In this research article, we investigate the stability of a complex dynamical system involving coupled rigid bodies consisting of three equal masses joined by three rigid rods of equal lengths, hinged at each of their bases. The system is free to oscillate in the vertical plane. We obtained the equation of motion using the generalized coordinates and the Euler-Lagrange equations. We then proceeded to study the stability of the dynamical systems using the Jacobian linearization method and subsequently confirmed our result by phase portrait analysis. Finally, we performed MathCAD simulation of the resulting ordinary differential equations, describing the dynamics of the system and obtained the graphical profiles for each generalized coordinates representing the angles measured with respect to the vertical axis. It is discovered that the coupled rigid pendulum gives rise to irregular oscillations with ever increasing amplitude. Furthermore, the resulting phase portrait analysis depicted spiral sources for each of the oscillating masses showing that the system under investigation is unstable.
The dynamics of coupled bodies and oscillators is significant in mechanics, engineering, electronics as well as biological systems. They are mostly represented as nonlinear dynamical systems [
Chutiphon [
Maliki and Nwoba [
Maliki and Okereke [
We consider the problem of analyzing the dynamics of a triple pendulum as shown in
From the figure below we choose θ 1 , θ 2 and θ 3 as our generalized coordinates.
To compute the Lagrangean of the system, we first compute the total kinetic energy (K.E) and potential energy (P.E) of the system. However, we require the following expressions.
x 1 = l 1 sin θ 1 (1)
y 1 = − l 1 cos θ 1 (2)
x 2 = l 1 sin θ 1 + l 2 sin θ 2 (3)
y 2 = − l 1 cos θ 1 − l 2 cos θ 2 (4)
x 3 = l 1 sin θ 1 + l 2 sin θ 2 + l 3 sin θ 3 (5)
y 3 = − l 1 cos θ 1 − l 2 cos θ 2 − l 3 cos θ 3 (6)
Differentiating the above coordinates with respect to time, we get;
x ˙ 1 = l 1 θ ˙ 1 cos θ 1 (7)
y ˙ 1 = l 1 θ ˙ 1 sin θ 1 (8)
x ˙ 2 = l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 (9)
y ˙ 2 = l 1 θ ˙ 1 sin θ 1 + l 2 θ ˙ 2 sin θ 2 (10)
x ˙ 3 = l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 + l 3 θ ˙ 3 cos θ 3 (11)
y ˙ 3 = l 1 θ ˙ 1 sin θ 1 + l 2 θ ˙ 2 sin θ 2 + l 3 θ ˙ 3 sin θ 3 (12)
The square of the velocities for each mass is given by;
v 1 2 = x ˙ 2 1 + y ˙ 2 1 = ( l 1 θ ˙ 1 cos θ 1 ) 2 + ( l 1 θ ˙ 1 sin θ 1 ) 2 = l 1 2 θ ˙ 2 1 (13)
Similarly,
v 2 2 = x ˙ 2 2 + y ˙ 2 2 = ( l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 ) 2 + ( l 1 θ ˙ 1 sin θ 1 + l 2 θ ˙ 2 sin θ 2 ) 2 = l 1 2 θ ˙ 2 1 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) (14)
Finally,
v 3 2 = x ˙ 2 3 + y ˙ 2 3 = ( l 1 θ ˙ 1 cos θ 1 + l 2 θ ˙ 2 cos θ 2 + l 3 θ ˙ 3 cos θ 3 ) 2 + ( l 1 θ ˙ 1 sin θ 1 + l 2 θ ˙ 2 sin θ 2 + l 3 θ ˙ 3 sin θ 3 ) 2
After some algebraic manipulations, we get
v 3 2 = l 1 2 θ ˙ 2 1 + l 2 2 θ ˙ 2 2 + l 3 2 θ ˙ 2 3 + 2 [ l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) + l 1 l 3 θ ˙ 1 θ ˙ 3 cos ( θ 1 − θ 3 ) + l 2 l 3 θ ˙ 2 θ ˙ 3 cos ( θ 2 − θ 3 ) ] (15)
The total kinetic energy of the pendulum is then:
T = 1 2 m 1 v 1 2 + 1 2 m 2 v 2 2 + 1 2 m 3 v 3 2 = 1 2 m 1 l 1 2 θ ˙ 2 1 + 1 2 m 2 [ l 1 2 θ ˙ 2 1 + l 2 2 θ ˙ 2 2 + 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) ] + 1 2 m 3 ( l 1 2 θ ˙ 2 1 + l 2 2 θ ˙ 2 2 + l 3 2 θ ˙ 2 3 + 2 [ l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) + l 1 l 3 θ ˙ 1 θ ˙ 3 cos ( θ 1 − θ 3 ) + l 2 l 3 θ ˙ 2 θ ˙ 3 cos ( θ 2 − θ 3 ) ] ) (16)
The total potential energy of the pendulum is the sum of the potential energy of each mass;
V = m 1 g y 1 + m 2 g y 2 + m 3 g y 3
⇒ V = − m 1 g l 1 cos θ 1 + m 2 g ( − l 1 cos θ 1 − l 2 cos θ 2 ) + m 3 g ( − l 1 cos θ 1 − l 2 cos θ 2 − l 3 cos θ 3 ) (18)
Recall that the Lagrangean is given by L = T − V .
