Notes for Oscillating Systems (Part 1)
Physics Part IA Cambridge University, 2012-2013
Notes for Oscillating Systems (Part 1) Undamped simple harmonic motion: This is an ideal case of simple harmonic motion in which the only force acting on an object is a restoring force towards equilibrium: By Newton’s Second Law:
x m
∴ ́x + Equilibrium position
Letting
−kx=m x́
k x =0 m
Which is the equation of motion for an
2
k /m=ω0 such that the equation of motion is not specific to the mass/spring
system: 2
∴ ́x +ω 0 x=0 Which is the equation of motion for undamped SHM in general. The solutions to this differential equation can be expressed in two forms: Complex exponential form: Let
αt
2 αt
x= A e ∴ ́x = A α e
Real (physical) sinusoidal form:
:
∴ A e αt ( α 2 +ω 20) =0, A e αt ≠ 0 ∴ α =± iω 0 ∴ x= A e ℜ A +B Im Re
ϕ a0
a0
i ω0 t
−i ω 0 t
+B e
Physics Part IA Cambridge University, 2012-2013
ϕ A−B (¿)i ℜ¿
Due to the equivalence of exponentials and sinusoids:
x=( A+ B ) cos ω 0 t + ( A−B ) isin ω 0 t : A
Where
B
and
are in general
complex.
ω0
We are only interested in the real part of this, so we are looking for the real parts of
A + B and ( A−B ) i :
defines the period of the motion and
is thus called the ‘angular frequency’:
a0 =¿
If
amplitude and
ϕ=¿
phase,
then:
2π 2π T = → ω 0= ω0 T
When
t=0 :
x= A+ B
∴ ω 0=2 πυ
∴ ℜ { A+ B } =a0 cos ϕ Where
υ is the oscillatory frequency of
the motion and
T
When
is the oscillatory
t=π /2 ω0
period.
:
x=( A−B ) i ∴ ℜ {( A+ B ) i }=a0 sin ϕ
∴ ℜ { x }=a0 cos ϕ cos ω0 t+ a0 sin ϕ sin ω0 t ∴ ℜ { x }=a0 cos(ω 0 t + ϕ) Or
ℜ { x }=C cos ω0 t+ D sin ω 0 t
a0 =√ C 2 + D2 , ϕ=tan −1
−D C
Mechanical systems displaying undamped SHM: All mechanical systems displaying SHM have the defining equation
ω20
is defined differently for each:
́x + ω20 x=0 , but
Physics Part IA Cambridge University, 2012-2013 Simple pendulum:
−F T =m ́s =ml θ́
θ s ́s l mg FT
∴−mg sinθ=ml θ́ As long as
∴−gθ=l θ́
́ g x=0 ∴ θ+ l ∴ ω 20=
g l
∴ ω 20=
τ I
θ FT G Torsional pendulum:
́ G=I θ=−τθ
́ τ x=0 ∴ θ+ I
θ is small, sin θ ≈ θ :
Physics Part IA Cambridge University, 2012-2013
Velocity, acceleration and phasor diagrams: Phasor diagrams represent the phase relationships between displacement, velocity and acceleration for an oscillating system: Displacement:
Velocity:
x=a 0 cos (ω 0 t+ ϕ) ́x =
Phase relationship:
Acceleration:
d ( x ) =−a0 ω0 sin(ω 0 t + ϕ) dt
Phase relationship:
x=a 0 cos (ω 0 t+ ϕ)
(
́x =a 0 cos ω 0 t + ϕ+
2 ́x = d 2 ( x )=−a0 ω20 cos (ω 0 t+ϕ) dt
Phase relationship:
π 2
)
́x =a 0 cos ( ω0 t + ϕ+ π )
The successive phase difference between displacement, velocity and acceleration is therefore
π /2 :
Argand plane:
a0 e
Phasor diagram:
i (ω0 t +ϕ )
a0
Im
a0 ω0 t a0 cos (ω0 t+ϕ)
a0
a0 ϕ
ω0 t
Re
a0 cos (ω0 t+ϕ)
a0 ϕ
x
Physics Part IA Cambridge University, 2012-2013 Velocity: Argand plane:
Phasor diagram:
Im Re
a0 ω 0
a0
(
a0 ω 0 cos ω 0 t + ϕ+
a0 ω 0 e
(
i ω0 t +ϕ +
π 2
π 2
)
π 2
)
)
a0 ́x a0 ω 0
a0
(
a0 ω 0 cos ω 0 t + ϕ+
Physics Part IA Cambridge University, 2012-2013 Acceleration: Argand plane:
Phasor diagram:
a0 ω 20 cos ( ω0 t + ϕ ) ́x Re Im 2
a0 ω 0
a0 a0 ω 20 cos ( ω0 t + ϕ ) a0 ω 20 e
i (ω 0 t +ϕ )
a0
Superposition of simple harmonic motion: The equation of motion for SHM is linear, hence solutions of this equation can be superimposed to give another solution: This can be proven for the general case of superimposing two simple harmonic oscillations of the same frequency;
SHM 1
and
SHM 2
:
Physics Part IA Cambridge University, 2012-2013
SHM 1 : x́1+ ω0 x 1=0, SHM 2 : x́2+ ω0 x 2=0 Where
x 1+ x 2=x ∴ x́1+ x́2= x́
:
λ1 SHM 1+ λ2 SHM 2= λ1 ( x́1 +ω 0 x 1 ) + λ 2 ( x́2 +ω0 x 2 )
¿ ( λ1 + λ 2 ) ́x + ( λ 1+ λ2 ) ω0 x ¿ λ3 ́x + λ3 ω0 x The main use of phasor diagrams is to superimpose SHMs of either common or different frequency.
