RECAP

• Small oscillations in Bound System

• Periodic motion • Simple Harmonic Motion (Description, Examples of SHM) • Example Problem 4.15

• Example Problem 10.1

Total energy (potential +kinetic) is a constant for an undamped oscillator

22 mv2

1kx

2

1KUE

Energy of an oscillator

Total Energy

tkAtAmE 22222 sin2

1cos

2

1

tAmtAmE 222222 sin2

1cos

2

1

222

2

1

2

1kAAmE Expected

Work with harmonic oscillator : Time average values of sin(wt) and sin2(wt) over one cycle of oscillation

0 1 2 3 4 5 6

-1.0

-0.5

0.0

0.5

1.0

Sin

()

(radian)

0sin t

Area above the axis equals the area below the axis

0 1 2 3 4 5 60.0

0.2

0.4

0.6

0.8

1.0

(radian)

Sin

2(

)

2

1sin2 t

Mathematically

2

1)(sin

2sin

2

0

22

dttt

By symmetry

0cos t 2

1cos2 t

Question : What is the time averaged value of P.E. or K.E. over one period?

222

4

1sin

2

1. kAtkAEK

222

4

1cos

2

1. kAtkAEP

Similarly

.... EPEK

Next Topic

• Complex Numbers • Damped harmonic oscillator Equation of motion and solution Lightly damped Heavily damped Critically damped • Energy of the Damped Oscillator • Quality factor

Complex numbers are represented by z = x + iy

x is the real part and y is the imaginary part

Graphical representation of complex numbers

y

x

A

A cos

A sin

Imaginary axis

Real axis

z = x + iy = A (cos + i sin )

z = x + iy = A (cos + i sin )

z = A ei

y

x

A

xy - complex plane

vector of length A makes an angle with the real axis

Geometrically what is the meaning?

φ)tAcos(ωx 0

What is the use of complex numbers in harmonic oscillator?

Sol. of a SHM

φ)t(ω sinAωx 00

φ)tcos(ωAωx 0

2

0

To simplify the calculations

φ)tAcos(ωx 0

How to represent in complex form ?

φ)t sin(ωAy 0

Consider the imaginary component

φ)t sin(ωA i φ)t(ω cos A Z 00

φ)t(ω i 0e A Z

Calculation becomes simpler

Real part represents the equ. of SHM

φ)t(ω i 0e A Z

φ)t(ω i

00eiω A Z

φ)t(ω sinAωx 00 Real part

φ)t(ω i 2

00eω A Z

φ)tcos(ωAωx 0

2

0 Real part

Harmonic Oscillator : System is displaced from Equilibrium position (Experiences a restoring force)

Damping : Any effect that tends to reduce the amplitude of oscillations

Mechanics : Friction is one such damping effect

Damped harmonic oscillator

Consider the effect of friction on the harmonic oscillator Special form of friction force : Viscous force Forces arise : Object moves through a fluid

bvf

b depends on Shape of the mass Medium through which it is moving

Assuming a frictional force

Total force on the mass = Fspring+ f

m

bvkxF

0

0

2

0

xxx

xm

kx

m

bx

xbkxxm

m

b

m

k2

0

In complex form

02

0 xωxγx

How to solve ?

Companion equation

02

0 yωyγy

02

0 zωzγz

Solution will be of the form αtezz 0

Substituting the solution into the original equation:

02

0

2

0 )ωαγ(αez αt

02

0 zωzγz

2

0

2

ω4

γ

2

γα

Most general solution will be

tα

B

tα

A21 ezezz

zA and zB are constants 1 and 2 are the two roots

02

0

2

0 )ωαγ(αez αt

2

0

2

2142

ωγγ

α ,

tα

B

tα

A ezezz 21

Three possibilities

2

o

2

ω4

γ

2

o

2

ω4

γ

2

o

2

ω4

γ

Case (i)

Case (ii)

Case (iii)

Lightly damped and Oscillatory damped SHM

Heavily damped system

Critically damped system

2

o

2

ω4

γ Case (i)