∴ L = 1 2 m 1 l 1 2 θ ˙ 2 1 + 1 2 m 2 l 1 2 θ ˙ 2 1 + 1 2 m 2 l 2 2 θ ˙ 2 2 + 1 2 m 3 l 1 2 θ ˙ 2 1 + 1 2 m 3 l 2 2 θ ˙ 2 2 + 1 2 m 3 l 3 2 θ ˙ 2 3 + m 2 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) + m 3 l 1 l 2 θ ˙ 1 θ ˙ 2 cos ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 cos ( θ 1 − θ 3 ) + m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 cos ( θ 2 − θ 3 ) + m 2 g l 1 cos θ 1 + m 2 g l 2 cos θ 2 + m 3 g l 1 cos θ 1 + m 3 g l 2 cos θ 2 + m 3 g l 3 cos θ 3 (19)
We now employ the E-L equations to obtain the equations of motion, i.e.;
d d t ( ∂ L ∂ θ ˙ j ) − ∂ L ∂ θ j = 0 , j = 1 , 2 , 3 (20)
For the first generalized coordinate j = 1, we have;
∂ L ∂ θ ˙ 1 = m 1 l 1 2 θ ˙ 1 + m 2 l 1 2 θ ˙ 1 + m 2 l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) + m 3 l 1 2 θ ˙ 1 + m 3 l 1 l 2 θ ˙ 2 cos ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 3 cos ( θ 1 − θ 3 )
⇒ ∂ L ∂ θ ˙ 1 = l 1 2 θ ˙ 1 ( m 1 + m 2 + m 3 ) + l 1 l 2 θ ˙ 2 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 3 cos ( θ 1 − θ 3 )
∴ d d t ( ∂ L ∂ θ ˙ 1 ) = l 1 2 θ ¨ 1 ( m 1 + m 2 + m 3 ) + l 1 l 2 θ ˙ 2 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 3 cos ( θ 1 − θ 3 )
Furthermore
∂ L ∂ θ 1 = − m 2 l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − m 3 l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) − m 1 g l 1 sin θ 1 − m 2 g l 1 sin θ 1 − m 3 g l 1 sin θ 1
⇒ ∂ L ∂ θ 1 = − l 1 l 2 θ ˙ 1 θ ˙ 2 ( m 2 + m 3 ) sin ( θ 1 − θ 2 ) − m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) − g l 1 sin θ 1 ( m 1 + m 2 + m 3 )
∴ d d t ( ∂ L ∂ θ ˙ 1 ) − ∂ L ∂ θ 1 = l 1 2 θ ¨ 1 ( m 1 + m 2 + m 3 ) + l 1 l 2 θ ˙ 2 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 3 cos ( θ 1 − θ 3 ) + l 1 l 2 θ ˙ 1 θ ˙ 2 ( m 2 + m 3 ) sin ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) + g l 1 sin θ 1 ( m 1 + m 2 + m 3 ) = 0 (21)
For the second generalized coordinate j = 2 in (1.20), we have;
∂ L ∂ θ ˙ 2 = m 2 l 2 2 θ ˙ 2 + m 2 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + m 3 l 2 2 θ ˙ 2 + m 3 l 1 l 2 θ ˙ 1 cos ( θ 1 − θ 2 ) + m 3 l 2 l 3 θ ˙ 3 cos ( θ 2 − θ 3 ) = l 2 2 θ ˙ 2 ( m 2 + m 3 ) + l 1 l 2 θ ˙ 1 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 2 l 3 θ ˙ 3 cos ( θ 2 − θ 3 )
⇒ d d t ( ∂ L ∂ θ ˙ 2 ) = l 2 2 θ ¨ 2 ( m 2 + m 3 ) + l 1 l 2 θ ˙ 1 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 2 l 3 θ ˙ 3 cos ( θ 2 − θ 3 )
