Maple 12 is used to compute the transfer function of a simple mechanical translation system using differential equations, Laplace transforms , matrices and vectors.

2. Mechanical Translations Systems
Solution
First, notice that this is the same system as used in Example 2.17 of Reference [1], however here we
have no friction between the masses and the ground. Secondly, we will use the common symbols for
the mechanical components and not the fancy ones introduced by Nise.
(a)
First draw the FBD's of the two masses, see the figure below.
u k2$ x1 K x2 k2$ x2 K x1
.. .. m2 k3$x2
m1 m1$x1 m2$x2
k1$x1 . . . .
b$ x1 K x2 b$ x2 K 1
x
n
Using FDB's of the two masses and Newton's second law ( >F = 0 ) we can obtain the following
k= 1
k
equations of motion:
FBD mass 1 : Keep m2 still , move m1 in positive direction and evaluate the forces acting on m1
.. . .
m1$x1 Cb$ x1 Kx2 Ck1$x1 Ck2$ x1 Kx2 = u
FBD mass 2 : Keep m1 still , move m2 in positive direction and evaluate the forces acting on m2
.. . .
m2$x2 Cb$ x2 Kx1 Ck3$x2 Ck2$ x2 Kx1 = 0
Simplifying and rearranging we obtain
2
d x1 d x1 d x2
m1$ Cb$ C k1 Ck2 $x1 = b$ Ck2$x2 Cu
d t2 dt dt
2
d x2 d x2 d x1
m2$ 2
Cb$ C k2 Ck3 $x2 = b$ Ck2$x1
dt dt dt
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3. Mechanical Translations Systems
Taking the laplace transform of these two equations assuming zero initial conditions, we have
m1$s2 Cb$s Ck1 Ck2 $X1 = b$s Ck2 $X2 CU
m2$s2 Cb$s Ck2 Ck3 $X2 = b$s Ck2 $X1
We solve these two equations for X1 , X2 with Maple's solve and we find
X1 s m2 s2 Cb s Ck2 Ck3
G1 s = =
m1$s2 Cb$s Ck1 Ck2 $ m2$s2 Cb$s Ck2 Ck3 K b$s Ck2
U s 2
X2 s b s Ck2
G2 s = =
m1$s Cb$s Ck1 Ck2 $ m2$s2 Cb$s Ck2 Ck3 K b$s Ck2
U s 2 2
Look behind the button, to see how it is done in Maple. We also could have used Cramer's rule, as done
in Example 2.17, but having Maple at hand this is only a waste of time!
(b)
The two not simplified equations of motion were:
.. . .
m1$x1 Cb$ x1 Kx2 Ck1$x1 Ck2$ x1 Kx2 = u
.. . .
m2$x2 Cb$ x2 Kx1 Ck3$x2 Ck2$ x2 Kx1 = 0
Taking the Laplace transform and rearranging for matrix-vector form we have
m1$s2 Cb$s Ck1 Ck2 $X1 C K Kk2 $X2 = U
s$b
K Kk2 $X1 C m2$s2 Cb$s Ck2 Ck3 $X2 = 0
s$b
The matrix-vector form A.X = B in the s-domain now is
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4. Mechanical Translations Systems
m1$s2 Cb$s Ck1 Ck2 K Kk2
s$b X1 u
. =
K Kk2
s$b m2$s2 Cb$s Ck2 Ck3 X2 0
Solutions can be found using matrix inverse and is found to be the same as before:
X = AK $B
1
Look behind the button to see how this result is achieved.
Example 2 (Skill-Assessment Exercise 2.8)
Problem
Find the transfer function G s = X2 s /F s for the translational mechanical system shown in Fig. 2.21
of Reference [1].
Solution
First, rename the component symbols as follows:
M1 = m1 , M2 = m2 = 1 kg
fv = bi , i = 1 ..4 = 1 Ns$mK1
i
K = k = 1 N$mK1
f t =F t N
To see if the force F is in the t-domain or in the s-domain use F t respectively F s .
Next, draw the FDB's and write the following equations of motion
.. . . . . . .
Mass 1 : m1$x1 C b1$ x1 K 2 C b2$ x1 K 2 C b3$ x1 K 2 Ck$ x1 Kx2 = F s
x x x
.. . . . . . . .
Mass 2 : m2$x2 C b1$ x2 K 1 C b2$ x2 K 1 C b3$ x2 K 1 Cb4$x2 Ck$ x2 Kx1 = 0
x x x
Filling in component values we have
.. . .
x1 C3$ x1 K 2 C x1 Kx2 = F s
x
.. . . .
x2 C 3$ x2 K 1 Cx2 C x2 Kx1 = 0
x
Then, take the Laplace transform of these equations and rearrange
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5. Mechanical Translations Systems
s2 C3$s C1 $X1 K 3$s C1 $X2 = F s
K 3$s C1 $X1 C s2 C4$s C1 $X2 = 0
Finally, we could find the required transfer function by solving these equations with Cramer's rule ( see
solution on CD-ROM) or Maple's solve command. Here, however, we will use the matrix-vector
notation and Maple.
O restart :
O with LinearAlgebra :
s2 C3$s C1 K 3$s C1
O Ad :
K 3$s C1 s2 C4$s C1
F
O Bd :
0
O X d AK .B
1
2
s C4 s C1 F
s s3 C7 s2 C5 s C1
X :=
3 s C1 F
s s C7 s2 C5 s C1
3
O Θ1 d X1 : Θ2 d X2 :
Required transfer function
Θ2
O Gd
F
3 s C1
G :=
s s C7 s2 C5 s C1
3
End of Mechanical Translation System
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