PHY F111

Mechanics, Oscillations and Waves

InstructorsDr. V. Satya Narayana MurthyDr. Asrarul HaqueBITS Pilani Hyderabad Campus

Part - I

Oscillations and Waves

V. Satya Narayana MurthyA217BITS Pilani Hyderabad CampusHyderabad

Topics to be covered

Kleppner & KolenkowCh 10 Harmonic oscillator

A P FrenchCh 3 Free vibrations of a physical system

Oscillations involving massive springs

Ch 4 Forced vibrations and resonanceThe power absorbed by a driven oscillator

Ch 2 Super position of periodic motions

Ch 5 Coupled oscillators & normal modes

Ch 6 Normal modes of continuous systems

Free vibrations of a stretched stringSuperposition of modes on a stringForced harmonic vibrations of a

stretched string

Ch 7 Progressive wavesNormal modes and travelling wavesProgressive wavesDispersion, phase and group velocityThe energy in a mechanical wave

The Harmonic Oscillator

Kleppner & Kolenkow (CH 10)A P French (CH - 3 & 4)

Topics to be covered

Periodic motions

Simple harmonic motion

Damped harmonic oscillator

Forced harmonic oscillator

The power absorbed by a driven oscillator

Oscillations involving massive springs

In every day life we come across various

things that move

The motion of physical systems can be

classified into 2 broad categories

1 Translational motion

2 Vibrational motion

Periodic motions

Vibratory / Oscillatory motions

SHM

Periodic motions

A movement that repeats with periodicity

Ex:

The pattern that repeats may be simple or complicated

Vibratory / oscillatory motion

What is the difference between oscillatory and vibratory motion?

A body in periodic motion moves back and forth over the same path

In oscillation time taken to complete one cycle is constant, in vibration it may not beOscillations occur in physical or biological systemsVibrations occur in mechanical systems

Every oscillatory motion is periodic

but every periodic motion need not be oscillatory

EX. the oscillations of a pendulumthe vibrations of a string of a guitar

Uniform circular motion is a periodic motion, but it is not oscillatory

Simple Harmonic Motion / Sinusoidal Motion

Simple periodic motion

In many systems a small displacement (x) from the equilibrium position sets up SHM

Restoring force = -kx

Where k is a constant (stiffness or spring constant)

SHM No friction force

ldl

x

In equilibrium net force acting on the mass, F = -k dl = Mg

Now M is displaced from equilibrium position by a distance x

MgMg

Total net force acting on M is F net = -k (dl + x) Mg = -kx

Vertical spring mass system

Fnet

The equation for the motion of SHM is:

)1( 0xmkx

kxxm

How to solve?

kxF

The solution for this equation is of the form:tCcosx o

Another possible solution is : tBsinx o

Therefore the most general solution will be

)2( tCcostBsinx oo

)3( )tAcos(x 0

Equ. (2) can be written in convenient form as

Rotating vector representation

SHM can be represented as geometric projection of uniform circular motion

Acosx If we take counter clock wise direction as +vethen = 0t +

)3( )tAcos(x 0

The value of is determined from the value of x at t=0

0

tCcostBsinx oo )2(

)tAcos(x 0 )3(

AcosC

AsinB

or

where

)tcos(Ax 02

0

)4( 0xx 20

Comparing eq. (1) & (4)

)5(mk

0

)tAcos(x 0 )3(

0xmkx )1(

Examples of SHMSimple pendulum

l

mmgcos

s

mgsin

mg

For small angular displacementssin

lS

l S

Eq. Of motion is

mgsinSm

(6) 0 gsinS

and

0 SlgS )6(

lg

0

kxxm

Oscillations of floating bodies

When a body is in equilibrium,the weight is balanced by thebuoyant force

xgxm A mAg

0

mDisplace the body from its equilibrium position by an amount xthe extra buoyancy force is given by:

mg

Fb

x

Harmonic oscillations of an LC circuit

LC

Ksource

voltage across capacitor

2

2

dtqdL

dtdiLv

voltage across inductor Cqv

qLCq

0qLC1q

LC1

0

Nomenclature

)tAcos(x 0 )3(

x = instantaneous displacement of theparticle at time tA = amplitude (maximum displacement)o = angular frequency = 2/T = phase factor or phase angle

A and are fixed by initial conditions

Let at t = 0, the position of the mass is x(0)and its velocity is v(0)

cos)0( Ax sin)0( 0Av

)0()0(tan

0

1

xv

2

0

2 )0()0(

vxA

How to calculate A and ?