Also,
∂ L ∂ θ 2 = − m 2 l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − m 3 l 1 l 2 θ ˙ 1 θ ˙ 2 sin ( θ 1 − θ 2 ) − m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) − m 2 g l 2 sin θ 2 − m 2 g l 2 sin θ 2
⇒ ∂ L ∂ θ 2 = − l 1 l 2 θ ˙ 1 θ ˙ 2 ( m 2 + m 3 ) sin ( θ 1 − θ 2 ) − m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) − g l 2 sin θ 2 ( m 2 + m 3 )
∴ d d t ( ∂ L ∂ θ ˙ 2 ) − ∂ L ∂ θ 2 = l 2 2 θ ¨ 2 ( m 2 + m 3 ) + l 1 l 2 θ ˙ 1 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 2 l 3 θ ˙ 3 cos ( θ 2 − θ 3 ) + l 1 l 2 θ ˙ 1 θ ˙ 2 ( m 2 + m 3 ) sin ( θ 1 − θ 2 ) + m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) + g l 2 sin θ 2 ( m 2 + m 3 ) = 0 (22)
For the third generalized coordinate j = 3 in (20), we have;
∂ L ∂ θ ˙ 3 = m 3 l 3 2 θ ˙ 3 + m 3 l 1 l 3 θ ˙ 1 cos ( θ 1 − θ 3 ) + m 3 l 2 l 3 θ ˙ 2 cos ( θ 2 − θ 3 )
⇒ d d t ( ∂ L ∂ θ ˙ 3 ) = m 3 l 3 2 θ ¨ 3 + m 3 l 1 l 3 θ ˙ 1 cos ( θ 1 − θ 3 ) + m 3 l 2 l 3 θ ˙ 2 cos ( θ 2 − θ 3 )
Furthermore,
∂ L ∂ θ 3 = − m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) − m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) − m 3 g l 3 sin θ 3
⇒ d d t ( ∂ L ∂ θ ˙ 3 ) − ∂ L ∂ θ 3 = m 3 l 3 2 θ ¨ 3 + m 3 l 1 l 3 θ ˙ 1 cos ( θ 1 − θ 3 ) + m 3 l 2 l 3 θ ˙ 2 cos ( θ 2 − θ 3 ) + m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) + m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) + m 3 g l 3 sin θ 3 = 0 (23)
Therefore, the equations of motion are;
l 1 2 θ ¨ 1 ( m 1 + m 2 + m 3 ) + l 1 l 2 θ ˙ 2 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 3 cos ( θ 1 − θ 3 ) + l 1 l 2 θ ˙ 1 θ ˙ 2 ( m 2 + m 3 ) sin ( θ 1 − θ 2 ) + m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) + g l 1 ( m 1 + m 2 + m 3 ) sin θ 1 = 0 (24)
l 2 2 θ ¨ 2 ( m 2 + m 3 ) + l 1 l 2 θ ˙ 1 ( m 2 + m 3 ) cos ( θ 1 − θ 2 ) + m 3 l 2 l 3 θ ˙ 3 cos ( θ 2 − θ 3 ) + l 1 l 2 θ ˙ 1 2 ( m 2 + m 3 ) sin ( θ 1 − θ 2 ) + m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) + g l 2 ( m 2 + m 3 ) sin θ 2 = 0 (25)
m 3 l 3 2 θ ¨ 3 + m 3 l 1 l 3 θ ˙ 1 cos ( θ 1 − θ 3 ) + m 3 l 2 l 3 θ ˙ 2 cos ( θ 2 − θ 3 ) + m 3 l 1 l 3 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) + m 3 l 2 l 3 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) + m 3 g l 3 sin θ 3 = 0 (26)
Assuming equal masses and equal lengths of rods, i.e., l 1 = l 2 = l 3 = l , m 1 = m 2 = m 3 = m .
