MA 1506Mathematics II

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MA 1506 Mathematics II

Tutorial 4The harmonic oscillator

Groups: B03 & B08February 15, 2012

Ngo Quoc AnhDepartment of Mathematics

National University of Singapore

MA 1506Mathematics II

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Question 1: Equilibrium points for ODEs

Equilibrium solutions of ODEs

A solution x of a given ODE x = f(x) is said to be anequilibrium solution if x is constant.

If xE is an equilibrium solution of x = f(x) thenxE = f(xE). Consequently, f(xE) = 0. In other words,finding equilibrium solution is to find solutions of f(x) = 0.

Since coshx = ex+e−x

2 > 0, the equation x = coshxadmits no equilibrium solution.

By solving cosx = 0, we conclude that equilibrium forthe equation x = cosx are π

2 + kπ where k ∈ Z.

Similarly, by solving tan(sinx) = 0, we conclude thatequilibrium for the equation x = tan(sinx) are kπwhere k ∈ Z.

Since the behavior of a solution x depends strongly on f , weneed to study f at points those are near equilibrium points.

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Question 1: Equilibrium points for ODEs

To achieve that goal, we simply use Taylor series. Assumef ′(xE) 6= 0. Note that when x is close enough to xE, thereholds

f(x) ≈ f(xE) + f ′(xE)(x− xE).

Therefore, near the equilibrium point xE, there is nodifference between x = f(x) and its approximated ODE

x− f ′(xE)x = −f ′(xE)xE + f(xE).

Concerning the above approximated ODE, since xE is aparticular solution, we can easily see that

x = xh + xE︸︷︷︸constant

.

Since xE is constant, it contributes nothing to the stabilityof the equilibrium xE. It turns out that we need to study thestability of the so-called linearized equation

x = f ′(xE)x.

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Question 1: Equilibrium points for ODEs

Up to constants, x = f ′(xE)x is equivalent to either x = x(xE is therefore unstable) or x = −x (xE is now stable).

The full stability of items (ii) and (iii) can be describedthrough the following phase portraits.

MA 1506Mathematics II

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Question 1: Equilibrium points for ODEs

For item (i), the phase por-trait looks like the picture onthe right.

Keep in mind that the x-axisrepresents x while the y-axisrepresents x. Thus, equilib-rium points belong to the x-axis.

The meaning of this phaseportrait is the following: Each

point of the plane represents an initial value. Given such apoint, a solution x(t) to the ODE exists. Accordingly, thereis a curve representing (x(t), x(t)) when t varies.

To identify equilibrium points, we look for points on thex-axis those have arrows concentrating in a smallneighborhood. To draw the above picture, we can use Maple.

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Question 2: RLC circuitBy the Kirchhoff voltage law, there holds

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 = Ld

dti(t), VC =

Q

C, i(t) =

d

dtQ(t)

we have

Ri(t) + Ld

dti(t) +

Q

C= V (t).

Making use of i(t) = dQdt , it leads to a 2nd order differential

equation for Q given by

d2

dt2Q(t) +

R

L

d

dtQ(t) +

1

LCQ(t) =

v(t)

L.

In addition, it is given that i(0) = 0 and Q(0) = 0.

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Question 2: RLC circuit

Since R = 5 Ω, L = 0.05H, C = 4× 10−4 F , andV (t) = 200 cos(100t)V , we arrive at

d2

dt2Q(t) + 100

d

dtQ(t) + 50000Q(t) = 4000 cos(100t).

The characteristic equation, λ2 + 100λ+ 5× 104 = 0, hastwo 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.

Since the function on the right, r(t), is of the form”polynomial × cosine”, we can use the method ofundetermined coefficients. Precisely, we shall find

Qp = A cos(100t) +B sin(100t).

Substituting and equalizing, we get that

Qp =16

170cos(100t) +

4

170sin(100t).

