Download Exercises Lecture 15 Harmonic Oscillators

AM1
Mathematical Analysis 1
Oct. 2011 – Feb. 2012
Exercises Lecture 15
Date: January 27
Harmonic Oscillators
In classical mechanics, a harmonic oscillator is a system that, when displaced from its equilibrium
position, experiences a restoring force, F~ , proportional to the displacement, ~x:
F~ = −k~x
(15.1)
where k is a positive constant, known as spring constant.
Consider now a one-dimensional harmonic oscillator, like the one in Figure 15.1, now (15.1) becomes
F = −kx and by Newton’s Law the force can be expressed as the mass of the oscillator times its
acceleration
F = ma = −kx
The acceleration is given by the second derivative of the displacement x, then we arrive to the
second order linear differential equation
mẍ(t) + kx(t) = 0
(15.2)
Equation (15.2) can be rewritten in the following standard form
r
2
ẍ(t) + ω x(t) = 0
where ω =
k
m
(15.3)
The solution of (15.3) is of the form
x(t) = A sin(ωt) + B cos(ωt)
(15.4)
in fact,
ẍ(t) = −Aω 2 sin(ωt) − Bω 2 cos(ωt)
= −ω 2 (A sin(ωt) + B cos(ωt))
= −ω 2 x(t)
Now define
C=
p
A2 + B 2
ϕ = arctan
15-1
B
A
15-2
Figure 15.1: An example of harmonic oscillator, given by a mass-spring system.
That definition gives A = C cos ϕ and B = C sin ϕ, thus
x(t) = C cos ϕ sin(ωt) + C sin ϕ cos(ωt)
= C sin(ωt + ϕ)
= C sin(2πνt + ϕ)
2π
= C sin
t+ϕ
T
r
1
k
1
ω=
where ν =
2π
2π m
r
2π
m
where T =
= 2π
ω
k
(15.5)
In this form the coefficients take the following names
• T is the period of the oscillations
• ν is the frequency of the oscillations
• ϕ is the initial phase
• C is the amplitude of the oscillations
Damped Harmonic Oscillators
In real oscillators, friction, or damping, slows the motion of the system. In many cases the frictional
force F~f is modeled as being proportional to the velocity ~v of the system:
F~f = −c~v
15-3
Figure 15.2: An example of damped harmonic oscillator, given by a mass-spring-damper system.
where c is called the viscous damping coefficient. Figure 15.2 gives a sketch of a damped-like system.
Now the total force acting onto the system is given by
F~ = −k~x − cv
(15.6)
Since the velocity is the first derivative of the displacement, in the case of a one-dimensional system
the analogous of (15.2) is
mẍ(t) + cẋ(t) + kx(t) = 0
(15.7)
which can be conveniently rewritten as
r
ẍ(t) + 2ζω0 ẋ(t) + ω02 x(t) = 0
where ω0 =
k
,
m
c
ζ= √
2 mk
(15.8)
• ω0 is called the natural frequency of oscillation, because it corresponds to the frequency of
the undamped model
• ζ is called the damping ratio and determines the behavior of the solution
For the system to be physically consistent the viscous friction coefficient, c, must be a positive
number (we will return to this point later), thus ζ > 0.
Equation (15.8) is a second order linear ODE
1
of the form
ẍ(t) + aẋ(t) + bx(t) = 0
(15.9)
This kind of equation, depending on the discriminant
∆ = a2 − 4b = 4ω02 (ζ 2 − 1)
has the following solution
at
x(t) = e− 2 [Au1 (t) + Bu2 (t)] = e−ω0 ζ [Au1 (t) + Bu2 (t)]
where
1
ODE is a TLA for Ordinary Differential Equation; a TLA is a TLA for Three Letter Acronym.
(15.10)
15-4
• if ∆ = 0, i.e. ζ = 1 then
u1 (t) = 1
u2 (t) = t
• if ∆ > 0, i.e. ζ > 1 then
u1 (t) = eκt
u2 (t) = e−κt
p
√
where κ = 21 ∆ = ω0 ζ 2 − 1. Note that |κ| < |ω0 ζ|.
