MA 1506
Mathematics II
Tutorials
MA 1506 Mathematics II
Tutorial 4
Ngo Quoc Anh
The harmonic oscillator
Question 1
Groups: B03 & B08
Question 2
February 15, 2012
Question 3
Question 4
Ngo Quoc Anh
Department of Mathematics
National University of Singapore
1/13
Question 1: Equilibrium points for ODEs
Equilibrium solutions of ODEs
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A solution x of a given ODE ẍ = f (x) is said to be an
equilibrium solution if x is constant.
If xE is an equilibrium solution of ẍ = f (x) then
ẍE = f (xE ). Consequently, f (xE ) = 0. In other words,
finding equilibrium solution is to find solutions of f (x) = 0.
ex +e−x
Since cosh x =
> 0, the equation ẍ = cosh x
2
admits no equilibrium solution.
Question 1
Question 2
Question 3
Question 4
By solving cos x = 0, we conclude that equilibrium for
the equation ẍ = cos x are π2 + kπ where k ∈ Z.
Similarly, by solving tan(sin x) = 0, we conclude that
equilibrium for the equation ẍ = tan(sin x) are kπ
where k ∈ Z.
Since the behavior of a solution x depends strongly on f , we
need to study f at points those are near equilibrium points.
2/13
Question 1: Equilibrium points for ODEs
To achieve that goal, we simply use Taylor series. Assume
f 0 (xE ) 6= 0. Note that when x is close enough to xE , there
holds
f (x) ≈ f (xE ) + f 0 (xE )(x − xE ).
Therefore, near the equilibrium point xE , there is no
difference between ẍ = f (x) and its approximated ODE
0
0
ẍ − f (xE )x = −f (xE )xE + f (xE ).
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Question 1
Question 2
Question 3
Question 4
Concerning the above approximated ODE, since xE is a
particular solution, we can easily see that
x = xh +
xE .
|{z}
constant
Since xE is constant, it contributes nothing to the stability
of the equilibrium xE . It turns out that we need to study the
stability of the so-called linearized equation
ẍ = f 0 (xE )x.
3/13
Question 1: Equilibrium points for ODEs
Up to constants, ẍ = f 0 (xE )x is equivalent to either ẍ = x
(xE is therefore unstable) or ẍ = −x (xE is now stable).
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The full stability of items (ii) and (iii) can be described
through the following phase portraits.
Question 1
Question 2
Question 3
Question 4
4/13
Question 1: Equilibrium points for ODEs
For item (i), the phase portrait looks like the picture on
the right.
Keep in mind that the x-axis
represents x while the y-axis
represents ẋ. Thus, equilibrium points belong to the xaxis.
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Question 1
Question 2
Question 3
Question 4
The meaning of this phase
portrait is the following: Each
point of the plane represents an initial value. Given such a
point, a solution x(t) to the ODE exists. Accordingly, there
is a curve representing (x(t), ẋ(t)) when t varies.
To identify equilibrium points, we look for points on the
x-axis those have arrows concentrating in a small
neighborhood. To draw the above picture, we can use Maple.
5/13
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Question 2: RLC circuit
By the Kirchhoff voltage law, there holds
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VR + VL + VC = V (t)
where VR , VL , and VC are the voltages across R,
L, and C respectively and V (t) is the time varying
voltage from the source. Using the constitutive equations,
VR = Ri(t),
VL = L
d
i(t),
dt
we have
Ri(t) + L
VC =
Q
,
C
i(t) =
d
Q(t)
dt
Question 1
Question 2
Question 3
Question 4
d
Q
i(t) +
= V (t).
dt
C
Making use of i(t) = dQ
dt , it leads to a 2nd order differential
equation for Q given by
d2
R d
1
v(t)
Q(t) +
Q(t) +
Q(t) =
.
dt2
L dt
LC
L
In addition, it is given that i(0) = 0 and Q(0) = 0.
6/13
Question 2: RLC circuit
Since R = 5 Ω, L = 0.05 H, C = 4 × 10−4 F , and
V (t) = 200 cos(100t) V , we arrive at
d
d2
Q(t) + 100 Q(t) + 50000Q(t) = 4000 cos(100t).
2
dt
dt
4
The characteristic equation, λ2 + 100λ
√ + 5 × 10 = 0, has
two complex roots λ1,2 = −50 ± 50 19i. Consequently, the
homogeneous equation has a general solution given by
√
√
Qh = (c1 cos(50 19t) + c2 sin(50 19t)e−50t .
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Question 1
Question 2
Question 3
Question 4
Since the function on the right, r(t), is of the form
”polynomial × cosine”, we can use the method of
undetermined coefficients. Precisely, we shall find
Qp = A cos(100t) + B sin(100t).
