Application: Two mass, three spring system
Two mass-three spring system
m1 , m2 > 0, two masses
Math 216
Differential Equations
k1 , k2 , k3 > 0, spring elasticity
x1 (t), x2 (t), displacement of m1 , m2 from equilibrium.
Positive is right, negative left
No friction or external forces
Kenneth Harris
[email protected]
Department of Mathematics
University of Michigan
November 3, 2008
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Application: Two mass, three spring system
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Math 216 Differential Equations
Forces on mass m2
Apply Newton’s law to mass m1 .
Apply Newton’s law to mass m2 .
m2 x200 = −k2 (x2 − x1 ) − k3 x2
m1 x100 = −k1 x1 + k2 (x2 − x1 )
= k2 x1 − (k2 + k3 )x2 .
= −(k1 + k2 )x1 + k2 x2 .
k2 Hx2 -x1 L
-k2 Hx2 -x1 L
k2 Hx2 -x1 L
-k1 x1
-k3 x2
-k2 Hx2 -x1 L
-k3 x1
m1
m1
m2
m2
0<x1 <x2
0<x2 <x1
0<x1 <x2
0<x2 <x1
Hk2 stretchedL
Hk2 stretchedL
Hk2 compressedL
Hk1 stretchedL
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Application: Two mass, three spring system
Forces on mass m1
-k1 x1
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Hk3 compressedL
Hk1 stretchedL
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Hk3 compressedL
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Application: Two mass, three spring system
Application: Two mass, three spring system
Two mass-three spring system: Equations
First method for solving second order system
Two mass, three spring system as a system of equations:
First of three ways to solve the second-order system:
m1 x100 = −(k1 + k2 )x1 + k2 x2
m1 x100 = −(k1 + k2 )x1 + k2 x2
m2 x200 = k2 x1 − (k2 + k3 )x2 .
m2 x200 = k2 x1 − (k2 + k3 )x2 .
As a matrix equation (dividing by m1 , m2 ):
 (k +k )

k2
 
 00 
− 1m1 2
m1
x1
 x1

 
 =




x200
x2
(k2 +k3 )
k2
− m2
m2
Method 1. Rewrite using differential operators:
m1 D 2 + (k1 + k2 )I x1 − k2 x2 = 0
−k2 x1 + m2 D 2 + (k2 + k3 )I x2 = 0.
Solve using Method of Elimination from Section 4.2.
The equation has the form
00
x = Ax.
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Application: Two mass, three spring system
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Application: Two mass, three spring system
Second method for solving second order system
Third method for solving second order system
Second of three ways to solve the second-order system:
m1 x100
m2 x200
Third of three ways to solve the second-order system:
= −(k1 + k2 )x1 + k2 x2
m1 x100 = −(k1 + k2 )x1 + k2 x2
= k2 x1 − (k2 + k3 )x2 .
m2 x200 = k2 x1 − (k2 + k3 )x2 .
Method 2. Rewrite as a system of four equations and four unknowns
(Section 4.1):
y10
Method 3. Rewrite as a 2 × 2 matrix equation (Section 5.1):
= y2
y20
=
y30
=
y40
=

2)
− (k1m+k
1


x00 = 

k2
(k1 + k2 )
y1 +
y3
−
m1
m1
y4
k2
(k2 + k3 )
y1 −
y3
m1
m2
k2
m2
k2
m1
3)
− (k2m+k
2



