

Topic 2 damped oscillation

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