Damped Simple

Harmonic
Motion
Dr. Farah Aziz PHY-1201
 Damped Oscillations
Ideally, in SHM the total energy remained constant and the displacement followed a sine curve, for
an infinite time.
In practice, some energy is always dissipated by a resistive or viscous force
Frictional or resistive force Here r is the constant of
• is proportional to the velocity. proportionality
• acts in the direction opposite to that of the velocity (dimensions: force per unit
velocity)
𝐹𝑟𝑒𝑠 = −𝑟𝑥

By Newton’s Law
𝑑2 𝑥
𝑚𝑥 = −𝑠𝑥 − 𝑟𝑥 𝑥=
𝑑𝑡 2
𝑑𝑥
𝑚𝑥 + 𝑟𝑥 + 𝑠𝑥 = 0 𝑥=
𝑑𝑡

Equation of motion of a damped Simple

Harmonic Oscillator
• The energy is not constant in this case
• The energy is lost due to this extra term 𝑟𝑥
Dr. Farah Aziz PHY-1201
 Solution of Equation of Motion
The solutions of the equation of motion of a damped harmonic oscillator are given as
1 2
𝑟 𝑟2 𝑠
2𝑚 𝑡+ 4𝑚2 −𝑚 𝑡
𝑥1 = 𝐶1 𝑒 Here 𝑪𝟏 and 𝑪𝟐 are the
constants
or
1 2
𝑟 𝑟2 𝑠
𝑡− − 𝑡
2𝑚 4𝑚2 𝑚
𝑥2 = 𝐶2 𝑒

or the superposition of two solutions 𝑥 = 𝑥1 + 𝑥2

1 2 1 2
𝑟 𝑟2 𝑠 𝑟 𝑟2 𝑠
𝑡+ − 𝑡 𝑡− − 𝑡
2𝑚 4𝑚2 𝑚 2𝑚 4𝑚2 𝑚
𝑥 = 𝐶1 𝑒 + 𝐶2 𝑒

𝑟2 𝑠
The bracket − can be positive, zero or negative depending on the relative
4𝑚2 𝑚
magnitude of the two terms inside it.

Dr. Farah Aziz PHY-1201

 Types of Damping
1) Heavy damping
𝑟2 𝑠
Bracket +ve >
4𝑚2 𝑚
𝑟2
damping resistance term dominates the
4𝑚2
𝑠
stiffness term 𝑚, and heavy damping results in a
dead beat system.

2) Critical damping
𝑟2 𝑠
Bracket zero =𝑚
4𝑚2

The balance between the two terms results in a

critically damped sysrtem.

Case 1) and 2) do not show oscillatory behaviour

Dr. Farah Aziz PHY-1201
 Types of Damping
3) Light damping
𝑟2 𝑠
Bracket -ve <
4𝑚2 𝑚
𝑟2 𝑠
When < 𝑚 the damping is light, and this gives oscillatory damped simple harmonic
4𝑚2
motion.
1 2 the square root of a negative
𝑟2 𝑠
• ± −𝑚 is an imaginary quantity number is an imaginary quantity
4𝑚2
−𝟏 = 𝒊
1 2 1 2 1 2
𝑟2 𝑠 𝑠 𝑟2 𝑠 𝑟2
± − = ± −1 − = ±𝒊 −
4𝑚2 𝑚 𝑚 4𝑚2 𝑚 4𝑚2
1 2 1 2
𝑟𝑡 𝑠 𝑟2 𝑟𝑡 𝑠 𝑟2
−2𝑚+𝒊 𝑚− 𝑡 −2𝑚−𝒊 𝑚− 𝑡
4𝑚2 4𝑚2
𝑥 = 𝐶1 𝑒 + 𝐶2 𝑒