The equations become respectively;
3 l 2 θ ¨ 1 m + 2 l 2 θ ˙ 2 m cos ( θ 1 − θ 2 ) + m l 2 θ ˙ 3 cos ( θ 1 − θ 3 ) + 2 l 2 θ ˙ 1 θ ˙ 2 m sin ( θ 1 − θ 2 ) + m l 2 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) + 3 g l m sin θ 1 = 0 (27)
2 l 2 θ ¨ 2 m + 2 l 2 θ ˙ 1 m cos ( θ 1 − θ 2 ) + m l 2 θ ˙ 3 cos ( θ 2 − θ 3 ) + 2 l 2 θ ˙ 1 2 m sin ( θ 1 − θ 2 ) + m l 2 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) + 2 g l m sin θ 2 = 0 (28)
m l 2 θ ¨ 3 + m l 2 θ ˙ 1 cos ( θ 1 − θ 3 ) + m l 2 θ ˙ 3 cos ( θ 2 − θ 3 ) + m l 2 θ ˙ 1 θ ˙ 3 sin ( θ 1 − θ 3 ) + m l 2 θ ˙ 2 θ ˙ 3 sin ( θ 2 − θ 3 ) + m g l sin θ 3 = 0 (29)
We assume the generalized coordinates θ 1 , θ 2 , θ 3 representing the angular displacements are small so that;
sin θ i ≈ θ i , s i n ( θ i − θ j ) ≈ θ i − θ j , cos ( θ i − θ j ) ≈ 1 − 1 2 ( θ i − θ j ) 2 , ∀ i ≠ j
The equations of motion for the coupled rigid body then become;
3 θ ¨ 1 + 2 θ ˙ 2 [ 1 − 1 2 ( θ 1 − θ 2 ) 2 ] + θ ˙ 3 [ 1 − 1 2 ( θ 1 − θ 3 ) 2 ] + 2 θ ˙ 1 θ ˙ 2 ( θ 1 − θ 2 ) + θ ˙ 1 θ ˙ 3 ( θ 1 − θ 3 ) + 3 g l θ 1 = 0
⇒ 3 θ ¨ 1 + 2 θ ˙ 2 + θ ˙ 3 − θ ˙ 2 ( θ 1 − θ 2 ) 2 − 1 2 θ ˙ 3 ( θ 1 − θ 3 ) 2 + 2 θ ˙ 1 θ ˙ 2 ( θ 1 − θ 2 ) + θ ˙ 1 θ ˙ 3 ( θ 1 − θ 3 ) + 3 g l θ 1 = 0 (30)
Following the above procedure, we have for Equation (28)
2 θ ¨ 2 + 2 θ ˙ 1 [ 1 − 1 2 ( θ 1 − θ 2 ) 2 ] + θ ˙ 3 [ 1 − 1 2 ( θ 2 − θ 3 ) 2 ] + 2 θ ˙ 1 2 ( θ 1 − θ 2 ) + θ ˙ 2 θ ˙ 3 ( θ 2 − θ 3 ) + 2 g l θ 2 = 0
⇒ 2 θ ¨ 2 + 2 θ ˙ 1 + θ ˙ 3 − 1 2 θ ˙ 1 ( θ 1 − θ 2 ) 2 − 1 2 θ ˙ 3 ( θ 2 − θ 3 ) 2 + 2 θ ˙ 1 2 ( θ 1 − θ 2 ) + θ ˙ 2 θ ˙ 3 ( θ 2 − θ 3 ) + 2 g l θ 2 = 0 (31)
Similarly for Equation (29) we have;
θ ¨ 3 + θ ˙ 1 [ 1 − 1 2 ( θ 1 − θ 3 ) 2 ] + θ ˙ 3 [ 1 − 1 2 ( θ 2 − θ 3 ) 2 ] + θ ˙ 1 θ ˙ 3 ( θ 1 − θ 3 ) + θ ˙ 2 θ ˙ 3 ( θ 2 − θ 3 ) + g l θ 3 = 0