Energy of an oscillatorThe total energy (potential +kinetic) is a constant for an undamped oscillator

22 mv21kx

21KUE

tsinAm21

tcoskA21E 0

22200

22

2kA21E

The individual values of P.E and K.E will vary with time

Time average values

f(t)

t1 t2

fWhat is time average?

dttf2

1

t

t

What is area under the curve between t 1 and t2 ?

dttfttf2

1

t

t12 dttftt

1f2

1

t

t12or

Examples

0 1 2 3 4 5 6-1.0

-0.5

0.0

0.5

1.0

Sin

()

(radian)

0sin t

0 1 2 3 4 5 60.0

0.2

0.4

0.6

0.8

1.0

(radian)Si

n2(

)

21sin2 t

21)(sin

2sin

2

0

22

dtttMathematically

What is the time average values of P.E. or K.E. over one period?

22241sin

21.. kAtkAEP

22241cos

21.. kAtkAEK

.... EPEK

Time average value

When friction is present, this is no longer true

Calculating 0 or T from ESpring mass system

m

22 xm21kx

21KUE

Since E is constant 0dtdE

0xmkx

mk

0

km2T

Simple pendulum

ms

l-y

y

m

l

22

222

y2lysylsl

For small

sy

2lsy

2

mgymv21E 2

22

dtdsv

22 sl

mg21sm

21E

2lsy

dtdsv

2

22

Since E is constant 0dtdE

0dtdE

0slgs

lg

0

gl2T

mgymv21E 2

Simple pendulum

E interms of

l cos

ms

my

ldtdl

dtdsv

22

2 mgl21

dtdml

21E

0dtdE

0lg

lg

0

For small

2lcos1ly

2

Recap

Periodic motionVibratory / Oscillatory motionEqu. of motion for different SHOHow to guess a solution for second order differential equ. having constant coefficientsTime average values of KE & PE

Today's topics

Complex numbersDamped harmonic oscillator

Equ. of motionLightly dampedHeavily dampedCritically damped

EnergyQuality factor

Complex numbers

)tAcos(x 0 Sol. of a SHM

)t( sinAx 00

)tcos(Ax 02

0

To simplify the calculations we use complex numbers

What is the use of complex numbers in harmonic oscillator?

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?

O X

Y

) ( i) ( i21 112 e eAAz t

2 1 1

Add vector of length A2at angle (2- 1) to A1 1

Turn it by an angle (t+ 1)

O XA1

A212

1 t

A1

12

A2

)tAcos(x 0

How to represent in complex form ?

)t sin(Ay 0

Consider the imaginary component

)t sin(A i )t( cos A Z 00

)t( i 0e A Z Calculation becomes simpler

Real part represents the equ. of SHM

Real part represents the equ. of SHM

)t( i 0e A Z

)t( i 0

0ei A Z

)t( sinAx 00 Real part

)t( i 20

0e A Z

)tcos(Ax 02

0 Real part

Damped Harmonic OscillatorSHM No friction forceWhat is the effect of friction on the harmonic oscillator?

Assume a special form of friction force viscous force velocity f = - bv

Condition: Viscous force arises when an object moves through a fluid at speeds which are not so large to cause turbulence

b = coefficient of damping force

Total force acting on m is F = Fspring + f

bvkxF

xbkxxm

0xxx

0xmkx

mbx

20

Equ. of motion is

In complex form

020 xxx

How to solve ?