Light Damping or Under Damping

22

4o

is imaginary

1

22

o iω2

γ

4

γωi

2

γα

2

0

2

2142

ωγγ

α ,

General solution :

tα

2

tα

121 ezezz

1

22

o iω2

γ

4

γωi

2

γα

tiω

2

tiω

1

γt/2 11 ezezez

tiω

2

tiω

1

γt/2 11 ezezez

Solution to the differential equation :

tCsinωtBcosωex 11

γt/2

Real part of the solution is

φtωA(t)cosφtωcosAex 112

γt

or

Solution is oscillatory, but with a reduced frequency and time varying (exponentially decaying) amplitude

4

γωω

22

o1

2

o

2

ω4

γ

01 ωω

φtωA(t)φtωAexγt

112 coscos

2

o

2

ω4

γ Case (ii)

Heavy Damping or Over Damping

2

o

2

ω4

γ is real

2

2

o

γ

4ω1

2

γ

2

γα

Both roots are negative

2

0

2

ω4

γ

2

γα

Represents non-oscillatory behavior

Actual displacement : Initial conditions

tαtα 21 BeAex

Real part of the solution is

tαtαezezz 21

21

Solution is

2

o

2

ω4

γ Case (iii) Critical Damping

tγ/2Cex

2

γα

Sol. is

2

0

2

ω4

γ

2

γα

tγ/2Cex

Sol. for 2nd order differential eqn. should have two independent constants which are to be fixed by the initial conditions

Sol. is

Solution is incomplete. Why?

Solution will be of the form

teBtAx )2/(

Energy considerations

Why the amplitude must decrease with time?

From work-energy theorem

tUtKtE

fW Work done by friction

frictionW0EtE

)(

)0(

tx

x

f fdxW

Work done by friction

t

fvdt0

t

dtbv0

2

0Frictional force dissipates energy

E(t) decreases with time

We can find how E(t) versus time

2

2

1..

dt

dxmEK

tAext

12 cos

)cos(2

)sin( 1

1

1

2

1

ttAevdt

dx t

)cos(2

)sin( 1

1

1

2

1

ttAevdt

dx t

)sin( 1

2

1

tAevdt

dx t

can be neglected

12ω

γ

1

2

o

2

ω4

γ 01 ωω

Therefore

)(sin2

1

2

11

222

1

2 teAmmvtK t

)(cos2

1

2

11

222 tekAkxtU t

)(cos)(sin2

11

2

1

22

1

2

tktmeA

tUtKtE

t

Light damping condition Again using

mk /2

0

2

1

tekAtE 2

2

1

)(cos)(sin2

11

2

1

22

1

2

tktmeA

tUtKtE

t

At t = 0 2

02

1kAE

In general teEtE 0

0 1 2 3 4 5 6 7

0

1

2

3

4

5E

time(s)

tekAtE 2

2

1

teEtE 0

00 368.0 E

e

EE

When 1

Max Energy

Characteristic time

b

m

1

Time constant (τ) : The decay is characterized by the time required for the energy to drop to 1/e times its initial value :

Quality factor

The damping can be specified by a dimensionless parameter Q

radianperdissipatedenergy

oscillatortheinstoredenergyQ

Rate of change of energy γEeγEdt

dE γt

0

Energy dissipated in a time T is γETTdt

dE

E(t)

Time to oscillate thru 2 radians 1

2

Time to oscillate thru one radian 1

1

Energy dissipated per radian is

1ω

γE

γ

ω

γ

ω

ω

γE

EQ 01

1

Q≫1 : Lightly damped

Q≪1 : Heavily damped

A3 220.00 157.

A#3/Bb

3 233.08 148.

B3 246.94 140.

C4 261.63 132.

C#4/Db

4 277.18 124.

D4 293.66 117.

D#4/Eb

4 311.13 111.

E4 329.63 105.

F4 349.23 98.8

F#4/Gb

4 369.99 93.2

G4 392.00 88.0

G#4/Ab

4 415.30 83.1

A4 440.00

Middle C

Note Freq(Hz) Musical note

A

A musician’s tuning fork rings at A above middle C, 440 Hz. A sound level meter indicates that the sound intensity decreases by a factor of 5 in 4 s. What is the Q of the tuning fork?

Example 10.2

54

0

0

0 eE

eE54 e 14.0

4

5ln

5ln4

s

700

4.0

44021

Q

Sound Intensity α Energy of oscillation

γt

0eEtE