Chapt 3 French

8/18/2019 Chapt 3 French

1/61

Lectures on

Oscillations and WavesBy

D. D. PantAssociate Professor

Department of Physics

BITS PilaniChamber No. 3258

Email Id [email protected]

Mobile No. 09950425605

mailto:[email protected]:[email protected]


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Text Book:

Vibrations and Wavesby

A. P. FrenchReference Book:

Waves and Oscillations

by

N. K. Bajaj


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Oscillations and Waves

Why s tudy o sc i llat ions and waves?

– Almost al l phy sical s i tuat ions involve periodic or

osc i l latory behavior

• Motion of the planets• Stable mechanical systems

• Electrical systems

• Fundamental forces

–Per iod ic mo t ion in con t inuous media

• Wave propagation

• Electromagnetic radiation (light/optics)

• Matter particles are “waves”.


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The world is full of oscillatory motions

-A child on a swing-A guitar string being played

-Swinging pendulum of wall clock

-Atoms in molecules or in solid lattice

-Air molecules as a sound wave passes by

-Radio waves, microwaves and visible light are

oscillating magnetic and electric field vectors


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Isaac Newton(1642-1727)

Early Studies of Oscillations

Robert Hooke(1635-1703)

Christian Huygens(1623-1697)


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Periodic motion:- Any motion that

repeats itself in equal intervals oftime.

Oscillatory motion:- If a particlemoves back and forth over the

same path.

Harmonic motion:- Oscillatory

motions which can be expressed in

terms of sine and cosine functions.


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Simple Harmonic Oscillators

A simple pendulum

A mass fixed to a wall via a spring

A frictionless U tube containing liquid

A hydrometer floating in a liquid

An inductor connected across a

capacitor carrying a charge q


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Floating objects

Example I : The up-down motion of a partiallyimmersed solid

x

Equilibrium Position Pushed down by x

F

xg A Force Buoyancy Additional F


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Equation of motion of the body is :

x g Adt xd m 2

2

Simple Harmonic motion withm

g A


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yy

2yg

L

M)y(U

M : Total mass of liquid

L : Total length of the water column

2yM

2

1KE


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Energy conservation :

0dt

dE

0yyL

Mg2yyM

0yL

g2y

SHM of frequency :

L

g2

22 ygL

MyM

2

1E


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Example III

Prob. 6.17 ( K & K): A rod of length l and mass m, pivoted at one end, isheld by a spring at its midpoint and a spring at its

far end, both pulling in opposite directions. The

springs have spring constant k, and at equilibrium

their pull is perpendicular to the rod. Find thefrequency of small oscillations about the

equilibrium position.

l

g

m

k

2

3

4

15


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tsin&tcos

0equationSHMof Solution 2..

x x

The two independent solutions are :

The most general solution of SHM equation is :

)sin()cos()( t Bt At x

(A & B are arbitrary constants)


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Any arbitrary initial condition on position and

velocity can be accommodated within the solution

of above kind, with appropriate values of A & B

Example :

Suppose the initial (t = 0) position and initial

velocity are respectively. Obtain thesolution.

00 v&x

Bv;Ax00

tsinv

tcosx)t(x 00


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The solution of a linear differential equation with

constant coefficient is an exponential function :

t jt j eC eC t x 21)(

t peC t x )(

Substituting this into the eq.

j p

So the most general complex solution is :

)tCos(A

)tCos(C2CC )()(

t jt j

ee

complexare& 21 C C

022

2

xdt

xd

We get


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The complex solution :

)( t je A

)(

t je A z

A

t

is thus, a rotating vector of fixed length ,

rotating counter-clockwise, with an angular

velocity

A


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)( t je A z

)tcos(Ax

The SHM is the projection of the vector on the x-

axis.

)sin( t A y


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Prob. 3.19

Mass m connected to two

springs on frictionless

horizontal table. Spring

constant k and unstretched

lengths of springs 0

x

y

(k) (k)

xk 2

dt

xdm

2

2

m

k 2

x

a) Eq. of motion along x

F = - 2kx


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y

2

22/1

2

2

2

y

1

y

1

)(2

y)( 0

2

00

yk 2F 0y

b) Eq. of motion along y : y

yk 2dt

yd

m 0

2

2

2/1

0y

m

k 2


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x

y

2/1

0y

x

T

T

t At x x

cos)( 0

t At y y cos)( 0

c) Ratio of periods along x & y

d) x & y as functions of time if and mass

starts from rest

0

00 A y x

00

A,A

at t = 0

x

y


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Damped Simple Harmonic Motion

1. Pendulum with air drag

In addition to the restoring force, there is adamping force, always opposing the motion

of the oscillator

2. U-tube with viscous liquid

The damping force is usually proportional tothe velocity of the oscillator :

dt

dx bv bF

damp


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Damped Simple Harmonic Motion