To convert into complex form use the companion equation

020 yyy

020 zzz

The solution will be of the form, tezz 0

Substituting the solution back into the original equation gives us:

0202

0 )(ezt

020 zzz

20

2

4

2

The most general solution will be:

tB

tA

21 ezezz

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

0202

0 )(ezt

20

2

21 42

,

tB

tA ezezz 21

2o

2

4 2o

2

4 2o

2

4

Case (i) Case (ii) Case (iii)

2o

2

4 Case (i) Light Damping

orUnder Damping

22

4 o

is imaginary

1

22o i2

4

i2

ti2ti1t/2 11 ezezez The solution to the differential equation is:

tCsintBcosex 11t/2 Real part of x is

tA(t)costcosAex 112t

or

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

4

2

2o1

2o

2

4

01

tA(t)tAext

112 coscos

2o

2

4 Case (ii)

Heavy Dampingor

Over Damping

2o

2

4 is real

2

2o

412

2

Both roots are negative

This represents non-oscillatory behaviorThe actual displacement will depend upon the initial conditions

tt 21 BeAex

Real part of the solution is

tt ezezz 21 21 Solution is

01

2o

2

4

2o

2

4 Case (iii) Critical Damping

t/2Cex

The sol. to a 2nd order differential equ.should have two independent constantswhich are to be fixed by the initial conditions

2

Sol. is

The solution is incomplete Why?

solution will be of the form teBtAx )2/(

teBtAx )2/(tt 21 BeAex tAext

12 cos

Air Thickoil Water

Light Heavy Critical

Energy of a Damped Harmonic Oscillator

frictionW0EtE From work energy theorem

Wfriction = work done by the friction forcefrom time 0 to t

f = -bvopposes the motion

x(t)

x(0)f fdxW

t

0

2dtbv 0 Friction force dissipates energy

E(t) decreases with time

22

21

21 xmkxK (t)U (t)E (t)

For the lightly damped oscillator)tt/2)cos(Aexp(x 1

)t(t)sinexp(2mAmv

21K(t) 1

221

22

can be neglected

)tcos(2

)tsin(Aev 11

1

t2

1

12

1

2o

2

4 01

)t(cosekA21kx

21tU 1

2t22

)t(kcos)t(sinmeA21

tE

12

122

1t2

For light dampingmk

202

1

t2ekA21tE

At t=0 20 kA2

1E

In general t0eEtE

0 1 2 3 4 5 6 7

0

1

2

3

4

5E

time(s)

The decay is characterized by a time , damping time, during which the energy falls to e-1 of its initial value

t0eEtE 00 0.368E

eEtE

When 1

Time constant

Recap

Complex numbersDamped harmonic oscillatorEqu. of motion xbkxxm

20

2

21 42

,

tB

tA ezezz 21

2o

2

4 2o

2

4 2

o

2

4

Case (i) Case (ii) Case (iii)

teBtAx )2/(tt 21 BeAex tAext

12 cos

Light Heavy Critical

t0eEtE

0 1 2 3 4 5 6 7

0

1

2

3

4

5

E

time(s)

00 0.368EeEtE

1

Energy of a lightly damped harmonic oscillator

Time constant ()

Todays topics

Q factor of DHO

Forced Harmonic oscillator

Undamped FHOEqu. Of motionSolutionResonance+ve and ve aspects

Quality factorThe damping can be specified by a dimensionless parameter Q

radianperdissipatedenergyoscillatortheinstoredenergyQ

Rate of change of energy EeEdtdE t

0

Energy dissipated in a time T is ETTdtdE

E(t)

T = 2 / 1 oscillates through 2 radians

Energy dissipated per radian is 1

E

EEQ 01

1

Light damping Q>>1Heavy damping Q is lowUndamped oscillator Q is infinite

ETTdtdE

Energy dissipated in 2 radians

01 Q

1

In an experiment, a paperweight suspendedfrom a hefty rubber band had a period of1.2 s and the amplitude of oscillationdecreased by a factor of 2 after threeperiods. What is the estimated Q of thesystem?

2t

AeA(t)

Solution

1.8

(0)

AeAe2

ln21.8

10.39s

131.2*0.39

2

Q 1

10.3 In an undamped free harmonic oscillator the motion is given by x = A sin0t. The displacement is maximum exactly midway between the zero crossings. In a damped oscillator the motion is no longer sinusoidal and the maximum is advanced before the midpoint of zero crossings. Show that the maximum is advanced by a phase angle given approximately by = 1/2Q

Forced Harmonic OscillatorUndamped Forced Oscillator

Equ. of motion of a SHO

tcosFkxxm 0

kxxm

Driving force

tcosFkxxm 0

How to solve?

