Topic 2 damped oscillation

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Topic 1-2 Damped SHM UEEP1033 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

tm

s

m

rGex

qtGex

mrt

pt

2/1

2

22/

4sinh

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)

Þ 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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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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Methods of Describing the Damping of an OscillatorLogarithmic 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:

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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(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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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

Science

Topic 2 damped oscillation