x

Solve using the (extended) Eigenvalue Method of Section 5.3.
Solve using Eigenvalue Method of Section 5.2.
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Extended eigenvalue method to second-order systems
Extended eigenvalue method to second-order systems
Extended eigenvalue method
Mechanical systems
Solve. x00 = Ax, where A has real constant components.
Solution. If λ = α2 is an eigenvalue and v is an eigenvector,
veαt is a solution to x00 = Ax.
Application. Mass-spring problems are of the form x00 = Ax where
where the eigenvalues are negative.
Guess a solution: x(t) = veαt .
Substitute into x00 = Ax:
since veαt
α2 veαt
00
= α2 veαt ,
Let α2 = −ω 2 and v an associated (real-valued) eigenvector. Then
= Aveαt
x(t) = veiωt = v(cos ωt + i sin ωt)
Cancel eαt
is a solution to x00 = Ax.
2
α v = Av
Real Solutions. The real and imaginary parts
Answer. x(t) = veαt is a solution when
α2 is an eigenvalue and
v an associated eigenvector of A.
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Math 216 Differential Equations
x1 (t) = v cos ωt
x2 (t) = v sin ωt
are linearly independent real-valued solutions to x00 = Ax .
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Extended eigenvalue method to second-order systems
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Example 1
Second-order homogeneous linear systems
Example 1
Problem. Find a general solution to the mass-spring system:
m1 = m2 = 1, two masses
The following is useful for mechanical systems.
k1 = k3 = 4, k2 = 6, spring elasticity
Theorem
Suppose an n × n matrix A has distinct negative eigenvalues
−ω12 , −ω22 , . . . , −ωn2 with associated real eigenvectors v1 , v2 , . . . , vn .
x1 (t), x2 (t), displacement of m1 , m2 from equilibrium.
No friction or outside forces
The second-order equations are
Then a general solution to x00 = Ax is
x(t) =
n
X
x100 = −10x1 + 6x2
x200 = 6x1 − 10x2 .
(aj cos ωj t + bj sin ωj t)
j=1
Equivalently, as a matrix equation:
00 x1
−10
6
x1
=
x200
6
−10 x2
where ai and bi are parameters.
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Example 1
Example 1
Example 1 continued
Example 1 continued
Compute eigenvalues.
−10 − λ
6
= λ2 + 20λ + 64 = (λ + 16)(λ + 4)
6
−10 − λ
Compute eigenvector for λ = −16. These are solutions to
0
−10 + 16
6
x1
=
0
6
−10 + 16 x2
So, λ = −16, −4.
This reduces to a single equation
6x1 + 6x2 = 0
There are four linearly independent real solutions
v1 cos 4t,
v1 sin 4t,
v2 cos 2t,
or
x1 = −x2 .
v2 sin 2t
−1
Eigenvector. v1 =
.
1
where v1 is an eigenvector for λ = −16 and
v2 is an eigenvector for λ = −4.
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Example 1
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Example 1
Example 1 continued
Example 1 concluded
General Solution. The mass-spring system
Compute eigenvector for λ = −4. These are solutions to
−10 + 4
6
x1
0
=
0
6
−10 + 4 x2
x100 = −10x1 + 6x2
x200 = 6x1 − 10x2 .
has a general solution (as a vector equation)
−1
−1
1
1
~x (t) = a1
cos 4t + a2
sin 4t + b1
cos 2t + b2
sin 2t
1
1
1
1
This reduces to a single equation
−6x1 + 6x2 = 0
or
x1 = x2 .
and as a scalar system:
1
Eigenvector. v2 =
.
1
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x1 (t) = −(a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t)
x2 (t) = (a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t)
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Example 1
Example 1
Example 1 analysis
Example 1 analysis
Analysis. The general solution
The displacements x1 of mass m1 and x2 of mass m2 :
x1 (t) = −(a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t)
x2 (t) =
(a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t)
x1 (t)
= −c1 cos(4t − α1 ) + c2 cos(2t − α2 )
x2 (t)
= c1 cos(4t − α1 ) + c2 cos(2t − α2 )
A linear combination of two natural modes of oscillation:
the natural frequencies ω1 = 4 and ω2 = 2.
can be simplified to
x1 (t) = −c1 cos(4t − α1 ) + c2 cos(2t − α2 )
The natural mode ω2 = 2
x2 (t) = c1 cos(4t − α1 ) + c2 cos(2t − α2 )
where
q
c1 = a12 + a22 , tan α1 =
q
c2 = b12 + b22 , tan α1 =
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a2
a1 ,
x1 (t)
= c2 cos(2t − α2 )
x2 (t)
= c2 cos(2t − α2 )
a free oscillation (no damping) in which the masses move in
synchrony in the same direction and with the same frequency (ω2 = 2)
and equal amplitudes of oscillation (c2 ).
b2
b1
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Example 1
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Forced oscillations
Example 1 analysis
Mass-spring systems with external forces
The natural mode ω1 = 4
x1 (t)
=
−c1 cos(4t − α1 )
x2 (t)
=
c1 cos(4 − 1)
Suppose we apply external forces F1 to mass m1 and F2 to mass m2 in
the mass-spring system. We now have a nonhomogeneous system
a free oscillation (no damping) in which the masses move in synchrony in the
opposite directions and with the same frequency (ω1 = 4) and equal
amplitudes of oscillation (c1 ).
m1 x100 (t) = −(k1 + k2 )x1 (t) + k2 x2 (t) + F1 (t)
m2 x200 (t) = k2 x1 (t) − (k2 + k3 )x2 (t) + F2 (t)
x
t
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Forced oscillations
Forced oscillations
Mechanical systems with forced oscillations
First method for solving second order system
Forced oscillations. We are interested in periodic external forces
applied to the masses (F0 is a constant vector)
First of two ways to solve the second-order system:
F
F0 cos ωt = 0 cos ωt
F1
m1 x100 = −(k1 + k2 )x1 + k2 x2 + F0 cos ωt
m2 x200 = k2 x1 − (k2 + k3 )x2 + F1 cos ωt
As a system of equations:
m1 x100
= −(k1 + k2 )x1 + k2 x2 + F0 cos ωt
m2 x200
= k2 x1 − (k2 + k3 )x2 + F1 cos ωt
Method 1. Rewrite using differential operators:
m1 D + (k1 + k2 )I x1 − k2 x2 = F0 cos ωt
−k2 x1 + m2 D + (k2 + k3 )I x2 = F1 cos ωt
As a matrix equation
2 (k +k )
− 1m 2
1
6
4 5=6
6
4
x200
k2
2
x100
3
m2
3
k2
m1
− (k2m+k3 )
2 3 2 3
F0
7 x1
74 5 4 5
+
cos ωt,
7
5
x2
F1
Solve using Method of Elimination from Section 4.2.
2
Note the form x00 = Ax + F0 cos ωt.
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Forced oscillations
Solve the second-order system
x00 = Ax + F0 cos ωt
m1 x100 = −(k1 + k2 )x1 + k2 x2 + F0 cos ωt
m2 x200 = k2 x1 − (k2 + k3 )x2 + F1 cos ωt
Guess a particular solution of the form xp = c cos ωt.
Substitute xp where
x00p = −ω 2 c cos ωt
Method 2. Rewrite as a 2 × 2 matrix equation (Section 5.1):
k2
m2
3)
− (k2m+k
2
−ω 2 c cos ωt