Try the solution

tcos A x

RHS of equ. has cos t

LHS of equ. must also have cos t

tcosFkxxm 0

tcos A x

Equ. of motion

solution

tFtAkm o coscos2

2mkFA o

22

1

o

o

mFA

The solution is tmFx

cos1

220

0

Incomplete solution ???

No arbitrary constants

Must able to specify x0 and v0

Complete solution is

)tBcos(cos

1mFx 022

0

0

t

Steady state solution

General solution ofundamped oscillator

020 xx

Resonance22

1

o

o

mFA

0A

= 0 A is finite

A 0Resonance

0 20 40 60 80 100

-0.010

-0.005

0.000

0.005

0.010

0

A

221

o

o

mFA

00;A

00;A

The displacement is opposite to the directionof the force!There is a phase difference of between thedisplacement and the applied force

0;A

-ve A ?

The phenomenon of resonance has both +veand ve aspects+ve aspectsSmall driving force gives large amplitudeTuning radios to the desired frequency

-wave ovenfood with no water content cannot be heated

Applied -wave frequency is equal to the H2O molecules (non zero dipole moment) natural frequency

-ve aspects

To reduce response at resonance dissipative friction force is needed - Forced Damped Harmonic Oscillator

Recap

Q 01

tcosFkxxm 0

solution tmFx

cos1

220

0

Forced Undamped oscillator

Q factor of DHO

Complete solution is

)tBcos(cos

1mFx 022

0

0

t

Steady state solution

General solution ofundamped oscillator

020 xx

221

o

o

mFA

0 20 40 60 80 100

-0.010

-0.005

0.000

0.005

0.010

0

A

-ve A ?

displacement is opposite to the direction ofthe force!There is a phase difference of between thedisplacement and the applied force

tcosFkxxm 0

Todays topics

Forced damped harmonic oscillator

Equ. Of motionSolutionResonanceEnergyQuality factor

Forced Damped Harmonic Oscillator

Undamped FHO drivingspring FFF

tcosFkxxm 0

SHM springFF kxxm

Actual motion is the superposition of oscillations at two frequencies and 0

Transient behaviorTransient behavior

drivingviscousspring FFFF Damped FHO

tcosFbv -kxxm 0

In the initial stage transient state exists

After a sufficiently long time the natural oscillations dies out because of the damping force Now the oscillator oscillates at the frequency of the driving force Steady state

tmFxxx oo cos

2

Will x =A cos t satisfy this differential equ.?No!

The velocity term gives sin t

tcosFbv -kxxm 0

tcosmFx

mkx

mb x 0

tmFxxx oo cos

2

How to find the solution?

Write the above equation in complex form

tiemFzzz 020

Solution will be of the form z = zo eit

Real part of z = zo eit gives the solution to Forced damped harmonic oscillator

Substituting z = zo eit in complex equation

imFz

emFiez titi

220

00

020

20

1

)(

i1

mFz

220

00

In polar form

220

1

21

22220

0*00

0

tan

)()(1

Re

mFzzR

z i

22220

2200

)()(i(

mF

)

Real part

The complete solution is z = zo eit

titi-i ReeRez

)cos( tRx

1/22222oo

1mFRA

22

1tan

o

Phase difference betweenthe driving force and the displacement

1/22222oo

1mFRA

A is constant for a given frequency

0dtdA

21

20m 2Q11

At = max

For light damping, A is maximum for = oand the amplitude at resonance is:

o

oo m

FA )(

The behavior of A and as functions of ,depends on the ratio / o

1/22222oo

1mFRA

10

1

Q 01 1/22222o

o

1mFA

21

20m 2Q11

As increases, the maximum amplitude occurs at a frequency less than the resonant frequency