⇒ θ ¨ 3 + θ ˙ 1 + θ ˙ 3 − 1 2 θ ˙ 1 ( θ 1 − θ 3 ) 2 − 1 2 θ ˙ 3 ( θ 2 − θ 3 ) 2 + θ ˙ 1 θ ˙ 3 ( θ 1 − θ 3 ) + θ ˙ 2 θ ˙ 3 ( θ 2 − θ 3 ) + g l θ 3 = 0 (32)
The model equations for the given problem are summarized as;
{ θ ¨ 1 + 2 3 θ ˙ 2 + 1 3 θ ˙ 3 − 1 3 θ ˙ 2 ( θ 1 − θ 2 ) 2 − 1 6 θ ˙ 3 ( θ 1 − θ 3 ) 2 + 2 3 θ ˙ 1 θ ˙ 2 ( θ 1 − θ 2 ) + 1 3 θ ˙ 1 θ ˙ 3 ( θ 1 − θ 3 ) + g l θ 1 = 0 θ ¨ 2 + θ ˙ 1 + 1 2 θ ˙ 3 − 1 4 θ ˙ 1 ( θ 1 − θ 2 ) 2 − 1 4 θ ˙ 3 ( θ 2 − θ 3 ) 2 + θ ˙ 1 2 ( θ 1 − θ 2 ) + 1 2 θ ˙ 2 θ ˙ 3 ( θ 2 − θ 3 ) + g l θ 2 = 0 θ ¨ 3 + θ ˙ 1 + θ ˙ 3 − 1 2 θ ˙ 1 ( θ 1 − θ 3 ) 2 − 1 2 θ ˙ 3 ( θ 2 − θ 3 ) 2 + θ ˙ 1 θ ˙ 3 ( θ 1 − θ 3 ) + θ ˙ 2 θ ˙ 3 ( θ 2 − θ 3 ) + g l θ 3 = 0 (33)
For the purpose of stability analysis we must vectorize the above coupled system of differential equations.
Let u 1 = θ 1 , u 2 = θ 2 , u 3 = θ 3 and u ˙ 1 = u 4 , u ˙ 2 = u 5 , u ˙ 3 = u 6 , hence u ¨ 1 = u ˙ 4 , u ¨ 2 = u ˙ 5 , u ¨ 3 = u ˙ 6 Thus the vectorized system of equations for the coupled pendulums is;
u ˙ 1 = u 4 (34)
u ˙ 2 = u 5 (35)
u ˙ 3 = u 6 (36)
u ˙ 4 = − 2 3 u 5 − 1 3 u 6 + 1 3 u 5 ( u 1 − u 2 ) 2 + 2 6 u 6 ( u 1 − u 3 ) 2 − 2 3 u 4 u 5 ( u 1 − u 2 ) − 1 3 u 4 u 6 ( u 1 − u 3 ) − g l u 1 (37)
u ˙ 5 = − u 4 − 1 2 u 6 + 1 4 u 4 ( u 1 − u 2 ) 2 + 1 4 u 6 ( u 2 − u 3 ) 2 − u 4 2 ( u 1 − u 2 ) − 1 2 u 5 u 6 ( u 2 − u 3 ) − g l u 2 (38)
u ˙ 6 = − u 4 − u 5 + 1 2 u 4 ( u 1 − u 3 ) 2 + 1 2 u 6 ( u 2 − u 3 ) 2 − u 4 u 6 ( u 1 − u 3 ) − u 5 u 6 ( u 2 − u 3 ) − g l u 3 (39)
We shall investigate the stability of the pendulum at the critical point, where u ˙ 1 = u ˙ 2 = u ˙ 3 = u ˙ 4 = u ˙ 5 = u ˙ 6 = 0 . This implies that u 4 = 0 , u 5 = 0 , u 6 = 0 and by substitution our critical point is ( 0 , 0 , 0 , 0 , 0 , 0 ) .
