Damped
Oscillations
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Damped Oscillations
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(Free) Damped Oscillations
The equation of motion is
Let us now find out the solution
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Try a solution
In the equation
Substitution yields
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The equation
has the roots
and
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Situation-1:Underdamped
or
let us call
then the roots are
then the general solution
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General solution: Underdamped
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Different Initial Conditions
Case-1.Released from extremity
At t  0 : x  a, x  0
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Underdamped Oscillations
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an example :
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Phase Comparison
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Logarithmic Decrement
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How to describe the damping of
an Oscillator
 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 time
What 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.
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Damped Vibration
Logarithmic Decrement (δ)
Amplitude of nth Oscillation: An = A0e-βnT
This measures the rate at which the oscillation dies away
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Damped Vibration
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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Damped Vibration
Example: LCR in series
Find charge on the
capacitor at time t.
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Damped Vibration
Example: LCR in series
Find charge on the
capacitor at time t.
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Damped Vibration
Conductor
Example:
Torsion constant
Mass
Resistance
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Uniform
magnetic
field B
Square coil
Side = a
Damped Vibration
E.M.F.
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Flux change:
Damped Vibration
Current:
Force:
Torque:
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Damped Vibration
Relaxation time:
Moment of inertia:
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Damped Vibration
a problem
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Different Initial Conditions
Case-2. Impulsed at equilibrium
General solution: Underdamped
At x  0 speed  v0
x(t) 
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v0

. sin(  t)
 t
e
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Situation-2: Overdamped
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General solution: Overdamped
Case-1. Released from extremity
    2  02
 t
x(t)  a0e
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cosh( t   )
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General solution: Overdamped
Case-2. Impulsed at equilibrium
At x  0 speed  v0
    
2
2
0
x(t) 
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v0

e  t sinh ( t)
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General solution: Overdamped
Case-3. position xo : velocity vo
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High damping
1   2 and  2  2
2
0
 02 
   t
 2 
x(t)  A1e
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 2 t
 A2e
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High damping
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Situation-3: Critically damped
Identical roots -
1  2  
General solution
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General solution: Critically damped
Case-1. Released from extremity
At t  0 : x  a, x  0
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General solution: Critically damped
Case-2. Impulsed at equilibrium
At x  0 speed  v0
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Critically damped
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Comparison
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Comparison
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Comparison
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