10

21

20m 2Q11

1/22222o

o

1mFA

22

1tan

o

0 20 40 60 80 100

-0.010

-0.005

0.000

0.005

0.010

0

A

Undamped FHO Damped FHO

1/22222oo

1mFA

22

1tan

o

221

o

o

mFA

EnergyFor steady state motion amplitude is constant in time

tAx cos

tAv sin

)sin(21

21)( 222 tAmmvtK

)(cos21

21)( 222 tkAkxtU

22

41 AmK

22

41 AkU

2222

222 )(41

o

oo

mFE

)(mA41E 20

22

22

41 AmK

22

41 AkU

Steady state

Light Damping

Resonance curve or lorentzian

222

2/1

81

o

o

mFE maximum

height 24

Falls to one half maximum

2

20 2

20

FWHM = 2(-0) = 2

Forced damped harmonic oscillator

tcosFbv -kxxm 0

Recap

Equ. Of motion

Steady state solution

1/22222oo

1mFRA

22

1tan

o

tAx cos

gA is constant for a given frequency

21

20m 2Q11

1/22222oo

1mFRA

22

1tan

o

Steady state 22

2

2/1

81

o

o

mFE

Todays topics

Power absorbed by an oscillatorSimilarity between the power and energy curvesQ factor calculation from resonance curvesOscillations involving massive springs

Power absorbed by an oscillator

How to maintain the amplitude of a forced harmonic oscillator constantly?

Rate at which energy is supplied to a driven oscillator to maintain its amplitude constantly is

FvdtdxF

dtdwP

tcos A x 221

o

o

mFAwhere

t sinA x v

t2 sin2AF - t cos t sinAF P 00 Fv

t cos F F 0 Driving force

&

Undamped FHO

t2 sin2AF P 0 Fv

t2 sin P

0 P

Energy is fed into the system in one half cycle and is taken out again during next half cycle

Damped FHO

2/12222 ])()[(1

o

o

mFA

)(t tsincosAFFv P 0

tAx cos

tAv sin

)sin(cos P 0 ttAFFv

tcos)sinAF(

tcostsin)cosAF(2

0

0

Average value is zero

sinAF21P 0

sin21 0 AFP

2222022

0

1m

F21 P

For light damping 0 =

2

20

20

2

1mF

81 P

220

20

/21

mF

81 P

222

2/1

81

o

o

mFEAverage energy is

Resonance curve or lorentzian

maximum height 2

4

Falls to one half maximum

2

20 2

20

220

20

/21

mF

81 P

222

2/1

81

o

o

mFE

E or P

20

0- +

= Full width at half maximum / resonance width

half maximum2E

Width of the curve

2P or

Quality factor

curve resonance ofwidth frequency frequency resonance

oQ

Gives the frequency selective property of an oscillator

Sharpness of resonance curve means the system will not respond unlessdriven very near its resonance frequency

Q = 10 is more selective

Response in time vs response in frequency

50 100 150

0.00

0.02

0.04

0.06

0.08

0.10

0.12

E

FWHM

P

Oscillators which are very frequencyselective also have weak damping.

So such an oscillator does not recover from adisturbance or does not respond quickly.

The damping time and the resonance curvewidth obey 1

1

Oscillations involving massive springs

mk M

Total energy is a constant

2

21 kxU

E = K + U = Constant

K = K spring + K mass

l0

dtdE

How to calculate the KE of the spring ?

What is the frequency of oscillation?

The spring oscillations are not so large that they cause the spring coils to bump into each otherStretching force is same at all points along the springAll the points in the spring undergo displacements proportional to their distances from fixed end Static extensionVelocity is the same for all the elements of the spring

Assumptions

l/3 l/3 l/3m

mx

32x

3x

l

ms ds

dsl

MdM

2

21 (dM)dvdK

dtdx

lsdv

xls

Displacement of ds

22

21

21

dtdx

lsds

lM(dM)dvdK

dssdtdx

lMdK 2

2

32

l

spring dssdtdx

lMK

0

22

32

2

6

dtdxMK spring

E = PE spring + KE spring + KE mass

222

21

621

dtdxm

dtdxMkxE

0dtdE

3Mm

k

3Mm

k

Suppose m = 0

Mk3

The above calculation is not exact Why?Because of the assumptions (i) Extension of the spring is proportional

to the distance from the fixed end(ii) Velocity (dx/dt) is the same for all the

elements of the spring

Is only an approximation

It will hold if M

10.7 Find the driving frequency for whichthe velocity of a forced damped oscillator isexactly in phase with the driving force.

10.10 A small cuckoo clock has a pendulum25 cm long with a mass of 10 g and aperiod of 1s. The clock is powered by a 200g weight which falls 2 m between dailywindings. The amplitude of the swings is0.2 rad. What is the Q of the clock? Howlong would the clock run if it were poweredby a battery with 1 J capacity?