Equations (34)-(39) can, for convenience) be written simply as;
u ˙ 1 = f 1 ( u 1 , ⋯ , u 6 ) ⋮ u ˙ 6 = f 6 ( u 1 , ⋯ , u 6 ) (40)
where the f i ( u 1 , ⋯ , u 6 ) , i = 1 , ⋯ , 6 represent the RHS of the system.
The Jacobian of the system is written
J f ( u ) = ( ∂ f i ∂ u j ) 6 × 6 (41)
Computing the entries and evaluating at the critical point, we get
J f = ( 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 − g l 0 0 0 − 2 3 − 1 3 0 − g l 0 − 1 0 − 1 2 0 0 − g l − 1 − 1 0 )
The eigenvalues of the matrix J are given by | J f − λ I | = 0 .
⇒ | − λ 0 0 1 0 0 0 − λ 0 0 1 0 0 0 − λ 0 0 1 − g l 0 0 − λ − 2 3 − 1 3 0 − g l 0 − 1 − λ − 1 2 0 0 − g l − 1 − 1 − λ | = 0
This simplifies to give;
λ 6 + ( − 3 α − 3 2 ) λ 4 + 2 3 λ 3 + ( 3 α 2 − 3 2 α ) λ − α 3 = 0 (42)
where α = − g / l . For simplicity we take l = 1 , and naturally g = 9.8 .
We now employ the POLYROOT algorithm in MathCad [
Define a vector v of the coefficients beginning with the constant term, i.e.:
v = [ − α 3 0 ( 3 α 2 + 3 2 α ) 2 3 ( − 3 α − 3 2 ) 0 1 ] T (43)
∴ polyroots ( v ) = [ − 0.703 − 3.051 i − 0.703 + 3.051 i 0.282 + 3.118 i 0.282 − 3.118 i 0.421 − 3.102 i 0.421 + 3.102 i ] (44)
Clearly not all the eigenvalues have negative real parts, we therefore conclude that the critical point of the system is unstable.
In MathCAD we define the vector of derivatives, viz.;
Additional arguments for the ODE solver are
t 0 : = 0 Initial value of independent variable
t 1 : = 10 End value of independent variable
Y 0 : = ( 0.01 0 0 0 0 0 ) Vector of initial function values
n u m : = 1 × 10 3 Number of solution values on [t0, t1]
Solution MatrixS 1 : = Rkadapt ( Y 0 , t 0 , t 1 , n u m , D )
t : = S 1 〈 0 〉 Independent variable values
u 1 : = S 1 〈 1 〉 First solution function values
u 2 : = S 1 〈 2 〉 Second solution function values
u 3 : = S 1 〈 3 〉 Third solution function values
u 4 : = S 1 〈 4 〉 Fourth solution function values
u 5 : = S 1 〈 5 〉 Fifth solution function values
u 6 : = S 1 〈 6 〉 Sixth solution function values
The graph of θ 1 ( t ) (
0 | 1 | 2 | 3 | 4 | 5 | 6 | ||
---|---|---|---|---|---|---|---|---|
0 | 0 | 0.01 | 0 | 0 | 0 | 0 | 0 | |
S1 = | 1 | 0.01 | 9.995 × 10−3 | 1.631 × 10−8 | 1.637 × 10−8 | −9.799 × 10−4 | 4.891 × 10−6 | 4.915 × 10−6 |
2 | 0.01 | 9.98 × 10−3 | 1.131 × 10−7 | 1.313 × 10−7 | −1.959 × 10−3 | 1.952 × 10−5 | 1.972 × 10−5 | |
3 | 0.01 | 9.956 × 10−3 | 4.39 × 10−7 | 4.439 × 10−7 | −2.936 × 10−3 | 4.381 × 10−5 | 4.448 × 10−5 | |
4 | 0.01 | 9.922 × 10−3 | 1.038 × 10−6 | 1.054 × 10−6 | −3.911 × 10−3 | 7.768 × 10−5 | 7.924 × 10−5 | |
