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Damped Oscillations

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Let us now find Let us now find out the solutionout the solution

The equation The equation of motion is of motion is

(Free) Damped (Free) Damped OscillationsOscillations

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Try a Try a solutionsolution

In the In the equationequation

SubstitutiSubstitution yieldson yields

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The The equationequation

has the has the rootsroots

anandd

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Situation-Situation-

11::UnderdampedUnderdampedoror

then the then the roots areroots are

let us let us callcall

then the then the general general solutionsolution

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General solution: General solution: UnderdampedUnderdamped

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Case-1Case-1.Released .Released from extremityfrom extremity

0:0At xa, xt

Different Initial Different Initial ConditionsConditions

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Underdamped OscillationsUnderdamped Oscillations

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an an examplexampl

e :e :

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Phase Comparison Phase Comparison

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Logarithmic Logarithmic DecrementDecrement

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What is the rate of amplitude dying ? Logarithmic decrement What is the time taken by amplitude to decay to 1/e (=0.368) times of its original value ? Relaxation timeWhat is the rate of energy decaying to 1/e (=0.368) times of its original value ? Quality Factor

The time for a natural decay process to reach zero is theoretically infinite. Measurement in terms of the fraction e-1 of the original value is a very common procedure in Physics.

How to describe the damping of an Oscillator

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Logarithmic Decrement (δ)Amplitude of nth Oscillation: An = A0e-βnT

This measures the rate at which the oscillation dies away

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Relaxation time (τ)Amplitude : A = A0e-βt ; at t=0, A=A0

(1/e)A0 = A0e-βτ

Quality factor (Q)Energy : ½k(Amplitude)2 ; E=E0e-2βt

(1/e)E0 = E0e-2β(Δt) ; Δt = 1/2β Q = ω´Δt = ω´/2β = π/δ

Quality factor is defined as the angle in radians through which the damped system oscillates as its energy decays to e-1 of its original energy.

Show that Q = 2π (Energy stored in system/Energy lost per cycle)

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Example: LCR in series

Find charge on the capacitor at time t.

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Example: LCR in series

Find charge on the capacitor at time t.

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Example:

Mass

Resistance

Conductor

Square coil Side = a

Uniform magnetic field B

Torsion constant

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Flux change:E.M.F.

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Current: Force:

Torque:

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Relaxation time:

Moment of inertia:

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a a problemproblem

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General solution: General solution: UnderdampedUnderdamped

Case-2.Case-2. Impulsed Impulsed at equilibriumat equilibrium

0 speed 0At vx

Different Initial Different Initial ConditionsConditions

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Situation-2Situation-2: : Overdamped Overdamped

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General solution: General solution: OverdampedOverdamped

Case-1.Case-1. Released Released from extremityfrom extremity 2

02

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General solution: General solution: OverdampedOverdamped

Case-2.Case-2. Impulsed Impulsed at equilibriumat equilibrium

20

2

0 speed 0At vx

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General solution: General solution: OverdampedOverdamped

Case-3.Case-3. position position xxo o

: velocity : velocity vvoo

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High dampingHigh damping

tt

eAeAtx

22

21

20

)(

2 and 2 22

1 0

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High dampingHigh damping

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Situation-3Situation-3: : CCritically ritically damped damped

General General solutionsolution

21Identical Identical roots -roots -

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General solution: General solution: Critically dampedCritically damped

Case-1.Case-1. Released Released from extremityfrom extremity 0:0At xa, x t

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General solution: General solution: Critically dampedCritically damped

Case-2.Case-2. Impulsed Impulsed at equilibriumat equilibrium 0 speed 0At vx

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Critically Critically dampeddamped

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ComparisonComparison

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ComparisonComparison

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ComparisonComparison