• if ∆ < 0, i.e. ζ < 1 then
u1 (t) = sin(κt)
p
√
where κ = 21 −∆ = ω0 1 − ζ 2
u2 (t) = cos(κt)
Remark 15.1. Provided that the viscous frictional coefficient c is positive, we have that ζ > 0 and
the solution x(t) tends to 0 as t → +∞ in any of the cases stated above. Whereas if c < 0 then we
have that ζ < 0 and any of the solutions stated above diverges for t → +∞.
Let us now study those three cases in detail.
Under-critical damping
We speak of under-critical regime whenever ζ < 1. In this situation the solution of the system is
h
p
p
i
x(t) = e−ω0 ζt A sin ω0 1 − ζ 2 t + B cos ω0 1 − ζ 2 t
(15.11)
Setting C =
√
A2 + B 2 and ϕ = arctan BA this solution can be conveniently rewritten as
h
p
i
x(t) = e−ω0 ζt C sin ω0 1 − ζ 2 t + ϕ
(15.12)
= e−ω0 ζ t [C sin (ωd t + ϕ)]
where ωd = ω0
p
the number T =
(15.13)
1 − ζ 2 is the damped frequency of the system. We can rewrite (15.13) as
2π
−ω0 ζt
e
C sin
t+ϕ
T
2π
ωd
is called the pseudo-period because in the limit of null damping we obtain (15.5).
Figures 15.3–15.5 present the solution x(t) and its derivative ẋ(t) (the velocity of the system) for
some values of ζ in the under-critical case.
Remark 15.2. In the limit for t → +∞ the solution x(t) tends to 0, this behavior follows from
the fact that ζ > 0. Figure 15.6 shows the solution for the case of an under-critically damped
oscillator with negative friction coefficient (that is to say c < 0, i.e. ζ < 0). This situation in
physically unacceptable, because the system is not slowed at all by the friction; on the contrary, it
is accelerated proportionally to its current speed.
15-5
√
Figure 15.3: Position and velocity for an under-critically damped harmonic oscillator, ζ =
2
10 .
Figure 15.4: Position and velocity for an under-critically damped harmonic oscillator, ζ =
√
3 2
16 .
15-6
√
Figure 15.5: Position and velocity for an under-critically damped harmonic oscillator, ζ =
2
2
Figure 15.6: Position and velocity for an under-critically damped harmonic oscillator, with viscous
friction coefficient c < 0.
15-7
Over-critical damping
We speak of under-critical regime whenever ζ > 1. In this situation the solution of the system is
√
√
h
i
2
2
x(t) = e−ω0 ζt Aeω0 ζ −1 t + Be−ω0 ζ −1 t
(15.14)
Remark 15.3. Moreover, the solution x(t) for an over-critically damped oscillator tends to zero
without oscillating.
Figure 15.7
√ presents the solution x(t) and its derivative for the case of a over-damped oscillator,
with ζ = 2.
Critical damping
We speak of under-critical regime whenever ζ = 1. In this situation the solution of the system is
x(t) = e−ω0 t [A + Bt]
(15.15)
Figure 15.8
√ presents the solution x(t) and its derivative for the case of a over-damped oscillator,
with ζ = 2.
Remark 15.4. The solution x(t) for an over-critically damped oscillator tends to zero as fast as
possible without oscillating. What means “as fast as possible”? It means that a critically-damped
oscillator slows down to its equilibrium position faster than any over-critically damped oscillator.
With respect to the same initial position, x(0), an velocity, ẋ(0), the convergence in the case
of critical damping is faster than that in the case of over-critical damping simply because the
convergence rate in (15.15) is given by the product of a converging exponential times a linearly
growing function, while in (15.14) by the product of converging times a diverging exponential.
Figures 15.7–15.8 and figures 15.9–15.10 confront in pair the convergence of the over-critical and
the critical damped case for two different set of initial data x(0), ẋ(0).