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Question 2: RLC circuit

In particular, we have

Q(t) =(c1 cos(50√

19t) + c2 sin(50√

19t)e−50t

+16

170cos(100t) +

4

170sin(100t).

It leaves out to determine c1 and c2. Thanks to the giveninitial data, we get that

0 = c1 +16

170, c1 = − 16

170

0 =80

17+ 50√

19c2 +40

17, c2 = −12

√19

1615.

Hence, the formula for i(t), which is ddtQ(t), is given by

i(t) =(− 680

289cos(50

√19) +

27889√

19

5491sin(50

√19))e−50t

+16

170cos(100t) +

4

170sin(100t).

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Question 3: Maximum of the amplitude response function

When solving the following ODE

mx+ bx+ kx = F0 cos(αt)

one eventually gets

x(t) = xh(t)︸ ︷︷ ︸TRANSIENT0

+1

m

F0√(ω2 − α2)2 + b2

m2α2︸ ︷︷ ︸A(α)

cos(αt− γ)

as a general solution where

ω =

√k

m, tan γ =

bα

k −mα2.

It is well-known that the behavior of A(α) depends on thefrequency α and the friction constant b. To find itsmaximum value, we simply differentiate A w.r.t α to find itscritical points.

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Question 3: Maximum of the amplitude response function

It’s not hard to find that A′(α) is nothing but

− F0

2m

((ω2 − α2)

2+

b2

m2α2

)− 32 [

2(ω2 − α2)(−2α) + 2αb2

m2︸ ︷︷ ︸4α

(b2

2m2−ω2+α2)

].

Consequently, A′(α) = 0 is equivalent to

α2 = ω2 − b2

2m2.

There are two cases

If b <√

2mω, then A′ = 0 hasa unique solution. Hence, A hasa maximum value given by

Aresonance =2mF0

b√

4m2ω2 − b2.

If b >√

2mω, then A′ < 0.

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Question 4Gravity F=Mg

M(mass of the tanker)

Buoyancy F=ρAdg

ρ(density of seawater)

ρobjectTanker

d

(horizontal cross-section area)A

Gravity F=Mg

M(mass of the tanker)

Buoyancy F=ρA(d+x)g

ρ(density of seawater)

object

(horizontal cross-section area)

Tanker

A

dx(t)

At rest, since gravity = buoyancy, we find that ρAdg = Mg.Therefore, the draught of the ship is

d =M

ρA.

Taking the downwards direction to be positive, while gravityremains unchanged, the new buoyancy is now −ρA(d+ x)g.

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Question 4

The net force is now Mg − ρA(d+x)g. In view of the Newton secondlaw, we get that

Mx = Mg − ρA(d+ x)g.

Thanks to the formula for d, we sim-ply have

x = −ρAgM

x.

Gravity F=Mg

M(mass of the tanker)

Buoyancy F=ρA(d+x)g

ρ(density of seawater)

object

(horizontal cross-section area)

Tanker

A

dx(t)

Wave F=cos(αt)0

Friction F=-bx

H-x(t)

Suppose we now have wave and friction forces. The newequation reads as following

Mx = Mg − ρA(d+ x)g − bx+ F0 cos(αt),

which is, by ρAdg = Mg,

Mx+ bx+ ρAxg = F0 cos(αt).

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Question 4

It is well-known that

x(t) = xh(t)︸ ︷︷ ︸TRANSIENT0

+F0

M

cos(αt− γ)√(ω2 − α2)2 + b2

M2α2, ω =

√ρAg

M.

Since the ship eventually bobs at the same frequency as thewaves, that is α, resonance could occurs. In particular, thereis a danger if

max of the amplitude of x(t) > H.

Using Q3, there are two cases

If b >√

2Mω =√

2ρAMg, there is no resonance. Inother words, the amplitude of x(t) always decreases in t.

If b <√

2ρAMg, the amplitude of x(t) achieves itsmaximum

2MF0

b√

4M2ω2 − b2=

2MF0

b√

4ρAMg − b2

which needs to be less than H.