topic 2 damped oscillation
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Oscillation and waves lecture notesTRANSCRIPT
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Topic 2: Damped Oscillation
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Taking C as a constant length:
The equation of motion become:
So:
Or:
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Solving the quadratic equation in gives:
The displacement can now be expressed as:
the sum of both these terms:
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
= 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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Case 1. Heavy Damping
Let: and
If now F = C1 + C2 and G = C1 C2
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Hyperbolic sine:
Hyperbolic cosine:
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
• 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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
If x = 0 at t = 0
0
0)0sinh0cosh(0
F
GFe p
tm
s
m
rGex
qtGex
mrt
pt
2/1
2
22/
4sinh
sinh
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Case 1: Heavy Damping
It will return to zero displacement quite slowly without oscillating about its equilibrium position
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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:
Case 2: Critical Damping
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
At maximum displacement:
At this time the displacement is:
Case 2: Critical Damping
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Case 2: Critical Damping
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Case 3: Damped Simple Harmonic Motion
when
= imaginary quantity
So the displacement is:
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
The bracket has the dimensions of inverse time (i.e. of frequency)
Þ the behaviour of the displacement x is oscillatory with a new frequency
= frequency of ideal simple harmonic motion
Case 3: Damped Simple Harmonic Motion
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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
Case 3: Damped Simple Harmonic Motion
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
SHM with new frequency:
Amplitude A is modified by e-rt/2m (which decays with time)
Case 3: Damped Simple Harmonic Motion
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Case 3: Damped Simple Harmonic Motion
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Summary
solution
solution
: heavy damped
: critically damped
: damped SHM
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
: heavy damped
Summary
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
: critically damped
Summary
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
: damped SHM
Amplitude A is modified by e-rt/2m (which decays with time)
Summary
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Logarithmic Decrement,
= 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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Methods of Describing the Damping of an OscillatorLogarithmic Decrement,
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
then one period later the amplitude is given by
If the period of oscillation is where = 2/
- logarithmic decrement
where:
Methods of Describing the Damping of an OscillatorLogarithmic Decrement,
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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
Methods of Describing the Damping of an OscillatorLogarithmic Decrement,
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Q-value of a Damped Simple Harmonic Oscillator
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
• 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
Q-value of a Damped Simple Harmonic Oscillator
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
If r is small, then Q is very large
to a very close approximation:
which is a constant of the damped system
Q-value of a Damped Simple Harmonic Oscillator
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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
Q-value of a Damped Simple Harmonic Oscillator
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
If
the energy lost per cycle is:
where (the period of oscillation)
Q-value of a Damped Simple Harmonic Oscillator
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
Solution
(a) Period is approximately that of undamped oscillator:
s 34.085
25.022
k
mT
5.0ln2/ln 2/ mrte mrt
(b) Amplitude of oscillation: Ae-rt/2m = 0.5A
s 95.4070.0
)6931.0(25.025.0ln2
r
mt
Example
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
(c) Mechanical energy at time t is
mrtekA /2
2
1
s 48.2070.0
)6931.0)(25.0(2
1ln
2
1ln
2
1
2
1
2
1 2/2
r
mt
m
rt
kAekA mrt
Solution
Example
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Topic 1-2 Damped SHM UEEP1033 Oscillations and Waves
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