5 | 0.01 | 9.878 × 10−3 | 2.024 × 10−6 | 2.062 × 10−6 | −4.882 × 10−3 | 1.21 × 10−4 | 1.24 × 10−4 | |
6 | 0.01 | 9.824 × 10−3 | 3.489 × 10−6 | 3.569 × 10−6 | −5.849 × 10−3 | 1.736 × 10−4 | 1.789 × 10−4 | |
7 | 0.01 | 9.761 × 10−3 | 5.527 × 10−6 | 5.674 × 10−6 | −6.811 × 10−3 | 2.354 × 10−4 | 2.438 × 10−4 | |
8 | 0.01 | 9.688 × 10−3 | 8.228 × 10−6 | 8.478 × 10−6 | −7.767 × 10−3 | 3.062 × 10−4 | 3.187 × 10−4 | |
9 | 0.01 | 9.605 × 10−3 | 1.168 × 10−5 | 1.208 × 10−5 | −8.716 × 10−3 | 3.859 × 10−4 | 4.036 × 10−4 | |
10 | 0.1 | 9.514 × 10−3 | 1.597 × 10−5 | 1.658 × 10−5 | −9.657 × 10−3 | 4.741 × 10−4 | 4.983 × 10−4 | |
11 | 0.11 | 9.412 × 10−3 | 2.119 × 10−5 | 2.208 × 10−5 | −0.011 | 5.708 × 10−4 | 6.029 × 10−4 | |
12 | 0.12 | 9.302 × 10−3 | 2.742 × 10−5 | 2.867 × 10−5 | −0.012 | 6.756 × 10−4 | 7.172 × 10−4 | |
13 | 0.13 | 9.182 × 10−3 | 3.473 × 10−5 | 3.645 × 10−5 | −0.012 | 7.884 × 10−4 | 8.41 × 10−4 | |
14 | 0.14 | 9.053 × 10−3 | 4.321 × 10−5 | 4.552 × 10−5 | −0.013 | 9.088 × 10−4 | 9.743 × 10−4 | |
15 | 0.15 | 8.916 × 10−3 | 5.597 × 10−5 | 5.597 × 10−5 | ∙∙∙ | 1.037 × 10−4 | ∙∙∙ |
θ ˙ 1 ( t ) = u 4 , θ ˙ 2 ( t ) = u 5 and θ ˙ 3 ( t ) = u 6 . These velocities are clearly oscillatory and irregular with increasing amplitude.
Figures 3(a)-(c) are respectively the phase portrait for θ ˙ 1 ( t ) = u 4 against θ 1 ( t ) = u 1 , θ ˙ 2 ( t ) = u 5 against θ 2 ( t ) = u 2 and θ ˙ 3 ( t ) = u 6 against θ 3 ( t ) = u 3 . For each of the phase portraits we obtain a spiral source.
Coupled oscillators in general, are useful in the study of vibrations and as such computing the specific modes of vibration for oscillating systems is very important from a practical point of view, particularly in engineering. The study of coupled systems is useful in mechanics, electronics as well as biological systems. It is also pertinent for the biomechanical analysis of animals, humans or robotic systems. An early application of mathematical modeling in biological systems was first pioneered by Van der Pol in 1928. The Van der Pol oscillator [
Our research work has clearly demonstrated the power of mathematical modeling, where the behaviour of a complex mechanical system can be captured in the form of a system of ordinary differential equations. Furthermore, with versatile mathematical software such as MathCAD, the system can be analyzed in detailed graphical format to give a deep understanding of the inherent dynamics as well as its stability properties.
Maliki, O.S. and Anozie, V.O. (2018) On the Stability Analysis of a Coupled Rigid Body. Applied Mathematics, 9, 210-222. https://doi.org/10.4236/am.2018.93016