Physics 105
Classical Mechanics
Lecture 7
October 9, 2009
R.P. Johnson
Damped Oscillator with Square Wave Forcing Function
n := 1 , 3 .. 9
b :=
n
T := 2 ⋅ π
4
ω := n ⋅
2⋅ π
n
n⋅ π
Fa( t) :=
T is the period of the driving function
Frequency harmonics 1, 3, 5, etc
T
−1 if t ≥ −π ∧ t ≤ 0
Exact square wave driving function, defined here over
a single period of T=2pi.
otherwise
1 if t > 0 ∧ t ≤ π
0 otherwise
Square wave driving function (truncated Fourier series). Note that
the constant term and cosine terms are zero, as F is an odd
function..
∑ ( n ( n ))
F( t) :=
b ⋅ sin ω ⋅ t
n
ω0 := 1.5
β := .5
m := 1
δ := atan2ω0 − ω

n
2
( n)
b
D :=
n
2
Oscillator parameters
, 2 ⋅ ω ⋅ β
n

n
2
ω 2 − ω 2 + 4⋅ ω 2⋅ β2
( n)
 0 ( n) 
(
)
s( n , t) := D ⋅ sin ω ⋅ t − δ 
n
x ( t) :=
1
m
n
n
∑ (s(n , t))
⋅

n
Phase of the response for each harmonic

Amplitude of the response for each harmonic
Fourier components of the
solution
Solution for the
steady-state response
Input Square Wave
1
F( t)
Fa( t)
0
−1
−1
− 0.5
0
t
π
0.5
1
Input Square Wave Fourier Components
b1⋅ sin( ω 1 t) 1
b3⋅ sin( ω 3 t)
b5⋅ sin( ω 5 t)
b7⋅ sin( ω 7 t) 0
b9⋅ sin( ω 9 t)
Fa( t)
−1
−1
− 0.5
0
0.5
1
t
π
Fourier Amplitude of the Response
0.8
0.6
Dn
0.4
0.2
0
0
2
4
6
n
8
Oscillator Steady-State Response
1
0.5
x( t)
0
− 0.5
−1
−1
− 0.5
0
0.5
1
t
π
Oscillator Response Fourier Components
1
s( 1 , t) 0.5
s( 3 , t)
s( 5 , t)
0
s( 7 , t)
− 0.5
−1
−1
− 0.5
0
t
π
0.5
1
Damped Oscillator with Saw-Tooth Wave Forcing Function
n := 1 , 2 .. 9
b :=
( −1 )
n
T := 2 ⋅ π
n+ 1
ω := n ⋅
2⋅ π
n
n⋅ π
t
Fa( t) :=
T is the period of the driving function
Frequency harmonics 1, 3, 5, etc
T
if t ≥ −π ∧ t ≤ π
2π
Exact saw tooth driving function, defined here over a
single period of T=2pi.
0 otherwise
∑ (bn⋅sin(ωn⋅t))
F( t) :=
Square wave driving function (truncated Fourier series). Note that
the constant term and cosine terms are zero, as F is an odd
function.
n
ω0 := 2.5
β := .5
m := 1
δ := atan2ω0 − ω

n
2
( n)
b
D :=
n
2
Oscillator parameters
, 2 ⋅ ω ⋅ β
n

n
2
ω 2 − ω 2 + 4⋅ ω 2⋅ β2
( n)
 0 ( n) 
(
)
s( n , t) := D ⋅ sin ω ⋅ t − δ 
n
x ( t) :=
1
m
n
n
∑ (s(n , t))
⋅

n
Phase of the response for each harmonic

Amplitude of the response for each harmonic
Fourier components of the
solution
Solution for the
steady-state response
Input Saw Tooth
1
F( t)
Fa( t)
0
−1
−1
− 0.5
0
t
π
0.5
1
Input Saw Tooth Fourier Components
b1⋅ sin( ω 1 t) 1
b2⋅ sin( ω 2 t)
b3⋅ sin( ω 3 t)
b4⋅ sin( ω 4 t) 0
b5⋅ sin( ω 5 t)
Fa( t)
−1
−1
− 0.5
0
0.5
1
t
π
Fourier Amplitude of the Response
0.1
0.05
Dn
0
− 0.05
− 0.1
0
2
4
6
n
8
Oscillator Steady-State Response
0.1
0
x( t)
− 0.1
− 0.2
−1
− 0.5
0
0.5
1
t
π
Oscillator Response Fourier Components
0.1
s( 1 , t) 0.05
s( 3 , t)
s( 5 , t)
0
s( 7 , t)
− 0.05
− 0.1
−1
− 0.5
0
t
π
0.5
1
Oscillator Steady-State Response
0.1
0
x( t)
− 0.1
− 0.2
−1
0
1
t
π
2
3