 
F0


 x +   cos ωt

F1
−ω 2 c = Ac + F0 .
New problem. We want a solution c to
A + ω 2 I c = −F0 .
Guess a particular solution of the form xp = c cos ωt,
and determine the values of the parameter c.
Math 216 Differential Equations
= Ac cos ωt + F0 cos ωt
and cancel the common cos ωt
which is of the form x00 = Ax + F0 cos ωt.
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Second method for solving second order system
Second of two ways to solve the second-order system:
k2
m1
November 3, 2008
Forced oscillations
Second method for solving second order system

2)
− (k1m+k
1


x00 = 

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Forced oscillations
Example 2
Second method for solving second order system
Example 2
Solve for unknown c
Problem. Find a general solution to the mass-spring system of
Example 1 when:
2
A + ω I c = −F0 .
Observation. As long as −ω 2 is not an eigenvalue for A, then A + ω 2 I
is invertible.
External forces: F1 = 30 cos t, F2 = 60 cos t
The second-order equations are
Solution. If −ω 2 is not an eigenvalue for A, then
A + ω 2 I c = −F0 .
x100 = −10x1 + 6x2 + 30 cos t
x200 = 6x1 − 10x2 + 60 cos t
has a unique solution.
Observation. If −ω 2 is an eigenvalue for A, then we have resonance.
In this case, we can use techniques of Section 5.6 which generalize
the Method of Undetermined Coefficients and Method of Variation of
Parameters to vector systems.
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Equivalently, as a matrix equation:
00 x1
−10
6
x1
30 cos t
=
+
x200
6
−10 x2
60 cos t
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Example 2
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Example 2
Example 2 continued
Example 2 continued
Compute a particular solution for
Solve the system of equations
00 x1
−10
6
x1
30 cos t
=
+
x200
6
−10 x2
60 cos t
−10 = −3c1 + 2c2
−20 = 2c1 − 3c2
Guess. xp = c cos t.
Let ω = 1. Since −ω 2 = −1 is not an eigenvalue for the system
(λ = −16, −4), there is no resonance.
Substitute the guess xp = c cos t into the equation.
c1
−10
6
c1
30
−
cos t =
cos t +
cos t
c2
6
−10 c2
60
14
xp =
cos t
16
x1 (t) = −(a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t) + 14 cos t
−10 + 1
6
c1
30
=−
6
−10 + 1 c2
60
Math 216 Differential Equations
Particular solution
General solution. xc + xp , where xc is from Example 1:
This reduces to
Kenneth Harris (Math 216)
So, c1 = 14 and c2 = 16.
x2 (t) = (a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t) + 16 cos t
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Example 2
Example 2
Example 2 continued
Example 2 continued
Problem. Find the motion when the two masses begin at the
equillibrium position at rest. The initial conditions are
Solve.
x1 (0) = 0,
x10 (0) = 0,
x2 (0) = 0,
0 = −a1 + b1 + 14
x20 (0) = 0
0 = −4a2 + 2b2
The general solution and derivatives are
0 = a1 + b1 + 16
x1 (t)
=
−(a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t) + 14 cos t
x10 (t)
=
−(−4a1 sin 4t + 4a2 cos t) + (−2b1 sin 2t + 2b2 cos 2t) − 16 sin t
x2 (t) =
(a1 cos 4t + a2 sin 4t) + (b1 cos 2t + b2 sin 2t) + 16 cos t
x20 (t) =
(−4a1 sin 4t + 4a2 cos 4t) + (−2b1 sin 2t + 2b2 cos 2t) − 14 sin t
0 = 4a2 + 2b2
The solutions are
a1 = −1 b1 = −15 a2 = b2 = 0
Substitute.
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0
= −a1 + b1 + 14
0
= −4a2 + 2b2
0
= a1 + b1 + 16
0
= 4a2 + 2b2
Math 216 Differential Equations
Solution. The general solution and derivatives are
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x1 (t)
=
cos 4t − 15 cos 2t + 14 cos t
x2 (t)
=
− cos 4t − 15 cos 2t + 16 cos t
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