Equation of motion :

dt

dx bxk

dt

xdm

2

2

Or, 0xdt

dx

dt

xd 202

2

frequencynaturalorfrequencyundampedcalledisIt

absentisdampingwhenfrequencyangularisand 0m

k

frequencyof dimensionhaswhere

m

b


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Each value of this term describes particular

type of motion

negativeorzero positive, becan4

root termSquare 20

2

t

eC t x

20

2

42

)(


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Real exponential functions means

Non Oscillatory Motion

For example Pendulum inside thick syrup

Most general solution is

CCex(t) 212-

qt qt t

ee

Case 1: Heavily Damped or Over Damped

Motion

q2

0

2

4 writeusLet

2

0

2

4

or square root term is +ve

Or damping force > restoring force


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i) Initial conditions :Pendulum released from rest

0)0(;)0(i.e. 0 x x x

qt t qt

eqeqeq

xt x

224

)( 20

t

x(t)


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ii) With the initial conditions :

0v)0(x;0)0(x

t qt qt

eeeq

vt x

20

2)(

t

x(t)


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The most general solution is

t2e)tBA()t(x

Case 2: Critical Damping

2

0

2

4 i.e.

This is the limiting case of behaviour of case 1

as q changes from +ve to –ve value.

square root term is zero i.e. q = 0


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i) With the initial conditions :

0)0(x;x)0(x 0

t2

0 et21x)t(x

t

x(t)


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ii) With the initial conditions :

0v)0(x;0)0(x

t2

0

etv)t(x

t

x(t)


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Applications of Critical Damping Mechanism

In many systems, quick damping is desirable

to bring the system to a quick stop.

i) Needle in meters such as ammeter,

voltmeter etc.


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ii) Door closers :

Out of the two non-oscillatory damping – over

damping and critical damping – it is the latterthat brings the system back to equilibrium

quicker


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t

x(t) Critical Damping

Over Damping

0)t(x

)t(xim

od

cd

t


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motiondampedof frequencyangularis

4 Where

22

0

j p

p

2

i.e.

quantitycomplexaisSo

Case 3: Damped Simple Harmonic Motion

2

0

2

4

or square root term is -veOr damping force < restoring force


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The most general complex solution :

t jt j

t

e Be Aet z

2

)(

The most general real solution :

)tcos(eA)t(xt

20

This is a SHM with decaying amplitude

conditioninitialfromobtainedareand0

A


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Rotating vector representation of Damped SHO

A lit d f th ill t d ith ti


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Energy of the oscillator also decays with time as

t

e Amt Ak E

2

0

2

0

2

2

1

)(2

1

t

0 eE

4

),(2

2

0

2

0

t2

0

eA)t(A

Amplitude of the oscillator decays with time as

Frequency of damped oscillator is less than theundamped oscillator


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22


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1

22

2

2

2

2

0

2

m

E

x

m

E

x

This is an equation of an ellipse

P h i lli


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Path is an ellipse

x

x m

E2

2m

E2


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Q lit F t Q l


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Quality Factor or Q – valueIt describes the charactertics of Damped

Harmonic Motion

T2 timein thisradiansof Number

1 t

1

00

e E e E E t

Q

It is defined as the number of radians through

which damped oscillator oscillates as its energy

decays to e-1 of its initial value.

)t(x


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)t(x

t

)t(x

t

t

)t(x

1

2

3

Quality Factor


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4

Oscillator 4 is a better quality oscillator than

1 -3 , even though its amplitude decreases

faster


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The true quality of a damped SHO is not

measured by how long it lives (time in

which the amplitude drops substantially),but rather, by how many cycles of

oscillations it completes in this lifetime.


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Amplitude after n cycles :

Q

n

0n eAA

Energy after n cycles :

Q

n2

0n eEE

cycle perlostenergy

systeminstoredenergy2 Q

Q is also obtained from following energy relation

P b 3 16


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Prob. 3.16 According to classical electro-magnetic theory an accelerated electron

radiates energy at the rate :

3

22

c

aeK P

229106 C m N K

e = electron chargec = speed of light

a = instantaneous acceleration

) If l ill i l


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a) If an electron were oscillating along astraight line :

tsinAx

how much energy would it radiate in one cycle?

Ans : tsinAxa

2

tsin

c

AeK P

2

3

242

3

232T

0

cyclec

AeK dtPE


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b) What is the Q of this oscillator?

cycle perlostenergy

systeminstoredenergy2 Q

2

3

232

322

Ae2K

Am2

Ke

mcc

Q

c) After how many oscillations will the


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c) After how many oscillations, will theenergy be down to half the initial value?

2

EeEE 0Q

n2

0n

2n2Qn

d) Putting for the typical optical


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opt Ke

mcQ 2

3

115

opt

opt sec100.4c2

810

sec107.12nQn2

T 82/12/1

d) Putting for the typical opticalfrequency, find Q and the half life

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