Notes for Oscillating Systems (Part 1) Undamped simple harmonic motion: This is an ideal case of simple harmonic motion in which the only force acting on an object is a restoring force towards equilibrium: By Newton’s Second Law:
x m
∴ ́x + Equilibrium position
Letting
−kx=m x́
k x =0 m
Which is the equation of motion for an
2
k /m=ω0 such that the equation of motion is not specific to the mass/spring
system: 2
∴ ́x +ω 0 x=0 Which is the equation of motion for undamped SHM in general. The solutions to this differential equation can be expressed in two forms: Complex exponential form: Let
αt
2 αt
x= A e ∴ ́x = A α e
Real (physical) sinusoidal form:
:
∴ A e αt ( α 2 +ω 20) =0, A e αt ≠ 0 ∴ α =± iω 0 ∴ x= A e ℜ A +B Im Re
ϕ a0
a0
i ω0 t
−i ω 0 t
+B e
Physics Part IA Cambridge University, 2012-2013
ϕ A−B (¿)i ℜ¿
Due to the equivalence of exponentials and sinusoids:
x=( A+ B ) cos ω 0 t + ( A−B ) isin ω 0 t : A
Where
B
and
are in general
complex.
ω0
We are only interested in the real part of this, so we are looking for the real parts of
A + B and ( A−B ) i :
defines the period of the motion and
is thus called the ‘angular frequency’:
a0 =¿
If
amplitude and
ϕ=¿
phase,
then:
2π 2π T = → ω 0= ω0 T
When
t=0 :
x= A+ B
∴ ω 0=2 πυ
∴ ℜ { A+ B } =a0 cos ϕ Where
υ is the oscillatory frequency of
the motion and
T
When
is the oscillatory
t=π /2 ω0
period.
:
x=( A−B ) i ∴ ℜ {( A+ B ) i }=a0 sin ϕ
∴ ℜ { x }=a0 cos ϕ cos ω0 t+ a0 sin ϕ sin ω0 t ∴ ℜ { x }=a0 cos(ω 0 t + ϕ) Or
ℜ { x }=C cos ω0 t+ D sin ω 0 t
a0 =√ C 2 + D2 , ϕ=tan −1
−D C
Mechanical systems displaying undamped SHM: All mechanical systems displaying SHM have the defining equation
ω20
is defined differently for each:
́x + ω20 x=0 , but
Physics Part IA Cambridge University, 2012-2013 Simple pendulum:
−F T =m ́s =ml θ́
θ s ́s l mg FT
∴−mg sinθ=ml θ́ As long as
∴−gθ=l θ́
́ g x=0 ∴ θ+ l ∴ ω 20=
g l
∴ ω 20=
τ I
θ FT G Torsional pendulum:
́ G=I θ=−τθ
́ τ x=0 ∴ θ+ I
θ is small, sin θ ≈ θ :
Physics Part IA Cambridge University, 2012-2013
Velocity, acceleration and phasor diagrams: Phasor diagrams represent the phase relationships between displacement, velocity and acceleration for an oscillating system: Displacement:
Velocity:
x=a 0 cos (ω 0 t+ ϕ) ́x =
Phase relationship:
Acceleration:
d ( x ) =−a0 ω0 sin(ω 0 t + ϕ) dt
Phase relationship:
x=a 0 cos (ω 0 t+ ϕ)
(
́x =a 0 cos ω 0 t + ϕ+
2 ́x = d 2 ( x )=−a0 ω20 cos (ω 0 t+ϕ) dt
Phase relationship:
π 2
)
́x =a 0 cos ( ω0 t + ϕ+ π )
The successive phase difference between displacement, velocity and acceleration is therefore
π /2 :
Argand plane:
a0 e
Phasor diagram:
i (ω0 t +ϕ )
a0
Im
a0 ω0 t a0 cos (ω0 t+ϕ)
a0
a0 ϕ
ω0 t
Re
a0 cos (ω0 t+ϕ)
a0 ϕ
x
Physics Part IA Cambridge University, 2012-2013 Velocity: Argand plane:
Phasor diagram:
Im Re
a0 ω 0
a0
(
a0 ω 0 cos ω 0 t + ϕ+
a0 ω 0 e
(
i ω0 t +ϕ +
π 2
π 2
)
π 2
)
)
a0 ́x a0 ω 0
a0
(
a0 ω 0 cos ω 0 t + ϕ+
Physics Part IA Cambridge University, 2012-2013 Acceleration: Argand plane:
Phasor diagram:
a0 ω 20 cos ( ω0 t + ϕ ) ́x Re Im 2
a0 ω 0
a0 a0 ω 20 cos ( ω0 t + ϕ ) a0 ω 20 e
i (ω 0 t +ϕ )
a0
Superposition of simple harmonic motion: The equation of motion for SHM is linear, hence solutions of this equation can be superimposed to give another solution: This can be proven for the general case of superimposing two simple harmonic oscillations of the same frequency;
SHM 1
and
SHM 2
:
Physics Part IA Cambridge University, 2012-2013
SHM 1 : x́1+ ω0 x 1=0, SHM 2 : x́2+ ω0 x 2=0 Where
x 1+ x 2=x ∴ x́1+ x́2= x́
:
λ1 SHM 1+ λ2 SHM 2= λ1 ( x́1 +ω 0 x 1 ) + λ 2 ( x́2 +ω0 x 2 )
¿ ( λ1 + λ 2 ) ́x + ( λ 1+ λ2 ) ω0 x ¿ λ3 ́x + λ3 ω0 x The main use of phasor diagrams is to superimpose SHMs of either common or different frequency.
