Forced harmonic motion – the damped and driven harmonic oscillator
Restoring force: Fs = –kx
External driver
F0 cos ω f t
Damping force: Fd = –cv
Driving force: Ff = F0 cos ωft
Net force: Fnet = –kx –cv + F0 cos ωft
Newton’s
2nd
law:
m
d 2x
dt
2
= − kx − b
k
= Ω2
m
Define:
dx
+ F0 cos ω f t
dt
m
c
= 2γ
m
F
x + 2 γx + Ω 2 x = 0 cos ω f t
m
Equation of motion, 2nd order differential equation:
Assume the oscillator is underdamped, γ < Ω.
Important (angular) frequencies:
ωf
Driving frequency, driving period
k
m
Ω=
Tf =
2π
ωf
Natural frequency of the undamped oscillator
a = Ω 2 − γ 2
Frequency of the underdamped oscillator
( )
To find the steady state solution for x, consider the
driving force the real part of a complex driving force.
F0 cos ω f t = Re e
Find the complex solution of the new differential
equation, the real and imaginary parts of the solution will
solve the differential equation separately.
x(t ) = A e
Try a solution of the form
and insert into the
differential equation:
i (ω f t −ϕ)
⇒ x (t ) = iω f A e
= Ae
F iω t
x + 2 γx + Ω 2 x = 0 e f
m
iω f t − iϕ
iω f t − i ϕ
e
iω f t
e
⇒ x(t ) = −ω2f A e
iω f t − iϕ
e
(− ω2f + i2γω f + Ω2 )Aeiω t e−iϕ = Fm0 eiω t
(Ω2 − ω2f )A + i2γω f A = Fm0 cos ϕ + i Fm0 sin ϕ
f
f
Equate real and imaginary parts:
(
)
F
Ω − ω2f A = 0 cos ϕ
2
m
F0
2 γω f A = sin ϕ
m
Take the ratio of the two equations to
find an expression for the tangent of the
phase angle. Square each of the
equations and add to find an equation
for the amplitude A. See next pages for
the results.
Amplitude of oscillations:
A=
F0 / m
( Ω2 − ω2f ) + ( 2γω f )2
2
Ω =1, γ = 0.2, F0/m = 1
3
Amplitude
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
ωf / Ω
2
2.5
3
2.5
3
Phase difference between oscillator and applied force:
2 γω f
Ω 2 − ω2f
Ω = 1, γ = 0.2, F0/m = 1
3
2.5
Phase difference
tan ϕ =
2
1.5
1
0.5
0
0
0.5
1
1.5
ωf / Ω
2