15-8
Figure 15.7: Position and velocity for an over-critically damped harmonic oscillator, ζ =
Figure 15.8: Position and velocity for an critically damped harmonic oscillator
√
2
15-9
Figure 15.9: Position and velocity for an over-critically damped harmonic oscillator, ζ =
Figure 15.10: Position and velocity for an critically damped harmonic oscillator
√
2
15-10
Driven Harmonic Oscillators
Driven harmonic oscillators are damped oscillators further affected by an externally applied force
F (t). We will consider the special case of a sinusoidal driving force, that is
F (t) = q sin(ωt + θ)
(15.16)
Then the ODE describing the behavior of this new model becomes
ẍ(t) + 2ζω0 ẋ(t) + ω02 x(t) =
q
sin(ωt + θ)
m
(15.17)
The general solution of (15.17) is
x(t) = x0 (t) + x∗ (t)
where x0 (t) is the solution of (15.8), the homogeneous equation, whilst x∗ (t) is the so-called particular integral.
Remark 15.5. Note that x0 (t) will tend to 0 for x → ∞. Therefore the solution of (15.17) for
large t’s will have the following behavior, x(t) = x∗ (t) + o(1).
Let us look for a special solution of (15.17) of the form
x∗ (t) = p sin(ωt + φ)
(15.18)
Substituting (15.18) into (15.17) we obtain
q
p ω02 − ω 2 sin(ωt + φ) + 2pζω0 ω cos(ωt + φ) =
sin(ωt + θ)
m
(15.19)
which gives
p ω02 − ω 2 [sin(ωt) cos φ + cos(ωt) sin φ] + 2pζω0 ω [cos(ωt) cos φ − sin(ωt) sin φ] =
q
[sin(ωt) cos θ + cos(ωt) sin θ]
(15.20)
m
or equivalently
n o
q
p ω02 − ω 2 cos φ − 2ζω0 ω sin φ − cos θ sin(ωt)+
m
o
n q
2
2
p ω0 − ω sin φ + 2ζω0 ω cos φ − sin θ cos(ωt) = 0
m
So the following system arises
( p ω02 − ω 2 sin φ + 2ζω0 ω cos φ
p ω02 − ω 2 cos φ − 2ζω0 ω sin φ
=
=
q
m
q
m
sin θ
cos θ
(15.21)
(15.22)
Our unknown are p and θ, the parameters of the guessed solution x∗ (t). To find p raise to the
square both sides of both equations, then sum side-by-side:
( 2
q2
2
p2 ω02 − ω 2 sin φ + 2ζω0 ω cos φ
=m
2 sin θ
(15.23)
2
2
q
2
p2 ω02 − ω 2 cos φ − 2ζω0 ω sin φ
=m
2 cos θ
15-11
=⇒ p2
h
ω02 − ω 2
2
i
q2
+ 4ζ 2 ω02 ω 2 = 2
m
(15.24)
From (15.24) we see that
p=
q
A(ω)q
q
=
m
m
ω02 − ω 2
1
2
+ 4ζ 2 ω02 ω 2
To determine φ take (15.22) and divide side-by-side the first equality by the second. We obtain
ω02 − ω 2 sin φ + 2ζω0 ω cos φ
tan θ =
ω02 − ω 2 cos φ − 2ζω0 ω sin φ
sin φ +
=
cos φ −
2ζω0 ω
cos φ
ω02 −ω 2
2ζω0 ω
sin φ
ω02 −ω 2
2ζω0 ω
ω02 −ω 2
2ζω0 ω
tan φ
ω02 −ω 2
tan φ +
=
1−
(15.25)
0ω
Now let us set ϑ = arctan ω2ζω
2 −ω 2 . From (15.25) we have
0
tan φ + tan ϑ
1 − tan φ tan ϑ
= tan(φ + ϑ)
tan θ =
and therefore
φ=θ−
2ζω0 ω
ω02 − ω 2
(15.26)