Substituting and equalizing, we get that
Qp =
16
4
cos(100t) +
sin(100t).
170
170
7/13
Question 2: RLC circuit
In particular, we have
√
√
Q(t) =(c1 cos(50 19t) + c2 sin(50 19t)e−50t
4
16
cos(100t) +
sin(100t).
+
170
170
It leaves out to determine c1 and c2 . Thanks to the given
initial data, we get that
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Question 1
Question 2
Question 3
16
16
0 = c1 +
,
c1 = −
170
170
√
√
80
40
12 19
0=
+ 50 19c2 + ,
c2 = −
.
17
17
1615
Hence, the formula for i(t), which is
Question 4
d
dt Q(t),
is given by
√
680
√
√ 27889 19
i(t) = −
cos(50 19) +
sin(50 19) e−50t
289
5491
16
4
+
cos(100t) +
sin(100t).
170
170
8/13
Question 3: Maximum of the amplitude response function
When solving the following ODE
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mẍ + bẋ + kx = F0 cos(αt)
one eventually gets
x(t) =
x (t)
| h{z }
TRANSIENT&0
+
1
F0
q
m (ω 2 − α2 )2 +
{z
|
Question 1
b2 2
α
m2
A(α)
cos(αt − γ)
}
Question 2
Question 3
Question 4
as a general solution where
r
k
bα
ω=
, tan γ =
.
m
k − mα2
It is well-known that the behavior of A(α) depends on the
frequency α and the friction constant b. To find its
maximum value, we simply differentiate A w.r.t α to find its
critical points.
9/13
Question 3: Maximum of the amplitude response function
It’s not hard to find that A0 (α) is nothing but
− 3
b2 2 2 h
b2 i
F0
2
2 2
(ω − α ) + 2 α
2(ω 2 − α2 )(−2α) + 2α 2 .
−
2m
m
m}
|
{z
4α
Consequently, A0 (α) = 0 is equivalent to
α2 = ω 2 −
b2
.
2m2
b2
−ω 2 +α2
2m2
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Question 1
Question 2
Question 3
Question 4
There are two cases
√
If b < 2mω, then A0 = 0 has
a unique solution. Hence, A has
a maximum value given by
2mF0
Aresonance = √
.
b 4m2 ω 2 − b2
√
If b > 2mω, then A0 < 0.
10/13
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Question 4
Gravity F=Mg
Gravity F=Mg
M
(mass of the tanker)
d
ρ
M
(mass of the tanker)
ρobject
Tanker
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object
d Tanker
A
x(t)
(horizontal cross-section area)
(density of seawater)
ρ
Question 1
Question 2
A
(horizontal cross-section area)
(density of seawater)
Buoyancy F=ρAdg
Question 3
Question 4
Buoyancy F=ρA(d+x)g
At rest, since gravity = buoyancy, we find that ρAdg = M g.
Therefore, the draught of the ship is
d=
M
.
ρA
Taking the downwards direction to be positive, while gravity
remains unchanged, the new buoyancy is now −ρA(d + x)g.
11/13
Question 4
The net force is now M g − ρA(d +
x)g. In view of the Newton second
law, we get that
Gravity F=Mg
M
(mass of the tanker)
M ẍ = M g − ρA(d + x)g.
Thanks to the formula for d, we simply have
ρAg
x.
M
Wave F=cos(αt)
0
H-x(t)
object
Question 1
Question 2
x(t)
(density of seawater)
ẍ = −
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d Tanker
ρ
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A
(horizontal cross-section area)
Question 3
Question 4
Friction F=-bx
Buoyancy F=ρA(d+x)g
Suppose we now have wave and friction forces. The new
equation reads as following
M ẍ = M g − ρA(d + x)g − bẋ + F0 cos(αt),
which is, by ρAdg = M g,
M ẍ + bẋ + ρAxg = F0 cos(αt).
12/13
MA 1506
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Question 4
It is well-known that
cos(αt − γ)
F0
, ω=
x(t) =
xh (t)
+ q
| {z }
M (ω 2 − α2 )2 + b2 α2
M2
TRANSIENT&0
r
ρAg
.
M
Since the ship eventually bobs at the same frequency as the
waves, that is α, resonance could occurs. In particular, there
is a danger if
max of the amplitude of x(t) > H.
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Question 1
Question 2
Question 3
Question 4
Using Q3, there are two cases
√
√
If b > 2M ω = 2ρAM g, there is no resonance. In
other words, the amplitude of x(t) always decreases in t.
√
If b < 2ρAM g, the amplitude of x(t) achieves its
maximum
2M F0
2M F0
√
= p
2
2
2
b 4M ω − b
b 4ρAM g − b2
which needs to be less than H.
13/13