𝑠 𝑟2
𝜔′ = −
𝑚 4𝑚2

Dr. Farah Aziz PHY-1201

 Damped Harmonic oscillator
𝑠 𝑟2
With 𝜔′ = − 4𝑚2 the solution looks like
𝑚
𝑟𝑡 𝑟𝑡
−2𝑚+𝒊𝜔′𝑡 −2𝑚−𝒊𝜔′𝑡
𝑥 = 𝐶1 𝑒 + 𝐶2 𝑒
𝑟𝑡
−2𝑚
𝑥=𝑒 𝐶1 𝑒 𝒊𝜔′𝑡 + 𝐶2 𝑒 −𝒊𝜔′𝑡 𝑨 −𝑨
We choose 𝑪𝟏 = 𝒆𝒊𝝋 and 𝑪𝟏 = 𝒆−𝒊𝝋 , where
−
𝑟𝑡 𝐴 𝒊𝜑+𝒊𝜔′𝑡 𝐴 −𝒊𝜑−𝒊𝜔′𝑡 𝟐𝒊 𝟐𝒊
𝑥= 𝑒 2𝑚 𝑒 − 𝑒 𝐴 and 𝜑are constants to be found by using initial
2𝒊 2𝒊 conditions (at 𝑡 = 0)
𝑟𝑡 𝑒𝒊 𝜔′ 𝒕+𝜑 − 𝑒 −𝒊 𝜔′ 𝒕+𝜑
−2𝑚 𝑒 𝒊𝜽 −𝑒 −𝒊𝜽
𝑥= 𝐴𝑒 As = sin 𝜽
2𝒊
2𝒊
𝑟𝑡
−2𝑚
𝑥= 𝐴𝑒 sin 𝜔′ 𝑡 + 𝜑

𝒓𝒕
−𝟐𝒎
𝒙= 𝑨𝒆 𝒔𝒊𝒏 𝝎′ 𝒕 + 𝝋

New Amplitude which decays exponentially New reduced Angular Frequency

𝑟𝑡
−
with time due to 𝑒 2𝑚 term

Dr. Farah Aziz PHY-1201

 Damped Harmonic Oscillator
𝑟𝑡
−2𝑚
𝑥 = 𝐴𝑒 𝑠𝑖𝑛 𝜔′ 𝑡 + 𝜑

New reduced frequency 𝑠

For no damping, 𝜔 = , hence
𝑠 𝑟2 𝑚
𝜔′ = − 4𝑚2 𝝎′ < 𝝎
𝑚

New Modified Amplitude For no damping, amplitude is 𝐴,

𝑟𝑡
−2𝑚 The new amplitude 𝐴′ decays exponentially
𝐴′ = 𝐴𝑒 with time
If 𝑥 = 0, at 𝑡 = 0 then 𝜑 = 0

The presence of the force term 𝑟𝑥 in the

equation of motion introduces a loss of energy
which causes the amplitude of oscillation to
𝑟𝑡
−
decay with time as 𝑒 2𝑚

Dr. Farah Aziz PHY-1201

 Quality factor of a damped SHO
Quality factor or Q value measures the rate at which the energy of a simple
harmonic oscillator decays.

The decay of the Amplitude with time is given as As the Energy is given by the
𝒓𝒕 1
− square of the amplitude 𝐸 = 𝑠𝑎2
𝑨 = 𝑨𝟎 𝒆 𝟐𝒎 2
As the energy is proprtional to the square of So energy decay ∝ square of amplitude
amplitude, decay
𝑟𝑡
−2𝑚
hence by squaring • Amplitude decreases as𝑒
𝑟𝑡 2 𝑟𝑡
2 −2𝑚 −𝑚
𝐴 = 2
𝐴0 𝑒 • Energy decreases as 𝑒
So the Energy decay is
The energy is lost due to the extra term 𝑟𝑥
𝒓𝒕
−
𝑬= 𝑬𝟎 𝒆 𝒎 The larger the value of r the
more rapid the decay of the
Here, amplitude and energy.
𝐴0 = Amplitude at 𝑡 = 0
𝐸0 = Energy at 𝑡 = 0

Dr. Farah Aziz PHY-1201

 Quality factor of a damped SHO
𝑚
• time for the energy 𝐸 to decay to 𝐸0 𝑒 −1 is given by 𝑡 = sec
𝑟
𝑚 𝑚
• during 𝑡 = sec the oscillator will have vibrated through 𝜔′ 𝑡(𝑜𝑟𝜔′ ) vibrations.
𝑟 𝑟

The quality factor 𝑸 is then given as

• After what time the Energy
𝑸=
𝒎
𝝎′ 𝒓 decays to 𝐸0 𝑒 −1 ?
𝑟𝑡
Put 𝐸0 𝑒 −𝑚 = 𝐸0 𝑒 −1
𝑟𝑡
The quality factor 𝑸 is the number of radians − = −1
𝑚
𝑚
through which the damped system oscillates as its 𝑡=
𝑟
energy decays to 𝑬𝟎 𝒆−𝟏

𝑠 𝑟2 • No. of vibrations?
If r is small, then Q is very large and ≫ 𝑛𝑜 𝑜𝑓 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛𝑠
𝑚 4𝑚2 𝜔=
𝑡
𝑠 𝑟2 𝑠 𝑛𝑜. 𝑜𝑓 𝑣𝑖𝑏𝑟𝑎𝑡𝑖𝑜𝑛𝑠 = 𝜔𝑡
So that 𝜔′ = − ≈
𝑚 4𝑚2 𝑚
′
𝜔 ≈𝜔
So that approximately 𝒎
𝑸 = 𝝎𝟎
𝒓

