Lecture 19

Classical Mechanics (PHL 103)

Day 19

March 01, 2012

Asoka Biswas

Q(uality) factor

03-01-2012 PHL103 2

sin 'tx t Ce t

It is defined as energy stored in the oscillator

energy dissipated per unit radianQ

It is a unit-less number and specifies the degree of damping.

We wish to calculate the Q-factor of an under-damped simple harmonic oscillator.

Thus the velocity is given by

' cos 'tx t Ce t using '

2 2

0'

Q(uality) factor

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The total mechanical energy is given by

2 2

2 22 2 2 2 2

22

1 1

2 2

sin ' ' cos '2 2

2

t t

t

E t kx mx

C Cke t m e t

Ce k

At t = 0, the energy of the system is 210

2E kC

So we can write 20 tE t E e

putting 0' k m

Q(uality) factor

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Thus, the rate of change of energy is

22 0 2tdE

E e Edt

Energy dissipated in a short time interval t is

2 dE

E t E tdt

For one radian, t = 1/ and hence, the Q-factor becomes,

0'

2 / ' 22

E

EQ

Energy stored in the system at any instant t

You may discover a different-looking definition of Q-factor as

energy stored in the oscillator2 2

2 / ' 2 energy dissipated per cycle

E EQ

E ET

Some Q-factors

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Systems Q-factors

Earth (in earthquake) 250-1400

Piano string 3000

Crystal in digital watch 104

Micro-wave cavity 104

Excited atom 107

Neutron star 1012

Excited Fe57 nucleus 3 x 1012

An undamped oscillator

Forced harmonic oscillator

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Due to damping force, the energy gets dissipated.

To maintain the oscillation, one needs to supply energy to make up for the loss.

To do this, one applies a time-dependent force.

The equation of motion becomes:

2 ( )mx m x kx F t

We consider a force that is sinusoidal: 0( ) sinF t F t

The equation becomes:

020

sin2

F tx x x

m


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The complete solution of this equation is a sum of the transient solution and the steady state solution.

The particular solution which does not die down with time. We call this the steady-state solution.

The transient solution is obtained by putting F(t) = 0. This solution dies down with time.

For sinusoidal driving, the system oscillates sinusoidally at the driving frequency at long times.


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For steady-state, assume a trial solution:

sinAx t

We get from the equation:

2 2 00 sin 2 cos sinF

A t A t tm

Comparing the coefficients of and cos t sin t

2 2 00 cosF

Am

02 sinF

Am

Solve to get:

0

2 2 22 2 2 2

00

2tan ;

4

F mA


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The steady-state solution thus becomes:

02

2 2 2 2

0

sin

4ss

F mx t t

The particle performs forced oscillation at a frequency equal to the driving frequency with the amplitude and phase as a function of damping and driving frequency

The transient solution is given by

1 2t t ttrx t e Ae A e where 2 2

0

The final solution is given by:

tr ssx t x t x t


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2 202

tan

Variation of phase :

If increases from zero to 0, the phase varies from zero to /2.

If increases from 0 to much larger values, the phase varies from /2 to .

=0 =0.1 =0.2

=0.5

If is zero, the phase abruptly changes from zero to , at the frequency =0

The driving frequency becomes equal to the natural frequency


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Variation of amplitude A:

There exists a frequency, at which the amplitude A becomes maximum, i.e. 2

20; 0

dA d A

d d

0

22 2 2 2

0 4

F mA

Using this condition, 2 2 20 2

The resonant frequency becomes: 2 2

res 0 2

The amplitude becomes maximum at = res and equals

0 0

max2 24 2 2 200

24 4 2

F m F mA

Ex.: In mechanical system, stiffness (~) of the rubber (or spring) is so chosen that the resonant frequency becomes far from the running frequency of the motor


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= 0

increasing

res

A(

)


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Variation of velocity:

0

22 2 2 2

0

cos cos

4

F mx t t A t

The magnitude of the velocity will be maximum, if the denominator is minimum

This happens if 0

The maximum value of velocity becomes 0

max2

Fx

m

The velocity and the amplitude become maximum at different frequencies

The phase difference between the applied force and the velocity is = - /2 2 2

0tan cot2

The force is in phase with the velocity at the resonance 0


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Variation of energy

cosx t A t sinx t A t We have found and

So, the total mechanical energy is

2 2 2 2 2 21 1

cos sin2 2

E t mx kx A m t k t

The time-averaged energy can be written as

2 2 2 2 201 1

44E t A m k mA

or

2 2 2

0 0

22 2 2 2

0

1

4 4

FE t

m

0

22 2 2 2

0 4

F mA

using


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For small damping: The energy profile becomes:

Full-Width-at-Half-Maximum (FWHM): 2

For 0 , the energy becomes

0

E t

2 2 2

0 0

22 2 2 2

0

2 2 2

0 0 0

2 22 2 2 2

0 0 0 0

1

4 4

21 1 1

4 4 84

FE t

m

F F

m m

Resonance curve

max2E E

This exhibits a Lorentzian profile

Q-factor revisited

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2

0

2

0 0

4 2Q

The quality factor becomes:

The Q-factor is given by the ratio between average energy stored in the system and that dissipated per cycle.

2 2 2

2 20

0

2

1 1

2 22

42

mx m x

QT m x

2 2 2 2 21

2x A x

using

Instantaneous rate of the dissipation of energy

2 2 2

velocity damping force

2

2 cos

x m x

m A t

So, the average energy dissipated per unit cycle is 22T m x

Clearly, for small , the quality factor will be very large, leading to sharpness of the resonance curve.

2T

Maintaining oscillation

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Energy issues:

The applied force supplies energy to the system to maintain the oscillation. This means that rate of this supply has to be equal to the rate of dissipation at the steady state.

Instantaneous rate of the supply of energy

0

velocity external force

cos sinA t F t

Time-averaging the above rate, 0

1sin

2W A F

2 2

0 sin

4W

mD

F

Time-averaged rate of dissipation of energy is 2 2 22D m x m A

Putting,

0

22 2 2 2

0 4

F mA

2 2

0

2tan

and

we can write, 0 sin 2A F m

Therefore

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