Dr. Farah Aziz PHY-1201

 Energy dissipaton due to damping
presence of the resistive force reduces the amplitude of oscillation with time as energy
is dissipated.
• The total energy is not a constant anymore
• Both amplitude and energy are reduced with time
𝑑𝐸 dE
≠0 dt
Should not be zero but sth negative because E decreases
𝑑𝑡
𝑑𝐸 𝑑 1 1 2 1 1
= 2
𝑚𝑥 + 𝑠𝑥 As E = 𝑚𝑥 2 + 𝑠𝑥 2
2 2
𝑑𝑡 𝑑𝑡 2 2
𝑑𝐸 1 1
= 𝑚2𝑥𝑥 𝑥 + 𝑠2𝑥𝑥
𝑑𝑡 2 2
𝑑𝐸
= 𝑥 𝑚𝑥 + 𝑠𝑥
𝑑𝑡
𝑑𝐸 𝑚𝑥 + −𝑟𝑥 + 𝑠𝑥 = 0
= 𝑥 −𝑟𝑥
𝑑𝑡 𝑚𝑥 + 𝑠𝑥 = −𝑟𝑥
𝒅𝑬
= −𝒓𝒙𝟐
𝒅𝒕
rate of doing work against the frictional
force (dimensions of power)
Dr. Farah Aziz PHY-1201
 Differences between an ideal and
Damped oscillator
Ideal SHM Damped SHM
𝑚𝑥 + 𝑠𝑥 = 0 𝑚𝑥 + 𝑟𝑥 + 𝑠𝑥 = 0

𝑟𝑡
𝑥 = 𝐴 𝑠𝑖𝑛 𝜔𝑡 + 𝜑 −
𝑥= 𝐴𝑒 2𝑚 𝑠𝑖𝑛 𝜔′ 𝑡 + 𝜑

𝑠
𝜔= 𝑠 𝑟2
𝑚 𝜔′ = −
𝑚 4𝑚2

𝑑𝐸 𝑑𝐸
=0 ≠0
𝑑𝑡 𝑑𝑡

Dr. Farah Aziz PHY-1201

 Damped SHM in an Electrical Circuit
The force equation in the mechanical oscillator is replaced by the voltage equation in the
electrical circuit of inductance, resistance and capacitance

The sum of the voltages around the circuit is given from

Kirchhoff ’s law
𝑑𝐼 𝑞
𝐿 + 𝑅𝐼 + = 0
𝑑𝑡 𝐶
𝑞
𝐿𝑞 + 𝑅𝑞 + = 0 Voltage equation for
𝐶 the damped electrical oscillator

By comparison with the mechanical case, the solution is then given as

1 2
𝑅𝑡 𝑅2 1
− ± − 𝑡
2𝐿 4𝐿2 𝐿𝐶
𝑞 = 𝑞0 𝑒
𝑹𝒕
or 𝒒 = 𝒒𝟎 𝒆−𝟐𝑳 𝒔𝒊𝒏 𝝎′ 𝒕 + 𝝋

1 𝑅2
which, for > 2 gives oscillatory behaviour at a
𝐿𝐶 4𝐿
frequency
1 𝑅2
𝜔′ = − 2
𝐿𝐶 4𝐿
Dr. Farah Aziz PHY-1201
Damped Mechanical Oscillator Damped Electrical Oscillator

Displacement 𝒙 Charge 𝒒
Velocity 𝒙 Current 𝒒
Stiffness 𝒔 Compliance 𝟏 𝑪
Mass 𝒎 Inductance 𝑳
Friction 𝒓 Resistance 𝑹
to

reduced Angular Frequency

𝒔 𝒓𝟐 𝟏 𝑹𝟐
− −
𝒎 𝟒𝒎𝟐 𝑳𝑪 𝟒𝑳𝟐

modified Amplitude
𝒓𝒕 𝑹𝒕
− −
𝒙 = 𝑨𝒆 𝟐𝒎 𝒒 = 𝒒𝟎 𝒆 𝟐𝑳

Equation of motion
𝟏
𝒎𝒙 + 𝒓𝒙 + 𝒔𝒙 = 𝟎 𝐋𝒒 + 𝑹𝒒 + 𝒒 = 𝟎
𝑪
(Force equation) (Voltage equation)

Dr. Farah Aziz PHY-1201